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PAGE 1 Elemen ts of Abstract and Linear Algebra E. H. Connell PAGE 2 iiE.H. Connell Departmen t of Mathematics Univ ersit y of Miami P .O. Bo x 249085 Coral Gables, Florida 33124 USA ec@math.miami.eduMathematical Sub ject Classications (1991): 1201, 1301, 1501, 1601, 2001 c r 1999 E.H. Connell Marc h 20, 2004 PAGE 3 iii In tro duction In 1965 I rst taugh t an undergraduate course in abstract algebra. It w as fun to teac h b ecause the material w as in teresting and the class w as outstanding. Fiv e of those studen ts later earned a Ph.D. in mathematics. Since then I ha v e taugh t the course ab out a dozen times from v arious texts. Ov er the y ears I dev elop ed a set of lecture notes and in 1985 I had them t yp ed so they could b e used as a text. They no w app ear (in mo died form) as the rst v e c hapters of this b o ok. Here w ere some of m y motiv es at the time. 1) T o ha v e something as short and inexp ensiv e as p ossible. In m y exp erience, studen ts lik e short b o oks. 2) T o a v oid all inno v ation. T o organize the material in the most simpleminded straigh tforw ard manner. 3) T o order the material linearly T o the exten t p ossible, eac h section should use the previous sections and b e used in the follo wing sections. 4) T o omit as man y topics as p ossible. This is a foundational course, not a topics course. If a topic is not used later, it should not b e included. There are three go o d reasons for this. First, linear algebra has top priorit y It is b etter to go forw ard and do more linear algebra than to stop and do more group and ring theory Second, it is more imp ortan t that studen ts learn to organize and write pro ofs themselv es than to co v er more sub ject matter. Algebra is a p erfect place to get started b ecause there are man y \easy" theorems to pro v e. There are man y routine theorems stated here without pro ofs, and they ma y b e considered as exercises for the studen ts. Third, the material should b e so fundamen tal that it b e appropriate for studen ts in the ph ysical sciences and in computer science. Zillions of studen ts tak e calculus and co okb o ok linear algebra, but few tak e abstract algebra courses. Something is wrong here, and one thing wrong is that the courses try to do to o m uc h group and ring theory and not enough matrix theory and linear algebra. 5) T o oer an alternativ e for computer science ma jors to the standard discrete mathematics courses. Most of the material in the rst four c hapters of this text is co v ered in v arious discrete mathematics courses. Computer science ma jors migh t b enet b y seeing this material organized from a purely mathematical viewp oin t. PAGE 4 iv Ov er the y ears I used the v e c hapters that w ere t yp ed as a base for m y algebra courses, supplemen ting them as I sa w t. In 1996 I wrote a sixth c hapter, giving enough material for a full rst y ear graduate course. This c hapter w as written in the same \st yle" as the previous c hapters, i.e., ev erything w as righ t do wn to the n ub. It h ung together prett y w ell except for the last t w o sections on determinan ts and dual spaces. These w ere indep enden t topics stuc k on at the end. In the academic y ear 199798 I revised all six c hapters and had them t yp ed in LaT eX. This is the p ersonal bac kground of ho w this b o ok came ab out. It is dicult to do an ything in life without help from friends, and man y of m y friends ha v e con tributed to this text. My sincere gratitude go es esp ecially to Marilyn Gonzalez, Lourdes Robles, Marta Alpar, John Zw eib el, Dmitry Gokhman, Brian Co omes, Huseyin Ko cak, and Sh ulim Kaliman. T o these and all who con tributed, this b o ok is fondly dedicated. This b o ok is a surv ey of abstract algebra with emphasis on linear algebra. It is in tended for studen ts in mathematics, computer science, and the ph ysical sciences. The rst three or four c hapters can stand alone as a one semester course in abstract algebra. Ho w ev er they are structured to pro vide the bac kground for the c hapter on linear algebra. Chapter 2 is the most dicult part of the b o ok b ecause groups are written in additiv e and m ultiplicativ e notation, and the concept of coset is confusing at rst. After Chapter 2 the b o ok gets easier as y ou go along. Indeed, after the rst four c hapters, the linear algebra follo ws easily Finishing the c hapter on linear algebra giv es a basic one y ear undergraduate course in abstract algebra. Chapter 6 con tin ues the material to complete a rst y ear graduate course. Classes with little bac kground can do the rst three c hapters in the rst semester, and c hapters 4 and 5 in the second semester. More adv anced classes can do four c hapters the rst semester and c hapters 5 and 6 the second semester. As bare as the rst four c hapters are, y ou still ha v e to truc k righ t along to nish them in one semester. The presen tation is compact and tigh tly organized, but still somewhat informal. The pro ofs of man y of the elemen tary theorems are omitted. These pro ofs are to b e pro vided b y the professor in class or assigned as homew ork exercises. There is a nontrivial theorem stated without pro of in Chapter 4, namely the determinan t of the pro duct is the pro duct of the determinan ts. F or the prop er ro w of the course, this theorem should b e assumed there without pro of. The pro of is con tained in Chapter 6. The Jordan form should not b e considered part of Chapter 5. It is stated there only as a reference for undergraduate courses. Finally Chapter 6 is not written primarily for reference, but as an additional c hapter for more adv anced courses. PAGE 5 v This text is written with the con viction that it is more eectiv e to teac h abstract and linear algebra as one coheren t discipline rather than as t w o separate ones. T eac hing abstract algebra and linear algebra as distinct courses results in a loss of synergy and a loss of momen tum. Also with this text the professor do es not extract the course from the text, but rather builds the course up on it. I am con vinced it is easier to build a course from a base than to extract it from a big b o ok. Because after y ou extract it, y ou still ha v e to build it. The bare b ones nature of this b o ok adds to its rexibilit y b ecause y ou can build whatev er course y ou w an t around it. Basic algebra is a sub ject of incredible elegance and utilit y but it requires a lot of organization. This b o ok is m y attempt at that organization. Ev ery eort has b een extended to mak e the sub ject mo v e rapidly and to mak e the ro w from one topic to the next as seamless as p ossible. The studen t has limited time during the semester for serious study and this time should b e allo cated with care. The professor pic ks whic h topics to assign for serious study and whic h ones to \w a v e arms at". The goal is to sta y fo cused and go forw ard, b ecause mathematics is learned in hindsigh t. I w ould ha v e made the b o ok shorter, but I did not ha v e an y more time. When using this text, the studen t already has the outline of the next lecture, and eac h assignmen t should include the study of the next few pages. Study forw ard, not just bac k. A few min utes of preparation do es w onders to lev erage classro om learning, and this b o ok is in tended to b e used in that manner. The purp ose of class is to learn, not to do transcription w ork. When studen ts come to class cold and sp end the p erio d taking notes, they participate little and learn little. This leads to a dead class and also to the bad psyc hology of \O K ; I am here, so teac h me the sub ject." Mathematics is not taugh t, it is learned, and man y studen ts nev er learn ho w to learn. Professors should giv e more direction in that regard. Unfortunately mathematics is a dicult and hea vy sub ject. The st yle and approac h of this b o ok is to mak e it a little ligh ter. This b o ok w orks b est when view ed ligh tly and read as a story I hop e the studen ts and professors who try it, enjo y it. E. H. Connell Departmen t of Mathematics Univ ersit y of Miami Coral Gables, FL 33124 ec@math.miami.edu PAGE 6 viOutlineChapter 1 Bac kground and F undamen tals of Mathematics Sets, Cartesian pro ducts 1 Relations, partial orderings, Hausdor maximalit y principle, 3 equiv alence relations F unctions, bijections, strips, solutions of equations, 5 righ t and left in v erses, pro jections Notation for the logic of mathematics 13 In tegers, subgroups, unique factorization 14 Chapter 2 Groups Groups, scalar m ultiplication for additiv e groups 19 Subgroups, order, cosets 21 Normal subgroups, quotien t groups, the in tegers mo d n 25 Homomorphisms 27 P erm utations, the symmetric groups 31 Pro duct of groups 34 Chapter 3 Rings Rings 37 Units, domains, elds 38 The in tegers mo d n 40 Ideals and quotien t rings 41 Homomorphisms 42 P olynomial rings 45 Pro duct of rings 49 The Chinese remainder theorem 50 Characteristic 50 Bo olean rings 51 Chapter 4 Matrices and Matrix Rings Addition and m ultiplication of matrices, in v ertible matrices 53 T ransp ose 56 T riangular, diagonal, and scalar matrices 56 Elemen tary op erations and elemen tary matrices 57 Systems of equations 59 PAGE 7 vii Determinan ts, the classical adjoin t 60 Similarit y trace, and c haracteristic p olynomial 64 Chapter 5 Linear Algebra Mo dules, submo dules 68 Homomorphisms 69 Homomorphisms on R n 71 Cosets and quotien t mo dules 74 Pro ducts and copro ducts 75 Summands 77 Indep endence, generating sets, and free basis 78 Characterization of free mo dules 79 Uniqueness of dimension 82 Change of basis 83 V ector spaces, square matrices o v er elds, rank of a matrix 85 Geometric in terpretation of determinan t 90 Linear functions appro ximate dieren tiable functions lo cally 91 The transp ose principle 92 Nilp oten t homomorphisms 93 Eigen v alues, c haracteristic ro ots 95 Jordan canonical form 96 Inner pro duct spaces, GramSc hmidt orthonormalization 98 Orthogonal matrices, the orthogonal group 102 Diagonalization of symmetric matrices 103 Chapter 6 App endix The Chinese remainder theorem 108 Prime and maximal ideals and UFD s 109 Splitting short exact sequences 114 Euclidean domains 116 Jordan blo c ks 122 Jordan canonical form 123 Determinan ts 128 Dual spaces 130 PAGE 8 viii 1 2 3 4 5 6 7 8 9 11 10 Abstract algebra is not only a ma jor sub ject of science, but it is also magic and fun. Abstract algebra is not all w ork and no pla y and it is certainly not a dull b o y See, for example, the neat card tric k on page 18. This tric k is based, not on sleigh t of hand, but rather on a theorem in abstract algebra. An y one can do it, but to understand it y ou need some group theory And b efore b eginning the course, y ou migh t rst try y our skills on the famous (some w ould sa y infamous) tile puzzle. In this puzzle, a frame has 12 spaces, the rst 11 with n um b ered tiles and the last v acan t. The last t w o tiles are out of order. Is it p ossible to slide the tiles around to get them all in order, and end again with the last space v acan t? After giving up on this, y ou can study p erm utation groups and learn the answ er! PAGE 9 Chapter 1 Bac kground and F undamen tals of MathematicsThis c hapter is fundamen tal, not just for algebra, but for all elds related to mathematics. The basic concepts are pro ducts of sets, partial orderings, equiv alence relations, functions, and the in tegers. An equiv alence relation on a set A is sho wn to b e simply a partition of A in to disjoin t subsets. There is an emphasis on the concept of function, and the prop erties of surjectiv e, injectiv e, and bijectiv e. The notion of a solution of an equation is cen tral in mathematics, and most prop erties of functions can b e stated in terms of solutions of equations. In elemen tary courses the section on the Hausdor Maximalit y Principle should b e ignored. The nal section giv es a pro of of the unique factorization theorem for the in tegers. Notation Mathematics has its o wn univ ersally accepted shorthand. The sym b ol 9 means \there exists" and 9 means \there exists a unique". The sym b ol 8 means \for eac h" and ) means \implies". Some sets (or collections) are so basic they ha v e their o wn proprietary sym b ols. Fiv e of these are listed b elo w. N = Z + = the set of p ositiv e in tegers = f 1 ; 2 ; 3 ; ::: g Z = the ring of in tegers = f :::; 2 ; 1 ; 0 ; 1 ; 2 ; ::: g Q = the eld of rational n um b ers = f a=b : a; b 2 Z ; b 6 = 0 g R = the eld of real n um b ers C = the eld of complex n um b ers = f a + bi : a; b 2 R g ( i 2 = 1) Sets Supp ose A; B ; C ; ... are sets. W e use the standard notation for in tersection and union. A \ B = f x : x 2 A and x 2 B g = the set of all x whic h are elemen ts 1 PAGE 10 2 Bac kground Chapter 1 of A and B A [ B = f x : x 2 A or x 2 B g = the set of all x whic h are elemen ts of A or B An y set called an index set is assumed to b e nonv oid. Supp ose T is an index set and for eac h t 2 T A t is a set. [ t 2 T A t = f x : 9 t 2 T with x 2 A t g \ t 2 T A t = f x : if t 2 T ; x 2 A t g = f x : 8 t 2 T ; x 2 A t g Let ; b e the n ull set. If A \ B = ; then A and B are said to b e disjoint Denition Supp ose eac h of A and B is a set. The statemen t that A is a subset of B ( A B ) means that if a is an elemen t of A then a is an elemen t of B That is, a 2 A ) a 2 B : If A B w e ma y sa y A is con tained in B or B con tains A Exercise Supp ose eac h of A and B is a set. The statemen t that A is not a subset of B means Theorem (De Morgan's la ws) Supp ose S is a set. If C S (i.e., if C is a subset of S ), let C 0 the complemen t of C in S b e dened b y C 0 = S C = f x 2 S : x 62 C g Then for an y A; B S ( A \ B ) 0 = A 0 [ B 0 and ( A [ B ) 0 = A 0 \ B 0 Cartesian Pro ducts If X and Y are sets, X Y = f ( x; y ) : x 2 X and y 2 Y g In other w ords, the Cartesian pro duct of X and Y is dened to b e the set of all ordered pairs whose rst term is in X and whose second term is in Y Example R R = R 2 = the plane. PAGE 11 Chapter 1 Bac kground 3 Denition If eac h of X 1 ; :::; X n is a set, X 1 X n = f ( x 1 ; :::; x n ) : x i 2 X i for 1 i n g = the set of all ordered n tuples whose i th term is in X i Example R R = R n = real n space. Question Is ( R R 2 ) = ( R 2 R ) = R 3 ? Relations If A is a nonv oid set, a nonv oid subset R A A is called a r elation on A If ( a; b ) 2 R w e sa y that a is related to b and w e write this fact b y the expression a b Here are sev eral prop erties whic h a relation ma y p ossess. 1) If a 2 A then a a (rerexiv e) 2) If a b; then b a (symmetric) 2 0 ) If a b and b a then a = b (an tisymmetric) 3) If a b and b c; then a c (transitiv e) Denition A relation whic h satises 1), 2 0 ), and 3) is called a p artial or dering In this case w e write a b as a b Then 1) If a 2 A then a a 2 0 ) If a b and b a then a = b 3) If a b and b c; then a c Denition A line ar or dering is a partial ordering with the additional prop ert y that, if a; b 2 A then a b or b a Example A = R with the ordinary ordering, is a linear ordering. Example A = all subsets of R 2 with a b dened b y a b is a partial ordering. Hausdor Maximalit y Principle (HMP) Supp ose S is a nonv oid subset of A and is a relation on A This denes a relation on S If the relation satises an y of the prop erties 1), 2), 2 0 ), or 3) on A; the relation also satises these prop erties when restricted to S In particular, a partial ordering on A denes a partial ordering PAGE 12 4 Bac kground Chapter 1 on S Ho w ev er the ordering ma y b e linear on S but not linear on A The HMP is that an y linearly ordered subset of a partially ordered set is con tained in a maximal linearly ordered subset. Exercise Dene a relation on A = R 2 b y ( a; b ) ( c; d ) pro vided a c and b d: Sho w this is a partial ordering whic h is linear on S = f ( a; a ) : a < 0 g : Find at least t w o maximal linearly ordered subsets of R 2 whic h con tain S One of the most useful applications of the HMP is to obtain maximal monotonic collections of subsets. Denition A collection of sets is said to b e monotonic if, giv en an y t w o sets of the collection, one is con tained in the other. Corollary to HMP Supp ose X is a nonv oid set and A is some nonv oid collection of subsets of X and S is a sub collection of A whic h is monotonic. Then 9 a maximal monotonic sub collection of A whic h con tains S Pro of Dene a partial ordering on A b y V W i V W ; and apply HMP The HMP is used t wice in this b o ok. First, to sho w that innitely generated v ector spaces ha v e free bases, and second, in the App endix, to sho w that rings ha v e maximal ideals (see pages 87 and 109). In eac h of these applications, the maximal monotonic sub collection will ha v e a maximal elemen t. In elemen tary courses, these results ma y b e assumed, and th us the HMP ma y b e ignored. Equiv alence Relations A relation satisfying prop erties 1), 2), and 3) is called an e quivalenc e r elation Exercise Dene a relation on A = Z b y n m i n m is a m ultiple of 3. Sho w this is an equiv alence relation. Denition If is an equiv alence relation on A and a 2 A w e dene the e quivalenc e class con taining a b y cl ( a ) = f x 2 A : a x g PAGE 13 Chapter 1 Bac kground 5 Theorem 1) If b 2 cl ( a ) then cl ( b ) = cl ( a ) : Th us w e ma y sp eak of a subset of A b eing an equiv alence class with no men tion of an y elemen t con tained in it. 2) If eac h of U; V A is an equiv alence class and U \ V 6 = ; then U = V 3) Eac h elemen t of A is an elemen t of one and only one equiv alence class. Denition A p artition of A is a collection of disjoin t nonv oid subsets whose union is A In other w ords, a collection of nonv oid subsets of A is a partition of A pro vided an y a 2 A is an elemen t of one and only one subset of the collection. Note that if A has an equiv alence relation, the equiv alence classes form a partition of A Theorem Supp ose A is a nonv oid set with a partition. Dene a relation on A b y a b i a and b b elong to the same subset of the partition. Then is an equiv alence relation, and the equiv alence classes are just the subsets of the partition. Summary There are t w o w a ys of viewing an equiv alence relation  one is as a relation on A satisfying 1), 2), and 3), and the other is as a partition of A in to disjoin t subsets. Exercise Dene an equiv alence relation on Z b y n m i n m is a m ultiple of 3. What are the equiv alence classes? Exercise Is there a relation on R satisfying 1), 2), 2 0 ) and 3) ? That is, is there an equiv alence relation on R whic h is also a partial ordering? Exercise Let H R 2 b e the line H = f ( a; 2 a ) : a 2 R g Consider the collection of all translates of H ; i.e., all lines in the plane with slop e 2. Find the equiv alence relation on R 2 dened b y this partition of R 2 F unctions Just as there are t w o w a ys of viewing an equiv alence relation, there are t w o w a ys of dening a function. One is the \in tuitiv e" denition, and the other is the \graph" or \ordered pairs" denition. In either case, domain and r ange are inheren t parts of the denition. W e use the \in tuitiv e" denition b ecause ev ery one thinks that w a y PAGE 14 6 Bac kground Chapter 1 Denition If X and Y are (nonv oid) sets, a function or mapping or map with domain X and range Y is an ordered triple ( X ; Y ; f ) where f assigns to eac h x 2 X a w ell dened elemen t f ( x ) 2 Y The statemen t that ( X ; Y ; f ) is a function is written as f : X Y or X f Y Denition The gr aph of a function ( X ; Y ; f ) is the subset X Y dened b y = f ( x; f ( x )) : x 2 X g : The connection b et w een the \in tuitiv e" and \graph" viewp oin ts is giv en in the next theorem. Theorem If f : X Y then the graph X Y has the prop ert y that eac h x 2 X is the rst term of one and only one ordered pair in Con v ersely if is a subset of X Y with the prop ert y that eac h x 2 X is the rst term of one and only ordered pair in then 9 f : X Y whose graph is The function is dened b y \ f ( x ) is the second term of the ordered pair in whose rst term is x ." Example Identity functions Here X = Y and f : X X is dened b y f ( x ) = x for all x 2 X : The iden tit y on X is denoted b y I X or just I : X X Example Constant functions Supp ose y 0 2 Y Dene f : X Y b y f ( x ) = y 0 for all x 2 X Restriction Giv en f : X Y and a nonv oid subset S of X dene f j S : S Y b y ( f j S )( s ) = f ( s ) for all s 2 S Inclusion If S is a nonv oid subset of X dene the inclusion i : S X b y i ( s ) = s for all s 2 S: Note that inclusion is a restriction of the iden tit y Comp osition Giv en W f X g Y dene g f : W Y b y ( g f )( x ) = g ( f ( x )). Theorem (The asso ciativ e la w of comp osition) If V f W g X h Y then h ( g f ) = ( h g ) f : This ma y b e written as h g f PAGE 15 Chapter 1 Bac kground 7 Denitions Supp ose f : X Y 1) If T Y the inverse image of T is a subset of X f 1 ( T ) = f x 2 X : f ( x ) 2 T g 2) If S X the image of S is a subset of Y f ( S ) = f f ( s ) : s 2 S g = f y 2 Y : 9 s 2 S with f ( s ) = y g 3) The image of f is the image of X i.e., image ( f ) = f ( X ) = f f ( x ) : x 2 X g = f y 2 Y : 9 x 2 X with f ( x ) = y g 4) f : X Y is surje ctive or onto pro vided image ( f ) = Y i.e., the image is the range, i.e., if y 2 Y f 1 ( y ) is a nonv oid subset of X 5) f : X Y is inje ctive or 11 pro vided ( x 1 6 = x 2 ) ) f ( x 1 ) 6 = f ( x 2 ) ; i.e., if x 1 and x 2 are distinct elemen ts of X then f ( x 1 ) and f ( x 2 ) are distinct elemen ts of Y 6) f : X Y is bije ctive or is a 11 c orr esp ondenc e pro vided f is surjectiv e and injectiv e. In this case, there is function f 1 : Y X with f 1 f = I X : X X and f f 1 = I Y : Y Y Note that f 1 : Y X is also bijectiv e and ( f 1 ) 1 = f Examples 1) f : R R dened b y f ( x ) = sin( x ) is neither surjectiv e nor injectiv e. 2) f : R [ 1 ; 1] dened b y f ( x ) = sin ( x ) is surjectiv e but not injectiv e. 3) f : [0 ; = 2] R dened b y f ( x ) = sin ( x ) is injectiv e but not surjectiv e. 4) f : [0 ; = 2] [0 ; 1] dened b y f ( x ) = sin ( x ) is bijectiv e. ( f 1 ( x ) is written as arcsin( x ) or sin 1 ( x ) : ) 5) f : R (0 ; 1 ) dened b y f ( x ) = e x is bijectiv e. ( f 1 ( x ) is written as ln( x ) : ) Note There is no suc h thing as \the function sin( x )." A function is not dened unless the domain and range are sp ecied. PAGE 16 8 Bac kground Chapter 1 Exercise Sho w there are natural bijections from ( R R 2 ) to ( R 2 R ) and from ( R 2 R ) to R R R : These three sets are disjoin t, but the bijections b et w een them are so natural that w e sometimes iden tify them. Exercise Supp ose X is a set with 6 elemen ts and Y is a nite set with n elemen ts. 1) There exists an injectiv e f : X Y i n 2) There exists a surjectiv e f : X Y i n 3) There exists a bijectiv e f : X Y i n Pigeonhole Principle Supp ose X is a nite set with m elemen ts, Y is a nite set with n elemen ts, and f : X Y is a function. 1) If m = n; then f is injectiv e i f is surjectiv e i f is bijectiv e. 2) If m > n; then f is not injectiv e. 3) If m < n; then f is not surjectiv e. If y ou are placing 6 pigeons in 6 holes, and y ou run out of pigeons b efore y ou ll the holes, then y ou ha v e placed 2 pigeons in one hole. In other w ords, in part 1) for m = n = 6, if f is not surjectiv e then f is not injectiv e. Of course, the pigeonhole principle do es not hold for innite sets, as can b e seen b y the follo wing exercise. Exercise Sho w there is a function f : Z + Z + whic h is injectiv e but not surjectiv e. Also sho w there is one whic h is surjectiv e but not injectiv e. Exercise Supp ose f : [ 2 ; 2] R is dened b y f ( x ) = x 2 Find f 1 ( f ([1 ; 2])). Also nd f ( f 1 ([3 ; 5])). Exercise Supp ose f : X Y is a function, S X and T Y Find the relationship b et w een S and f 1 ( f ( S )). Sho w that if f is injectiv e, S = f 1 ( f ( S )). Also nd the relationship b et w een T and f ( f 1 ( T )). Sho w that if f is surjectiv e, T = f ( f 1 ( T )). Strips W e no w dene the v ertical and horizon tal strips of X Y If x 0 2 X ; f ( x 0 ; y ) : y 2 Y g = ( x 0 Y ) is called a vertic al strip If y 0 2 Y ; f ( x; y 0 ) : x 2 X g = ( X y 0 ) is called a horizontal strip PAGE 17 Chapter 1 Bac kground 9 Theorem Supp ose S X Y The subset S is the graph of a function with domain X and range Y i eac h v ertical strip in tersects S in exactly one p oin t. This is just a restatemen t of the prop ert y of a graph of a function. The purp ose of the next theorem is to restate prop erties of functions in terms of horizon tal strips. Theorem Supp ose f : X Y has graph Then 1) Eac h horizon tal strip in tersects in at least one p oin t i f is 2) Eac h horizon tal strip in tersects in at most one p oin t i f is 3) Eac h horizon tal strip in tersects in exactly one p oin t i f is Solutions of Equations No w w e restate these prop erties in terms of solutions of equations. Supp ose f : X Y and y 0 2 Y Consider the equation f ( x ) = y 0 Here y 0 is giv en and x is considered to b e a \v ariable". A solution to this equation is an y x 0 2 X with f ( x 0 ) = y 0 Note that the set of all solutions to f ( x ) = y 0 is f 1 ( y 0 ). Also f ( x ) = y 0 has a solution i y 0 2 image( f ) i f 1 ( y 0 ) is nonv oid. Theorem Supp ose f : X Y 1) The equation f ( x ) = y 0 has at least one solution for eac h y 0 2 Y i f is 2) The equation f ( x ) = y 0 has at most one solution for eac h y 0 2 Y i f is 3) The equation f ( x ) = y 0 has a unique solution for eac h y 0 2 Y i f is Righ t and Left In v erses One w a y to understand functions is to study righ t and left in v erses, whic h are dened after the next theorem. Theorem Supp ose X f Y g W are functions. 1) If g f is injectiv e, then f is injectiv e. PAGE 18 10 Bac kground Chapter 1 2) If g f is surjectiv e, then g is surjectiv e. 3) If g f is bijectiv e, then f is injectiv e and g is surjectiv e. Example X = W = f p g Y = f p; q g f ( p ) = p and g ( p ) = g ( q ) = p Here g f is the iden tit y but f is not surjectiv e and g is not injectiv e. Denition Supp ose f : X Y is a function. A left in v erse of f is a function g : Y X suc h that g f = I X : X X A righ t in v erse of f is a function h : Y X suc h that f h = I Y : Y Y Theorem Supp ose f : X Y is a function. 1) f has a righ t in v erse i f is surjectiv e. An y suc h righ t in v erse m ust b e injectiv e. 2) f has a left in v erse i f is injectiv e. An y suc h left in v erse m ust b e surjectiv e. Corollary Supp ose eac h of X and Y is a nonv oid set. Then 9 an injectiv e f : X Y i 9 a surjectiv e g : Y X : Also a function from X to Y is bijectiv e i it has a left in v erse and a righ t in v erse i it has a left and righ t in v erse. Note The Axiom of Choice is not discussed in this b o ok. Ho w ev er, if y ou w ork ed 1) of the theorem ab o v e, y ou unkno wingly used one v ersion of it. F or completeness, w e state this part of 1) again. The Axiom of Choice If f : X Y is surjectiv e, then f has a righ t in v erse h That is, for eac h y 2 Y it is p ossible to c ho ose an x 2 f 1 ( y ) and th us to dene h ( y ) = x Note It is a classical theorem in set theory that the Axiom of Choice and the Hausdor Maximalit y Principle are equiv alen t. Ho w ev er in this text w e do not go that deeply in to set theory F or our purp oses it is assumed that the Axiom of Choice and the HMP are true. Exercise Supp ose f : X Y is a function. Dene a relation on X b y a b if f ( a ) = f ( b ) : Sho w this is an equiv alence relation. If y b elongs to the image of f then f 1 ( y ) is an equiv alence class and ev ery equiv alence class is of this form. In the next c hapter where f is a group homomorphism, these equiv alence classes will b e called cosets. PAGE 19 Chapter 1 Bac kground 11 Pro jections If X 1 and X 2 are nonv oid sets, w e dene the pro jection maps 1 : X 1 X 2 X 1 and 2 : X 1 X 2 X 2 b y i ( x 1 ; x 2 ) = x i Theorem If Y ; X 1 and X 2 are nonv oid sets, there is a 11 corresp ondence b et w een f functions f : Y X 1 X 2 g and f ordered pairs of functions ( f 1 ; f 2 ) where f 1 : Y X 1 and f 2 : Y X 2 g Pro of Giv en f dene f 1 = 1 f and f 2 = 2 f Giv en f 1 and f 2 dene f : Y X 1 X 2 b y f ( y ) = ( f 1 ( y ) ; f 2 ( y )). Th us a function from Y to X 1 X 2 is merely a pair of functions from Y to X 1 and Y to X 2 This concept is displa y ed in the diagram b elo w. It is summarized b y the equation f = ( f 1 ; f 2 ). X 1 X 2 X 1 X 2 Y ? @ @ @ @ @ R f 1 f 2 f 1 2 One nice thing ab out this concept is that it w orks ne for innite Cartesian pro ducts. Denition Supp ose T is an index set and for eac h t 2 T X t is a nonv oid set. Then the pr o duct Y t 2 T X t = Q X t is the collection of all sequences f x t g t 2 T = f x t g where x t 2 X t F ormally these sequences are functions from T to S X t with eac h ( t ) in X t and written as ( t ) = x t : If T = f 1 ; 2 ; : : : ; n g then f x t g is the ordered n tuple ( x 1 ; x 2 ; : : : ; x n ) : If T = Z + then f x t g is the sequence ( x 1 ; x 2 ; : : : ) : F or an y T and an y s in T the pr oje ction map s : Q X t X s is dened b y s ( f x t g ) = x s Theorem If Y is an y nonv oid set, there is a 11 corresp ondence b et w een f functions f : Y Q X t g and f sequences of functions f f t g t 2 T where f t : Y X t g Giv en f the sequence f f t g is dened b y f t = t f : Giv en f f t g f is dened b y f ( y ) = f f t ( y ) g PAGE 20 12 Bac kground Chapter 1 A Calculus Exercise Let A b e the collection of all functions f : [0 ; 1] R whic h ha v e an innite n um b er of deriv ativ es. Let A 0 A b e the sub collection of those functions f with f (0) = 0. Dene D : A 0 A b y D ( f ) = d f =dx Use the mean v alue theorem to sho w that D is injectiv e. Use the fundamen tal theorem of calculus to sho w that D is surjectiv e. Exercise This exercise is not used elsewhere in this text and ma y b e omitted. It is included here for studen ts who wish to do a little more set theory Supp ose T is a nonv oid set. 1) If Y is a nonv oid set, dene Y T to b e the collection of all functions with domain T and range Y Sho w that if T and Y are nite sets with m and n elemen ts, then Y T has n m elemen ts. In particular, when T = f 1 ; 2 ; 3 g ; Y T = Y Y Y has n 3 elemen ts. Sho w that if n 3, the subset of Y f 1 ; 2 ; 3 g of all injectiv e functions has n ( n 1)( n 2) elemen ts. These injectiv e functions are called p erm utations on Y tak en 3 at a time. If T = N then Y T is the innite pro duct Y Y That is, Y N is the set of all innite sequences ( y 1 ; y 2 ; : : : ) where eac h y i 2 Y F or an y Y and T let Y t b e a cop y of Y for eac h t 2 T : Then Y T = Y t 2 T Y t 2) Supp ose eac h of Y 1 and Y 2 is a nonv oid set. Sho w there is a natural bijection from ( Y 1 Y 2 ) T to Y T 1 Y T 2 (This is the fundamen tal prop ert y of Cartesian pro ducts presen ted in the t w o previous theorems.) 3) Dene P ( T ), the p o w er set of T to b e the collection of all subsets of T (including the n ull set). Sho w that if T is a nite set with m elemen ts, P ( T ) has 2 m elemen ts. 4) If S is an y subset of T dene its c haracteristic function S : T f 0 ; 1 g b y letting S ( t ) b e 1 when t 2 S and b e 0 when t 2 j S Dene : P ( T ) f 0 ; 1 g T b y ( S ) = S Dene : f 0 ; 1 g T P ( T ) b y ( f ) = f 1 (1). Sho w that if S T then ( S ) = S and if f : T f 0 ; 1 g then ( f ) = f Th us is a bijection and = 1 P ( T ) f 0 ; 1 g T 5) Supp ose r : T f 0 ; 1 g T is a function and sho w that it cannot b e surjectiv e. If t 2 T denote r ( t ) b y r ( t ) = f t : T f 0 ; 1 g Dene f : T f 0 ; 1 g b y f ( t ) = 0 if f t ( t ) = 1, and f ( t ) = 1 if f t ( t ) = 0. Sho w that f is not in the image of r and th us r cannot b e surjectiv e. This sho ws that if T is an innite set, then the set f 0 ; 1 g T represen ts a \higher order of innit y than T ". 6) An innite set Y is said to b e c ountable if there is a bijection from the p ositiv e PAGE 21 Chapter 1 Bac kground 13 in tegers N to Y : Sho w Q is coun table but the follo wing three collections are not. i) P ( N ), the collection of all subsets of N ii) f 0 ; 1 g N the collection of all functions f : N f 0 ; 1 g iii) The collection of all sequences ( y 1 ; y 2 ; : : : ) where eac h y i is 0 or 1. W e kno w that ii) and iii) are equal and there is a natural bijection b et w een i) and ii). W e also kno w there is no surjectiv e map from N to f 0 ; 1 g N i.e., f 0 ; 1 g N is uncoun table. Finally sho w there is a bijection from f 0 ; 1 g N to the real n um b ers R (This is not so easy T o start with, y ou ha v e to decide what the real n um b ers are.) Notation for the Logic of Mathematics Eac h of the w ords \Lemma", \Theorem", and \Corollary" means \true statemen t". Supp ose A and B are statemen ts. A theorem ma y b e stated in an y of the follo wing w a ys: Theorem Hyp othesis Statemen t A Conclusion Statemen t B Theorem Supp ose A is true. Then B is true. Theorem If A is true, then B is true. Theorem A ) B ( A implies B ). There are t w o w a ys to pro v e the theorem  to supp ose A is true and sho w B is true, or to supp ose B is false and sho w A is false. The expressions \ A B ", \ A is equiv alen t to B ", and \ A is true i B is true ha v e the same meaning (namely that A ) B and B ) A ). The imp ortan t thing to remem b er is that though ts and expressions ro w through the language. Mathematical sym b ols are shorthand for phrases and sen tences in the English language. F or example, \ x 2 B means \ x is an elemen t of the set B ." If A is the statemen t \ x 2 Z + and B is the statemen t \ x 2 2 Z + ", then \ A ) B "means \If x is a p ositiv e in teger, then x 2 is a p ositiv e in teger". Mathematical Induction is based up on the fact that if S Z + is a nonv oid subset, then S con tains a smallest elemen t. PAGE 22 14 Bac kground Chapter 1 Theorem Supp ose P ( n ) is a statemen t for eac h n = 1 ; 2 ; ::: Supp ose P (1) is true and for eac h n 1, P ( n ) ) P ( n + 1) : Then for eac h n 1, P ( n ) is true. Pro of If the theorem is false, then 9 a smallest p ositiv e in teger m suc h that P ( m ) is false. Since P ( m 1) is true, this is imp ossible. Exercise Use induction to sho w that, for eac h n 1 ; 1 + 2 + + n = n ( n + 1) = 2. The In tegers In this section, lo w er case letters a; b; c; ::: will represen t in tegers, i.e., elemen ts of Z Here w e will establish the follo wing three basic prop erties of the in tegers. 1) If G is a subgroup of Z then 9 n 0 suc h that G = n Z 2) If a and b are in tegers, not b oth zero, and G is the collection of all linear com binations of a and b then G is a subgroup of Z and its p ositiv e generator is the greatest common divisor of a and b 3) If n 2, then n factors uniquely as the pro duct of primes. All of this will follo w from long division, whic h w e no w state formally Euclidean Algorithm Giv en a; b with b 6 = 0, 9 m and r with 0 r < j b j and a = bm + r : In other w ords, b divides a \ m times with a remainder of r ". F or example, if a = 17 and b = 5, then m = 4 and r = 3, 17 = 5( 4) + 3. Denition If r = 0, w e sa y that b divides a or a is a multiple of b This fact is written as b j a Note that b j a the rational n um b er a=b is an in teger 9 m suc h that a = bm a 2 b Z Note An ything (except 0) divides 0. 0 do es not divide an ything. 1 divides an ything If n 6 = 0, the set of in tegers whic h n divides is n Z = f nm : m 2 Z g = f :::; 2 n; n; 0 ; n; 2 n; ::: g : Also n divides a and b with the same remainder i n divides ( a b ). Denition A nonv oid subset G Z is a sub gr oup pro vided ( g 2 G ) g 2 G ) and ( g 1 ; g 2 2 G ) ( g 1 + g 2 ) 2 G ). W e sa y that G is closed under negation and closed under addition. PAGE 23 Chapter 1 Bac kground 15 Theorem If n 2 Z then n Z is a subgroup. Th us if n 6 = 0, the set of in tegers whic h n divides is a subgroup of Z The next theorem states that ev ery subgroup of Z is of this form. Theorem Supp ose G Z is a subgroup. Then 1) 0 2 G 2) If g 1 and g 2 2 G then ( m 1 g 1 + m 2 g 2 ) 2 G for all in tegers m 1 ; m 2 3) 9 nonnegativ e in teger n suc h that G = n Z : In fact, if G 6 = f 0 g and n is the smallest p ositiv e in teger in G then G = n Z Pro of Since G is nonv oid, 9 g 2 G No w ( g ) 2 G and th us 0 = g + ( g ) b elongs to G and so 1) is true. P art 2) is straigh tforw ard, so consider 3). If G 6 = 0, it m ust con tain a p ositiv e elemen t. Let n b e the smallest p ositiv e in teger in G If g 2 G g = nm + r where 0 r < n: Since r 2 G it m ust b e 0, and g 2 n Z : No w supp ose a; b 2 Z and at least one of a and b is nonzero. Theorem Let G b e the set of all linear com binations of a and b i.e., G = f ma + nb : m; n 2 Z g : Then 1) G con tains a and b 2) G is a subgroup. In fact, it is the smallest subgroup con taining a and b It is called the subgroup generated b y a and b 3) Denote b y ( a; b ) the smallest p ositiv e in teger in G: By the previous theorem, G = ( a; b ) Z and th us ( a; b ) j a and ( a; b ) j b: Also note that 9 m; n suc h that ma + nb = ( a; b ) : The in teger ( a; b ) is called the gr e atest c ommon divisor of a and b 4) If n is an in teger whic h divides a and b then n also divides ( a; b ). Pro of of 4) Supp ose n j a and n j b i.e., supp ose a; b 2 n Z Since G is the smallest subgroup con taining a and b n Z ( a; b ) Z ; and th us n j ( a; b ). Corollary The follo wing are equiv alen t. 1) a and b ha v e no common divisors, i.e., ( n j a and n j b ) ) n = 1. PAGE 24 16 Bac kground Chapter 1 2) ( a; b ) = 1 ; i.e., the subgroup generated b y a and b is all of Z 3) 9 m; n 2 Z with ma + nb = 1. Denition If an y one of these three conditions is satised, w e sa y that a and b are r elatively prime This next theorem is the basis for unique factorization. Theorem If a and b are relativ ely prime with a not zero, then a j bc ) a j c Pro of Supp ose a and b are relativ ely prime, c 2 Z and a j bc Then there exist m; n with ma + nb = 1, and th us mac + nbc = c No w a j mac and a j nbc Th us a j ( mac + nbc ) and so a j c Denition A prime is an in teger p > 1 whic h do es not factor, i.e., if p = ab then a = 1 or a = p The rst few primes are 2, 3, 5, 7, 11, 13, 17,... Theorem Supp ose p is a prime. 1) If a is an in teger whic h is not a m ultiple of p then ( p; a ) = 1 : In other w ords, if a is an y in teger, ( p; a ) = p or ( p; a ) = 1. 2) If p j ab then p j a or p j b 3) If p j a 1 a 2 a n then p divides some a i : Th us if eac h a i is a prime, then p is equal to some a i Pro of P art 1) follo ws immediately from the denition of prime. No w supp ose p j ab If p do es not divide a then b y 1), ( p; a ) = 1 and b y the previous theorem, p m ust divide b Th us 2) is true. P art 3) follo ws from 2) and induction on n The Unique F actorization Theorem Supp ose a is an in teger whic h is not 0,1, or 1. Then a ma y b e factored in to the pro duct of primes and, except for order, this factorization is unique. That is, 9 a unique collection of distinct primes p 1 ; p 2 ; :::; p k and p ositiv e in tegers s 1 ; s 2 ; :::; s k suc h that a = p s 1 1 p s 2 2 p s k k Pro of F actorization in to primes is ob vious, and uniqueness follo ws from 3) in the theorem ab o v e. The p o w er of this theorem is uniqueness, not existence. PAGE 25 Chapter 1 Bac kground 17 No w that w e ha v e unique factorization and part 3) ab o v e, the picture b ecomes transparen t. Here are some of the basic prop erties of the in tegers in this ligh t. Theorem (Summary) 1) Supp ose j a j > 1 has prime factorization a = p s 1 1 p s k k Then the only divisors of a are of the form p t 1 1 p t k k where 0 t i s i for i = 1 ; :::; k 2) If j a j > 1 and j b j > 1, then ( a; b ) = 1 i there is no common prime in their factorizations. Th us if there is no common prime in their factorizations, 9 m; n with ma + nb = 1 ; and also ( a 2 ; b 2 ) = 1. 3) Supp ose j a j > 1 and j b j > 1. Let f p 1 ; : : : ; p k g b e the union of the distinct primes of their factorizations. Th us a = p s 1 1 p s k k where 0 s i and b = p t 1 1 p t k k where 0 t i Let u i b e the minim um of s i and t i Then ( a; b ) = p u 1 1 p u k k : F or example (2 3 5 11 ; 2 2 5 4 7) = 2 2 5. 3 0 ) Let v i b e the maxim um of s i and t i Then c = p v 1 1 p v k k is the le ast (p ositiv e) c ommon multiple of a and b: Note that c is a m ultiple of a and b; and if n is a m ultiple of a and b; then n is a m ultiple of c Finally if a and b are p ositiv e, their least common m ultiple is c = ab= ( a; b ) ; and if in addition a and b are relativ ely prime, then their least common m ultiple is just their pro duct. 4) There is an innite n um b er of primes. (Pro of: Supp ose there w ere only a nite n um b er of primes p 1 ; p 2 ; :::; p k : Then no prime w ould divide ( p 1 p 2 p k + 1).) 5) Supp ose c is an in teger greater than 1. Then p c is rational i p c is an in teger. In particular, p 2 and p 3 are irrational. (Pro of: If p c is rational, 9 p ositiv e in tegers a and b with p c = a=b and ( a; b ) = 1. If b > 1, then it is divisible b y some prime, and since cb 2 = a 2 this prime will also app ear in the prime factorization of a This is a con tradiction and th us b = 1 and p c is an in teger.) (See the fth exercise b elo w.) Exercise Find (180,28), i.e., nd the greatest common divisor of 180 and 28, i.e., nd the p ositiv e generator of the subgroup generated b y f 180,28 g Find in tegers m and n suc h that 180 m + 28 n = (180 ; 28). Find the least common m ultiple of 180 and 28, and sho w that it is equal to (180 28) = (180 ; 28). PAGE 26 18 Bac kground Chapter 1 Exercise W e ha v e dened the greatest common divisor (gcd) and the least common m ultiple (lcm) of a pair of in tegers. No w supp ose n 2 and S = f a 1 ; a 2 ; ::; a n g is a nite collection of in tegers with j a i j > 1 for 1 i n Dene the gcd and the lcm of the elemen ts of S and dev elop their prop erties. Express the gcd and the lcm in terms of the prime factorizations of the a i When is the lcm of S equal to the pro duct a 1 a 2 a n ? Sho w that the set of all linear com binations of the elemen ts of S is a subgroup of Z and its p ositiv e generator is the gcd of the elemen ts of S Exercise Sho w that the gcd of S = f 90 ; 70 ; 42 g is 2, and nd in tegers n 1 ; n 2 ; n 3 suc h that 90 n 1 + 70 n 2 + 42 n 3 = 2 : Also nd the lcm of the elemen ts of S Exercise Sho w that if eac h of G 1 ; G 2 ; :::; G m is a subgroup of Z then G 1 \ G 2 \ \ G m is also a subgroup of Z No w let G = (90 Z ) \ (70 Z ) \ (42 Z ) and nd the p ositiv e in teger n with G = n Z Exercise Sho w that if the n th ro ot of an in teger is a rational n um b er, then it itself is an in teger. That is, supp ose c and n are in tegers greater than 1. There is a unique p ositiv e real n um b er x with x n = c Sho w that if x is rational, then it is an in teger. Th us if p is a prime, its n th ro ot is an irrational n um b er. Exercise Sho w that a p ositiv e in teger is divisible b y 3 i the sum of its digits is divisible b y 3. More generally let a = a n a n 1 : : : a 0 = a n 10 n + a n 1 10 n 1 + + a 0 where 0 a i 9. No w let b = a n + a n 1 + + a 0 and sho w that 3 divides a and b with the same remainder. Although this is a straigh tforw ard exercise in long division, it will b e more transparen t later on. In the language of the next c hapter, it sa ys that [ a ] = [ b ] in Z 3 Card T ric k Ask friends to pic k out sev en cards from a dec k and then to select one to lo ok at without sho wing it to y ou. T ak e the six cards face do wn in y our left hand and the selected card in y our righ t hand, and announce y ou will place the selected card in with the other six, but they are not to kno w where. Put y our hands b ehind y our bac k and place the selected card on top, and bring the sev en cards in fron t in y our left hand. Ask y our friends to giv e y ou a n um b er b et w een one and sev en (not allo wing one). Supp ose they sa y three. Y ou mo v e the top card to the b ottom, then the second card to the b ottom, and then y ou turn o v er the third card, lea ving it face up on top. Then rep eat the pro cess, mo ving the top t w o cards to the b ottom and turning the third card face up on top. Con tin ue un til there is only one card face do wn, and this will b e the selected card. Magic? Sta y tuned for Chapter 2, where it is sho wn that an y nonzero elemen t of Z 7 has order 7. PAGE 27 Chapter 2 GroupsGroups are the cen tral ob jects of algebra. In later c hapters w e will dene rings and mo dules and see that they are sp ecial cases of groups. Also ring homomorphisms and mo dule homomorphisms are sp ecial cases of group homomorphisms. Ev en though the denition of group is simple, it leads to a ric h and amazing theory Ev erything presen ted here is standard, except that the pro duct of groups is giv en in the additiv e notation. This is the notation used in later c hapters for the pro ducts of rings and mo dules. This c hapter and the next t w o c hapters are restricted to the most basic topics. The approac h is to do quic kly the fundamen tals of groups, rings, and matrices, and to push forw ard to the c hapter on linear algebra. This c hapter is, b y far and ab o v e, the most dicult c hapter in the b o ok, b ecause group op erations ma y b e written as addition or m ultiplication, and also the concept of coset is confusing at rst. Denition Supp ose G is a nonv oid set and : G G G is a function. is called a binary op er ation and w e will write ( a; b ) = a b or ( a; b ) = a + b Consider the follo wing prop erties. 1) If a; b; c 2 G then a ( b c ) = ( a b ) c If a; b; c 2 G then a + ( b + c ) = ( a + b ) + c: 2) 9 e = e G 2 G suc h that if a 2 G 9 0 =0 G 2 G suc h that if a 2 G e a = a e = a: 0 + a = a +0 = a 3) If a 2 G 9 b 2 G with a b = b a = e If a 2 G; 9 b 2 G with a + b = b + a = 0 ( b is written as b = a 1 ). ( b is written as b = a ). 4) If a; b 2 G then a b = b a: If a; b 2 G then a + b = b + a Denition If prop erties 1), 2), and 3) hold, ( G; ) is said to b e a gr oup If w e write ( a; b ) = a b w e sa y it is a multiplic ative group. If w e write ( a; b ) = a + b 19 PAGE 28 20 Groups Chapter 2 w e sa y it is an additive group. If in addition, prop ert y 4) holds, w e sa y the group is ab elian or c ommutative Theorem Let ( G; ) b e a m ultiplicativ e group. (i) Supp ose a; c; c 2 G Then a c = a c ) c = c: Also c a = c a ) c = c: In other w ords, if f : G G is dened b y f ( c ) = a c then f is injectiv e. Also f is bijectiv e with f 1 giv en b y f 1 ( c ) = a 1 c (ii) e is unique, i.e., if e 2 G satises 2), then e = e: In fact, if a; b 2 G then ( a b = a ) ) ( b = e ) and ( a b = b ) ) ( a = e ) : Recall that b is an iden tit y in G pro vided it is a righ t and left iden tit y for an y a in G: Ho w ev er, group structure is so rigid that if 9 a 2 G suc h that b is a righ t iden tit y for a then b = e Of course, this is just a sp ecial case of the cancellation la w in (i). (iii) Ev ery righ t in v erse is an in v erse, i.e., if a b = e then b = a 1 Also if b a = e then b = a 1 : Th us in v erses are unique. (iv) If a 2 G; then ( a 1 ) 1 = a (v) The m ultiplication a 1 a 2 a 3 = a 1 ( a 2 a 3 ) = ( a 1 a 2 ) a 3 is w elldened. In general, a 1 a 2 a n is w ell dened. (vi) If a; b 2 G; ( a b ) 1 = b 1 a 1 : Also ( a 1 a 2 a n ) 1 = a 1 n a 1 n 1 a 1 1 (vii) Supp ose a 2 G: Let a 0 = e and if n > 0 ; a n = a a ( n times) and a n = a 1 a 1 ( n times). If n 1 ; n 2 ; :::; n t 2 Z then a n 1 a n 2 a n t = a n 1 + + n t : Also ( a n ) m = a nm Finally if G is ab elian and a; b 2 G then ( a b ) n = a n b n Exercise W rite out the ab o v e theorem where G is an additiv e group. Note that part (vii) states that G has a scalar m ultiplication o v er Z This means that if a is in G and n is an in teger, there is dened an elemen t an in G This is so basic, that w e state it explicitly Theorem Supp ose G is an additiv e group. If a 2 G let a 0 =0 and if n > 0, let an = ( a + + a ) where the sum is n times, and a ( n ) = ( a ) + ( a ) + ( a ), PAGE 29 Chapter 2 Groups 21 whic h w e write as ( a a a ) : Then the follo wing prop erties hold in general, except the rst requires that G b e ab elian. ( a + b ) n = an + bn a ( n + m ) = an + am a ( nm ) = ( an ) m a 1 = a Note that the plus sign is used am biguously  sometimes for addition in G and sometimes for addition in Z In the language used in Chapter 5, this theorem states that an y additiv e ab elian group is a Z mo dule. (See page 71.) Exercise Supp ose G is a nonv oid set with a binary op eration ( a; b ) = a b whic h satises 1), 2) and [ 3 0 ) If a 2 G 9 b 2 G with a b = e ]. Sho w ( G; ) is a group, i.e., sho w b a = e: In other w ords, the group axioms are stronger than necessary If ev ery elemen t has a righ t in v erse, then ev ery elemen t has a t w o sided in v erse. Exercise Supp ose G is the set of all functions from Z to Z with m ultiplication dened b y comp osition, i.e., f g = f g Note that G satises 1) and 2) but not 3), and th us G is not a group. Sho w that f has a righ t in v erse in G i f is surjectiv e, and f has a left in v erse in G i f is injectiv e (see page 10). Also sho w that the set of all bijections from Z to Z is a group under comp osition. Examples G = R G = Q or G = Z with ( a; b ) = a + b is an additiv e ab elian group. Examples G = R 0 or G = Q 0 with ( a; b ) = ab is a m ultiplicativ e ab elian group. G = Z 0 with ( a; b ) = ab is not a group. G = R + = f r 2 R : r > 0 g with ( a; b ) = ab is a m ultiplicativ e ab elian group. Subgroups Theorem Supp ose G is a m ultiplicativ e group and H G is a nonv oid subset satisfying 1) if a; b 2 H then a b 2 H and 2) if a 2 H then a 1 2 H PAGE 30 22 Groups Chapter 2 Then e 2 H and H is a group under m ultiplication. H is called a sub gr oup of G Pro of Since H is nonv oid, 9 a 2 H By 2), a 1 2 H and so b y 1), e 2 H The asso ciativ e la w is immediate and so H is a group. Example G is a subgroup of G and e is a subgroup of G These are called the impr op er subgroups of G Example If G = Z under addition, and n 2 Z then H = n Z is a subgroup of Z By a theorem in the section on the in tegers in Chapter 1, ev ery subgroup of Z is of this form (see page 15). This is a k ey prop ert y of the in tegers. Exercises Supp ose G is a m ultiplicativ e group. 1) Let H b e the c enter of G i.e., H = f h 2 G : g h = h g for all g 2 G g Sho w H is a subgroup of G 2) Supp ose H 1 and H 2 are subgroups of G: Sho w H 1 \ H 2 is a subgroup of G 3) Supp ose H 1 and H 2 are subgroups of G with neither H 1 nor H 2 con tained in the other. Sho w H 1 [ H 2 is not a subgroup of G 4) Supp ose T is an index set and for eac h t 2 T H t is a subgroup of G Sho w \ t 2 T H t is a subgroup of G 5) F urthermore, if f H t g is a monotonic collection, then [ t 2 T H t is a subgroup of G 6) Supp ose G = f all functions f : [0 ; 1] R g Dene an addition on G b y ( f + g )( t ) = f ( t ) + g ( t ) for all t 2 [0 ; 1]. This mak es G in to an ab elian group. Let K b e the subset of G comp osed of all dieren tiable functions. Let H b e the subset of G comp osed of all con tin uous functions. What theorems in calculus sho w that H and K are subgroups of G ? What theorem sho ws that K is a subset (and th us subgroup) of H ? Order Supp ose G is a m ultiplicativ e group. If G has an innite n um b er of PAGE 31 Chapter 2 Groups 23 elemen ts, w e sa y that o ( G ), the or der of G is innite. If G has n elemen ts, then o ( G ) = n Supp ose a 2 G and H = f a i : i 2 Z g H is an ab elian subgroup of G called the sub gr oup gener ate d by a W e dene the or der of the element a to b e the order of H i.e., the order of the subgroup generated b y a Let f : Z H b e the surjectiv e function dened b y f ( m ) = a m Note that f ( k + l ) = f ( k ) f ( l ) where the addition is in Z and the m ultiplication is in the group H W e come no w to the rst real theorem in group theory It sa ys that the elemen t a has nite order i f is not injectiv e, and in this case, the order of a is the smallest p ositiv e in teger n with a n = e Theorem Supp ose a is an elemen t of a m ultiplicativ e group G and H = f a i : i 2 Z g If 9 distinct in tegers i and j with a i = a j then a has some nite order n In this case H has n distinct elemen ts, H = f a 0 ; a 1 ; : : : ; a n 1 g and a m = e i n j m In particular, the order of a is the smallest p ositiv e in teger n with a n = e and f 1 ( e ) = n Z Pro of Supp ose j < i and a i = a j Then a i j = e and th us 9 a smallest p ositiv e in teger n with a n = e This implies that the elemen ts of f a 0 ; a 1 ; :::; a n 1 g are distinct, and w e m ust sho w they are all of H If m 2 Z the Euclidean algorithm states that 9 in tegers q and r with 0 r < n and m = nq + r Th us a m = a nq a r = a r and so H = f a 0 ; a 1 ; :::; a n 1 g and a m = e i n j m Later in this c hapter w e will see that f is a homomorphism from an additiv e group to a m ultiplicativ e group and that, in additiv e notation, H is isomorphic to Z or Z n Exercise W rite out this theorem for G an additiv e group. T o b egin, supp ose a is an elemen t of an additiv e group G and H = f ai : i 2 Z g Exercise Sho w that if G is a nite group of ev en order, then G has an o dd n um b er of elemen ts of order 2. Note that e is the only elemen t of order 1. Denition A group G is cyclic if 9 an elemen t of G whic h generates G Theorem If G is cyclic and H is a subgroup of G then H is cyclic. Pro of Supp ose G = f a i : i 2 Z g is a cyclic group and H is a subgroup of G If H = e then H is cyclic, so supp ose H 6 = e No w there is a smallest p ositiv e in teger m with a m 2 H If t is an in teger with a t 2 H then b y the Euclidean algorithm, m divides t and th us a m generates H Note that in the case G has nite order n i.e., G = f a 0 ; a 1 ; : : : ; a n 1 g then a n = e 2 H and th us the p ositiv e in teger m divides n In either case, w e ha v e a clear picture of the subgroups of G: Also note that this theorem w as pro v ed on page 15 for the additiv e group Z PAGE 32 24 Groups Chapter 2 Cosets Supp ose H is a subgroup of a group G It will b e sho wn b elo w that H partitions G in to righ t cosets. It also partitions G in to left cosets, and in general these partitions are distinct. Theorem If H is a subgroup of a m ultiplicativ e group G then a b dened b y a b i a b 1 2 H is an equiv alence relation. If a 2 G cl ( a ) = f b 2 G : a b g = f h a : h 2 H g = H a: Note that a b 1 2 H i b a 1 2 H If H is a subgroup of an additiv e group G then a b dened b y a b i ( a b ) 2 H is an equiv alence relation. If a 2 G cl ( a ) = f b 2 G : a b g = f h + a : h 2 H g = H + a: Note that ( a b ) 2 H i ( b a ) 2 H Denition These equiv alence classes are called right c osets If the relation is dened b y a b i b 1 a 2 H then the equiv alence classes are cl ( a ) = aH and they are called left c osets H is a left and righ t coset. If G is ab elian, there is no distinction b et w een righ t and left cosets. Note that b 1 a 2 H i a 1 b 2 H In the theorem ab o v e, H is used to dene an equiv alence relation on G and th us a partition of G W e no w do the same thing a dieren t w a y W e dene the righ t cosets directly and sho w they form a partition of G: Y ou migh t nd this easier. Theorem Supp ose H is a subgroup of a m ultiplicativ e group G If a 2 G dene the righ t coset con taining a to b e H a = f h a : h 2 H g : Then the follo wing hold. 1) H a = H i a 2 H 2) If b 2 H a; then H b = H a; i.e., if h 2 H then H ( h a ) = ( H h ) a = H a 3) If H c \ H a 6 = ; ; then H c = H a 4) The righ t cosets form a partition of G i.e., eac h a in G b elongs to one and only one righ t coset. 5) Elemen ts a and b b elong to the same righ t coset i a b 1 2 H i b a 1 2 H Pro of There is no b etter w a y to dev elop facilit y with cosets than to pro v e this theorem. Also write this theorem for G an additiv e group. Theorem Supp ose H is a subgroup of a m ultiplicativ e group G PAGE 33 Chapter 2 Groups 25 1) An y t w o righ t cosets ha v e the same n um b er of elemen ts. That is, if a; b 2 G f : H a H b dened b y f ( h a ) = h b is a bijection. Also an y t w o left cosets ha v e the same n um b er of elemen ts. Since H is a righ t and left coset, an y t w o cosets ha v e the same n um b er of elemen ts. 2) G has the same n um b er of righ t cosets as left cosets. The function F dened b y F ( H a ) = a 1 H is a bijection from the collection of righ t cosets to the left cosets. The n um b er of righ t (or left) cosets is called the index of H in G 3) If G is nite, o ( H ) (index of H ) = o ( G ) and so o ( H ) j o ( G ) : In other w ords, o ( G ) =o ( H ) = the n um b er of righ t cosets = the n um b er of left cosets. 4) If G is nite, and a 2 G then o ( a ) j o ( G ) : (Pro of: The order of a is the order of the subgroup generated b y a and b y 3) this divides the order of G .) 5) If G has prime order, then G is cyclic, and an y elemen t (except e ) is a generator. (Pro of: Supp ose o ( G ) = p and a 2 G a 6 = e: Then o ( a ) j p and th us o ( a ) = p: ) 6) If o ( G ) = n and a 2 G then a n = e (Pro of: a o ( a ) = e and n = o ( a ) ( o ( G ) =o ( a )) : ) Exercisesi) Supp ose G is a cyclic group of order 4, G = f e; a; a 2 ; a 3 g with a 4 = e Find the order of eac h elemen t of G: Find all the subgroups of G ii) Supp ose G is the additiv e group Z and H = 3 Z : Find the cosets of H iii) Think of a circle as the in terv al [0 ; 1] with end p oin ts iden tied. Supp ose G = R under addition and H = Z : Sho w that the collection of all the cosets of H can b e though t of as a circle. iv) Let G = R 2 under addition, and H b e the subgroup dened b y H = f ( a; 2 a ) : a 2 R g : Find the cosets of H : (See the last exercise on p 5.) Normal Subgroups W e w ould lik e to mak e a group out of the collection of cosets of a subgroup H In PAGE 34 26 Groups Chapter 2 general, there is no natural w a y to do that. Ho w ev er, it is easy to do in case H is a normal subgroup, whic h is describ ed b elo w. Theorem If H is a subgroup of a group G then the follo wing are equiv alen t. 1) If a 2 G then aH a 1 = H 2) If a 2 G then aH a 1 H 3) If a 2 G then aH = H a 4) Ev ery righ t coset is a left coset, i.e., if a 2 G 9 b 2 G with H a = bH Pro of 1) ) 2) is ob vious. Supp ose 2) is true and sho w 3). W e ha v e ( aH a 1 ) a H a so aH H a Also a ( a 1 H a ) aH so H a aH : Th us aH = H a 3) ) 4) is ob vious. Supp ose 4) is true and sho w 3). H a = bH con tains a so bH = aH b ecause a coset is an equiv alence class. Th us aH = H a Finally supp ose 3) is true and sho w 1). Multiply aH = H a on the righ t b y a 1 Denition If H satises an y of the four conditions ab o v e, then H is said to b e a normal subgroup of G: (This concept go es bac k to Ev ariste Galois in 1831.) Note F or an y group G G and e are normal subgroups. If G is an ab elian group, then ev ery subgroup of G is normal. Exercise Sho w that if H is a subgroup of G with index 2, then H is normal. Exercise Sho w the in tersection of a collection of normal subgroups of G is a normal subgroup of G: Sho w the union of a monotonic collection of normal subgroups of G is a normal subgroup of G Exercise Let A R 2 b e the square with v ertices ( 1 ; 1) ; (1 ; 1) ; (1 ; 1), and ( 1 ; 1), and G b e the collection of all \isometries" of A on to itself. These are bijections of A on to itself whic h preserv e distance and angles, i.e., whic h preserv e dot pro duct. Sho w that with m ultiplication dened as comp osition, G is a m ultiplicativ e group. Sho w that G has four rotations, t w o rerections ab out the axes, and t w o rerections ab out the diagonals, for a total of eigh t elemen ts. Sho w the collection of rotations is a cyclic subgroup of order four whic h is a normal subgroup of G Sho w that the rerection ab out the x axis together with the iden tit y form a cyclic subgroup of order t w o whic h is not a normal subgroup of G Find the four righ t cosets of this subgroup. Finally nd the four left cosets of this subgroup. PAGE 35 Chapter 2 Groups 27 Quotien t Groups Supp ose N is a normal subgroup of G and C and D are cosets. W e wish to dene a coset E whic h is the pro duct of C and D If c 2 C and d 2 D dene E to b e the coset con taining c d i.e., E = N ( c d ). The coset E do es not dep end up on the c hoice of c and d This is made precise in the next theorem, whic h is quite easy Theorem Supp ose G is a m ultiplicativ e group, N is a normal subgroup, and G= N is the collection of all cosets. Then ( N a ) ( N b ) = N ( a b ) is a w ell dened m ultiplication (binary op eration) on G= N and with this m ultiplication, G= N is a group. Its iden tit y is N and ( N a ) 1 = ( N a 1 ). F urthermore, if G is nite, o ( G= N ) = o ( G ) =o ( N ). Pro of Multiplication of elemen ts in G= N is m ultiplication of subsets in G ( N a ) ( N b ) = N ( aN ) b = N ( N a ) b = N ( a b ) : Once m ultiplication is w ell dened, the group axioms are immediate. Exercise W rite out the ab o v e theorem for G an additiv e group. In the additiv e ab elian group R = Z ; determine those elemen ts of nite order. Example Supp ose G = Z under +, n > 1, and N = n Z Z n the gr oup of inte gers mo d n is dened b y Z n = Z =n Z : If a is an in teger, the coset a + n Z is denoted b y [ a ]. Note that [ a ] + [ b ] = [ a + b ], [ a ] = [ a ], and [ a ] = [ a + nl ] for an y in teger l An y additiv e ab elian group has a scalar m ultiplication o v er Z and in this case it is just [ a ] m = [ am ]. Note that [ a ] = [ r ] where r is the remainder of a divided b y n and th us the distinct elemen ts of Z n are [0] ; [1] ; :::; [ n 1]. Also Z n is cyclic b ecause eac h of [1] and [ 1] = [ n 1] is a generator. W e already kno w that if p is a prime, an y nonzero elemen t of Z p is a generator, b ecause Z p has p elemen ts. Theorem If n > 1 and a is an y in teger, then [ a ] is a generator of Z n i ( a; n ) = 1. Pro of The elemen t [ a ] is a generator i the subgroup generated b y [ a ] con tains [1] i 9 an in teger k suc h that [ a ] k = [1] i 9 in tegers k and l suc h that ak + nl = 1. Exercise Sho w that a p ositiv e in teger is divisible b y 3 i the sum of its digits is divisible b y 3. Note that [10] = [1] in Z 3 : (See the fth exercise on page 18.) Homomorphisms Homomorphisms are functions b et w een groups that comm ute with the group operations. It follo ws that they honor iden tities and in v erses. In this section w e list PAGE 36 28 Groups Chapter 2 the basic prop erties. Prop erties 11), 12), and 13) sho w the connections b et w een coset groups and homomorphisms, and should b e considered as the cornerstones of abstract algebra. As alw a ys, the studen t should rewrite the material in additiv e notation. Denition If G and G are m ultiplicativ e groups, a function f : G G is a homomorphism if, for all a; b 2 G f ( a b ) = f ( a ) f ( b ). On the left side, the group op eration is in G while on the righ t side it is in G The kernel of f is dened b y k er ( f ) = f 1 ( e ) = f a 2 G : f ( a ) = e g In other w ords, the k ernel is the set of solutions to the equation f ( x ) = e (If G is an additiv e group, k er ( f ) = f 1 (0 ) : ) Examples The constan t map f : G G dened b y f ( a ) = e is a homomorphism. If H is a subgroup of G the inclusion i : H G is a homomorphism. The function f : Z Z dened b y f ( t ) = 2 t is a homomorphism of additiv e groups, while the function dened b y f ( t ) = t + 2 is not a homomorphism. The function h : Z R 0 dened b y h ( t ) = 2 t is a homomorphism from an additiv e group to a m ultiplicativ e group. W e no w catalog the basic prop erties of homomorphisms. These will b e helpful later on in the study of ring homomorphisms and mo dule homomorphisms. Theorem Supp ose G and G are groups and f : G G is a homomorphism. 1) f ( e ) = e 2) f ( a 1 ) = f ( a ) 1 : The rst in v erse is in G and the second is in G 3) f is injectiv e k er( f ) = e 4) If H is a subgroup of G f ( H ) is a subgroup of G: In particular, image( f ) is a subgroup of G 5) If H is a subgroup of G f 1 ( H ) is a subgroup of G: F urthermore, if H is normal in G then f 1 ( H ) is normal in G 6) The k ernel of f is a normal subgroup of G 7) If g 2 G f 1 ( g ) is v oid or is a coset of k er( f ), i.e., if f ( g ) = g then f 1 ( g ) = N g where N = k er( f ). In other w ords, if the equation f ( x ) = g has a PAGE 37 Chapter 2 Groups 29 solution, then the set of all solutions is a coset of N = k er( f ). This is a k ey fact whic h is used routinely in topics suc h as systems of equations and linear dieren tial equations. 8) The comp osition of homomorphisms is a homomorphism, i.e., if h : G = G is a homomorphism, then h f : G = G is a homomorphism. 9) If f : G G is a bijection, then the function f 1 : G G is a homomorphism. In this case, f is called an isomorphism and w e write G G: In the case G = G f is also called an automorphism 10) Isomorphisms preserv e all algebraic prop erties. F or example, if f is an isomorphism and H G is a subset, then H is a subgroup of G i f ( H ) is a subgroup of G H is normal in G i f ( H ) is normal in G G is cyclic i G is cyclic, etc. Of course, this is somewhat of a copout, b ecause an algebraic prop ert y is one that, b y denition, is preserv ed under isomorphisms. 11) Supp ose H is a normal subgroup of G Then : G G=H dened b y ( a ) = H a is a surjectiv e homomorphism with k ernel H : F urthermore, if f : G G is a surjectiv e homomorphism with k ernel H then G=H G (see b elo w). 12) Supp ose H is a normal subgroup of G If H k er ( f ), then f : G=H G dened b y f ( H a ) = f ( a ) is a w elldened homomorphism making the follo wing diagram comm ute. G G G=H f ? > f Th us dening a homomorphism on a quotien t group is the same as dening a homomorphism on the n umerator whic h sends the denominator to e The image of f is the image of f and the k ernel of f is k er( f ) =H Th us if H = k er ( f ), f is injectiv e, and th us G=H image( f ). 13) Giv en an y group homomorphism f domain( f ) = k er ( f ) image( f ) : This is the fundamen tal connection b et w een quotien t groups and homomorphisms. PAGE 38 30 Groups Chapter 2 14) Supp ose K is a group. Then K is an innite cycle group i K is isomorphic to the in tegers under addition, i.e., K Z K is a cyclic group of order n i K Z n Pro of of 14) Supp ose G = K is generated b y some elemen t a Then f : Z K dened b y f ( m ) = a m is a homomorphism from an additiv e group to a m ultiplicativ e group. If o ( a ) is innite, f is an isomorphism. If o ( a ) = n k er( f ) = n Z and f : Z n K is an isomorphism. Exercise If a is an elemen t of a group G there is alw a ys a homomorphism from Z to G whic h sends 1 to a When is there a homomorphism from Z n to G whic h sends [1] to a ? What are the homomorphisms from Z 2 to Z 6 ? What are the homomorphisms from Z 4 to Z 8 ? Exercise Supp ose G is a group and g is an elemen t of G g 6 = e 1) Under what conditions on g is there a homomorphism f : Z 7 G with f ([1]) = g ? 2) Under what conditions on g is there a homomorphism f : Z 15 G with f ([1]) = g ? 3) Under what conditions on G is there an injectiv e homomorphism f : Z 15 G ? 4) Under what conditions on G is there a surjectiv e homomorphism f : Z 15 G ? Exercise W e kno w ev ery nite group of prime order is cyclic and th us ab elian. Sho w that ev ery group of order four is ab elian. Exercise Let G = f h : [0 ; 1] R : h has an innite n um b er of deriv ativ es g Then G is a group under addition. Dene f : G G b y f ( h ) = dh dt = h 0 Sho w f is a homomorphism and nd its k ernel and image. Let g : [0 ; 1] R b e dened b y g ( t ) = t 3 3 t + 4 : Find f 1 ( g ) and sho w it is a coset of k er ( f ). Exercise Let G b e as ab o v e and g 2 G Dene f : G G b y f ( h ) = h 00 + 5 h 0 + 6 t 2 h Then f is a group homomorphism and the dieren tial equation h 00 + 5 h 0 + 6 t 2 h = g has a solution i g lies in the image of f No w supp ose this equation has a solution and S G is the set of all solutions. F or whic h subgroup H of G is S an H coset? PAGE 39 Chapter 2 Groups 31 Exercise Supp ose G is a m ultiplicativ e group and a 2 G Dene f : G G to b e conjugation b y a i.e., f ( g ) = a 1 g a: Sho w that f is a homomorphism. Also sho w f is an automorphism and nd its in v erse. P erm utations Supp ose X is a (nonv oid) set. A bijection f : X X is called a p ermutation on X and the collection of all these p erm utations is denoted b y S = S ( X ). In this setting, v ariables are written on the left, i.e., f = ( x ) f Therefore the comp osition f g means \ f follo w ed b y g ". S ( X ) forms a m ultiplicativ e group under comp osition. Exercise Sho w that if there is a bijection b et w een X and Y there is an isomorphism b et w een S ( X ) and S ( Y ). Th us if eac h of X and Y has n elemen ts, S ( X ) S ( Y ), and these groups are called the symmetric groups on n elemen ts. They are all denoted b y the one sym b ol S n Exercise Sho w that o ( S n ) = n !. Let X = f 1 ; 2 ; :::; n g ; S n = S ( X ), and H = f f 2 S n : ( n ) f = n g Sho w H is a subgroup of S n whic h is isomorphic to S n 1 Let g b e an y p erm utation on X with ( n ) g = 1 : Find g 1 H g The next theorem sho ws that the symmetric groups are incredibly ric h and complex.Theorem (Ca yley's Theorem) Supp ose G is a m ultiplicativ e group with n elemen ts and S n is the group of all p erm utations on the set G Then G is isomorphic to a subgroup of S n Pro of Let h : G S n b e the function whic h sends a to the bijection h a : G G dened b y ( g ) h a = g a: The pro of follo ws from the follo wing observ ations. 1) F or eac h giv en a h a is a bijection from G to G 2) h is a homomorphism, i.e., h a b = h a h b 3) h is injectiv e and th us G is isomorphic to image( h ) S n The Symmetric Groups No w let n 2 and let S n b e the group of all p erm utations on f 1 ; 2 ; :::; n g The follo wing denition sho ws that eac h elemen t of S n ma y PAGE 40 32 Groups Chapter 2 b e represen ted b y a matrix. Denition Supp ose 1 < k n f a 1 ; a 2 ; :::; a k g is a collection of distinct in tegers with 1 a i n and f b 1 ; b 2 ; :::; b k g is the same collection in some dieren t order. Then the matrix a 1 a 2 ::: a k b 1 b 2 ::: b k represen ts f 2 S n dened b y ( a i ) f = b i for 1 i k and ( a ) f = a for all other a The comp osition of t w o p erm utations is computed b y applying the matrix on the left rst and the matrix on the righ t second. There is a sp ecial t yp e of p erm utation called a cycle F or these w e ha v e a sp ecial notation.Denition a 1 a 2 :::a k 1 a k a 2 a 3 :::a k a 1 is called a k cycle, and is denoted b y ( a 1 ; a 2 ; :::; a k ). A 2cycle is called a tr ansp osition The cycles ( a 1 ; :::; a k ) and ( c 1 ; :::; c ` ) are disjoint pro vided a i 6 = c j for all 1 i k and 1 j ` Listed here are eigh t basic prop erties of p erm utations. They are all easy except 4), whic h tak es a little w ork. Prop erties 9) and 10) are listed solely for reference. Theorem1) Disjoin t cycles comm ute. (This is ob vious.) 2) Ev ery noniden tit y p erm utation can b e written uniquely (except for order) as the pro duct of disjoin t cycles. (This is easy .) 3) Ev ery p erm utation can b e written (nonuniquely) as the pro duct of transp ositions. (Pro of: I = (1 ; 2)(1 ; 2) and ( a 1 ; :::; a k ) = ( a 1 ; a 2 )( a 1 ; a 3 ) ( a 1 ; a k ). ) 4) The parit y of the n um b er of these transp ositions is unique. This means that if f is the pro duct of p transp ositions and also of q transp ositions, then p is ev en i q is ev en. In this case, f is said to b e an even p erm utation. In the other case, f is an o dd p erm utation. 5) A k cycle is ev en (o dd) i k is o dd (ev en). F or example (1 ; 2 ; 3) = (1 ; 2)(1 ; 3) is an ev en p erm utation. 6) Supp ose f ; g 2 S n If one of f and g is ev en and the other is o dd, then g f is PAGE 41 Chapter 2 Groups 33 o dd. If f and g are b oth ev en or b oth o dd, then g f is ev en. (Ob vious.) 7) The map h : S n Z 2 dened b y h (ev en)= [0] and h (o dd)= [1] is a homomorphism from a m ultiplicativ e group to an additiv e group. Its k ernel (the subgroup of ev en p erm utations) is denoted b y A n and is called the alternating group. Th us A n is a normal subgroup of index 2, and S n = A n Z 2 8) If a; b; c and d are distinct in tegers in f 1 ; 2 ; : : : ; n g then ( a; b )( b; c ) = ( a; c; b ) and ( a; b )( c; d ) = ( a; c; d )( a; c; b ) : Since I = (1 ; 2 ; 3) 3 ; it follo ws that for n 3 ; ev ery ev en p erm utation is the pro duct of 3cycles. The follo wing parts are not included in this course. They are presen ted here merely for reference. 9) F or an y n 6 = 4, A n is simple, i.e., has no prop er normal subgroups. 10) S n can b e generated b y t w o elemen ts. In fact, f (1 ; 2) ; (1 ; 2 ; :::; n ) g generates S n (Of course there are subgroups of S n whic h cannot b e generated b y t w o elemen ts). Pro of of 4) It suces to pro v e if the pro duct of t transp ositions is the iden tit y I on f 1 ; 2 ; : : : ; n g then t is ev en. Supp ose this is false and I is written as t transp ositions, where t is the smallest o dd in teger this is p ossible. Since t is o dd, it is at least 3. Supp ose for con v enience the rst transp osition is ( a; n ). W e will rewrite I as a pro duct of transp ositions 1 2 t where ( n ) i = ( n ) for 1 i < t and ( n ) t 6 = n whic h will b e a con tradiction. This can b e done b y inductiv ely \pushing n to the righ t" using the equations b elo w. If a; b and c are distinct in tegers in f 1 ; 2 ; : : : ; n 1 g then ( a; n )( a; n ) = I ( a; n )( b; n ) = ( a; b )( a; n ), ( a; n )( a; c ) = ( a; c )( c; n ), and ( a; n )( b; c ) = ( b; c )( a; n ). Note that ( a; n )( a; n ) cannot o ccur here b ecause it w ould result in a shorter o dd pro duct. (No w y ou ma y solv e the tile puzzle on page viii.) Exercise1) W rite 1 2 3 4 5 6 7 6 5 4 3 1 7 2 as the pro duct of disjoin t cycles. W rite (1,5,6,7)(2,3,4)(3,7,1) as the pro duct of disjoin t cycles. W rite (3,7,1)(1,5,6,7)(2,3,4) as the pro duct of disjoin t cycles. Whic h of these p erm utations are o dd and whic h are ev en? PAGE 42 34 Groups Chapter 2 2) Supp ose ( a 1 ; : : : ; a k ) and ( c 1 ; : : : ; c ` ) are disjoin t cycles. What is the order of their pro duct? 3) Supp ose 2 S n : Sho w that 1 (1 ; 2 ; 3) = ((1) ; (2) ; (3) ) : This sho ws that conjugation b y is just a t yp e of relab eling. Also let = (4 ; 5 ; 6) and nd 1 (1 ; 2 ; 3 ; 4 ; 5) 4) Sho w that H = f 2 S 6 : (6) = 6 g is a subgroup of S 6 and nd its righ t cosets and its left cosets. 5) Let A R 2 b e the square with v ertices ( 1 ; 1) ; (1 ; 1), (1 ; 1), and ( 1 ; 1), and G b e the collection of all isometries of A on to itself. W e kno w from a previous exercise that G is a group with eigh t elemen ts. It follo ws from Ca yley's theorem that G is isomorphic to a subgroup of S 8 Sho w that G is isomorphic to a subgroup of S 4 6) If G is a m ultiplicativ e group, dene a new m ultiplication on the set G b y a b = b a In other w ords, the new m ultiplication is the old m ultiplication in the opp osite order. This denes a new group denoted b y G op the opp osite group. Sho w that it has the same iden tit y and the same in v erses as G and that f : G G op dened b y f ( a ) = a 1 is a group isomorphism. No w consider the sp ecial case G = S n The con v en tion used in this section is that an elemen t of S n is a p erm utation on f 1 ; 2 ; : : : ; n g with the v ariable written on the left. Sho w that an elemen t of S op n is a p erm utation on f 1 ; 2 ; : : : ; n g with the v ariable written on the righ t. (Of course, either S n or S op n ma y b e called the symmetric group, dep ending on p ersonal preference or con text.) Pro duct of Groups The pro duct of groups is usually presen ted for m ultiplicativ e groups. It is presen ted here for additiv e groups b ecause this is the form that o ccurs in later c hapters. As an exercise, this section should b e rewritten using m ultiplicativ e notation. The t w o theorems b elo w are transparen t and easy but quite useful. F or simplicit y w e rst consider the pro duct of t w o groups, although the case of innite pro ducts is only sligh tly more dicult. F or bac kground, read rst the t w o theorems on page 11. Theorem Supp ose G 1 and G 2 are additiv e groups. Dene an addition on G 1 G 2 b y ( a 1 ; a 2 ) + ( b 1 ; b 2 ) = ( a 1 + b 1 ; a 2 + b 2 ). This op eration mak es G 1 G 2 in to a group. Its \zero" is (0 1 ; 0 2 ) and ( a 1 ; a 2 ) = ( a 1 ; a 2 ). The pro jections 1 : G 1 G 2 G 1 PAGE 43 Chapter 2 Groups 35 and 2 : G 1 G 2 G 2 are group homomorphisms. Supp ose G is an additiv e group. W e kno w there is a bijection from f functions f : G G 1 G 2 g to f ordered pairs of functions ( f 1 ; f 2 ) where f 1 : G G 1 and f 2 : G G 2 g Under this bijection, f is a group homomorphism i eac h of f 1 and f 2 is a group homomorphism. Pro of It is transparen t that the pro duct of groups is a group, so let's pro v e the last part. Supp ose G; G 1 ; and G 2 are groups and f = ( f 1 ; f 2 ) is a function from G to G 1 G 2 No w f ( a + b ) = ( f 1 ( a + b ) ; f 2 ( a + b )) and f ( a ) + f ( b ) = ( f 1 ( a ) ; f 2 ( a )) + ( f 1 ( b ) ; f 2 ( b )) = ( f 1 ( a ) + f 1 ( b ) ; f 2 ( a ) + f 2 ( b )). An examination of these t w o equations sho ws that f is a group homomorphism i eac h of f 1 and f 2 is a group homomorphism.Exercise Supp ose G 1 and G 2 are groups. Sho w that G 1 G 2 and G 2 G 1 are isomorphic.Exercise If o ( a 1 ) = m and o ( a 2 ) = n; nd the order of ( a 1 ; a 2 ) in G 1 G 2 Exercise Sho w that if G is an y group of order 4, G is isomorphic to Z 4 or Z 2 Z 2 Sho w Z 4 is not isomorphic to Z 2 Z 2 : Sho w Z 12 is isomorphic to Z 4 Z 3 : Finally sho w that Z mn is isomorphic to Z m Z n i ( m; n ) = 1. Exercise Supp ose G 1 and G 2 are groups and i 1 : G 1 G 1 G 2 is dened b y i 1 ( g 1 ) = ( g 1 ; 0 2 ). Sho w i 1 is an injectiv e group homomorphism and its image is a normal subgroup of G 1 G 2 Usually G 1 is iden tied with its image under i 1 so G 1 ma y b e considered to b e a normal subgroup of G 1 G 2 Let 2 : G 1 G 2 G 2 b e the pro jection map dened in the Bac kground c hapter. Sho w 2 is a surjectiv e homomorphism with k ernel G 1 Therefore ( G 1 G 2 ) =G 1 G 2 as y ou w ould exp ect. Exercise Let R b e the reals under addition. Sho w that the addition in the pro duct R R is just the usual addition in analytic geometry Exercise Supp ose n > 2. Is S n isomorphic to A n G where G is a m ultiplicativ e group of order 2 ? One nice thing ab out the pro duct of groups is that it w orks ne for an y nite n um b er, or ev en an y innite n um b er. The next theorem is stated in full generalit y PAGE 44 36 Groups Chapter 2 Theorem Supp ose T is an index set, and for an y t 2 T G t is an additiv e group. Dene an addition on Y t 2 T G t = Q G t b y f a t g + f b t g = f a t + b t g This operation mak es the pro duct in to a group. Its \zero" is f 0 t g and f a t g = f a t g Eac h pro jection s : Q G t G s is a group homomorphism. Supp ose G is an additiv e group. Under the natural bijection from f functions f : G Q G t g to f sequences of functions f f t g t 2 T where f t : G G t g f is a group homomorphism i eac h f t is a group homomorphism. Finally the scalar m ultiplication on Q G t b y in tegers is giv en co ordinatewise, i.e., f a t g n = f a t n g Pro of The addition on Q G t is co ordinatewise. Exercise Supp ose s is an elemen t of T and s : Q G t G s is the pro jection map dened in the Bac kground c hapter. Sho w s is a surjectiv e homomorphism and nd its k ernel. Exercise Supp ose s is an elemen t of T and i s : G s Q G t is dened b y i s ( a ) = f a t g where a t = 0 if t 6 = s and a s = a: Sho w i s is an injectiv e homomorphism and its image is a normal subgroup of Q G t Th us eac h G s ma y b e considered to b e a normal subgroup of Q G t Exercise Let f : Z Z 30 Z 100 b e the homomorphism dened b y f ( m ) = ([4 m ] ; [3 m ]) : Find the k ernel of f : Find the order of ([4] ; [3]) in Z 30 Z 100 Exercise Let f : Z Z 90 Z 70 Z 42 b e the group homomorphism dened b y f ( m ) = ([ m ] ; [ m ] ; [ m ]). Find the k ernel of f and sho w that f is not surjectiv e. Let g : Z Z 45 Z 35 Z 21 b e dened b y g ( m ) = ([ m ] ; [ m ] ; [ m ]). Find the k ernel of g and determine if g is surjectiv e. Note that the gcd of f 45 ; 35 ; 21 g is 1. No w let h : Z Z 8 Z 9 Z 35 b e dened b y h ( m ) = ([ m ] ; [ m ] ; [ m ]). Find the k ernel of h and sho w that h is surjectiv e. Finally supp ose eac h of b; c; and d is greater than 1 and f : Z Z b Z c Z d is dened b y f ( m ) = ([ m ] ; [ m ] ; [ m ]). Find necessary and sucien t conditions for f to b e surjectiv e (see the rst exercise on page 18). Exercise Supp ose T is a nonv oid set, G is an additiv e group, and G T is the collection of all functions f : T G with addition dened b y ( f + g )( t ) = f ( t ) + g ( t ). Sho w G T is a group. F or eac h t 2 T let G t = G Note that G T is just another w a y of writing Y t 2 T G t : Also note that if T = [0 ; 1] and G = R the addition dened on G T is just the usual addition of functions used in calculus. (F or the ring and mo dule v ersions, see exercises on pages 44 and 69.) PAGE 45 Chapter 3 RingsRings are additiv e ab elian groups with a second op eration called m ultiplication. The connection b et w een the t w o op erations is pro vided b y the distributiv e la w. Assuming the results of Chapter 2, this c hapter ro ws smo othly This is b ecause ideals are also normal subgroups and ring homomorphisms are also group homomorphisms. W e do not sho w that the p olynomial ring F [ x ] is a unique factorization domain, although with the material at hand, it w ould b e easy to do. Also there is no men tion of prime or maximal ideals, b ecause these concepts are unnecessary for our dev elopmen t of linear algebra. These concepts are dev elop ed in the App endix. A section on Bo olean rings is included b ecause of their imp ortance in logic and computer science. Supp ose R is an additiv e ab elian group, R 6 = 0 and R has a second binary op eration (i.e., map from R R to R ) whic h is denoted b y m ultiplication. Consider the follo wing prop erties. 1) If a; b; c 2 R ( a b ) c = a ( b c ). (The asso ciativ e prop ert y of m ultiplication.) 2) If a; b; c 2 R a ( b + c ) = ( a b ) + ( a c ) and ( b + c ) a = ( b a ) + ( c a ). (The distributiv e la w, whic h connects addition and m ultiplication.) 3) R has a m ultiplicativ e iden tit y i.e., there is an elemen t 1 = 1 R 2 R suc h that if a 2 R a 1 = 1 a = a 4) If a; b 2 R a b = b a (The comm utativ e prop ert y for m ultiplication.) Denition If 1), 2), and 3) are satised, R is said to b e a ring If in addition 4) is satised, R is said to b e a c ommutative ring Examples The basic comm utativ e rings in mathematics are the in tegers Z the 37 PAGE 46 38 Rings Chapter 3 rational n um b ers Q the real n um b ers R and the complex n um b ers C It will b e sho wn later that Z n the in tegers mo d n has a natural m ultiplication under whic h it is a comm utativ e ring. Also if R is an y comm utativ e ring, w e will dene R [ x 1 ; x 2 ; : : : ; x n ], a p olynomical ring in n v ariables. No w supp ose R is an y ring, n 1, and R n is the collection of all n n matrices o v er R In the next c hapter, op erations of addition and m ultiplication of matrices will b e dened. Under these op erations, R n is a ring. This is a basic example of a noncomm utativ e ring. If n > 1, R n is nev er comm utativ e, ev en if R is comm utativ e. The next t w o theorems sho w that ring m ultiplication b eha v es as y ou w ould wish it to. They should b e w ork ed as exercises. Theorem Supp ose R is a ring and a; b 2 R 1) a 0 = 0 a = 0 : Since R 6 = 0 ; it follo ws that 1 6 = 0 2) ( a ) b = a ( b ) = ( a b ). Recall that, since R is an additiv e ab elian group, it has a scalar m ultiplication o v er Z (page 20). This scalar m ultiplication can b e written on the righ t or left, i.e., na = an and the next theorem sho ws it relates nicely to the ring m ultiplication. Theorem Supp ose a; b 2 R and n; m 2 Z 1) ( na ) ( mb ) = ( nm )( a b ). (This follo ws from the distributiv e la w and the previous theorem.) 2) Let n = n 1 F or example, 2 = 1 + 1 Then na = n a that is, scalar m ultiplication b y n is the same as ring m ultiplication b y n Of course, n ma y b e 0 ev en though n 6 = 0. Units Denition An elemen t a of a ring R is a unit pro vided 9 an elemen t a 1 2 R with a a 1 = a 1 a = 1 Theorem 0 can nev er b e a unit. 1 is alw a ys a unit. If a is a unit, a 1 is also a unit with ( a 1 ) 1 = a The pro duct of units is a unit with ( a b ) 1 = b 1 a 1 More PAGE 47 Chapter 3 Rings 39 generally if a 1 ; a 2 ; :::; a n are units, then their pro duct is a unit with ( a 1 a 2 a n ) 1 = a 1 n a 1 n 1 a 1 1 The set of all units of R forms a m ultiplicativ e group denoted b y R : Finally if a is a unit, ( a ) is a unit and ( a ) 1 = ( a 1 ). In order for a to b e a unit, it m ust ha v e a t w osided in v erse. It suces to require a left in v erse and a righ t in v erse, as sho wn in the next theorem. Theorem Supp ose a 2 R and 9 elemen ts b and c with b a = a c = 1 Then b = c and so a is a unit with a 1 = b = c Pro of b = b 1 = b ( a c ) = ( b a ) c = 1 c = c Corollary In v erses are unique. Domains and Fields In order to dene these t w o t yp es of rings, w e rst consider the concept of zero divisor. Denition Supp ose R is a comm utativ e ring. An elemen t a 2 R is called a zer o divisor pro vided it is nonzero and 9 a nonzero elemen t b with a b = 0 Note that if a is a unit, it cannot b e a zero divisor. Theorem Supp ose R is a comm utativ e ring and a 2 ( R 0 ) is not a zero divisor. Then ( a b = a c ) ) b = c: In other w ords, m ultiplication b y a is an injectiv e map from R to R : It is surjectiv e i a is a unit. Denition A domain (or inte gr al domain ) is a comm utativ e ring suc h that, if a 6 = 0 a is not a zero divisor. A eld is a comm utativ e ring suc h that, if a 6 = 0 a is a unit. In other w ords, R is a eld if it is comm utativ e and its nonzero elemen ts form a group under m ultiplication. Theorem A eld is a domain. A nite domain is a eld. Pro of A eld is a domain b ecause a unit cannot b e a zero divisor. Supp ose R is a nite domain and a 6 = 0 Then f : R R dened b y f ( b ) = a b is injectiv e and, b y the pigeonhole principle, f is surjectiv e. Th us a is a unit and so R is a eld. PAGE 48 40 Rings Chapter 3 Exercise Let C b e the additiv e ab elian group R 2 Dene m ultiplication b y ( a; b ) ( c; d ) = ( ac bd; ad + bc ). Sho w C is a comm utativ e ring whic h is a eld. Note that 1 = (1 ; 0) and if i = (0 ; 1), then i 2 = 1 Examples Z is a domain. Q R and C are elds. The In tegers Mo d n The concept of in tegers mo d n is fundamen tal in mathematics. It leads to a neat little theory as seen b y the theorems b elo w. Ho w ev er, the basic theory cannot b e completed un til the pro duct of rings is dened. (See the Chinese Remainder Theorem on page 50.) W e kno w from page 27 that Z n is an additiv e ab elian group. Theorem Supp ose n > 1. Dene a m ultiplication on Z n b y [ a ] [ b ] = [ ab ]. This is a w ell dened binary op eration whic h mak es Z n in to a comm utativ e ring. Pro of Since [ a + k n ] [ b + l n ] = [ ab + n ( a l + bk + k l n )] = [ ab ], the m ultiplication is w elldened. The ring axioms are easily v eried. Theorem Supp ose n > 1 and a 2 Z Then the follo wing are equiv alen t. 1) [ a ] is a generator of the additiv e group Z n 2) ( a; n ) = 1. 3) [ a ] is a unit of the ring Z n Pro of W e already kno w from page 27 that 1) and 2) are equiv alen t. Recall that if b is an in teger, [ a ] b = [ a ] [ b ] = [ ab ]. Th us 1) and 3) are equiv alen t, b ecause eac h sa ys 9 an in teger b with [ a ] b = [1]. Corollary If n > 1, the follo wing are equiv alen t. 1) Z n is a domain. 2) Z n is a eld. 3) n is a prime. Pro of W e already kno w 1) and 2) are equiv alen t, b ecause Z n is nite. Supp ose 3) is true. Then b y the previous theorem, eac h of [1], [2],...,[ n 1] is a unit, and th us 2) is true. No w supp ose 3) is false. Then n = ab where 1 < a < n 1 < b < n PAGE 49 Chapter 3 Rings 41 [ a ][ b ] = [0] ; and th us [ a ] is a zero divisor and 1) is false. Exercise List the units and their in v erses for Z 7 and Z 12 Sho w that ( Z 7 ) is a cyclic group but ( Z 12 ) is not. Sho w that in Z 12 the equation x 2 = 1 has four solutions. Finally sho w that if R is a domain, x 2 = 1 can ha v e at most t w o solutions in R (see the rst theorem on page 46). Subrings Supp ose S is a subset of a ring R The statemen t that S is a subring of R means that S is a subgroup of the group R 1 2 S and ( a; b 2 S ) a b 2 S ). Then clearly S is a ring and has the same m ultiplicativ e iden tit y as R Note that Z is a subring of Q Q is a subring of R and R is a subring of C Subrings do not pla y a role analogous to subgroups. That role is pla y ed b y ideals, and an ideal is nev er a subring (unless it is the en tire ring). Note that if S is a subring of R and s 2 S then s ma y b e a unit in R but not in S Note also that Z and Z n ha v e no prop er subrings, and th us o ccup y a sp ecial place in ring theory as w ell as in group theory Ideals and Quotien t Rings Ideals in ring theory pla y a role analagous to normal subgroups in group theory Denition A subset I of a ring R is a 8><>: leftrigh t 2 sided 9>=>; ideal pro vided it is a subgroup of the additiv e group R and if a 2 R and b 2 I then 8><>: a b 2 I b a 2 I a b and b a 2 I 9>=>; The w ord \ideal means \2sided ideal". Of course, if R is comm utativ e, ev ery righ t or left ideal is an ideal. Theorem Supp ose R is a ring. 1) R and 0 are ideals of R : These are called the impr op er ideals. 2) If f I t g t 2 T is a collection of righ t (left, 2sided) ideals of R then \ t 2 T I t is a righ t (left, 2sided) ideal of R : (See page 22.) PAGE 50 42 Rings Chapter 3 3) F urthermore, if the collection is monotonic, then [ t 2 T I t is a righ t (left, 2sided) ideal of R 4) If a 2 R ; I = aR is a righ t ideal. Th us if R is comm utativ e, aR is an ideal, called a princip al ide al Th us ev ery subgroup of Z is a principal ideal, b ecause it is of the form n Z 5) If R is a comm utativ e ring and I R is an ideal, then the follo wing are equiv alen t. i) I = R ii) I con tains some unit u iii) I con tains 1 Exercise Supp ose R is a comm utativ e ring. Sho w that R is a eld i R con tains no prop er ideals. The follo wing theorem is just an observ ation, but it is in some sense the b eginning of ring theory Theorem Supp ose R is a ring and I R is an ideal, I 6 = R Since I is a normal subgroup of the additiv e group R R =I is an additiv e ab elian group. Multiplication of cosets dened b y ( a + I ) ( b + I ) = ( ab + I ) is w elldened and mak es R =I a ring. Pro of ( a + I ) ( b + I ) = a b + aI + I b + I I a b + I Th us m ultiplication is w ell dened, and the ring axioms are easily v eried. The m ultiplicativ e iden tit y is (1 + I ). Observ ation If R = Z n > 1, and I = n Z ; the ring structure on Z n = Z =n Z is the same as the one previously dened. Homomorphisms Denition Supp ose R and R are rings. A function f : R R is a ring homomorphism pro vided 1) f is a group homomorphism 2) f (1 R ) = 1 R and 3) if a; b 2 R then f ( a b ) = f ( a ) f ( b ). (On the left, m ultiplication PAGE 51 Chapter 3 Rings 43 is in R while on the righ t m ultiplication is in R .) The kernel of f is the k ernel of f considered as a group homomorphism, namely k er ( f ) = f 1 (0 ). Here is a list of the basic prop erties of ring homomorphisms. Muc h of this w ork has already b een done b y the theorem in group theory on page 28. Theorem Supp ose eac h of R and R is a ring. 1) The iden tit y map I R : R R is a ring homomorphism. 2) The zero map from R to R is not a ring homomorphism (b ecause it do es not send 1 R to 1 R ). 3) The comp osition of ring homomorphisms is a ring homomorphism. 4) If f : R R is a bijection whic h is a ring homomorphism, then f 1 : R R is a ring homomorphism. Suc h an f is called a ring isomorphism In the case R = R f is also called a ring automorphism 5) The image of a ring homomorphism is a subring of the range. 6) The k ernel of a ring homomorphism is an ideal of the domain. In fact, if f : R R is a homomorphism and I R is an ideal, then f 1 ( I ) is an ideal of R 7) Supp ose I is an ideal of R I 6 = R and : R R =I is the natural pro jection, ( a ) = ( a + I ) : Then is a surjectiv e ring homomorphism with k ernel I F urthermore, if f : R R is a surjectiv e ring homomorphism with k ernel I then R =I R (see b elo w). 8) F rom no w on the w ord \homomorphism" means \ring homomorphism". Supp ose f : R R is a homomorphism and I is an ideal of R I 6 = R If I k er ( f ) ; then f : R =I R dened b y f ( a + I ) = f ( a ) PAGE 52 44 Rings Chapter 3 is a w elldened homomorphism making the follo wing diagram comm ute. R R R =I f ? > f Th us dening a homomorphism on a quotien t ring is the same as dening a homomorphism on the n umerator whic h sends the denominator to zero. The image of f is the image of f and the k ernel of f is k er( f ) =I : Th us if I = k er ( f ), f is injectiv e, and so R =I image ( f ). Pro of W e kno w all this on the group lev el, and it is only necessary to c hec k that f is a ring homomorphism, whic h is ob vious. 9) Giv en an y ring homomorphism f ; domain( f ) = k er ( f ) image( f ). Exercise Find a ring R with a prop er ideal I and an elemen t b suc h that b is not a unit in R but ( b + I ) is a unit in R =I Exercise Sho w that if u is a unit in a ring R then conjugation b y u is an automorphism on R : That is, sho w that f : R R dened b y f ( a ) = u 1 a u is a ring homomorphism whic h is an isomorphism. Exercise Supp ose T is a nonv oid set, R is a ring, and R T is the collection of all functions f : T R Dene addition and m ultiplication on R T p oin twise. This means if f and g are functions from T to R then ( f + g )( t ) = f ( t ) + g ( t ) and ( f g )( t ) = f ( t ) g ( t ). Sho w that under these op erations R T is a ring. Supp ose S is a nonv oid set and : S T is a function. If f : T R is a function, dene a function ( f ) : S R b y ( f ) = f : Sho w : R T R S is a ring homomorphism. Exercise No w consider the case T = [0 ; 1] and R = R Let A R [0 ; 1] b e the collection of all C 1 functions, i.e., A = f f : [0 ; 1] R : f has an innite n um b er of deriv ativ es g Sho w A is a ring. Notice that m uc h of the w ork has b een done in the previous exercise. It is only necessary to sho w that A is a subring of the ring R [0 ; 1] PAGE 53 Chapter 3 Rings 45 P olynomial Rings In calculus, w e consider real functions f whic h are p olynomials, f ( x ) = a 0 + a 1 x + + a n x n The sum and pro duct of p olynomials are again p olynomials, and it is easy to see that the collection of p olynomial functions forms a comm utativ e ring. W e can do the same thing formally in a purely algebraic setting. Denition Supp ose R is a comm utativ e ring and x is a \v ariable" or \sym b ol". The p olynomial ring R [ x ] is the collection of all p olynomials f = a 0 + a 1 x + + a n x n where a i 2 R Under the ob vious addition and m ultiplication, R [ x ] is a comm utativ e ring. The de gr e e of a nonzero p olynomial f is the largest in teger n suc h that a n 6 = 0 and is denoted b y n = deg ( f ) : If the top term a n = 1 then f is said to b e monic T o b e more formal, think of a p olynomial a 0 + a 1 x + as an innite sequence ( a 0 ; a 1 ; ::: ) suc h that eac h a i 2 R and only a nite n um b er are nonzero. Then ( a 0 ; a 1 ; ::: ) + ( b 0 ; b 1 ; ::: ) = ( a 0 + b 0 ; a 1 + b 1 ; ::: ) and ( a 0 ; a 1 ; ::: ) ( b 0 ; b 1 ; ::: ) = ( a 0 b 0 ; a 0 b 1 + a 1 b 0 ; a 0 b 2 + a 1 b 1 + a 2 b 0 ; ::: ). Note that on the righ t, the ring m ultiplication a b is written simply as ab as is often done for con v enience. Theorem If R is a domain, R [ x ] is also a domain. Pro of Supp ose f and g are nonzero p olynomials. Then deg ( f )+deg ( g ) = deg ( f g ) and th us f g is not 0 Another w a y to pro v e this theorem is to lo ok at the b ottom terms instead of the top terms. Let a i x i and b j x j b e the rst nonzero terms of f and g : Then a i b j x i + j is the rst nonzero term of f g Theorem (The Division Algorithm) Supp ose R is a comm utativ e ring, f 2 R [ x ] has degree 1 and its top co ecien t is a unit in R (If R is a eld, the top co ecien t of f will alw a ys b e a unit.) Then for an y g 2 R [ x ], 9 h; r 2 R [ x ] suc h that g = f h + r with r = 0 or deg ( r ) < deg ( f ). Pro of This theorem states the existence and uniqueness of p olynomials h and r W e outline the pro of of existence and lea v e uniqueness as an exercise. Supp ose f = a 0 + a 1 x + + a m x m where m 1 and a m is a unit in R F or an y g with deg( g ) < m set h = 0 and r = g F or the general case, the idea is to divide f in to g un til the remainder has degree less than m The pro of is b y induction on the degree of g Supp ose n m and the result holds for an y p olynomial of degree less than PAGE 54 46 Rings Chapter 3 n Supp ose g is a p olynomial of degree n No w 9 a monomial bx t with t = n m and deg( g f bx t ) < n By induction, 9 h 1 and r with f h 1 + r = ( g f bx t ) and deg( r ) < m: The result follo ws from the equation f ( h 1 + bx t ) + r = g Note If r = 0 w e sa y that f divides g : Note that f = x c divides g i c is a r o ot of g ; i.e., g ( c ) = 0 : More generally x c divides g with remainder g ( c ). Theorem Supp ose R is a domain, n > 0, and g ( x ) = a 0 + a 1 x + + a n x n is a p olynomial of degree n with at least one ro ot in R Then g has at most n ro ots. Let c 1 ; c 2 ; ::; c k b e the distinct ro ots of g in the ring R Then 9 a unique sequence of p ositiv e in tegers n 1 ; n 2 ; ::; n k and a unique p olynomial h with no ro ot in R so that g ( x ) = ( x c 1 ) n 1 ( x c k ) n k h ( x ). (If h has degree 0, i.e., if h = a n then w e sa y \all the ro ots of g b elong to R ". If g = a n x n w e sa y \all the ro ots of g are 0 ".) Pro of Uniqueness is easy so let's pro v e existence. The theorem is clearly true for n = 1. Supp ose n > 1 and the theorem is true for an y p olynomial of degree less than n No w supp ose g is a p olynomial of degree n and c 1 is a ro ot of g Then 9 a p olynomial h 1 with g ( x ) = ( x c 1 ) h 1 Since h 1 has degree less than n the result follo ws b y induction. Note If g is an y nonconstan t p olynomial in C [ x ], all the ro ots of g b elong to C i.e., C is an algebr aic al ly close d eld This is called The F undamen tal Theorem of Algebra, and it is assumed without pro of for this textb o ok. Exercise Supp ose g is a nonconstan t p olynomial in R [ x ]. Sho w that if g has o dd degree then it has a real ro ot. Also sho w that if g ( x ) = x 2 + bx + c then it has a real ro ot i b 2 4 c; and in that case b oth ro ots b elong to R Denition A domain T is a princip al ide al domain (PID) if, giv en an y ideal I 9 t 2 T suc h that I = tT : Note that Z is a PID and an y eld is PID. Theorem Supp ose F is a eld, I is a prop er ideal of F [ x ], and n is the smallest p ositiv e in teger suc h that I con tains a p olynomial of degree n Then I con tains a unique p olynomial of the form f = a 0 + a 1 x + + a n 1 x n 1 + x n and it has the prop ert y that I = f F [ x ] : Th us F [ x ] is a PID. F urthermore, eac h coset of I can b e written uniquely in the form ( c 0 + c 1 x + + c n 1 x n 1 + I ). Pro of This is a go o d exercise in the use of the division algorithm. Note this is similar to sho wing that a subgroup of Z is generated b y one elemen t (see page 15). PAGE 55 Chapter 3 Rings 47 Theorem Supp ose R is a subring of a comm utativ e ring C and c 2 C Then 9 homomorphism h : R [ x ] C with h ( x ) = c and h ( r ) = r for all r 2 R It is dened b y h ( a 0 + a 1 x + + a n x n ) = a 0 + a 1 c + + a n c n i.e., h sends f ( x ) to f ( c ). The image of h is the smallest subring of C con taining R and c This map h is called an evaluation map. The theorem sa ys that adding t w o p olynomials in R [ x ] and ev aluating is the same as ev aluating and then adding in C Also m ultiplying t w o p olynomials in R [ x ] and ev aluating is the same as ev aluating and then m ultiplying in C In street language the theorem sa ys y ou are free to send x wherev er y ou wish and extend to a ring homomorphism on R [ x ]. Exercise Let C = f a + bi : a; b 2 R g Since R is a subring of C there exists a homomorphism h : R [ x ] C whic h sends x to i and this h is surjectiv e. Sho w k er ( h ) = ( x 2 + 1) R [ x ] and th us R [ x ] = ( x 2 + 1) C : This is a go o d w a y to lo ok at the complex n um b ers, i.e., to obtain C ; adjoin x to R and set x 2 = 1. Exercise Z 2 [ x ] = ( x 2 + x + 1) has 4 elemen ts. W rite out the m ultiplication table for this ring and sho w that it is a eld. Exercise Sho w that, if R is a domain, the units of R [ x ] are just the units of R Th us if F is a eld, the units of F [ x ] are the nonzero constan ts. Sho w that [1] + [2] x is a unit in Z 4 [ x ]. In this c hapter w e do not pro v e F [ x ] is a unique factorization domain, nor do w e ev en dene unique factorization domain. The next denition and theorem are included merely for reference, and should not b e studied at this stage. Denition Supp ose F is a eld and f 2 F [ x ] has degree 1. The statemen t that g is an asso ciate of f means 9 a unit u 2 F [ x ] suc h that g = uf The statemen t that f is irr e ducible means that if h is a nonconstan t p olynomial whic h divides f then h is an asso ciate of f W e do not dev elop the theory of F [ x ] here. Ho w ev er, the dev elopmen t is easy b ecause it corresp onds to the dev elopmen t of Z in Chapter 1. The Division Algorithm corresp onds to the Euclidean Algorithm. Irreducible p olynomials corresp ond to prime in tegers. The degree function corresp onds to the absolute v alue function. One dierence is that the units of F [ x ] are nonzero constan ts, while the units of Z PAGE 56 48 Rings Chapter 3 are just 1. Th us the asso ciates of f are all cf with c 6 = 0 while the asso ciates of an in teger n are just n Here is the basic theorem. (This theory is dev elop ed in full in the App endix under the topic of Euclidean domains.) Theorem Supp ose F is a eld and f 2 F [ x ] has degree 1. Then f factors as the pro duct of irreducibles, and this factorization is unique up to order and asso ciates. Also the follo wing are equiv alen t. 1) F [ x ] = ( f ) is a domain. 2) F [ x ] = ( f ) is a eld. 3) f is irreducible. Denition No w supp ose x and y are \v ariables". If a 2 R and n; m 0, then a x n y m = a y m x n is called a monomial Dene an elemen t of R [ x ; y ] to b e an y nite sum of monomials. Theorem R [ x ; y ] is a comm utativ e ring and ( R [ x ])[ y ] R [ x ; y ] ( R [ y ])[ x ]. In other w ords, an y p olynomial in x and y with co ecien ts in R ma y b e written as a p olynomial in y with co ecien ts in R [ x ], or as a p olynomial in x with co ecien ts in R [ y ]. Side Commen t It is true that if F is a eld, eac h f 2 F [ x ; y ] factors as the pro duct of irreducibles. Ho w ev er F [ x ; y ] is not a PID. F or example, the ideal I = xF [ x; y ] + y F [ x; y ] = f f 2 F [ x; y ] : f (0 ; 0 ) = 0 g is not principal. If R is a comm utativ e ring and n 2, the concept of a p olynomial ring in n v ariables w orks ne without a hitc h. If a 2 R and v 1 ; v 2 ; :::; v n are nonnegativ e in tegers, then a x v 1 1 x v 2 2 x v n n is called a monomial. Order do es not matter here. Dene an elemen t of R [ x 1 ; x 2 ; :::; x n ] to b e an y nite sum of monomials. This giv es a comm utativ e ring and there is canonical isomorphism R [ x 1 ; x 2 ; :::; x n ] ( R [ x 1 ; x 2 ; :::; x n 1 ])[ x n ]. Using this and induction on n it is easy to pro v e the follo wing theorem. Theorem If R is a domain, R [ x 1 ; x 2 ; :::; x n ] is a domain and its units are just the units of R PAGE 57 Chapter 3 Rings 49 Exercise Supp ose R is a comm utativ e ring and f : R [ x; y ] R [ x ] is the ev aluation map whic h sends y to 0 This means f ( p ( x; y )) = p ( x; 0 ). Sho w f is a ring homomorphism whose k ernel is the ideal ( y ) = y R [ x; y ]. Use the fact that \the domain mo d the k ernel is isomorphic to the image" to sho w R [ x; y ] = ( y ) is isomorphic to R [ x ] : That is, if y ou adjoin y to R [ x ] and then factor it out, y ou get R [ x ] bac k. Pro duct of Rings The pro duct of rings w orks ne, just as do es the pro duct of groups. Theorem Supp ose T is an index set and for eac h t 2 T R t is a ring. On the additiv e ab elian group Y t 2 T R t = Q R t dene m ultiplication b y f r t g f s t g = f r t s t g Then Q R t is a ring and eac h pro jection s : Q R t R s is a ring homomorphism. Supp ose R is a ring. Under the natural bijection from f functions f : R Q R t g to f sequences of functions f f t g t 2 T where f t : R R t g f is a ring homomorphism i eac h f t is a ring homomorphism. Pro of W e already kno w f is a group homomorphism i eac h f t is a group homomorphism (see page 36). Note that f 1 t g is the m ultiplicativ e iden tit y of Q R t and f (1 R ) = f 1 t g i f t (1 R ) = 1 t for eac h t 2 T : Finally since m ultiplication is dened co ordinatewise, f is a ring homomorphism i eac h f t is a ring homomorphism. Exercise Supp ose R and S are rings. Note that R 0 is not a subring of R S b ecause it do es not con tain (1 R ; 1 S ). Sho w R 0 is an ideal and ( R S=R 0 ) S Supp ose I R and J S are ideals. Sho w I J is an ideal of R S and ev ery ideal of R S is of this form. Exercise Supp ose R and S are comm utativ e rings. Sho w T = R S is not a domain. Let e = (1 ; 0 ) 2 R S and sho w e 2 = e (1 e ) 2 = (1 e ), R 0 = eT and 0 S = (1 e ) T Exercise If T is an y ring, an elemen t e of T is called an idemp otent pro vided e 2 = e The elemen ts 0 and 1 are idemp oten ts called the trivial idemp oten ts. Supp ose T is a comm utativ e ring and e 2 T is an idemp oten t with 0 6 = e 6 = 1 Let R = eT and S = (1 e ) T Sho w eac h of the ideals R and S is a ring with iden tit y and f : T R S dened b y f ( t ) = ( et; (1 e ) t ) is a ring isomorphism. This sho ws that a comm utativ e ring T splits as the pro duct of t w o rings i it con tains a nontrivial idemp oten t. PAGE 58 50 Rings Chapter 3 The Chinese Remainder Theorem The natural map from Z to Z m Z n is a group homomorphism and also a ring homomorphism. If m and n are relativ ely prime, this map is surjectiv e with k ernel mn Z and th us Z mn and Z m Z n are isomorphic as groups and as rings. The next theorem is a classical generalization of this. (See exercise three on page 35.) Theorem Supp ose n 1 ; :::; n t are in tegers, eac h n i > 1, and ( n i ; n j ) = 1 for all i 6 = j Let f i : Z Z n i b e dened b y f i ( a ) = [ a ]. (Note that the brac k et sym b ol is used am biguously .) Then the ring homomorphism f = ( f 1 ; ::; f t ) : Z Z n 1 Z n t is surjectiv e. F urthermore, the k ernel of f is n Z where n = n 1 n 2 n t Th us Z n and Z n 1 Z n t are isomorphic as rings, and th us also as groups. Pro of W e wish to sho w that the order of f (1) is n and th us f (1) is a group generator, and th us f is surjectiv e. The elemen t f (1) m = ([1] ; ::; [1]) m = ([ m ] ; ::; [ m ]) is zero i m is a m ultiple of eac h of n 1 ; ::; n t Since their least common m ultiple is n the order of f (1) is n: (See the fourth exercise on page 36 for the case t = 3.) Exercise Sho w that if a is an in teger and p is a prime, then [ a ] = [ a p ] in Z p (F ermat's Little Theorem). Use this and the Chinese Remainder Theorem to sho w that if b is a p ositiv e in teger, it has the same last digit as b 5 Characteristic The follo wing theorem is just an observ ation, but it sho ws that in ring theory the ring of in tegers is a \cornerstone". Theorem If R is a ring, there is one and only one ring homomorphism f : Z R It is giv en b y f ( m ) = m 1 = m Th us the subgroup of R generated b y 1 is a subring of R isomorphic to Z or isomorphic to Z n for some p ositiv e in teger n Denition Supp ose R is a ring and f : Z R is the natural ring homomorphism f ( m ) = m 1 = m The nonnegativ e in teger n with k er ( f ) = n Z is called the char acteristic of R : Th us f is injectiv e i R has c haracteristic 0 i 1 has innite order. If f is not injectiv e, the c haracteristic of R is the order of 1 It is an in teresting fact that, if R is a domain, all the nonzero elemen ts of R ha v e the same order. (See page 23 for the denition of order.) PAGE 59 Chapter 3 Rings 51 Theorem Supp ose R is a domain. If R has c haracteristic 0, then eac h nonzero a 2 R has innite order. If R has nite c haracteristic n then n is a prime and eac h nonzero a 2 R has order n Pro of Supp ose R has c haracteristic 0, a is a nonzero elemen t of R and m is a p ositiv e in teger. Then ma = m a cannot b e 0 b ecause m ; a 6 = 0 and R is a domain. Th us o ( a ) = 1 No w supp ose R has c haracteristic n Then R con tains Z n as a subring, and th us Z n is a domain and n is a prime. If a is a nonzero elemen t of R na = n a = 0 a = 0 and th us o ( a ) j n and th us o ( a ) = n Exercise Sho w that if F is a eld of c haracteristic 0, F con tains Q as a subring. That is, sho w that the injectiv e homomorphism f : Z F extends to an injectiv e homomorphism f : Q F Bo olean Rings This section is not used elsewhere in this b o ok. Ho w ev er it ts easily here, and is included for reference. Denition A ring R is a Bo ole an ring if for eac h a 2 R a 2 = a i.e., eac h elemen t of R is an idemp oten t. Theorem Supp ose R is a Bo olean ring. 1) R has c haracteristic 2. If a 2 R ; 2 a = a + a = 0 ; and so a = a Pro of ( a + a ) = ( a + a ) 2 = a 2 + 2 a 2 + a 2 = 4 a: Th us 2 a = 0 : 2) R is comm utativ e. Pro of ( a + b ) = ( a + b ) 2 = a 2 + ( a b ) + ( b a ) + b 2 = a + ( a b ) ( b a ) + b: Th us a b = b a 3) If R is a domain, R Z 2 Pro of Supp ose a 6 = 0 : Then a (1 a ) = 0 and so a = 1 4) The image of a Bo olean ring is a Bo olean ring. That is, if I is an ideal of R with I 6 = R then ev ery elemen t of R =I is idemp oten t and th us R =I is a Bo olean ring. It follo ws from 3) that R =I is a domain i R =I is a eld i R =I Z 2 : (In the language of Chapter 6, I is a prime ideal i I is a maximal ideal i R =I Z 2 ). PAGE 60 52 Rings Chapter 3 Supp ose X is a nonv oid set. If a is a subset of X let a 0 = ( X a ) b e a complemen t of a in X No w supp ose R is a nonv oid collection of subsets of X Consider the follo wing prop erties whic h the collection R ma y p ossess. 1) a 2 R ) a 0 2 R 2) a; b 2 R ) ( a \ b ) 2 R : 3) a; b 2 R ) ( a [ b ) 2 R : 4) ; 2 R and X 2 R Theorem If 1) and 2) are satised, then 3) and 4) are satised. In this case, R is called a Bo ole an algebr a of sets Pro of Supp ose 1) and 2) are true, and a; b 2 R Then a [ b = ( a 0 \ b 0 ) 0 b elongs to R and so 3) is true. Since R is nonv oid, it con tains some elemen t a Then ; = a \ a 0 and X = a [ a 0 b elong to R and so 4) is true. Theorem Supp ose R is a Bo olean algebra of sets. Dene an addition on R b y a + b = ( a [ b ) ( a \ b ). Under this addition, R is an ab elian group with 0 = ; and a = a Dene a m ultiplication on R b y a b = a \ b Under this m ultiplication R b ecomes a Bo olean ring with 1 = X Exercise Let X = f 1 ; 2 ; :::; n g and let R b e the Bo olean ring of all subsets of X Note that o ( R ) = 2 n Dene f i : R Z 2 b y f i ( a ) = [1] i i 2 a Sho w eac h f i is a homomorphism and th us f = ( f 1 ; :::; f n ) : R Z 2 Z 2 Z 2 is a ring homomorphism. Sho w f is an isomorphism. (See exercises 1) and 4) on page 12.) Exercise Use the last exercise on page 49 to sho w that an y nite Bo olean ring is isomorphic to Z 2 Z 2 Z 2 ; and th us also to the Bo olean ring of subsets ab o v e. Note Supp ose R is a Bo olean ring. It is a classical theorem that 9 a Bo olean algebra of sets whose Bo olean ring is isomorphic to R So let's just supp ose R is a Bo olean algebra of sets whic h is a Bo olean ring with addition and m ultiplication dened as ab o v e. No w dene a b = a [ b and a ^ b = a \ b These op erations cup and cap are asso ciativ e, comm utativ e, ha v e iden tit y elemen ts, and eac h distributes o v er the other. With these t w o op erations (along with complemen t), R is called a Bo ole an algebr a R is not a group under cup or cap. An yw a y it is a classical fact that, if y ou ha v e a Bo olean ring (algebra), y ou ha v e a Bo olean algebra (ring). The adv an tage of the algebra is that it is symmetric in cup and cap. The adv an tage of the ring viewp oin t is that y ou can dra w from the ric h theory of comm utativ e rings. PAGE 61 Chapter 4 Matrices and Matrix Rings W e rst consider matrices in full generalit y i.e., o v er an arbitrary ring R Ho w ev er, after the rst few pages, it will b e assumed that R is comm utativ e. The topics, suc h as in v ertible matrices, transp ose, elemen tary matrices, systems of equations, and determinan t, are all classical. The highligh t of the c hapter is the theorem that a square matrix is a unit in the matrix ring i its determinan t is a unit in the ring. This c hapter concludes with the theorem that similar matrices ha v e the same determinan t, trace, and c haracteristic p olynomial. This will b e used in the next c hapter to sho w that an endomorphism on a nitely generated v ector space has a w elldened determinan t, trace, and c haracteristic p olynomial. Denition Supp ose R is a ring and m and n are p ositiv e in tegers. Let R m;n b e the collection of all m n matrices A = ( a i;j ) = 0BB@ a 1 ; 1 : : : a 1 ;n ... ... a m; 1 : : : a m;n 1CCA where eac h en try a i;j 2 R : A matrix ma y b e view ed as m n dimensional ro w v ectors or as n m dimensional column v ectors. A matrix is said to b e squar e if it has the same n um b er of ro ws as columns. Square matrices are so imp ortan t that they ha v e a sp ecial notation, R n = R n;n R n is dened to b e the additiv e ab elian group R R R T o emphasize that R n do es not ha v e a ring structure, w e use the \sum" notation, R n = R R R Our con v en tion is to write elemen ts of R n as column v ectors, i.e., to iden tify R n with R n; 1 If the elemen ts of R n are written as ro w v ectors, R n is iden tied with R 1 ;n 53 PAGE 62 54 Matrices Chapter 4 Addition of matrices T o \add" t w o matrices, they m ust ha v e the same n um b er of ro ws and the same n um b er of columns, i.e., addition is a binary op eration R m;n R m;n R m;n The addition is dened b y ( a i;j ) + ( b i;j ) = ( a i;j + b i;j ), i.e., the i; j term of the sum is the sum of the i; j terms. The follo wing theorem is just an observ ation. Theorem R m;n is an additiv e ab elian group. Its \zero" is the matrix 0 = 0 m;n all of whose terms are zero. Also ( a i;j ) = ( a i;j ). F urthermore, as additiv e groups, R m;n R mn Scalar m ultiplication An elemen t of R is called a sc alar A matrix ma y b e \m ultiplied" on the righ t or left b y a scalar. Righ t scalar m ultiplication is dened b y ( a i;j ) c = ( a i;j c ). It is a function R m;n R R m;n Note in particular that scalar m ultiplication is dened on R n Of course, if R is comm utativ e, there is no distinction b et w een righ t and left scalar m ultiplication. Theorem Supp ose A; B 2 R m;n and c; d 2 R : Then ( A + B ) c = Ac + B c A ( c + d ) = Ac + Ad A ( cd ) = ( Ac ) d and A 1 = A This theorem is en tirely transparen t. In the language of the next c hapter, it merely states that R m;n is a righ t mo dule o v er the ring R Multiplication of Matrices The matrix pro duct AB is dened i the n um b er of columns of A is equal to the n um b er of ro ws of B The matrix AB will ha v e the same n um b er of ro ws as A and the same n um b er of columns as B i.e., m ultiplication is a function R m;n R n;p R m;p The pro duct ( a i;j )( b i;j ) is dened to b e the matrix whose ( s; t ) term is a s; 1 b 1 ;t + + a s;n b n;t i.e., the dot pro duct of ro w s of A with column t of B Exercise Consider real matrices A = a b c d U = 2 0 0 1 V = 0 1 1 0 and W = 1 2 0 1 : Find the matrices AU; U A; AV ; V A; AW and W A PAGE 63 Chapter 4 Matrices 55 Denition The identity matrix I n 2 R n is the square matrix whose diagonal terms are 1 and whose odiagonal terms are 0 Theorem Supp ose A 2 R m;n 1) 0 p;m A = 0 p;n A 0 n;p = 0 m;p 2) I m A = A = AI n Theorem (The distributiv e la ws) ( A + B ) C = AC + B C and C ( A + B ) = C A + C B whenev er the op erations are dened. Theorem (The asso ciativ e la w for matrix m ultiplication) Supp ose A 2 R m;n ; B 2 R n;p and C 2 R p;q Then ( AB ) C = A ( B C ). Note that AB C 2 R m;q Pro of W e m ust sho w that the ( s; t ) terms are equal. The pro of in v olv es writing it out and c hanging the order of summation. Let ( x i;j ) = AB and ( y i;j ) = B C Then the ( s; t ) term of ( AB ) C is X i x s;i c i;t = X i X j a s;j b j;i c i;t = X i;j a s;j b j;i c i;t = X j a s;j X i b j;i c i;t = X j a s;j y j;t whic h is the ( s; t ) term of A ( B C ). Theorem F or eac h ring R and in teger n 1 ; R n is a ring. Pro of This elegan t little theorem is immediate from the theorems ab o v e. The units of R n are called invertible or nonsingular matrices. They form a group under m ultiplication called the gener al line ar gr oup and denoted b y GL n ( R ) = ( R n ) Exercise Recall that if A is a ring and a 2 A then aA is righ t ideal of A Let A = R 2 and a = ( a i;j ) where a 1 ; 1 = 1 and the other en tries are 0 Find aR 2 and R 2 a Sho w that the only ideal of R 2 con taining a is R 2 itself. Multiplication b y blo c ks Supp ose A; E 2 R n ; B ; F 2 R n;m ; C ; G 2 R m;n and D ; H 2 R m Then m ultiplication in R n + m is giv en b y A B C D E F G H = AE + B G AF + B H C E + D G C F + D H : PAGE 64 56 Matrices Chapter 4 T ransp ose Notation F or the r emainder of this chapter on matric es, supp ose R is a c ommutative ring Of course, for n > 1 ; R n is noncomm utativ e. T ransp ose is a function from R m;n to R n;m If A 2 R m;n ; A t 2 R n;m is the matrix whose ( i; j ) term is the ( j; i ) term of A So ro w i (column i ) of A b ecomes column i (ro w i ) of A t : If A is an n dimensional ro w v ector, then A t is an n dimensional column v ector. If A is a square matrix, A t is also square. Theorem 1) ( A t ) t = A 2) ( A + B ) t = A t + B t 3) If c 2 R ; ( Ac ) t = A t c 4) ( AB ) t = B t A t 5) If A 2 R n then A is in v ertible i A t is in v ertible. In this case ( A 1 ) t = ( A t ) 1 Pro of of 5) Supp ose A is in v ertible. Then I = I t = ( AA 1 ) t = ( A 1 ) t A t Exercise Characterize those in v ertible matrices A 2 R 2 whic h ha v e A 1 = A t Sho w that they form a subgroup of GL 2 ( R ). T riangular Matrices If A 2 R n then A is upp er (lower) triangular pro vided a i;j = 0 for all i > j (all j > i ). A is strictly upp er (lower) triangular pro vided a i;j = 0 for all i j (all j i ). A is diagonal if it is upp er and lo w er triangular, i.e., a i;j = 0 for all i 6 = j: Note that if A is upp er (lo w er) triangular, then A t is lo w er (upp er) triangular. Theorem If A 2 R n is strictly upp er (or lo w er) triangular, then A n = 0 Pro of The w a y to understand this is just m ultiply it out for n = 2 and n = 3 : The geometry of this theorem will b ecome transparen t later in Chapter 5 when the matrix A denes an R mo dule endomorphism on R n (see page 93). Denition If T is an y ring, an elemen t t 2 T is said to b e nilp otent pro vided 9 n suc h that t n = 0. In this case, (1 t ) is a unit with in v erse 1 + t + t 2 + + t n 1 Th us if T = R n and B is a nilp oten t matrix, I B is in v ertible. PAGE 65 Chapter 4 Matrices 57 Exercise Let R = Z Find the in v erse of 0B@ 1 2 3 0 1 4 0 0 1 1CA Exercise Supp ose A = 0BBBBBB@ a 1 a 2 0 0 a n 1CCCCCCA is a diagonal matrix, B 2 R m;n and C 2 R n;p Sho w that B A is obtained from B b y m ultiplying column i of B b y a i Sho w AC is obtained from C b y m ultiplying ro w i of C b y a i Sho w A is a unit in R n i eac h a i is a unit in R Scalar matrices A sc alar matrix is a diagonal matrix for whic h all the diagonal terms are equal, i.e., a matrix of the form cI n The map R R n whic h sends c to cI n is an injectiv e ring homomorphism, and th us w e ma y consider R to b e a subring of R n Multiplying b y a scalar is the same as m ultiplying b y a scalar matrix, and th us scalar matrices comm ute with ev erything, i.e., if B 2 R n ; ( cI n ) B = cB = B c = B ( cI n ) : Recall w e are assuming R is a comm utativ e ring. Exercise Supp ose A 2 R n and for eac h B 2 R n ; AB = B A Sho w A is a scalar matrix. F or n > 1, this sho ws ho w noncomm utativ e R n is. Elemen tary Op erations and Elemen tary Matrices Supp ose R is a comm utativ e ring and A is a matrix o v er R There are 3 t yp es of elemen tary ro w and column op erations on the matrix A A need not b e square. T yp e 1 Multiply ro w i b y some Multiply column i b y some unit a 2 R : unit a 2 R : T yp e 2 In terc hange ro w i and ro w j: In terc hange column i and column j: T yp e 3 Add a times ro w j Add a times column i to ro w i where i 6 = j and a to column j where i 6 = j and a is an y elemen t of R : is an y elemen t of R : PAGE 66 58 Matrices Chapter 4 Elemen tary Matrices Elemen tary matrices are square and in v ertible. There are three t yp es. They are obtained b y p erforming ro w or column op erations on the iden tit y matrix. T yp e 1 B = 0BBBBBBBB@ 1 1 0 a 1 0 1 1 1CCCCCCCCA where a is a unit in R T yp e 2 B = 0BBBBBBBB@ 1 0 1 1 1 1 0 1 1CCCCCCCCA T yp e 3 B = 0BBBBBBBB@ 1 1 a i;j 1 1 0 1 1 1CCCCCCCCA where i 6 = j and a i;j is an y elemen t of R : In t yp e 1, all the odiagonal elemen ts are zero. In t yp e 2, there are t w o nonzero odiagonal elemen ts. In t yp e 3, there is at most one nonzero odiagonal elemen t, and it ma y b e ab o v e or b elo w the diagonal. Exercise Sho w that if B is an elemen tary matrix of t yp e 1,2, or 3, then B is in v ertible and B 1 is an elemen tary matrix of the same t yp e. The follo wing theorem is handy when w orking with matrices. Theorem Supp ose A is a matrix. It need not b e square. T o p erform an elementary ro w (column) op eration on A p erform the op eration on an iden tit y matrix to obtain an elemen tary matrix B and m ultiply on the left (righ t). That is, B A = ro w op eration on A and AB = column op eration on A: (See the exercise on page 54.) PAGE 67 Chapter 4 Matrices 59 Exercise Supp ose F is a eld and A 2 F m;n 1) Sho w 9 in v ertible matrices B 2 F m and C 2 F n suc h that B AC = ( d i;j ) where d 1 ; 1 = = d t;t = 1 and all other en tries are 0 The in teger t is called the r ank of A: (See page 89 of Chapter 5.) 2) Supp ose A 2 F n is in v ertible. Sho w A is the pro duct of elemen tary matrices. 3) A matrix T is said to b e in r ow e chelon form if, for eac h 1 i < m the rst nonzero term of ro w ( i + 1) is to the righ t of the rst nonzero term of ro w i Sho w 9 an in v ertible matrix B 2 F m suc h that B A is in ro w ec helon form. 4) Let A = 3 11 0 4 and D = 3 11 1 4 W rite A and D as pro ducts of elemen tary matrices o v er Q Is it p ossible to write them as pro ducts of elemen tary matrices o v er Z ? F or 1), p erform ro w and column op erations on A to reac h the desired form. This sho ws the matrices B and C ma y b e selected as pro ducts of elemen tary matrices. P art 2) also follo ws from this pro cedure. F or part 3), use only ro w op erations. Notice that if T is in ro wec helon form, the n um b er of nonzero ro ws is the rank of T Systems of Equations Supp ose A = ( a i;j ) 2 R m;n and C = 0BBB@ c 1 c m 1CCCA 2 R m = R m; 1 The system a 1 ; 1 x 1 + + a 1 ;n x n = c 1 ... ... ... a m; 1 x 1 + + a m;n x n = c m of m equations in n unkno wns, can b e written as one matrix equation in one unkno wn, namely as ( a i;j ) 0BBB@ x 1 x n 1CCCA = 0BBB@ c 1 c m 1CCCA or AX = C PAGE 68 60 Matrices Chapter 4 Dene f : R n R m b y f ( D ) = AD Then f is a group homomorphism and also f ( D c ) = f ( D ) c for an y c 2 R : In the language of the next c hapter, this sa ys that f is an R mo dule homomorphism. The next theorem summarizes what w e already kno w ab out solutions of linear equations in this setting. Theorem 1) AX = 0 is called the homo gene ous e quation Its solution set is k er( f ). 2) AX = C has a solution i C 2 image( f ) : If D 2 R n is one solution, the solution set f 1 ( C ) is the coset D + k er ( f ) in R n (See part 7 of the theorem on homomorphisms in Chapter 2, page 28.) 3) Supp ose B 2 R m is in v ertible. Then AX = C and ( B A ) X = B C ha v e the same set of solutions. Th us w e ma y p erform an y ro w op eration on b oth sides of the equation and not c hange the solution set. 4) If m = n and A 2 R m is in v ertible, then AX = C has the unique solution X = A 1 C The geometry of systems of equations o v er a eld will not b ecome really transparen t un til the dev elopmen t of linear algebra in Chapter 5. Determinan ts The concept of determinan t is one of the most amazing in all of mathematics. The prop er dev elopmen t of this concept requires a study of m ultilinear forms, whic h is giv en in Chapter 6. In this section w e simply presen t the basic prop erties. F or eac h n 1 and eac h comm utativ e ring R determinan t is a function from R n to R : F or n = 1 ; j ( a ) j = a: F or n = 2 ; a b c d = ad bc Denition Let A = ( a i;j ) 2 R n If is a p erm utation on f 1 ; 2 ; :::; n g let sign( ) = 1 if is an ev en p erm utation, and sign( ) = 1 if is an o dd p erm utation. The determinant is dened b y j A j = X all sign ( ) a 1 ; (1) a 2 ; (2) a n; ( n ) Chec k that for n = 2, this agrees with the denition ab o v e. (Note that here w e are writing the p erm utation functions as ( i ) and not as ( i ) .) PAGE 69 Chapter 4 Matrices 61 F or eac h a 1 ; (1) a 2 ; (2) a n; ( n ) con tains exactly one factor from eac h ro w and one factor from eac h column. Since R is comm utativ e, w e ma y rearrange the factors so that the rst comes from the rst column, the second from the second column, etc. This means that there is a p erm utation on f 1 ; 2 ; : : : ; n g suc h that a 1 ; (1) a n; ( n ) = a (1) ; 1 a ( n ) ;n : W e wish to sho w that = 1 and th us sign( ) = sign( ) : T o reduce the abstraction, supp ose (2) = 5. Then the rst expression will con tain the factor a 2 ; 5 In the second expression, it will app ear as a (5) ; 5 and so (5) = 2. An yw a y is the in v erse of and th us there are t w o w a ys to dene determinan t. It follo ws that the determinan t of a matrix is equal to the determinan t of its transp ose. Theorem j A j = X all sign ( ) a 1 ; (1) a 2 ; (2) a n; ( n ) = X all sign ( ) a (1) ; 1 a (2) ; 2 a ( n ) ;n : Corollary j A j = j A t j Y ou ma y view an n n matrix A as a sequence of n column v ectors or as a sequence of n ro w v ectors. Here w e will use column v ectors. This means w e write the matrix A as A = ( A 1 ; A 2 ; : : : ; A n ) where eac h A i 2 R n; 1 = R n Theorem If t w o columns of A are equal, then j A j = 0 Pro of F or simplicit y assume the rst t w o columns are equal, i.e., A 1 = A 2 No w j A j = X all sign ( ) a (1) ; 1 a (2) ; 2 a ( n ) ;n and this summation has n terms and n is an ev en n um b er. Let r b e the transp osition whic h in terc hanges one and t w o. Then for an y a (1) ; 1 a (2) ; 2 a ( n ) ;n = a r (1) ; 1 a r (2) ; 2 a r ( n ) ;n This pairs up the n terms of the summation, and since sign( )= sign ( r ) ; these pairs cancel in the summation. Therefore j A j = 0 Theorem Supp ose 1 r n; C r 2 R n; 1 ; and a; c 2 R : Then j ( A 1 ; : : : ; A r 1 ; aA r + cC r ; A r +1 ; : : : ; A n ) j = a j ( A 1 ; : : : ; A n ) j + c j ( A 1 ; : : : ; A r 1 ; C r ; A r +1 ; : : : ; A n ) j Pro of This is immediate from the denition of determinan t and the distributiv e la w of m ultiplication in the ring R Summary Determinan t is a function d : R n R In the language used in the App endix, the t w o previous theorems sa y that d is an alternating m ultilinear form. The next t w o theorems sho w that alternating implies sk ewsymmetric (see page 129). PAGE 70 62 Matrices Chapter 4 Theorem In terc hanging t w o columns of A m ultiplies the determinan t b y min us one.Pro of F or simplicit y sho w that j ( A 2 ; A 1 ; A 3 ; : : : ; A n ) j = j A j W e kno w 0 = j ( A 1 + A 2 ; A 1 + A 2 ; A 3 ; : : : ; A n ) j = j ( A 1 ; A 1 ; A 3 ; : : : ; A n ) j + j ( A 1 ; A 2 ; A 3 ; : : : ; A n ) j + j ( A 2 ; A 1 ; A 3 ; : : : ; A n ) j + j ( A 2 ; A 2 ; A 3 ; : : : ; A n ) j Since the rst and last of these four terms are zero, the result follo ws. Theorem If is a p erm utation of (1 ; 2 ; : : : ; n ), then j A j = sign( ) j ( A (1) ; A (2) ; : : : ; A ( n ) ) j Pro of The p erm utation is the nite pro duct of transp ositions. Exercise Rewrite the four preceding theorems using ro ws instead of columns. The follo wing theorem is just a summary of some of the w ork done so far. Theorem Multiplying an y ro w or column of matrix b y a scalar c 2 R m ultiplies the determinan t b y c In terc hanging t w o ro ws or t w o columns m ultiplies the determinan t b y 1. Adding c times one ro w to another ro w, or adding c times one column to another column, do es not c hange the determinan t. If a matrix has t w o ro ws equal or t w o columns equal, its determinan t is zero. More generally if one ro w is c times another ro w, or one column is c times another column, then the determinan t is zero. There are 2 n w a ys to compute j A j ; expansion b y an y ro w or expansion b y an y column. Let M i;j b e the determinan t of the ( n 1) ( n 1) matrix obtained b y remo ving ro w i and column j from A: Let C i;j = ( 1) i + j M i;j : M i;j and C i;j are called the ( i; j ) minor and c ofactor of A: The follo wing theorem is useful but the pro of is a little tedious and should not b e done as an exercise. Theorem F or an y 1 i n; j A j = a i; 1 C i; 1 + a i; 2 C i; 2 + + a i;n C i;n F or an y 1 j n; j A j = a 1 ;j C 1 ;j + a 2 ;j C 2 ;j + + a n;j C n;j Th us if an y ro w or an y column is zero, the determinan t is zero. Exercise Let A = 0B@ a 1 a 2 a 3 b 1 b 2 b 3 c 1 c 2 c 3 1CA The determinan t of A is the sum of six terms. PAGE 71 Chapter 4 Matrices 63 W rite out the determinan t of A expanding b y the rst column and also expanding b y the second ro w. Theorem If A is an upp er or lo w er triangular matrix, j A j is the pro duct of the diagonal elemen ts. If A is an elemen tary matrix of t yp e 2, j A j = 1. If A is an elemen tary matrix of t yp e 3, j A j = 1. Pro of W e will pro v e the rst statemen t for upp er triangular matrices. If A 2 R 2 is an upp er triangular matrix, then its determinan t is the pro duct of the diagonal elemen ts. Supp ose n > 2 and the theorem is true for matrices in R n 1 Supp ose A 2 R n is upp er triangular. The result follo ws b y expanding b y the rst column. An elemen tary matrix of t yp e 3 is a sp ecial t yp e of upp er or lo w er triangular matrix, so its determinan t is 1. An elemen tary matrix of t yp e 2 is obtained from the iden tit y matrix b y in terc hanging t w o ro ws or columns, and th us has determinan t 1. Theorem (Determinan t b y blo c ks) Supp ose A 2 R n ; B 2 R n;m and D 2 R m Then the determinan t of A B O D is j A jj D j Pro of Expand b y the rst column and use induction on n The follo wing remark able theorem tak es some w ork to pro v e. W e assume it here without pro of. (F or the pro of, see page 130 of the App endix.) Theorem The determinan t of the pro duct is the pro duct of the determinan ts, i.e., if A; B 2 R n ; j AB j = j A jj B j Th us j AB j = j B A j and if C is in v ertible, j C 1 AC j = j AC C 1 j = j A j Corollary If A is a unit in R n ; then j A j is a unit in R and j A 1 j = j A j 1 Pro of 1 = j I j = j AA 1 j = j A jj A 1 j : One of the ma jor goals of this c hapter is to pro v e the con v erse of the preceding corollary Classical adjoin t Supp ose R is a comm utativ e ring and A 2 R n The classic al adjoint of A is ( C i;j ) t i.e., the matrix whose ( j; i ) term is the ( i; j ) cofactor. Before PAGE 72 64 Matrices Chapter 4 w e consider the general case, let's examine 2 2 matrices. If A = a b c d then ( C i;j ) = d c b a and so ( C i;j ) t = d b c a Then A ( C i;j ) t = ( C i;j ) t A = j A j 0 0 j A j = j A j I Th us if j A j is a unit in R ; A is in v ertible and A 1 = j A j 1 ( C i;j ) t In particular, if j A j = 1 ; A 1 = d b c a Here is the general case. Theorem If R is comm utativ e and A 2 R n then A ( C i;j ) t = ( C i;j ) t A = j A j I Pro of W e m ust sho w that the diagonal elemen ts of the pro duct A ( C i;j ) t are all j A j and the other elemen ts are 0 The ( s; s ) term is the dot pro duct of ro w s of A with ro w s of ( C i;j ) and is th us j A j (computed b y expansion b y ro w s ). F or s 6 = t the ( s; t ) term is the dot pro duct of ro w s of A with ro w t of ( C i;j ). Since this is the determinan t of a matrix with ro w s = ro w t the ( s; t ) term is 0 The pro of that ( C i;j ) t A = j A j I is similar and is left as an exercise. W e are no w ready for one of the most b eautiful and useful theorems in all of mathematics.Theorem Supp ose R is a comm utativ e ring and A 2 R n Then A is a unit in R n i j A j is a unit in R (Th us if R is a eld, A is in v ertible i j A j 6 = 0 .) If A is in v ertible, then A 1 = j A j 1 ( C i;j ) t Th us if j A j = 1 ; A 1 = ( C i;j ) t the classical adjoin t of A Pro of This follo ws immediately from the preceding theorem. Exercise Sho w that an y righ t in v erse of A is also a left in v erse. That is, supp ose A; B 2 R n and AB = I : Sho w A is in v ertible with A 1 = B and th us B A = I Similarit y Supp ose A; B 2 R n B is said to b e similar to A if 9 an in v ertible C 2 R n suc h that B = C 1 AC ; i.e., B is similar to A i B is a c onjugate of A Theorem B is similar to B PAGE 73 Chapter 4 Matrices 65 B is similar to A i A is similar to B If D is similar to B and B is similar to A then D is similar to A \Similarit y" is an equiv alence relation on R n Pro of This is a go o d exercise using the denition. Theorem Supp ose A and B are similar. Then j A j = j B j and th us A is in v ertible i B is in v ertible. Pro of Supp ose B = C 1 AC : Then j B j = j C 1 AC j = j AC C 1 j = j A j T race Supp ose A = ( a i;j ) 2 R n Then the tr ac e is dened b y trace ( A ) = a 1 ; 1 + a 2 ; 2 + + a n;n : That is, the trace of A is the sum of its diagonal terms. One of the most useful prop erties of trace is trace ( AB ) = trace( B A ) whenev er AB and B A are dened. F or example, supp ose A = ( a 1 ; a 2 ; :::; a n ) and B = ( b 1 ; b 2 ; :::; b n ) t Then AB is the scalar a 1 b 1 + + a n b n while B A is the n n matrix ( b i a j ). Note that trace ( AB ) = trace( B A ) : Here is the theorem in full generalit y Theorem Supp ose A 2 R m;n and B 2 R n;m Then AB and B A are square matrices with trace ( AB ) = trace( B A ). Pro of This pro of in v olv es a c hange in the order of summation. By denition, trace( AB ) = X 1 i m a i; 1 b 1 ;i + + a i;n b n;i = X 1 i m 1 j n a i;j b j;i = X 1 j n b j; 1 a 1 ;j + + b j;m a m;j = trace( B A ). Theorem If A; B 2 R n ; trace( A + B ) = trace( A ) + trace( B ) and trace( AB ) = trace( B A ). Pro of The rst part of the theorem is immediate, and the second part is a sp ecial case of the previous theorem. Theorem If A and B are similar, then trace( A ) = trace( B ). Pro of trace ( B ) = trace( C 1 AC ) = trace( AC C 1 ) = trace ( A ). PAGE 74 66 Matrices Chapter 4 Summary Determinan t and trace are functions from R n to R Determinan t is a m ultiplicativ e homomorphism and trace is an additiv e homomorphism. F urthermore j AB j = j B A j and trace( AB ) = trace ( B A ). If A and B are similar, j A j = j B j and trace( A ) = trace( B ). Exercise Supp ose A 2 R n and a 2 R : Find j aA j and trace( aA ). Characteristic p olynomials If A 2 R n the char acteristic p olynomial C P A ( x ) 2 R [ x ] is dened b y C P A ( x ) = j ( xI A ) j An y 2 R whic h is a ro ot of C P A ( x ) is called a char acteristic r o ot of A Theorem C P A ( x ) = a 0 + a 1 x + + a n 1 x n 1 + x n where trace( A ) = a n 1 and j A j = ( 1) n a 0 Pro of This follo ws from a direct computation of the determinan t. Theorem If A and B are similar, then they ha v e the same c haracteristic p olynomials.Pro of Supp ose B = C 1 AC C P B ( x ) = j ( xI C 1 AC ) j = j C 1 ( xI A ) C j = j ( xI A ) j = C P A ( x ). Exercise Supp ose R is a comm utativ e ring, A = a b c d is a matrix in R 2 and C P A ( x ) = a 0 + a 1 x + x 2 : Find a 0 and a 1 and sho w that a 0 I + a 1 A + A 2 = 0 i.e., sho w A satises its c haracteristic p olynomial. In other w ords, C P A ( A ) = 0 Exercise Supp ose F is a eld and A 2 F 2 : Sho w the follo wing are equiv alen t. 1) A 2 = 0 2) j A j = trace( A ) = 0 3) C P A ( x ) = x 2 4) 9 an elemen tary matrix C suc h that C 1 AC is strictly upp er triangular. Note This exercise is a sp ecial case of a more general theorem. A square matrix o v er a eld is nilp oten t i all its c haracteristic ro ots are 0 i it is similar to a strictly upp er triangular matrix. This remark able result cannot b e pro v ed b y matrix theory alone, but dep ends on linear algebra (see pages 93, 94, and 98). PAGE 75 Chapter 5 Linear Algebra The exalted p osition held b y linear algebra is based up on the sub ject's ubiquitous utilit y and ease of application. The basic theory is dev elop ed here in full generalit y i.e., mo dules are dened o v er an arbitrary ring R and not just o v er a eld. The elemen tary facts ab out cosets, quotien ts, and homomorphisms follo w the same pattern as in the c hapters on groups and rings. W e giv e a simple pro of that if R is a comm utativ e ring and f : R n R n is a surjectiv e R mo dule homomorphism, then f is an isomorphism. This sho ws that nitely generated free R mo dules ha v e a w ell dened dimension, and simplies some of the dev elopmen t of linear algebra. It is in this c hapter that the concepts ab out functions, solutions of equations, matrices, and generating sets come together in one unied theory After the general theory w e restrict our atten tion to v ector spaces, i.e., mo dules o v er a eld. The k ey theorem is that an y v ector space V has a free basis, and th us if V is nitely generated, it has a w ell dened dimension, and incredible as it ma y seem, this single in teger determines V up to isomorphism. Also an y endomorphism f : V V ma y b e represen ted b y a matrix, and an y c hange of basis corresp onds to conjugation of that matrix. One of the goals in linear algebra is to select a basis so that the matrix represen ting f has a simple form. F or example, if f is not injectiv e, then f ma y b e represen ted b y a matrix whose rst column is zero. As another example, if f is nilp oten t, then f ma y b e represen ted b y a strictly upp er triangular matrix. The theorem on Jordan canonical form is not pro v ed in this c hapter, and should not b e considered part of this c hapter. It is stated here in full generalit y only for reference and completeness. The pro of is giv en in the App endix. This c hapter concludes with the study of real inner pro duct spaces, and with the b eautiful theory relating orthogonal matrices and symmetric matrices. 67 PAGE 76 68 Linear Algebra Chapter 5 Denition Supp ose R is a ring and M is an additiv e ab elian group. The statemen t that M is a right Rmo dule means there is a scalar m ultiplication M R M satisfying ( a 1 + a 2 ) r = a 1 r + a 2 r ( m; r ) mr a ( r 1 + r 2 ) = ar 1 + ar 2 a ( r 1 r 2 ) = ( ar 1 ) r 2 a 1 = a for all a; a 1 ; a 2 2 M and r ; r 1 ; r 2 2 R The statemen t that M is a left Rmo dule means there is a scalar m ultiplication R M M satisfying r ( a 1 + a 2 ) = r a 1 + r a 2 ( r ; m ) r m ( r 1 + r 2 ) a = r 1 a + r 2 a ( r 1 r 2 ) a = r 1 ( r 2 a ) 1 a = a Note that the plus sign is used am biguously as addition in M and as addition in R Notation The fact that M is a righ t (left) R mo dule will b e denoted b y M = M R ( M = R M ). If R is comm utativ e and M = M R then left scalar m ultiplication dened b y r a = ar mak es M in to a left R mo dule. Th us for comm utativ e rings, w e ma y write the scalars on either side. In this text w e stic k to righ t R mo dules. Con v en tion Unless otherwise stated, it is assumed that R is a ring and the w ord \ R mo dule" (or sometimes just \mo dule") means \righ t R mo dule". Theorem Supp ose M is an R mo dule. 1) If r 2 R then f : M M dened b y f ( a ) = ar is a homomorphism of additiv e groups. In particular (0 M ) r = 0 M 2) If a 2 M a 0 R = 0 M 3) If a 2 M and r 2 R ; then ( a ) r = ( ar ) = a ( r ). Pro of This is a go o d exercise in using the axioms for an R mo dule. PAGE 77 Chapter 5 Linear Algebra 69 Submo dules If M is an R mo dule, the statemen t that a subset N M is a submo dule means it is a subgroup whic h is closed under scalar m ultiplication, i.e., if a 2 N and r 2 R then ar 2 N : In this case N will b e an R mo dule b ecause the axioms will automatically b e satised. Note that 0 and M are submo dules, called the impr op er submo dules of M Theorem Supp ose M is an R mo dule, T is an index set, and for eac h t 2 T N t is a submo dule of M 1) \ t 2 T N t is a submo dule of M 2) If f N t g is a monotonic collection, [ t 2 T N t is a submo dule. 3) + t 2 T N t = f all nite sums a 1 + + a m : eac h a i b elongs to some N t g is a submo dule. If T = f 1 ; 2 ; ::; n g then this submo dule ma y b e written as N 1 + N 2 + + N n = f a 1 + a 2 + + a n : eac h a i 2 N i g Pro of W e kno w from page 22 that v ersions of 1) and 2) hold for subgroups, and in particular for subgroups of additiv e ab elian groups. T o nish the pro ofs it is only necessary to c hec k scalar m ultiplication, whic h is immediate. Also the pro of of 3) is immediate. Note that if N 1 and N 2 are submo dules of M N 1 + N 2 is the smallest submo dule of M con taining N 1 [ N 2 Exercise Supp ose T is a nonv oid set, N is an R mo dule, and N T is the collection of all functions f : T N with addition dened b y ( f + g )( t ) = f ( t ) + g ( t ), and scalar m ultiplication dened b y ( f r )( t ) = f ( t ) r Sho w N T is an R mo dule. (W e kno w from the last exercise in Chapter 2 that N T is a group, and so it is only necessary to c hec k scalar m ultiplication.) This simple fact is quite useful in linear algebra. F or example, in 5) of the theorem b elo w, it is stated that Hom R ( M ; N ) forms an ab elian group. So it is only necessary to sho w that Hom R ( M ; N ) is a subgroup of N M : Also in 8) it is only necessary to sho w that Hom R ( M ; N ) is a submo dule of N M Homomorphisms Supp ose M and N are R mo dules. A function f : M N is a homomorphism (i.e., an R mo dule homomorphism) pro vided it is a group homomorphism and if a 2 M and r 2 R f ( ar ) = f ( a ) r On the left, scalar m ultiplication is in M and on the righ t it is in N The basic facts ab out homomorphisms are listed b elo w. Muc h PAGE 78 70 Linear Algebra Chapter 5 of this w ork has already b een done in the c hapter on groups (see page 28). Theorem 1) The zero map M N is a homomorphism. 2) The iden tit y map I : M M is a homomorphism. 3) The comp osition of homomorphisms is a homomorphism. 4) The sum of homomorphisms is a homomorphism. If f ; g : M N are homomorphisms, dene ( f + g ) : M N b y ( f + g )( a ) = f ( a ) + g ( a ). Then f + g is a homomorphism. Also ( f ) dened b y ( f )( a ) = f ( a ) is a homomorphism. If h : N P is a homomorphism, h ( f + g ) = ( h f ) + ( h g ). If k : P M is a homomorphism, ( f + g ) k = ( f k ) + ( g k ). 5) Hom R ( M ; N ) = Hom( M R ; N R ), the set of all homomorphisms from M to N forms an ab elian group under addition. Hom R ( M ; M ), with m ultiplication dened to b e comp osition, is a ring. 6) If a bijection f : M N is a homomorphism, then f 1 : N M is also a homomorphism. In this case f and f 1 are called isomorphisms A homomorphism f : M M is called an endomorphism An isomorphism f : M M is called an automorphism The units of the endomorphism ring Hom R ( M ; M ) are the automorphisms. Th us the automorphisms on M form a group under comp osition. W e will see later that if M = R n Hom R ( R n ; R n ) is just the matrix ring R n and the automorphisms are merely the in v ertible matrices. 7) If R is comm utativ e and r 2 R then g : M M dened b y g ( a ) = ar is a homomorphism. F urthermore, if f : M N is a homomorphism, f r dened b y ( f r )( a ) = f ( ar ) = f ( a ) r is a homomorphism. 8) If R is comm utativ e, Hom R ( M ; N ) is an R mo dule. 9) Supp ose f : M N is a homomorphism, G M is a submo dule, and H N is a submo dule. Then f ( G ) is a submo dule of N and f 1 ( H ) is a submo dule of M : In particular, image( f ) is a submo dule of N and k er ( f ) = f 1 (0 ) is a submo dule of M Pro of This is just a series of observ ations. PAGE 79 Chapter 5 Linear Algebra 71 Ab elian groups are Zmo dules On page 21, it is sho wn that an y additiv e group M admits a scalar m ultiplication b y in tegers, and if M is ab elian, the prop erties are satised to mak e M a Z mo dule. Note that this is the only w a y M can b e a Z mo dule, b ecause a 1 = a a 2 = a + a etc. F urthermore, if f : M N is a group homomorphism of ab elian groups, then f is also a Z mo dule homomorphism. Summary Additiv e ab elian groups are \the same things" as Z mo dules. While group theory in general is quite separate from linear algebra, the study of additiv e ab elian groups is a sp ecial case of the study of R mo dules. Exercise R mo dules are also Z mo dules and R mo dule homomorphisms are also Z mo dule homomorphisms. If M and N are Q mo dules and f : M N is a Z mo dule homomorphism, m ust it also b e a Q mo dule homomorphism? Homomorphisms on R n R n as an R mo dule On page 54 it w as sho wn that the additiv e ab elian group R m;n admits a scalar m ultiplication b y elemen ts in R The prop erties listed there w ere exactly those needed to mak e R m;n an R mo dule. Of particular imp ortance is the case R n = R R = R n; 1 (see page 53). W e b egin with the case n = 1. R as a righ t R mo dule Let M = R and dene scalar m ultiplication on the righ t b y ar = a r That is, scalar m ultiplication is just ring m ultiplication. This mak es R a righ t R mo dule denoted b y R R (or just R ). This is the same as the denition b efore for R n when n = 1. Theorem Supp ose R is a ring and N is a subset of R Then N is a submo dule of R R ( R R ) i N is a righ t (left) ideal of R Pro of The denitions are the same except expressed in dieren t language. Theorem Supp ose M = M R and f ; g : R M are homomorphisms with f (1 ) = g (1 ). Then f = g : F urthermore, if m 2 M 9 homomorphism h : R M with h (1 ) = m: In other w ords, Hom R ( R ; M ) M Pro of Supp ose f (1 ) = g (1 ). Then f ( r ) = f (1 r ) = f (1 ) r = g (1 ) r = g (1 r ) = g ( r ). Giv en m 2 M h : R M dened b y h ( r ) = mr is a homomorphism. Th us PAGE 80 72 Linear Algebra Chapter 5 ev aluation at 1 giv es a bijection from Hom R ( R ; M ) to M and this bijection is clearly a group isomorphism. If R is comm utativ e, it is an isomorphism of R mo dules. In the case M = R the ab o v e theorem states that m ultiplication on left b y some m 2 R denes a righ t R mo dule homomorphism from R to R and ev ery mo dule homomorphism is of this form. The elemen t m should b e though t of as a 1 1 matrix. W e no w consider the case where the domain is R n Homomorphisms on R n Dene e i 2 R n b y e i = 0BBBBBB@ 01 i 0 1CCCCCCA Note that an y 0BBBBBB@ r 1 r n 1CCCCCCA can b e written uniquely as e 1 r 1 + + e n r n The sequence f e 1 ; ::; e n g is called the c anonic al fr e e b asis or standar d b asis for R n Theorem Supp ose M = M R and f ; g : R n M are homomorphisms with f ( e i ) = g ( e i ) for 1 i n Then f = g F urthermore, if m 1 ; m 2 ; :::; m n 2 M 9 homomorphism h : R n M with h ( e i ) = m i for 1 i m The homomorphism h is dened b y h ( e 1 r 1 + + e n r n ) = m 1 r 1 + + m n r n Pro of The pro of is straigh tforw ard. Note this theorem giv es a bijection from Hom R ( R n ; M ) to M n = M M M and this bijection is a group isomorphism. W e will see later that the pro duct M n is an R mo dule with scalar m ultiplication dened b y ( m 1 ; m 2 ; ::; m n ) r = ( m 1 r ; m 2 r ; ::; m n r ). If R is comm utativ e so that Hom R ( R n ; M ) is an R mo dule, this theorem giv es an R mo dule isomorphism from Hom R ( R n ; M ) to M n This theorem rev eals some of the great simplicit y of linear algebra. It do es not matter ho w complicated the ring R is, or whic h R mo dule M is selected. An y R mo dule homomorphism from R n to M is determined b y its v alues on the basis, and an y function from that basis to M extends uniquely to a homomorphism from R n to M Exercise Supp ose R is a eld and f : R R M is a nonzero homomorphism. Sho w f is injectiv e. PAGE 81 Chapter 5 Linear Algebra 73 No w let's examine the sp ecial case M = R m and sho w Hom R ( R n ; R m ) R m;n : Theorem Supp ose A = ( a i;j ) 2 R m;n Then f : R n R m dened b y f ( B ) = AB is a homomorphism with f ( e i ) = column i of A Con v ersely if v 1 ; : : : ; v n 2 R m dene A 2 R m;n to b e the matrix with column i = v i Then f dened b y f ( B ) = AB is the unique homomorphism from R n to R m with f ( e i ) = v i Ev en though this follo ws easily from the previous theorem and prop erties of matrices, it is one of the great classical facts of linear algebra. Matrices o v er R giv e R mo dule homomorphisms! F urthermore, addition of matrices corresp onds to addition of homomorphisms, and m ultiplication of matrices corresp onds to comp osition of homomorphisms. These prop erties are made explicit in the next t w o theorems. Theorem If f ; g : R n R m are giv en b y matrices A; C 2 R m;n then f + g is giv en b y the matrix A + C Th us Hom R ( R n ; R m ) and R m;n are isomorphic as additiv e groups. If R is comm utativ e, they are isomorphic as R mo dules. Theorem If f : R n R m is the homomorphism giv en b y A 2 R m;n and g : R m R p is the homomorphism giv en b y C 2 R p;m then g f : R n R p is giv en b y C A 2 R p;n : That is, comp osition of homomorphisms corresp onds to m ultiplication of matrices. Pro of This is just the asso ciativ e la w of matrix m ultiplication, C ( AB ) = ( C A ) B The previous theorem rev eals where matrix m ultiplication comes from. It is the matrix whic h represen ts the comp osition of the functions. In the case where the domain and range are the same, w e ha v e the follo wing elegan t corollary Corollary Hom R ( R n ; R n ) and R n are isomorphic as rings. The automorphisms corresp ond to the in v ertible matrices. This corollary sho ws one w a y noncomm utativ e rings arise, namely as endomorphism rings. Ev en if R is comm utativ e, R n is nev er comm utativ e unless n = 1. W e no w return to the general theory of mo dules (o v er some giv en ring R ). PAGE 82 74 Linear Algebra Chapter 5 Cosets and Quotien t Mo dules After seeing quotien t groups and quotien t rings, quotien t mo dules go through without a hitc h. As b efore, R is a ring and mo dule means R mo dule. Theorem Supp ose M is a mo dule and N M is a submo dule. Since N is a normal subgroup of M the additiv e ab elian quotien t group M = N is dened. Scalar m ultiplication dened b y ( a + N ) r = ( ar + N ) is w elldened and giv es M = N the structure of an R mo dule. The natural pro jection : M M = N is a surjectiv e homomorphism with k ernel N F urthermore, if f : M M is a surjectiv e homomorphism with k er ( f ) = N then M = N M (see b elo w). Pro of On the group lev el, this is all kno wn from Chapter 2 (see pages 27 and 29). It is only necessary to c hec k the scalar m ultiplication, whic h is ob vious. The relationship b et w een quotien ts and homomorphisms for mo dules is the same as for groups and rings, as sho wn b y the next theorem. Theorem Supp ose f : M M is a homomorphism and N is a submo dule of M If N k er ( f ), then f : ( M = N ) M dened b y f ( a + N ) = f ( a ) is a w elldened homomorphism making the follo wing diagram comm ute. M M M = N f ? > f Th us dening a homomorphism on a quotien t mo dule is the same as dening a homomorphism on the n umerator that sends the denominator to 0 The image of f is the image of f and the k ernel of f is k er ( f ) = N Th us if N = k er ( f ), f is injectiv e, and th us ( M = N ) image( f ). Therefore for an y homomorphism f (domain( f )/k er( f )) image( f ). Pro of On the group lev el this is all kno wn from Chapter 2 (see page 29). It is only necessary to c hec k that f is a mo dule homomorphism, and this is immediate. PAGE 83 Chapter 5 Linear Algebra 75 Theorem Supp ose M is an R mo dule and K and L are submo dules of M i) The natural homomorphism K ( K + L ) =L is surjectiv e with k ernel K \ L: Th us ( K =K \ L ) ( K + L ) =L is an isomorphism. ii) Supp ose K L The natural homomorphism M =K M =L is surjectiv e with k ernel L=K : Th us ( M =K ) = ( L=K ) M =L is an isomorphism. Examples These t w o examples are for the case R = Z ; i.e., for ab elian groups. 1) M = Z K = 3 Z L = 5 Z K \ L = 15 Z K + L = Z K =K \ L = 3 Z = 15 Z Z = 5 Z = ( K + L ) =L 2) M = Z K = 6 Z L = 3 Z ( K L ) ( M =K ) = ( L=K ) = ( Z = 6 Z ) = (3 Z = 6 Z ) Z = 3 Z = M =L Pro ducts and Copro ducts Innite pro ducts w ork ne for mo dules, just as they do for groups and rings. This is stated b elo w in full generalit y although the studen t should think of the nite case. In the nite case something imp ortan t holds for mo dules that do es not hold for nonab elian groups or rings { namely the nite pro duct is also a copro duct. This mak es the structure of mo dule homomorphisms m uc h more simple. F or the nite case w e ma y use either the pro duct or sum notation, i.e., M 1 M 2 M n = M 1 M 2 M n Theorem Supp ose T is an index set and for eac h t 2 T M t is an R mo dule. On the additiv e ab elian group Y t 2 T M t = Q M t dene scalar m ultiplication b y f m t g r = f m t r g Then Q M t is an R mo dule and, for eac h s 2 T the natural pro jection s : Q M t M s is a homomorphism. Supp ose M is a mo dule. Under the natural 11 corresp ondence from f functions f : M Q M t g to f sequence of functions f f t g t 2 T where f t : M M t g f is a homomorphism i eac h f t is a homomorphism. Pro of W e already kno w from Chapter 2 that f is a group homomorphism i eac h f t is a group homomorphism. Since scalar m ultiplication is dened co ordinatewise, f is a mo dule homomorphism i eac h f t is a mo dule homomorphism. PAGE 84 76 Linear Algebra Chapter 5 Denition If T is nite, the copro duct and pro duct are the same mo dule. If T is innite, the c opr o duct or sum a t 2 T M t = M t 2 T M t = M t is the submo dule of Q M t consisting of all sequences f m t g with only a nite n um b er of nonzero terms. F or eac h s 2 T the inclusion homomorphisms i s : M s M t is dened b y i s ( a ) = f a t g where a t = 0 if t 6 = s and a s = a Th us eac h M s ma y b e considered to b e a submo dule of M t Theorem Supp ose M is an R mo dule. There is a 11 corresp ondence from f homomorphisms g : M t M g and f sequences of homomorphisms f g t g t 2 T where g t : M t M g Giv en g g t is dened b y g t = g i t Giv en f g t g g is dened b y g ( f m t g ) = X t g t ( m t ). Since there are only a nite n um b er of nonzero terms, this sum is w ell dened. F or T = f 1 ; 2 g the pro duct and sum prop erties are displa y ed in the follo wing comm utativ e diagrams. M 1 M 2 M 1 M 2 M 1 M 2 M 1 M 2 M M 1 i 1 2 i 2 f g f 1 f 2 g 1 g 2 ? 6 @ @ @ @ @ R @ @ @ @ @ I Theorem F or nite T the 11 corresp ondences in the ab o v e theorems actually pro duce group isomorphisms. If R is comm utativ e, they giv e isomorphisms of R mo dules. Hom R ( M ; M 1 M n ) Hom R ( M ; M 1 ) Hom R ( M ; M n ) and Hom R ( M 1 M n ; M ) Hom R ( M 1 ; M ) Hom R ( M n ; M ) Pro of Let's lo ok at this theorem for pro ducts with n = 2. All it sa ys is that if f = ( f 1 ; f 2 ) and h = ( h 1 ; h 2 ), then f + h = ( f 1 + h 1 ; f 2 + h 2 ). If R is comm utativ e, so that the ob jects are R mo dules and not merely additiv e groups, then the isomorphisms are mo dule isomorphisms. This sa ys merely that f r = ( f 1 ; f 2 ) r = ( f 1 r ; f 2 r ). PAGE 85 Chapter 5 Linear Algebra 77 Exercise Supp ose M and N are R mo dules. Sho w that M N is isomorphic to N M : No w supp ose A M ; B N are submo dules and sho w ( M N ) = ( A B ) is isomorphic to ( M = A ) ( N =B ). In particular, if a 2 R and b 2 R then ( R R ) = ( aR bR ) is isomorphic to ( R =aR ) ( R =bR ). F or example, the ab elian group ( Z Z ) = (2 Z 3 Z ) is isomorphic to Z 2 Z 3 These isomorphisms are transparen t and are used routinely in algebra without commen t (see Th 4, page 118). Exercise Supp ose R is a comm utativ e ring, M is an R mo dule, and n 1. Dene a function : Hom R ( R n ; M ) M n whic h is a R mo dule isomorphism. Summands One basic question in algebra is \When do es a mo dule split as the sum of t w o mo dules?". Before dening summand, here are t w o theorems for bac kground. Theorem Consider M 1 = M 1 0 as a submo dule of M 1 M 2 Then the pro jection map 2 : M 1 M 2 M 2 is a surjectiv e homomorphism with k ernel M 1 Th us ( M 1 M 2 ) = M 1 is isomorphic to M 2 : (See page 35 for the group v ersion.) This is exactly what y ou w ould exp ect, and the next theorem is almost as in tuitiv e. Theorem Supp ose K and L are submo dules of M and f : K L M is the natural homomorphism, f ( k ; l ) = k + l Then the image of f is K + L and the k ernel of f is f ( a; a ) : a 2 K \ L g Th us f is an isomorphism i K + L = M and K \ L = 0 In this case w e write K L = M This abuse of notation allo ws us to a v oid talking ab out \in ternal" and \external" direct sums. Denition Supp ose K is a submo dule of M The statemen t that K is a summand of M means 9 a submo dule L of M with K L = M According to the previous theorem, this is the same as there exists a submo dule L with K + L = M and K \ L = 0 If suc h an L exists, it need not b e unique, but it will b e unique up to isomorphism, b ecause L M =K : Of course, M and 0 are alw a ys summands of M Exercise Supp ose M is a mo dule and K = f ( m; m ) : m 2 M g M M Sho w K is a submo dule of M M whic h is a summand. Exercise R is a mo dule o v er Q and Q R is a submo dule. Is Q a summand of R ? With the material at hand, this is not an easy question. Later on, it will b e easy PAGE 86 78 Linear Algebra Chapter 5 Exercise Answ er the follo wing questions ab out ab elian groups, i.e., Z mo dules. (See the third exercise on page 35.) 1) Is 2 Z a summand of Z ? 2) Is 2 Z 4 a summand of Z 4 ? 3) Is 3 Z 12 a summand of Z 12 ? 4) Supp ose m; n > 1 : When is n Z mn a summand of Z mn ? Exercise If T is a ring, dene the cen ter of T to b e the subring f t : ts = st for all s 2 T g : Let R b e a comm utativ e ring and T = R n There is a exercise on page 57 to sho w that the cen ter of T is the subring of scalar matrices. Sho w R n is a left T mo dule and nd Hom T ( R n ; R n ). Indep endence, Generating Sets, and F ree Basis This section is a generalization and abstraction of the brief section Homomorphisms on R n These concepts w ork ne for an innite index set T b ecause linear com bination means nite linear com bination. Ho w ev er, to a v oid dizziness, the studen t should rst consider the case where T is nite. Denition Supp ose M is an R mo dule, T is an index set, and for eac h t 2 T s t 2 M Let S b e the sequence f s t g t 2 T = f s t g The statemen t that S is dep endent means 9 a nite n um b er of distinct elemen ts t 1 ; :::; t n in T and elemen ts r 1 ; ::; r n in R not all zero, suc h that the linear com bination s t 1 r 1 + + s t n r n = 0 Otherwise, S is indep endent Note that if some s t = 0 then S is dep enden t. Also if 9 distinct elemen ts t 1 and t 2 in T with s t 1 = s t 2 then S is dep enden t. Let S R b e the set of all linear com binations s t 1 r 1 + + s t n r n S R is a submo dule of M called the submo dule gener ate d b y S If S is indep enden t and generates M then S is said to b e a b asis or fr e e b asis for M In this case an y v 2 M can b e written uniquely as a linear com bination of elemen ts in S An R mo dule M is said to b e a fr e e R mo dule if it is zero or if it has a basis. The next t w o theorems are ob vious, except for the confusing notation. Y ou migh t try rst the case T = f 1 ; 2 ; :::; n g and R t = R n (see p 72). Theorem F or eac h t 2 T let R t = R R and for eac h c 2 T let e c 2 R t = M t 2 T R t b e e c = f r t g where r c = l and r t = 0 if t 6 = c Then f e c g c 2 T is a basis for R t called the c anonic al b asis or standar d b asis PAGE 87 Chapter 5 Linear Algebra 79 Theorem Supp ose N is an R mo dule and M is a free R mo dule with a basis f s t g Then 9 a 11 corresp ondence from the set of all functions g : f s t g N and the set of all homomorphisms f : M N Giv en g dene f b y f ( s t 1 r 1 + + s t n r n ) = g ( s t 1 ) r 1 + + g ( s t n ) r n Giv en f dene g b y g ( s t ) = f ( s t ). In other w ords, f is completely determined b y what it do es on the basis S and y ou are \free" to send the basis an y place y ou wish and extend to a homomorphism. Recall that w e ha v e already had the preceding theorem in the case S is the canonical basis for M = R n (p 72). The next theorem is so basic in linear algebra that it is used without commen t. Although the pro of is easy it should b e w ork ed carefully Theorem Supp ose N is a mo dule, M is a free mo dule with basis S = f s t g and f : M N is a homomorphism. Let f ( S ) b e the sequence f f ( s t ) g in N 1) f ( S ) generates N i f is surjectiv e. 2) f ( S ) is indep enden t in N i f is injectiv e. 3) f ( S ) is a basis for N i f is an isomorphism. 4) If h : M N is a homomorphism, then f = h i f j S = h j S Exercise Let ( A 1 ; ::; A n ) b e a sequence of n v ectors with eac h A i 2 Z n Sho w this sequence is linearly indep enden t o v er Z i it is linearly indep enden t o v er Q Is it true the sequence is linearly indep enden t o v er Z i it is linearly indep enden t o v er R ? This question is dicult un til w e learn more linear algebra. Characterization of F ree Mo dules An y free R mo dule is isomorphic to one of the canonical free R mo dules R t : This is just an observ ation, but it is a cen tral fact in linear algebra. Theorem A nonzero R mo dule M is free i 9 an index set T suc h that M M t 2 T R t : In particular, M has a nite free basis of n elemen ts i M R n Pro of If M is isomorphic to R t then M is certainly free. So no w supp ose M has a free basis f s t g Then the homomorphism f : M R t with f ( s t ) = e t sends the basis for M to the canonical basis for R t By 3) in the preceding theorem, f is an isomorphism. PAGE 88 80 Linear Algebra Chapter 5 Exercise Supp ose R is a comm utativ e ring, A 2 R n and the homomorphism f : R n R n dened b y f ( B ) = AB is surjectiv e. Sho w f is an isomorphism, i.e., sho w A is in v ertible. This is a k ey theorem in linear algebra, although it is usually stated only for the case where R is a eld. Use the fact that f e 1 ; ::; e n g is a free basis for R n The next exercise is routine, but still informativ e. Exercise Let R = Z A = 2 1 0 3 2 5 and f : Z 3 Z 2 b e the group homomorphism dened b y A Find a nontrivial linear com bination of the columns of A whic h is 0 Also nd a nonzero elemen t of k ernel( f ). If R is a comm utativ e ring, y ou can relate prop erties of R as an R mo dule to prop erties of R as a ring. Exercise Supp ose R is a comm utativ e ring and v 2 R v 6 = 0 1) v is indep enden t i v is 2) v is a basis for R i v generates R i v is Note that 2) here is essen tially the rst exercise for the case n = 1. That is, if f : R R is a surjectiv e R mo dule homomorphism, then f is an isomorphism. Relating these concepts to matrices The theorem stated b elo w giv es a summary of results w e ha v e already had. It sho ws that certain concepts ab out matrices, linear indep endence, injectiv e homomorphisms, and solutions of equations, are all the same  they are merely stated in dieren t language. Supp ose A 2 R m;n and f : R n R m is the homomorphism asso ciated with A i.e., f ( B ) = AB Let v 1 ; ::; v n 2 R m b e the columns of A i.e., f ( e i ) = v i = column i of A Let B = 0B@ b 1 : b n 1CA represen t an elemen t of R n and C = 0B@ c 1 : c m 1CA PAGE 89 Chapter 5 Linear Algebra 81 represen t an elemen t of R m Theorem 1) The elemen t f ( B ) is a linear com bination of the columns of A that is f ( B ) = f ( e 1 b 1 + + e n b n ) = v 1 b 1 + + v n b n : Th us the image of f is generated b y the columns of A (See b ottom of page 89.) 2) f v 1 ; ::; v n g generates R m i f is surjectiv e i (for an y C 2 R m AX = C has a solution). 3) f v 1 ; ::; v n g is indep enden t i f is injectiv e i AX = 0 has a unique solution i ( 9 C 2 R m suc h that AX = C has a unique solution). 4) f v 1 ; ::; v n g is a basis for R m i f is an isomorphism i (for an y C 2 R m AX = C has a unique solution). Relating these concepts to square matrices W e no w lo ok at the preceding theorem in the sp ecial case where n = m and R is a comm utativ e ring. So far in this c hapter w e ha v e just b een cataloging. No w w e pro v e something more substan tial, namely that if f : R n R n is surjectiv e, then f is injectiv e. Later on w e will pro v e that if R is a eld, injectiv e implies surjectiv e. Theorem Supp ose R is a comm utativ e ring, A 2 R n and f : R n R n is dened b y f ( B ) = AB Let v 1 ; ::; v n 2 R n b e the columns of A and w 1 ; ::; w n 2 R n = R 1 ;n b e the ro ws of A: Then the follo wing are equiv alen t. 1) f is an automorphism. 2) A is in v ertible, i.e., j A j is a unit in R 3) f v 1 ; ::; v n g is a basis for R n 4) f v 1 ; ::; v n g generates R n 5) f is surjectiv e. 2 t ) A t is in v ertible, i.e., j A t j is a unit in R 3 t ) f w 1 ; ::; w n g is a basis for R n PAGE 90 82 Linear Algebra Chapter 5 4 t ) f w 1 ; ::; w n g generates R n Pro of Supp ose 5) is true and sho w 2). Since f is on to, 9 u 1 ; :::; u n 2 R n with f ( u i ) = e i Let g : R n R n b e the homomorphism satisfying g ( e i ) = u i Then f g is the iden tit y No w g comes from some matrix D and th us AD = I This sho ws that A has a righ t in v erse and is th us in v ertible. Recall that the pro of of this fact uses determinan t, whic h requires that R b e comm utativ e (see the exercise on page 64). W e already kno w the rst three prop erties are equiv alen t, 4) and 5) are equiv alen t, and 3) implies 4). Th us the rst v e are equiv alen t. F urthermore, applying this result to A t sho ws that the last three prop erties are equiv alen t to eac h other. Since j A j = j A t j ; 2) and 2 t ) are equiv alen t. Uniqueness of Dimension There exists a ring R with R 2 R 3 as R mo dules, but this is of little in terest. If R is comm utativ e, this is imp ossible, as sho wn b elo w. First w e mak e a con v en tion. Con v en tion F or the r emainder of this chapter, R wil l b e a c ommutative ring Theorem If f : R m R n is a surjectiv e R mo dule homomorphism, then m n Pro of Supp ose k = n m is p ositiv e. Dene h : ( R m R k = R n ) R n b y h ( u; v ) = f ( u ). Then h is a surjectiv e homomorphism, and b y the previous section, also injectiv e. This is a con tradiction and th us m n Corollary If f : R m R n is an isomorphism, then m = n Pro of Eac h of f and f 1 is surjectiv e, so m = n b y the previous theorem. Corollary If f v 1 ; ::; v m g generates R n then m n Pro of The h yp othesis implies there is a surjectiv e homomorphism R m R n So this follo ws from the rst theorem. Lemma Supp ose M is a f.g. mo dule (i.e., a nite generated R mo dule). Then if M has a basis, that basis is nite. PAGE 91 Chapter 5 Linear Algebra 83 Pro of Supp ose U M is a nite generating set and S is a basis. Then an y elemen t of U is a nite linear com bination of elemen ts of S and th us S is nite. Theorem Supp ose M is a f.g. mo dule. If M has a basis, that basis is nite and an y other basis has the same n um b er of elemen ts. This n um b er is denoted b y dim( M ), the dimension of M : (By con v en tion, 0 is a free mo dule of dimension 0.) Pro of By the previous lemma, an y basis for M m ust b e nite. M has a basis of n elemen ts i M R n : The result follo ws b ecause R n R m i n = m Change of Basis Before c hanging basis, w e recall what a basis is. Previously w e dened generating, indep endence, and basis for sequences, not for collections. F or the concept of generating it matters not whether y ou use sequences or collections, but for indep endence and basis, y ou m ust use sequences. Consider the columns of the real matrix A = 2 3 2 1 4 1 If w e consider the column v ectors of A as a collection, there are only t w o of them, y et w e certainly don't wish to sa y the columns of A form a basis for R 2 In a set or collection, there is no concept of rep etition. In order to mak e sense, w e m ust consider the columns of A as an ordered triple of v ectors, and this sequence is dep enden t. In the denition of basis on page 78, basis is dened for sequences, not for sets or collections. Tw o sequences cannot b egin to b e equal unless they ha v e the same index set. Here w e follo w the classical con v en tion that an index set with n elemen ts will b e f 1 ; 2 ; ::; n g and th us a basis for M with n elemen ts is a sequence S = f u 1 ; ::; u n g or if y ou wish, S = ( u 1 ; ::; u n ) 2 M n Supp ose M is an R mo dule with a basis of n elemen ts. Recall there is a bijection : Hom R ( R n ; M ) M n dened b y ( h ) = ( h ( e 1 ) ; ::; h ( e n )) : No w h : R n M is an isomorphism i ( h ) is a basis for M Summary The p oin t of all this is that selecting a basis of n elemen ts for M is the same as selecting an isomorphism from R n to M and from this viewp oin t, c hange of basis can b e displa y ed b y the diagram b elo w. Endomorphisms on R n are represen ted b y square matrices, and th us ha v e a determinan t and trace. No w supp ose M is a f.g. free mo dule and f : M M is a homomorphism. In order to represen t f b y a matrix, w e m ust select a basis for M (i.e., an isomorphism with R n ). W e will sho w that this matrix is w ell dened up to similarit y and th us the determinan t, trace, and c haracteristic p olynomial of f are w elldened. PAGE 92 84 Linear Algebra Chapter 5 Denition Supp ose M is a free mo dule, S = f u 1 ; ::; u n g is a basis for M and f : M M is a homomorphism. The matrix A = ( a i;j ) 2 R n of f w.r.t. the basis S is dened b y f ( u i ) = u 1 a 1 ;i + + u n a n;i (Note that if M = R n and u i = e i A is the usual matrix asso ciated with f ). Theorem Supp ose T = f v 1 ; ::; v n g is another basis for M and B 2 R n is the matrix of f w.r.t. T Dene C = ( c i;j ) 2 R n b y v i = u 1 c 1 ;i + + u n c n;i Then C is in v ertible and B = C 1 AC ; i.e., A and B are similar. Therefore j A j = j B j trace( A )=trace( B ) ; and A and B ha v e the same c haracteristic p olynomial (see page 66 of c hapter 4). Con v ersely supp ose C = ( c i;j ) 2 R n is in v ertible. Dene T = f v 1 ; ::; v n g b y v i = u 1 c 1 ;i + + u n c n;i : Then T is a basis for M and the matrix of f w.r.t. T is B = C 1 AC In other w ords, conjugation of matrices corresp onds to c hange of basis. Pro of The pro of follo ws b y seeing that the follo wing diagram is comm utativ e. R n R n R n R n M M C C A B e i v i e i u i v i e i u i e i f ? ? @ @ @ @ @ R @ @ @ @ @ I I R @ @ @ I R @ @ @ R I The diagram also explains what it means for A to b e the matrix of f w.r.t. the basis S Let h : R n M b e the isomorphism with h ( e i ) = u i for 1 i n Then the matrix A 2 R n is the one determined b y the endomorphism h 1 f h : R n R n In other w ords, column i of A is h 1 ( f ( h ( e i ))). An imp ortan t sp ecial case is where M = R n and f : R n R n is giv en b y some matrix W Then h is giv en b y the matrix U whose i th column is u i and A = U 1 W U: In other w ords, W represen ts f w.r.t. the standard basis, and U 1 W U represen ts f w.r.t. the basis f u 1 ; ::; u n g Denition Supp ose M is a f.g. free mo dule and f : M M is a homomorphism. Dene j f j to b e j A j trace( f ) to b e trace( A ), and C P f ( x ) to b e C P A ( x ), where A is PAGE 93 Chapter 5 Linear Algebra 85 the matrix of f w.r.t. some basis. By the previous theorem, all three are w elldened, i.e., do not dep end up on the c hoice of basis. Exercise Let R = Z and f : Z 2 Z 2 b e dened b y f ( D ) = 3 3 0 1 D Find the matrix of f w.r.t. the basis ( 21 ; 31 !) Exercise Let L R 2 b e the line L = f ( r ; 2 r ) t : r 2 R g Sho w there is one and only one homomorphism f : R 2 R 2 whic h is the iden tit y on L and has f (( 1 ; 1) t ) = (1 ; 1) t Find the matrix A 2 R 2 whic h represen ts f with resp ect to the basis f (1 ; 2) t ; ( 1 ; 1) t g Find the determinan t, trace, and c haracteristic p olynomial of f Also nd the matrix B 2 R 2 whic h represen ts f with resp ect to the standard basis. Finally nd an in v ertible matrix C 2 R 2 with B = C 1 AC V ector Spaces So far in this c hapter w e ha v e b een dev eloping the theory of linear algebra in general. The previous theorem, for example, holds for an y comm utativ e ring R but it m ust b e assumed that the mo dule M is free. Endomorphisms in general will not ha v e a determinan t, trace, or c haracteristic p olynomial. W e no w fo cus on the case where R is a eld F and sho w that in this case, ev ery F mo dule is free. Th us an y nitely generated F mo dule will ha v e a w elldened dimension, and endomorphisms on it will ha v e w elldened determinan t, trace, and c haracteristic p olynomial. In this section, F is a eld. F mo dules ma y also b e called ve ctor sp ac es and F mo dule homomorphisms ma y also b e called line ar tr ansformations Theorem Supp ose M is an F mo dule and v 2 M Then v 6 = 0 i v is indep enden t. That is, if v 2 V and r 2 F v r = 0 implies v = 0 in M or r = 0 in F Pro of Supp ose v r = 0 and r 6 = 0 : Then 0 = ( v r ) r 1 = v 1 = v Theorem Supp ose M 6 = 0 is an F mo dule and v 2 M Then v generates M i v is a basis for M F urthermore, if these conditions hold, then M F F an y nonzero elemen t of M is a basis, and an y t w o elemen ts of M are dep enden t. PAGE 94 86 Linear Algebra Chapter 5 Pro of Supp ose v generates M Then v 6 = 0 and is th us indep enden t b y the previous theorem. In this case M F and an y nonzero elemen t of F is a basis, and an y t w o elemen ts of F are dep enden t. Theorem Supp ose M 6 = 0 is a nitely generated F mo dule. If S = f v 1 ; ::; v m g generates M then an y maximal indep enden t subsequence of S is a basis for M Th us an y nite indep enden t sequence can b e extended to a basis. In particular, M has a nite free basis, and th us is a free F mo dule. Pro of Supp ose, for notational con v enience, that f v 1 ; ::; v n g is a maximal indep enden t subsequence of S and n < i m It m ust b e sho wn that v i is a linear com bination of f v 1 ; ::; v n g Since f v 1 ; ::; v n ; v i g is dep enden t, 9 r 1 ; :::; r n ; r i not all zero, suc h that v 1 r 1 + + v n r n + v i r i = 0 Then r i 6 = 0 and v i = ( v 1 r 1 + + v n r n ) r 1 i Th us f v 1 ; ::; v n g generates S and th us all of M No w supp ose T is a nite indep enden t sequence. T ma y b e extended to a nite generating sequence, and inside that sequence it ma y b e extended to a maximal indep enden t sequence. Th us T extends to a basis. After so man y routine theorems, it is nice to ha v e one with real p o w er. It not only sa ys an y nite indep enden t sequence can b e extended to a basis, but it can b e extended to a basis inside an y nite generating set con taining it. This is one of the theorems that mak es linear algebra tic k. The k ey h yp othesis here is that the ring is a eld. If R = Z then Z is a free mo dule o v er itself, and the elemen t 2 of Z is indep enden t. Ho w ev er it certainly cannot b e extended to a basis. Also the niteness h yp othesis in this theorem is only for con v enience, as will b e seen momen tarily Since F is a comm utativ e ring, an y t w o bases of M m ust ha v e the same n um b er of elemen ts, and th us the dimension of M is w ell dened (see theorem on page 83). Theorem Supp ose M is an F mo dule of dimension n and f v 1 ; :::; v m g is an indep enden t sequence in M : Then m n and if m = n f v 1 ; ::; v m g is a basis. Pro of f v 1 ; ::; v m g extends to a basis with n elemen ts. The next theorem is just a collection of observ ations. Theorem Supp ose M and N are nitely generated F mo dules. PAGE 95 Chapter 5 Linear Algebra 87 1) M F n i dim( M ) = n 2) M N i dim ( M ) = dim ( N ) : 3) F m F n i n = m 4) dim ( M N ) = dim ( M ) + dim ( N ) : Here is the basic theorem for v ector spaces in full generalit y Theorem Supp ose M 6 = 0 is an F mo dule and S = f v t g t 2 T generates M 1) An y maximal indep enden t subsequence of S is a basis for M 2) An y indep enden t subsequence of S ma y b e extended to a maximal indep enden t subsequence of S and th us to a basis for M 3) An y indep enden t subsequence of M can b e extended to a basis for M In particular, M has a free basis, and th us is a free F mo dule. Pro of The pro of of 1) is the same as in the case where S is nite. P art 2) will follo w from the Hausdor Maximalit y Principle. An indep enden t subsequence of S is con tained in a maximal monotonic to w er of indep enden t subsequences. The union of these indep enden t subsequences is still indep enden t, and so the result follo ws. P art 3) follo ws from 2) b ecause an indep enden t sequence can alw a ys b e extended to a generating sequence. Theorem Supp ose M is an F mo dule and K M is a submo dule. 1) K is a summand of M i.e., 9 a submo dule L of M with K L = M 2) If M is f.g., then dim( K ) dim ( M ) and K = M i dim ( K ) = dim ( M ). Pro of Let T b e a basis for K Extend T to a basis S for M Then S T generates a submo dule L with K L = M : P art 2) follo ws from 1). Corollary Q is a summand of R In other w ords, 9 a Q submo dule V R with Q V = R as Q mo dules. (See exercise on page 77.) Pro of Q is a eld, R is a Q mo dule, and Q is a submo dule of R Corollary Supp ose M is a f.g. F mo dule, N is an F mo dule, and f : M N is a homomorphism. Then dim ( M ) = dim (k er ( f )) + dim(image ( f )). PAGE 96 88 Linear Algebra Chapter 5 Pro of Let K = k er ( f ) and L M b e a submo dule with K L = M Then f j L : L image( f ) is an isomorphism. Exercise Supp ose R is a domain with the prop ert y that, for R mo dules, ev ery submo dule is a summand. Sho w R is a eld. Exercise Find a free Z mo dule whic h has a generating set con taining no basis. Exercise The real v ector space R 2 is generated b y the sequence S = f ( ; 0) ; (2 ; 1) ; (3 ; 2) g : Sho w there are three maximal indep enden t subsequences of S and eac h is a basis for R 2 : (Ro w v ectors are used here just for con v enience.) The real v ector space R 3 is generated b y S = f (1 ; 1 ; 2) ; (1 ; 2 ; 1) ; (3 ; 4 ; 5) ; (1 ; 2 ; 0) g : Sho w there are three maximal indep enden t subsequences of S and eac h is a basis for R 3 : Y ou ma y use determinan t. Square matrices o v er elds This theorem is just a summary of what w e ha v e for square matrices o v er elds. Theorem Supp ose A 2 F n and f : F n F n is dened b y f ( B ) = AB Let v 1 ; ::; v n 2 F n b e the columns of A and w 1 ; ::; w n 2 F n = F 1 ;n b e the ro ws of A Then the follo wing are equiv alen t. 1) f v 1 ; ::; v n g is indep enden t, i.e., f is injectiv e. 2) f v 1 ; ::; v n g is a basis for F n i.e., f is an automorphism, i.e., A is in v ertible, i.e., j A j6 = 0 3) f v 1 ; ::; v n g generates F n i.e., f is surjectiv e. 1 t ) f w 1 ; ::; w n g is indep enden t. 2 t ) f w 1 ; ::; w n g is a basis for F n i.e., A t is in v ertible, i.e., j A t j6 = 0 3 t ) f w 1 ; ::; w n g generates F n PAGE 97 Chapter 5 Linear Algebra 89 Pro of Except for 1) and 1 t ), this theorem holds for an y comm utativ e ring R (See the section Relating these concepts to square matrices pages 81 and 82.) P arts 1) and 1 t ) follo w from the preceding section. Exercise Add to this theorem more equiv alen t statemen ts in terms of solutions of n equations in n unkno wns. Ov erview Supp ose eac h of X and Y is a set with n elemen ts and f : X Y is a function. By the pigeonhole principle, f is injectiv e i f is bijectiv e i f is surjectiv e. No w supp ose eac h of U and V is a v ector space of dimension n and f : U V is a linear transformation. It follo ws from the w ork done so far that f is injectiv e i f is bijectiv e i f is surjectiv e. This sho ws some of the simple and denitiv e nature of linear algebra. Exercise Let A = ( A 1 ; ::; A n ) b e an n n matrix o v er Z with column i = A i 2 Z n Let f : Z n Z n b e dened b y f ( B ) = AB and f : R n R n b e dened b y f ( C ) = AC : Sho w the follo wing are equiv alen t. (See the exercise on page 79.) 1) f : Z n Z n is injectiv e. 2) The sequence ( A 1 ; ::; A n ) is linearly indep enden t o v er Z 3) j A j 6 = 0. 4) f : R n R n is injectiv e. 5) The sequence ( A 1 ; ::; A n ) is linearly indep enden t o v er R Rank of a matrix Supp ose A 2 F m;n The ro w (column) rank of A is dened to b e the dimension of the submo dule of F n ( F m ) generated b y the ro ws (columns) of A Theorem If C 2 F m and D 2 F n are in v ertible, then the ro w (column) rank of A is the same as the ro w (column) rank of C AD Pro of Supp ose f : F n F m is dened b y f ( B ) = AB Eac h column of A is a v ector in the range F m and w e kno w from page 81 that eac h f ( B ) is a linear PAGE 98 90 Linear Algebra Chapter 5 com bination of those v ectors. Th us the image of f is the submo dule of F m generated b y the columns of A and its dimension is the column rank of A This dimension is the same as the dimension of the image of g f h : F n F m where h is an y automorphism on F n and g is an y automorphism on F m This pro v es the theorem for column rank. The theorem for ro w rank follo ws using transp ose. Theorem If A 2 F m;n the ro w rank and the column rank of A are equal. This n um b er is called the r ank of A and is min f m; n g Pro of By the theorem ab o v e, elemen tary ro w and column op erations c hange neither the ro w rank nor the column rank. By ro w and column op erations, A ma y b e c hanged to a matrix H where h 1 ; 1 = = h t;t = 1 and all other en tries are 0 (see the rst exercise on page 59). Th us ro w rank = t = column rank. Exercise Supp ose A has rank t Sho w that it is p ossible to select t ro ws and t columns of A suc h that the determined t t matrix is in v ertible. Sho w that the rank of A is the largest in teger t suc h that this is p ossible. Exercise Supp ose A 2 F m;n has rank t What is the dimension of the solution set of AX = 0 ? Denition If N and M are nite dimensional v ector spaces and f : N M is a linear transformation, the r ank of f is the dimension of the image of f If f : F n F m is giv en b y a matrix A then the rank of f is the same as the rank of the matrix A Geometric In terpretation of Determinan t Supp ose V R n is some nice subset. F or example, if n = 2, V migh t b e the in terior of a square or circle. There is a concept of the n dimensional v olume of V F or n = 1, it is length. F or n = 2, it is area, and for n = 3 it is \ordinary v olume". Supp ose A 2 R n and f : R n R n is the homomorphism giv en b y A The v olume of V do es not c hange under translation, i.e., V and V + p ha v e the same v olume. Th us f ( V ) and f ( V + p ) = f ( V ) + f ( p ) ha v e the same v olume. In street language, the next theorem sa ys that \ f m ultiplies v olume b y the absolute v alue of its determinan t". Theorem The n dimensional v olume of f ( V ) is j A j (the n dimensional v olume of V ). Th us if j A j = 1 ; f preserv es v olume. PAGE 99 Chapter 5 Linear Algebra 91 Pro of If j A j = 0, image( f ) has dimension < n and th us f ( V ) has n dimensional v olume 0. If j A j 6 = 0 then A is the pro duct of elemen tary matrices (see page 59) and for elemen tary matrices, the theorem is ob vious. The result follo ws b ecause the determinan t of the comp osition is the pro duct of the determinan ts. Corollary If P is the n dimensional parallelepip ed determined b y the columns v 1 ; :: ; v n of A; then the n dimensional v olume of P is j A j Pro of Let V = [0 ; 1] [0 ; 1] = f e 1 t 1 + + e n t n : 0 t i 1 g Then P = f ( V ) = f v 1 t 1 + + v n t n : 0 t i 1 g Linear functions appro ximate dieren tiable functions lo cally W e con tin ue with the sp ecial case F = R Linear functions arise naturally in business, science, and mathematics. Ho w ev er this is not the only reason that linear algebra is so useful. It is a cen tral fact that smo oth phenomena ma y b e appro ximated lo cally b y linear phenomena. Without this great simplication, the w orld of tec hnology as w e kno w it to da y w ould not exist. Of course, linear transformations send the origin to the origin, so they m ust b e adjusted b y a translation. As a simple example, supp ose h : R R is dieren tiable and p is a real n um b er. Let f : R R b e the linear transformation f ( x ) = h 0 ( p ) x Then h is appro ximated near p b y g ( x ) = h ( p ) + f ( x p ) = h ( p ) + h 0 ( p )( x p ). No w supp ose V R 2 is some nice subset and h = ( h 1 ; h 2 ) : V R 2 is injectiv e and dieren tiable. Dene the Jacobian b y J ( h )( x; y ) = @ h 1 @ x @ h 1 @ y @ h 2 @ x @ h 2 @ y and for eac h ( x; y ) 2 V let f ( x; y ) : R 2 R 2 b e the homomorphism dened b y J ( h )( x; y ). Then for an y ( p 1 ; p 2 ) 2 V h is appro ximated near ( p 1 ; p 2 ) (after translation) b y f ( p 1 ; p 2 ). The area of V is Z Z V 1 dxdy F rom the previous section w e kno w that an y homomorphism f m ultiplies area b y j f j The studen t ma y no w understand the follo wing theorem from calculus. (Note that if h is the restriction of a linear transformation from R 2 to R 2 this theorem is immediate from the previous section.) Theorem Supp ose the determinan t of J ( h )( x; y ) is nonnegativ e for eac h ( x; y ) 2 V Then the area of h ( V ) is Z Z V j J ( h ) j dxdy PAGE 100 92 Linear Algebra Chapter 5 The T ransp ose Principle W e no w return to the case where F is a eld (of arbitrary c haracteristic). F mo dules ma y also b e called ve ctor sp ac es and submo dules ma y b e called subsp ac es The study of R mo dules in general is imp ortan t and complex. Ho w ev er the study of F mo dules is short and simple { ev ery v ector space is free and ev ery subspace is a summand. The core of classical linear algebra is not the study of v ector spaces, but the study of homomorphisms, and in particular, of endomorphisms. One goal is to sho w that if f : V V is a homomorphism with some giv en prop ert y there exists a basis of V so that the matrix represen ting f displa ys that prop ert y in a prominen t manner. The next theorem is an illustration of this. Theorem Let F b e a eld and n b e a p ositiv e in teger. 1) Supp ose V is an n dimensional v ector space and f : V V is a homomorphism with j f j = 0 Then 9 a basis of V suc h that the matrix represen ting f has its rst ro w zero. 2) Supp ose A 2 F n has j A j = 0 Then 9 an in v ertible matrix C suc h that C 1 AC has its rst ro w zero. 3) Supp ose V is an n dimensional v ector space and f : V V is a homomorphism with j f j = 0. Then 9 a basis of V suc h that the matrix represen ting f has its rst column zero. 4) Supp ose A 2 F n has j A j = 0 Then 9 an in v ertible matrix D suc h that D 1 AD has its rst column zero. W e rst wish to sho w that these 4 statemen ts are equiv alen t. W e kno w that 1) and 2) are equiv alen t and also that 3) and 4) are equiv alen t b ecause c hange of basis corresp onds to conjugation of the matrix. No w supp ose 2) is true and sho w 4) is true. Supp ose j A j = 0 Then j A t j = 0 and b y 2) 9 C suc h that C 1 A t C has rst ro w zero. Th us ( C 1 A t C ) t = C t A ( C t ) 1 has rst ro w column zero. The result follo ws b y dening D = ( C t ) 1 : Also 4) implies 2). This is an example of the tr ansp ose principle Lo osely stated, it is that theorems ab out c hange of basis corresp ond to theorems ab out conjugation of matrices and theorems ab out the ro ws of a matrix corresp ond to theorems ab out the columns of a matrix, using transp ose. In the remainder of this c hapter, this will b e used without further commen t. PAGE 101 Chapter 5 Linear Algebra 93 Pro of of the theorem W e are free to select an y of the 4 parts, and w e select part 3). Since j f j = 0, f is not injectiv e and 9 a nonzero v 1 2 V with f ( v 1 ) = 0 No w v 1 is indep enden t and extends to a basis f v 1 ; ::; v n g : Then the matrix of f w.r.t this basis has rst column zero. Exercise Let A = 3 6 2 4 Find an in v ertible matrix C 2 R 2 so that C 1 AC has rst ro w zero. Also let A = 0B@ 0 0 0 1 3 4 2 1 4 1CA and nd an in v ertible matrix D 2 R 3 so that D 1 AD has rst column zero. Exercise Supp ose M is an n dimensional v ector space o v er a eld F k is an in teger with 0 < k < n and f : M M is an endomorphism of rank k Sho w there is a basis for M so that the matrix represen ting f has its rst n k ro ws zero. Also sho w there is a basis for M so that the matrix represen ting f has its rst n k columns zero. W ork these out directly without using the transp ose principle. Nilp oten t Homomorphisms In this section it is sho wn that an endomorphism f is nilp oten t i all of its c haracteristic ro ots are 0 i it ma y b e represen ted b y a strictly upp er triangular matrix. Denition An endomorphism f : V V is nilp oten t if 9 m with f m = 0 An y f represen ted b y a strictly upp er triangular matrix is nilp oten t (see page 56). Theorem Supp ose V is an n dimensional v ector space and f : V V is a nilp oten t homomorphism. Then f n = 0 and 9 a basis of V suc h that the matrix represen ting f w.r.t. this basis is strictly upp er triangular. Th us the c haracteristic p olynomial of f is C P f ( x ) = x n Pro of Supp ose f 6 = 0 is nilp oten t. Let t b e the largest p ositiv e in teger with f t 6 = 0 Then f t ( V ) f t 1 ( V ) f ( V ) V Since f is nilp oten t, all of these inclusions are prop er. Therefore t < n and f n = 0 Construct a basis for V b y starting with a basis for f t ( V ), extending it to a basis for f t 1 ( V ), etc. Then the matrix of f w.r.t. this basis is strictly upp er triangular. Note T o obtain a matrix whic h is strictly lo w er triangular, rev erse the order of the basis. PAGE 102 94 Linear Algebra Chapter 5 Exercise Use the transp ose principle to write 3 other v ersions of this theorem. Theorem Supp ose V is an n dimensional v ector space and f : V V is a homomorphism. Then f is nilp oten t i C P f ( x ) = x n : (See the exercise at the end of Chapter 4 for the case n = 2.) Pro of Supp ose C P f ( x ) = x n F or n = 1 this implies f = 0 so supp ose n > 1. Since the constan t term of C P f ( x ) is 0 the determinan t of f is 0 Th us 9 a basis of V suc h that the matrix A represen ting f has its rst column zero. Let B 2 F n 1 b e the matrix obtained from A b y remo ving its rst ro w and rst column. No w C P A ( x ) = x n = xC P B ( x ). Th us C P B ( x ) = x n 1 and b y induction on n B is nilp oten t and so 9 C suc h that C 1 B C is strictly upp er triangular. Then 0BBBBBB@ 1 0 0 0 C 1 0 1CCCCCCA 0BBBBBB@ 0 B 0 1CCCCCCA 0BBBBBB@ 1 0 0 0 C 0 1CCCCCCA = 0BBBBBB@ 0 0 C 1 B C 0 1CCCCCCA is strictly upp er triangular. Exercise Supp ose F is a eld, A 2 F 3 is a strictly lo w er triangular matrix of rank 2, and B = 0B@ 0 0 0 1 0 0 0 1 0 1CA Using conjugation b y elemen tary matrices, sho w there is an in v ertible matrix C so that C 1 AC = B No w supp ose V is a 3dimensional v ector space and f : V V is a nilp oten t endomorphism of rank 2. W e kno w f can b e represen ted b y a strictly lo w er triangular matrix. Sho w there is a basis f v 1 ; v 2 ; v 3 g for V so that B is the matrix represen ting f Also sho w that f ( v 1 ) = v 2 f ( v 2 ) = v 3 and f ( v 3 ) = 0 : In other w ords, there is a basis for V of the form f v ; f ( v ) ; f 2 ( v ) g with f 3 ( v ) = 0 Exercise Supp ose V is a 3dimensional v ector space and f : V V is a nilp oten t endomorphism of rank 1. Sho w there is a basis for V so that the matrix represen ting f is 0B@ 0 0 0 1 0 0 0 0 0 1CA PAGE 103 Chapter 5 Linear Algebra 95 Eigen v alues Our standing h yp othesis is that V is an n dimensional v ector space o v er a eld F and f : V V is a homomorphism. Denition An elemen t 2 F is an eigenvalue of f if 9 a nonzero v 2 V with f ( v ) = v An y suc h v is called an eigenve ctor E V is dened to b e the set of all eigen v ectors for (plus 0 ). Note that E = k er ( I f ) is a subspace of V The next theorem sho ws the eigen v alues of f are just the c haracteristic ro ots of f Theorem If 2 F then the follo wing are equiv alen t. 1) is an eigen v alue of f i.e., ( I f ) : V V is not injectiv e. 2) j ( I f ) j = 0 3) is a c haracteristic ro ot of f i.e., a ro ot of the c haracteristic p olynomial C P f ( x ) = j ( xI A ) j where A is an y matrix represen ting f Pro of It is immediate that 1) and 2) are equiv alen t, so let's sho w 2) and 3) are equiv alen t. The ev aluation map F [ x ] F whic h sends h ( x ) to h ( ) is a ring homomorphism (see theorem on page 47). So ev aluating ( xI A ) at x = and taking determinan t giv es the same result as taking the determinan t of ( xI A ) and ev aluating at x = : Th us 2) and 3) are equiv alen t. The nicest thing y ou can sa y ab out a matrix is that it is similar to a diagonal matrix. Here is one case where that happ ens. Theorem Supp ose 1 ; ::; k are distinct eigen v alues of f and v i is an eigen v ector of i for 1 i k : Then the follo wing hold. 1) f v 1 ; ::; v k g is indep enden t. 2) If k = n; i.e., if C P f ( x ) = ( x 1 ) ( x n ) ; then f v 1 ; ::; v n g is a basis for V The matrix of f w.r.t. this basis is the diagonal matrix whose ( i; i ) term is i Pro of Supp ose f v 1 ; ::; v k g is dep enden t. Supp ose t is the smallest p ositiv e in teger suc h that f v 1 ; ::; v t g is dep enden t, and v 1 r 1 + + v t r t = 0 is a nontrivial linear com bination. Note that at least t w o of the co ecien ts m ust b e nonzero. No w ( f t )( v 1 r 1 + + v t r t ) = v 1 ( 1 t ) r 1 + + v t 1 ( t 1 t ) r t 1 + 0 = 0 is a shorter PAGE 104 96 Linear Algebra Chapter 5 nontrivial linear com bination. This is a con tradiction and pro v es 1). P art 2) follo ws from 1) b ecause dim ( V ) = n Exercise Let A = 0 1 1 0 2 R 2 : Find an in v ertible C 2 C 2 suc h that C 1 AC is diagonal. Sho w that C cannot b e selected in R 2 : Find the c haracteristic p olynomial of A Exercise Supp ose V is a 3dimensional v ector space and f : V V is an endomorphism with C P f ( x ) = ( x ) 3 Sho w that ( f I ) has c haracteristic p olynomial x 3 and is th us a nilp oten t endomorphism. Sho w there is a basis for V so that the matrix represen ting f is 0B@ 0 0 1 0 0 1 1CA 0B@ 0 0 1 0 0 0 1CA or 0B@ 0 0 0 0 0 0 1CA W e could con tin ue and nally giv e an ad ho c pro of of the Jordan canonical form, but in this c hapter w e prefer to press on to inner pro duct spaces. The Jordan form will b e dev elop ed in Chapter 6 as part of the general theory of nitely generated mo dules o v er Euclidean domains. The next section is included only as a con v enien t reference. Jordan Canonical F orm This section should b e just skimmed or omitted en tirely It is unnecessary for the rest of this c hapter, and is not prop erly part of the ro w of the c hapter. The basic facts of Jordan form are summarized here simply for reference. The statemen t that a square matrix B o v er a eld F is a Jor dan blo ck means that 9 2 F suc h that B is a lo w er triangular matrix of the form B = 0BBBBBB@ 0 1 0 1 1CCCCCCA B giv es a homomorphism g : F m F m with g ( e m ) = e m and g ( e i ) = e i +1 + e i for 1 i < m Note that C P B ( x ) = ( x ) m and so is the only eigen v alue of B and B satises its c haracteristic p olynomial, i.e., C P B ( B ) = 0 PAGE 105 Chapter 5 Linear Algebra 97 Denition A matrix D 2 F n is in Jordan form if 9 Jordan blo c ks B 1 ; :: ; B t suc h that D = 0BBBBBBBB@ B 1 B 2 0 0 B t 1CCCCCCCCA : Supp ose D is of this form and B i 2 F n i has eigen v alue i Then n 1 + + n t = n and C P D ( x ) = ( x 1 ) n 1 ( x t ) n t Note that a diagonal matrix is a sp ecial case of Jordan form. D is a diagonal matrix i eac h n i = 1, i.e., i eac h Jordan blo c k is a 1 1 matrix. Theorem If A 2 F n the follo wing are equiv alen t. 1) 9 an in v ertible C 2 F n suc h that C 1 AC is in Jordan form. 2) 9 1 ; ::; n 2 F (not necessarily distinct) suc h that C P A ( x ) = ( x 1 ) ( x n ). (In this case w e sa y that all the eigen v alues of A b elong to F .) Theorem Jordan form (when it exists) is unique. This means that if A and D are similar matrices in Jordan form, they ha v e the same Jordan blo c ks, except p ossibly in dieren t order. The reader should use the transp ose principle to write three other v ersions of the rst theorem. Also note that w e kno w one sp ecial case of this theorem, namely that if A has n distinct eigen v alues in F then A is similar to a diagonal matrix. Later on it will b e sho wn that if A is a symmetric real matrix, then A is similar to a diagonal matrix. Let's lo ok at the classical case A 2 R n The complex n um b ers are algebraically closed. This means that C P A ( x ) will factor completely in C [ x ], and th us 9 C 2 C n with C 1 AC in Jordan form. C ma y b e selected to b e in R n i all the eigen v alues of A are real. Exercise Find all real matrices in Jordan form that ha v e the follo wing c haracteristic p olynomials: x ( x 2), ( x 2) 2 ( x 2)( x 3)( x 4), ( x 2)( x 3) 2 ( x 2) 2 ( x 3) 2 ( x 2)( x 3) 3 Exercise Supp ose D 2 F n is in Jordan form and has c haracteristic p olynomial a 0 + a 1 x + + x n : Sho w a 0 I + a 1 D + + D n = 0 i.e., sho w C P D ( D ) = 0 PAGE 106 98 Linear Algebra Chapter 5 Exercise (Ca yleyHamilton Theorem) Supp ose E is a eld and A 2 E n Assume the theorem that there is a eld F con taining E suc h that C P A ( x ) factors completely in F [ x ]. Th us 9 an in v ertible C 2 F n suc h that D = C 1 AC is in Jordan form. Use this to sho w C P A ( A ) = 0 : (See the second exercise on page 66.) Exercise Supp ose A 2 F n is in Jordan form. Sho w A is nilp oten t i A n = 0 i C P A ( x ) = x n : (Note ho w easy this is in Jordan form.) Inner Pro duct Spaces The t w o most imp ortan t elds for mathematics and science in general are the real n um b ers and the complex n um b ers. Finitely generated v ector spaces o v er R or C supp ort inner pro ducts and are th us geometric as w ell as algebraic ob jects. The theories for the real and complex cases are quite similar, and b oth could ha v e b een treated here. Ho w ev er, for simplicit y atten tion is restricted to the case F = R In the remainder of this c hapter, the p o w er and elegance of linear algebra b ecome transparen t for all to see. Denition Supp ose V is a real v ector space. An inner pr o duct (or dot pr o duct ) on V is a function V V R whic h sends ( u; v ) to u v and satises 1) ( u 1 r 1 + u 2 r 2 ) v = ( u 1 v ) r 1 + ( u 2 v ) r 2 for all u 1 ; u 2 ; v 2 V v ( u 1 r 1 + u 2 r 2 ) = ( v u 1 ) r 1 + ( v u 2 ) r 2 and r 1 ; r 2 2 R : 2) u v = v u for all u; v 2 V 3) u u 0 and u u = 0 i u = 0 for all u 2 V Theorem Supp ose V has an inner pro duct. 1) If v 2 V f : V R dened b y f ( u ) = u v is a homomorphism. Th us 0 v = 0. 2) Sc h w arz' inequalit y If u; v 2 V ; ( u v ) 2 ( u u )( v v ). Pro of of 2) Let a = p v v and b = p u u If a or b is 0, the result is ob vious. Supp ose neither a nor b is 0. No w 0 ( ua v b ) ( ua v b ) = ( u u ) a 2 2 ab ( u v )+ ( v v ) b 2 = b 2 a 2 2 ab ( u v ) + a 2 b 2 Dividing b y 2 ab yields 0 ab ( u v ) or j u v j ab PAGE 107 Chapter 5 Linear Algebra 99 Theorem Supp ose V has an inner pro duct. Dene the norm or length of a v ector v b y k v k = p v v : The follo wing prop erties hold. 1) k v k = 0 i v = 0 2) k v r k = k v k j r j 3) j u v j k u kk v k (Sc h w arz' inequalit y) 4) k u + v k k u k + k v k (The triangle inequalit y) Pro of of 4) k u + v k 2 = ( u + v ) ( u + v ) = k u k 2 + 2( u v ) + k v k 2 k u k 2 + 2 k u kk v k + k v k 2 = ( k u k + k v k ) 2 Denition An Inner Pro duct Space (IPS) is a real v ector space with an inner pro duct. Supp ose V is an IPS. A sequence f v 1 ; ::; v n g is ortho gonal pro vided v i v j = 0 when i 6 = j The sequence is orthonormal if it is orthogonal and eac h v ector has length 1, i.e., v i v j = i;j for 1 i; j n Theorem If S = f v 1 ; ::; v n g is an orthogonal sequence of nonzero v ectors in an IPS V ; then S is indep enden t. F urthermore ( v 1 k v 1 k ; ; v n k v n k ) is orthonormal. Pro of Supp ose v 1 r 1 + + v n r n = 0 Then 0 = ( v 1 r 1 + + v n r n ) v i = r i ( v i v i ) and th us r i = 0 : Th us S is indep enden t. The second statemen t is transparen t. It is easy to dene an inner pro duct, as is sho wn b y the follo wing theorem. Theorem Supp ose V is a real v ector space with a basis S = f v 1 ; ::; v n g Then there is a unique inner pro duct on V whic h mak es S an orthornormal basis. It is giv en b y the form ula ( v 1 r 1 + + v n r n ) ( v 1 s 1 + + v n s n ) = r 1 s 1 + + r n s n Con v en tion R n will b e assumed to ha v e the standar d inner pr o duct dened b y ( r 1 ; ::; r n ) t ( s 1 ; ::; s n ) t = r 1 s 1 + + r n s n : S = f e 1 ; ::; e n g will b e called the c anonic al or standar d orthonormal b asis (see page 72). The next theorem sho ws that this inner pro duct has an amazing geometry Theorem If u; v 2 R n u v = k u kk v k cos where is the angle b et w een u PAGE 108 100 Linear Algebra Chapter 5 and v Pro of Let u = ( r 1 ; ::; r n ) and v = ( s 1 ; ::; s n ). By the la w of cosines k u v k 2 = k u k 2 + k v k 2 2 k u kk v k cos So ( r 1 s 1 ) 2 + +( r n s n ) 2 = r 2 1 + + r 2 n + s 21 + + s 2n 2 k u kk v k cos Th us r 1 s 1 + + r n s n = k u kk v k cos Exercise This is a simple exercise to observ e that h yp erplanes in R n are cosets. Supp ose f : R n R is a nonzero homomorphism giv en b y a matrix A = ( a 1 ; ::; a n ) 2 R 1 ;n Then L = k er ( f ) is the set of all solutions to a 1 x 1 + + a n x n = 0 ; i.e., the set of all v ectors p erp endicular to A No w supp ose b 2 R and C = 0BBB@ c 1 c n 1CCCA 2 R n has f ( C ) = b Then f 1 ( b ) is the set of all solutions to a 1 x 1 + + a n x n = b whic h is the coset L + C and this the set of all solutions to a 1 ( x 1 c 1 ) + + a n ( x n c n ) = 0. GramSc hmidt orthonormalization Theorem (F ourier series) Supp ose W is an IPS with an orthonormal basis f w 1 ; ::; w n g : Then if v 2 W v = w 1 ( v w 1 ) + + w n ( v w n ). Pro of v = w 1 r 1 + + w n r n and v w i = ( w 1 r 1 + + w n r n ) w i = r i Theorem Supp ose W is an IPS, Y W is a subspace with an orthonormal basis f w 1 ; ::; w k g and v 2 W Y Dene the pr oje ction of v on to Y b y p ( v ) = w 1 ( v w 1 ) + + w k ( v w k ), and let w = v p ( v ). Then ( w w i ) = ( v w 1 ( v w 1 ) w k ( v w k )) w i = 0. Th us if w k +1 = w k w k then f w 1 ; ::; w k +1 g is an orthonormal basis for the subspace generated b y f w 1 ; ::; w k ; v g : If f w 1 ; ::; w k ; v g is already orthonormal, w k +1 = v Theorem (GramSc hmidt) Supp ose W is an IPS with a basis f v 1 ; ::; v n g Then W has an orthonormal basis f w 1 ; ::; w n g Moreo v er, an y orthonormal sequence in W extends to an orthonormal basis of W Pro of Let w 1 = v 1 k v 1 k Supp ose inductiv ely that f w 1 ; ::; w k g is an orthonormal basis for Y the subspace generated b y f v 1 ; ::; v k g Let w = v k +1 p ( v k +1 ) and PAGE 109 Chapter 5 Linear Algebra 101 w k +1 = w k w k Then b y the previous theorem, f w 1 ; ::; w k +1 g is an orthonormal basis for the subspace generated b y f w 1 ; ::; w k ; v k +1 g In this manner an orthonormal basis for W is constructed. Notice that this construction denes a function h whic h sends a basis for W to an orthonormal basis for W (see top ology exercise on page 103). No w supp ose W has dimension n and f w 1 ; ::; w k g is an orthonormal sequence in W Since this sequence is indep enden t, it extends to a basis f w 1 ; ::; w k ; v k +1 ; ::; v n g The pro cess ab o v e ma y b e used to mo dify this to an orthonormal basis f w 1 ; ::; w n g Exercise Let f : R 3 R b e the homomorphism dened b y the matrix (2,1,3). Find an orthonormal basis for the k ernel of f Find the pro jection of ( e 1 + e 2 ) on to k er( f ) : Find the angle b et w een e 1 + e 2 and the plane k er( f ). Exercise Let W = R 3 ha v e the standard inner pro duct and Y W b e the subspace generated b y f w 1 ; w 2 g where w 1 = (1 ; 0 ; 0) t and w 2 = (0 ; 1 ; 0) t W is generated b y the sequence f w 1 ; w 2 ; v g where v = (1 ; 2 ; 3) t : As in the rst theorem of this section, let w = v p ( v ), where p ( v ) is the pro jection of v on to Y and set w 3 = w k w k Find w 3 and sho w that for an y t with 0 t 1, f w 1 ; w 2 ; (1 t ) v + tw 3 g is a basis for W This is a k ey observ ation for an exercise on page 103 sho wing O ( n ) is a deformation retract of GL n ( R ). Isometries Supp ose eac h of U and V is an IPS. A homomorphism f : U V is said to b e an isometry pro vided it is an isomorphism and for an y u 1 ; u 2 in U ( u 1 u 2 ) U = ( f ( u 1 ) f ( u 2 )) V Theorem Supp ose eac h of U and V is an n dimensional IPS, f u 1 ; ::; u n g is an orthonormal basis for U and f : U V is a homomorphism. Then f is an isometry i f f ( u 1 ) ; ::; f ( u n ) g is an orthonormal sequence in V Pro of Isometries certainly preserv e orthonormal sequences. So supp ose T = f f ( u 1 ) ; ::; f ( u n ) g is an orthonormal sequence in V Then T is indep enden t and th us T is a basis for V and th us f is an isomorphism (see the second theorem on page 79). It is easy to c hec k that f preserv es inner pro ducts. W e no w come to one of the denitiv e theorems in linear algebra. It is that, up to isometry there is only one inner pro duct space for eac h dimension. PAGE 110 102 Linear Algebra Chapter 5 Theorem Supp ose eac h of U and V is an n dimensional IPS. Then 9 an isometry f : U V : In particular, U is isometric to R n with its standard inner pro duct. Pro of There exist orthonormal bases f u 1 ; ::; u n g for U and f v 1 ; ::; v n g for V By the rst theorem on page 79, there exists a homomorphism f : U V with f ( u i ) = v i ; and b y the previous theorem, f is an isometry Exercise Let f : R 3 R b e the homomorphism dened b y the matrix (2,1,3). Find a linear transformation h : R 2 R 3 whic h giv es an isometry from R 2 to k er ( f ). Orthogonal Matrices As noted earlier, linear algebra is not so m uc h the study of v ector spaces as it is the study of endomorphisms. W e no w wish to study isometries from R n to R n W e kno w from a theorem on page 90 that an endomorphism preserv es v olume i its determinan t is 1. Isometries preserv e inner pro duct, and th us preserv e angle and distance, and so certainly preserv e v olume. Theorem Supp ose A 2 R n and f : R n R n is the homomorphism dened b y f ( B ) = AB Then the follo wing are equiv alen t. 1) The columns of A form an orthonormal basis for R n ; i.e., A t A = I 2) The ro ws of A form an orthonormal basis for R n ; i.e., AA t = I 3) f is an isometry Pro of A left in v erse of a matrix is also a righ t in v erse (see the exercise on page 64). Th us 1) and 2) are equiv alen t b ecause eac h of them sa ys A is in v ertible with A 1 = A t No w f e 1 ; ::; e n g is the canonical orthonormal basis for R n and f ( e i ) is column i of A: Th us b y the previous section, 1) and 3) are equiv alen t. Denition If A 2 R n satises these three conditions, A is said to b e ortho gonal The set of all suc h A is denoted b y O ( n ), and is called the ortho gonal gr oup Theorem 1) If A is orthogonal, j A j = 1. 2) If A is orthogonal, A 1 is orthogonal. If A and C are orthogonal, AC is orthogonal. Th us O ( n ) is a m ultiplicativ e subgroup of GL n ( R ). PAGE 111 Chapter 5 Linear Algebra 103 3) Supp ose A is orthogonal and f is dened b y f ( B ) = AB Then f preserv es distances and angles. This means that if u; v 2 R n k u v k = k f ( u ) f ( v ) k and the angle b et w een u and v is equal to the angle b et w een f ( u ) and f ( v ). Pro of P art 1) follo ws from j A j 2 = j A j j A t j = j I j = 1. P art 2) is immediate, b ecause isometries clearly form a subgroup of the m ultiplicativ e group of all automorphisms. F or part 3) assume f : R n R n is an isometry Then k u v k 2 = ( u v ) ( u v ) = f ( u v ) f ( u v ) = k f ( u v ) k 2 = k f ( u ) f ( v ) k 2 The pro of that f preserv es angles follo ws from u v = k u kk v k cos Exercise Sho w that if A 2 O (2) has j A j = 1, then A = cos sin sin cos for some n um b er : (See the exercise on page 56.) Exercise (top ology) Let R n R n 2 ha v e its usual metric top ology This means a sequence of matrices f A i g con v erges to A i it con v erges co ordinatewise. Sho w GL n ( R ) is an op en subset and O ( n ) is closed and compact. Let h : GL n ( R ) O ( n ) b e dened b y GramSc hmidt. Sho w H : GL n ( R ) [0 ; 1] GL n ( R ) dened b y H ( A; t ) = (1 t ) A + th ( A ) is a deformation retract of GL n ( R ) to O ( n ). Diagonalization of Symmetric Matrices W e con tin ue with the case F = R Our goals are to pro v e that, if A is a symmetric matrix, all of its eigen v alues are real and that 9 an orthogonal matrix C suc h that C 1 AC is diagonal. As bac kground, w e rst note that symmetric is the same as selfadjoin t. Theorem Supp ose A 2 R n and u; v 2 R n : Then ( A t u ) v = u ( Av ). Pro of If y ; z 2 R n then the dot pro duct y z is the matrix pro duct y t z and matrix m ultiplication is asso ciativ e. Th us ( A t u ) v = ( u t A ) v = u t ( Av ) = u ( Av ). Denition Supp ose A 2 R n A is said to b e symmetric pro vided A t = A Note that an y diagonal matrix is symmetric. A is said to b e selfadjoint if ( Au ) v = u ( Av ) for all u; v 2 R n : The next theorem is just an exercise using the previous theorem. Theorem A is symmetric i A is selfadjoin t. PAGE 112 104 Linear Algebra Chapter 5 Theorem Supp ose A 2 R n is symmetric. Then 9 real n um b ers 1 ; ::; n (not necessarily distinct) suc h that C P A ( x ) = ( x 1 )( x 2 ) ( x n ) : That is, all the eigen v alues of A are real. Pro of W e kno w C P A ( x ) factors in to linears o v er C If = a + bi is a complex n um b er, its conjugate is dened b y = a bi If h : C C is dened b y h ( ) = then h is a ring isomorphism whic h is the iden tit y on R If w = ( a i;j ) is a complex matrix or v ector, its conjugate is dened b y w = ( a i;j ). Since A 2 R n is a real symmetric matrix, A = A t = A t No w supp ose is a complex eigen v alue of A and v 2 C n is an eigen v ector with Av = v Then ( v t v ) = ( v ) t v = ( Av ) t v = ( v t A ) v = v t ( A v ) = v t ( Av ) = v t ( v ) = ( v t v ). Th us = and 2 R Or y ou can dene a complex inner pro duct on C n b y ( w v ) = w t v The pro of then reads as ( v v ) = ( v v ) = ( Av v ) = ( v Av ) = ( v v ) = ( v v ) : Either w a y is a real n um b er. W e kno w that eigen v ectors b elonging to distinct eigen v alues are linearly indep enden t. F or symmetric matrices, w e sho w more, namely that they are p erp endicular. Theorem Supp ose A is symmetric, 1 ; 2 2 R are distinct eigen v alues of A; and Au = 1 u and Av = 2 v : Then u v = 0. Pro of 1 ( u v ) = ( Au ) v = u ( Av ) = 2 ( u v ). Review Supp ose A 2 R n and f : R n R n is dened b y f ( B ) = AB Then A represen ts f w.r.t. the canonical orthonormal basis. Let S = f v 1 ; ::; v n g b e another basis and C 2 R n b e the matrix with v i as column i Then C 1 AC is the matrix represen ting f w.r.t. S No w S is an orthonormal basis i C is an orthogonal matrix. Summary Represen ting f w.r.t. an orthonormal basis is the same as conjugating A b y an orthogonal matrix. Theorem Supp ose A 2 R n and C 2 O ( n ) : Then A is symmetric i C 1 AC is symmetric. Pro of Supp ose A is symmetric. Then ( C 1 AC ) t = C t A ( C 1 ) t = C 1 AC The next theorem has geometric and ph ysical implications, but for us, just the incredibilit y of it all will suce. PAGE 113 Chapter 5 Linear Algebra 105 Theorem If A 2 R n ; the follo wing are equiv alen t. 1) A is symmetric. 2) 9 C 2 O ( n ) suc h that C 1 AC is diagonal. Pro of By the previous theorem, 2) ) 1). Sho w 1) ) 2). Supp ose A is a symmetric 2 2 matrix. Let b e an eigen v alue for A and f v 1 ; v 2 g b e an orthonormal basis for R 2 with Av 1 = v 1 Then w.r.t this basis, the transformation determined b y A is represen ted b y b 0 d : Since this matrix is symmetric, b = 0. No w supp ose b y induction that the theorem is true for symmetric matrices in R t for t < n and supp ose A is a symmetric n n matrix. Denote b y 1 ; ::; k the distinct eigen v alues of A; k n If k = n the pro of is immediate, b ecause then there is a basis of eigen v ectors of length 1, and they m ust form an orthonormal basis. So supp ose k < n Let v 1 ; ::; v k b e eigen v ectors for 1 ; ::; k with eac h k v i k = 1. They ma y b e extended to an orthonormal basis v 1 ; ::; v n : With resp ect to this basis, the transformation determined b y A is represen ted b y 0BBBBBBBB@ 0BBB@ 1 k 1CCCA ( B ) (0) ( D ) 1CCCCCCCCA Since this is a symmetric matrix, B = 0 and D is a symmetric matrix of smaller size. By induction, 9 an orthogonal C suc h that C 1 D C is diagonal. Th us conjugating b y I 0 0 C mak es the en tire matrix diagonal. This theorem is so basic w e state it again in dieren t terminology If V is an IPS, a linear transformation f : V V is said to b e selfadjoin t pro vided ( u f ( v )) = ( f ( u ) v ) for all u; v 2 V Theorem If V is an n dimensional IPS and f : V V is a linear transformation, then the follo wing are equiv alen t. 1) f is selfadjoin t. 2) 9 an orthonormal basis f v 1 ; :::; v n g for V with eac h v i an eigen v ector of f PAGE 114 106 Linear Algebra Chapter 5 Exercise Let A = 2 2 2 2 Find an orthogonal C suc h that C 1 AC is diagonal. Do the same for A = 2 1 1 2 Exercise Supp ose A; D 2 R n are symmetric. Under what conditions are A and D similar? Sho w that, if A and D are similar, 9 an orthogonal C suc h that D = C 1 AC Exercise Supp ose V is an n dimensional real v ector space. W e kno w that V is isomorphic to R n Supp ose f and g are isomorphisms from V to R n and A is a subset of V Sho w that f ( A ) is an op en subset of R n i g ( A ) is an op en subset of R n This sho ws that V an algebraic ob ject, has a go dgiv en top ology Of course, if V has an inner pro duct, it automatically has a metric, and this metric will determine that same top ology Finally supp ose V and W are nitedimensional real v ector spaces and h : V W is a linear transformation. Sho w that h is con tin uous. Exercise Dene E : C n C n b y E ( A ) = e A = I + A + (1 = 2!) A 2 + This series con v erges and th us E is a w ell dened function. If AB = B A then E ( A + B ) = E ( A ) E ( B ). Since A and A comm ute, I = E (0 ) = E ( A A ) = E ( A ) E ( A ), and th us E ( A ) is in v ertible with E ( A ) 1 = E ( A ). F urthermore E ( A t ) = E ( A ) t and if C is in v ertible, E ( C 1 AC ) = C 1 E ( A ) C No w use the results of this section to pro v e the statemen ts b elo w. (F or part 1, assume the Jordan form, i.e., assume an y A 2 C n is similar to a lo w er triangular matrix.) 1) If A 2 C n then j e A j = e trace ( A ) Th us if A 2 R n j e A j = 1 i trace( A ) = 0. 2) 9 a nonzero matrix N 2 R 2 with e N = I 3) If N 2 R n is symmetric, then e N = I i N = 0 4) If A 2 R n and A t = A then e A 2 O ( n ). PAGE 115 Chapter 6 App endix The v e previous c hapters w ere designed for a y ear undergraduate course in algebra. In this app endix, enough material is added to form a basic rst y ear graduate course. Tw o of the main goals are to c haracterize nitely generated ab elian groups and to pro v e the Jordan canonical form. The st yle is the same as b efore, i.e., ev erything is righ t do wn to the n ub. The organization is mostly a linearly ordered sequence except for the last t w o sections on determinan ts and dual spaces. These are indep enden t sections added on at the end. Supp ose R is a comm utativ e ring. An R mo dule M is said to b e cyclic if it can b e generated b y one elemen t, i.e., M R =I where I is an ideal of R The basic theorem of this c hapter is that if R is a Euclidean domain and M is a nitely generated R mo dule, then M is the sum of cyclic mo dules. Th us if M is torsion free, it is a free R mo dule. Since Z is a Euclidean domain, nitely generated ab elian groups are the sums of cyclic groups { one of the jew els of abstract algebra. No w supp ose F is a eld and V is a nitely generated F mo dule. If T : V V is a linear transformation, then V b ecomes an F [ x ]mo dule b y dening v x = T ( v ). No w F [ x ] is a Euclidean domain and so V F [ x ] is the sum of cyclic mo dules. This classical and v ery p o w erful tec hnique allo ws an easy pro of of the canonical forms. There is a basis for V so that the matrix represen ting T is in Rational canonical form. If the c haracteristic p olynomial of T factors in to the pro duct of linear p olynomials, then there is a basis for V so that the matrix represen ting T is in Jordan canonical form. This alw a ys holds if F = C A matrix in Jordan form is a lo w er triangular matrix with the eigen v alues of T displa y ed on the diagonal, so this is a p o w erful concept. In the c hapter on matrices, it is stated without pro of that the determinan t of the pro duct is the pro duct of the determinan ts. A pro of of this, whic h dep ends up on the classication of certain t yp es of alternating m ultilinear forms, is giv en in this c hapter. The nal section giv es the fundamen tals of dual spaces. 107 PAGE 116 108 App endix Chapter 6 The Chinese Remainder Theorem On page 50 in the c hapter on rings, the Chinese Remainder Theorem w as pro v ed for the ring of in tegers. In this section this classical topic is presen ted in full generalit y Surprisingly the theorem holds ev en for noncomm utativ e rings. Denition Supp ose R is a ring and A 1 ; A 2 ; :::; A m are ideals of R Then the sum A 1 + A 2 + + A m is the set of all a 1 + a 2 + + a m with a i 2 A i The pr o duct A 1 A 2 A m is the set of all nite sums of elemen ts a 1 a 2 a m with a i 2 A i Note that the sum and pro duct of ideals are ideals and A 1 A 2 A m ( A 1 \ A 2 \ \ A m ). Denition Ideals A and B of R are said to b e c omaximal if A + B = R Theorem If A and B are ideals of a ring R then the follo wing are equiv alen t. 1) A and B are comaximal. 2) 9 a 2 A and b 2 B with a + b = 1 3) ( A ) = R =B where : R R =B is the pro jection. Theorem If A 1 ; A 2 ; :::; A m and B are ideals of R with A i and B comaximal for eac h i then A 1 A 2 A m and B are comaximal. Th us A 1 \ A 2 \ \ A m and B are comaximal. Pro of Consider : R R =B Then ( A 1 A 2 A m ) = ( A 1 ) ( A 2 ) ( A m ) = ( R =B )( R =B ) ( R =B ) = R =B Chinese Remainder Theorem Supp ose A 1 ; A 2 ; :::; A n are pairwise comaximal ideals of R with eac h A i 6 = R Then the natural map : R R = A 1 R = A 2 R = A n is a surjectiv e ring homomorphism with k ernel A 1 \ A 2 \ \ A n Pro of There exists a i 2 A i and b i 2 A 1 A 2 A i 1 A i +1 A n with a i + b i = 1 Note that ( b i ) = (0 ; ::; 0 ; 1 i ; 0 ; ::; 0). If ( r 1 + A 1 ; r 2 + A 2 ; :::; r n + A n ) is an elemen t of the range, it is the image of r 1 b 1 + r 2 b 2 + + r n b n = r 1 (1 a 1 ) + r 2 (1 a 2 ) + + r n (1 a n ). Theorem If R is comm utativ e and A 1 ; A 2 ; :::; A n are pairwise comaximal ideals of R then A 1 A 2 A n = A 1 \ A 2 \ \ A n Pro of for n = 2. Sho w A 1 \ A 2 A 1 A 2 9 a 1 2 A 1 and a 2 2 A 2 with a 1 + a 2 = 1 If c 2 A 1 \ A 2 then c = c ( a 1 + a 2 ) 2 A 1 A 2 PAGE 117 Chapter 6 App endix 109 Prime and Maximal Ideals and UFD s In the rst c hapter on bac kground material, it w as sho wn that Z is a unique factorization domain. Here it will b e sho wn that this prop ert y holds for an y principle ideal domain. Later on it will b e sho wn that ev ery Euclidean domain is a principle ideal domain. Th us ev ery Euclidean domain is a unique factorization domain. Denition Supp ose R is a comm utativ e ring and I R is an ideal. I is prime means I 6 = R and if a; b 2 R ha v e ab 2 I then a or b 2 I I is maximal means I 6 = R and there are no ideals prop erly b et w een I and R Theorem 0 is a prime ideal of R i R is 0 is a maximal ideal of R i R is Theorem Supp ose J R is an ideal, J 6 = R J is a prime ideal i R =J is J is a maximal ideal i R =J is Corollary Maximal ideals are prime. Pro of Ev ery eld is a domain. Theorem If a 2 R is not a unit, then 9 a maximal ideal I of R with a 2 I Pro of This is a classical application of the Hausdor Maximalit y Principle. Consider f J : J is an ideal of R con taining a with J 6 = R g This collection con tains a maximal monotonic collection f V t g t 2 T The ideal V = [ t 2 T V t do es not con tain 1 and th us is not equal to R Therefore V is equal to some V t and is a maximal ideal con taining a Note T o prop erly appreciate this pro of, the studen t should w ork the exercise in group theory at the end of this section (see page 114). Denition Supp ose R is a domain and a; b 2 R Then w e sa y a b i there exists a unit u with au = b: Note that is an equiv alence relation. If a b then a PAGE 118 110 App endix Chapter 6 and b are said to b e asso ciates Examples If R is a domain, the asso ciates of 1 are the units of R while the only asso ciate of 0 is 0 itself. If n 2 Z is not zero, then its asso ciates are n and n If F is a eld and g 2 F [ x ] is a nonzero p olynomial, then the asso ciates of g are all cg where c is a nonzero constan t. The follo wing theorem is elemen tary but it sho ws ho w asso ciates t in to the sc heme of things. An elemen t a divides b ( a j b ) if 9 c 2 R with ac = b Theorem Supp ose R is a domain and a; b 2 ( R 0 ). Then the follo wing are equiv alen t. 1) a b 2) a j b and b j a 3) aR = bR P arts 1) and 3) ab o v e sho w there is a bijection from the asso ciate classes of R to the principal ideals of R Th us if R is a PID, there is a bijection from the asso ciate classes of R to the ideals of R : If an elemen t of a domain generates a nonzero prime ideal, it is called a prime elemen t. Denition Supp ose R is a domain and a 2 R is a nonzero nonunit. 1) a is irr e ducible if it do es not factor, i.e., a = bc ) b or c is a unit. 2) a is prime if it generates a prime ideal, i.e., a j bc ) a j b or a j c Note If a is a prime and a j c 1 c 2 c n then a j c i for some i This follo ws from the denition and induction on n: If eac h c j is irreducible, then a c i for some i Note If a b then a is irreducible (prime) i b is irreducible (prime). In other w ords, if a is irreducible (prime) and u is a unit, then au is irreducible (prime). Note a is prime ) a is irreducible. This is immediate from the denitions. Theorem F actorization in to primes is unique up to order and asso ciates, i.e., if d = b 1 b 2 b n = c 1 c 2 c m with eac h b i and eac h c i prime, then n = m and for some p erm utation of the indices, b i and c ( i ) are asso ciates for ev ery i Note also 9 a unit u and primes p 1 ; p 2 ; : : : ; p t where no t w o are asso ciates and du = p s 1 1 p s 2 2 p s t t PAGE 119 Chapter 6 App endix 111 Pro of This follo ws from the notes ab o v e. Denition R is a factorization domain (FD) means that R is a domain and if a is a nonzero nonunit elemen t of R then a factors in to a nite pro duct of irreducibles. Denition R is a unique factorization domain (UFD) means R is a FD in whic h factorization is unique (up to order and asso ciates). Theorem If R is a UFD and a is a nonzero nonunit of R then a is irreducible a is prime. Th us in a UFD, elemen ts factor as the pro duct of primes. Pro of Supp ose R is a UFD, a is an irreducible elemen t of R and a j bc If either b or c is a unit or is zero, then a divides one of them, so supp ose eac h of b and c is a nonzero nonunit elemen t of R There exists an elemen t d with ad = bc Eac h of b and c factors as the pro duct of irreducibles and the pro duct of these pro ducts is the factorization of bc It follo ws from the uniqueness of the factorization of ad = bc that one of these irreducibles is an asso ciate of a and th us a j b or a j c Therefore the elemen t a is a prime. Theorem Supp ose R is a FD. Then the follo wing are equiv alen t. 1) R is a UFD. 2) Ev ery irreducible elemen t of R is prime, i.e., a irreducible a is prime. Pro of W e already kno w 1) ) 2). P art 2) ) 1) b ecause factorization in to primes is alw a ys unique. This is a rev ealing and useful theorem. If R is a FD, then R is a UFD i eac h irreducible elemen t generates a prime ideal. F ortunately principal ideal domains ha v e this prop ert y as seen in the next theorem. Theorem Supp ose R is a PID and a 2 R is nonzero nonunit. Then the follo wing are equiv alen t. 1) aR is a maximal ideal. 2) aR is a prime ideal, i.e., a is a prime elemen t. 3) a is irreducible. Pro of Ev ery maximal ideal is a prime ideal, so 1) ) 2). Ev ery prime elemen t is an irreducible elemen t, so 2) ) 3). No w supp ose a is irreducible and sho w aR is a maximal ideal. If I is an ideal con taining aR 9 b 2 R with I = bR Since b divides a the elemen t b is a unit or an asso ciate of a: This means I = R or I = aR PAGE 120 112 App endix Chapter 6 Our goal is to pro v e that a PID is a UFD. Using the t w o theorems ab o v e, it only remains to sho w that a PID is a FD. The pro of will not require that ideals b e principally generated, but only that they b e nitely generated. This turns out to b e equiv alen t to the prop ert y that an y collection of ideals has a \maximal" elemen t. W e shall see b elo w that this is a useful concept whic h ts naturally in to the study of unique factorization domains. Theorem Supp ose R is a comm utativ e ring. Then the follo wing are equiv alen t. 1) If I R is an ideal, 9 a nite set f a 1 ; a 2 ; :::; a n g R suc h that I = a 1 R + a 2 R + + a n R ; i.e., eac h ideal of R is nitely generated. 2) An y nonv oid collection of ideals of R con tains an ideal I whic h is maximal in the collection. This means if J is an ideal in the collection with J I then J = I (The ideal I is maximal only in the sense describ ed. It need not con tain all the ideals of the collection, nor need it b e a maximal ideal of the ring R .) 3) If I 1 I 2 I 3 ::: is a monotonic sequence of ideals, 9 t 0 1 suc h that I t = I t 0 for all t t 0 Pro of Supp ose 1) is true and sho w 3). The ideal I = I 1 [ I 2 [ : : : is nitely generated and 9 t 0 1 suc h that I t 0 con tains those generators. Th us 3) is true. No w supp ose 2) is true and sho w 1). Let I b e an ideal of R and consider the collection of all nitely generated ideals con tained in I By 2) there is a maximal one, and it m ust b e I itself, and th us 1) is true. W e no w ha v e 2) ) 1) ) 3), so supp ose 2) is false and sho w 3) is false. So there is a collection of ideals of R suc h that an y ideal in the collection is prop erly con tained in another ideal of the collection. Th us it is p ossible to construct a sequence of ideals I 1 I 2 I 3 : : : with eac h prop erly con tained in the next, and therefore 3) is false. (Actually this construction requires the Hausdor Maximalit y Principle or some form of the Axiom of Choice, but w e slide o v er that.) Denition If R satises these prop erties, R is said to b e No etherian or it is said to satisfy the asc ending chain c ondition This prop ert y is satised b y man y of the classical rings in mathematics. Ha ving three denitions mak es this prop ert y useful and easy to use. F or example, see the next theorem. Theorem A No etherian domain is a FD. In particular, a PID is a FD. Pro of Supp ose there is a nonzero nonunit elemen t that do es not factor as the nite pro duct of irreducibles. Consider all ideals dR where d do es not factor. Since R is No etherian, 9 a maximal one cR : The elemen t c m ust b e reducible, i.e., c = ab where neither a nor b is a unit. Eac h of aR and bR prop erly con tains cR and so eac h PAGE 121 Chapter 6 App endix 113 of a and b factors as a nite pro duct of irreducibles. This giv es a nite factorization of c in to irreducibles, whic h is a con tradiction. Corollary A PID is a UFD. So Z is a UFD and if F is a eld, F [ x ] is a UFD. Y ou see the basic structure of UFD s is quite easy It tak es more w ork to pro v e the follo wing theorems, whic h are stated here only for reference. Theorem If R is a UFD then R [ x 1 ; :::; x n ] is a UFD. Th us if F is a eld, F [ x 1 ; :::; x n ] is a UFD. (This theorem go es all the w a y bac k to Gauss.) If R is a PID, then the formal p o w er series R [[ x 1 ; :::; x n ]] is a UFD. Th us if F is a eld, F [[ x 1 ; :::; x n ]] is a UFD. (There is a UFD R where R [[ x ]] is not a UFD. See page 566 of Commutative A lgebr a b y N. Bourbaki.) Theorem Germs of analytic functions on C n form a UFD. Pro of See Theorem 6.6.2 of A n Intr o duction to Complex A nalysis in Sever al V ariables b y L. H ormander. Theorem Supp ose R is a comm utativ e ring. Then R is No etherian ) R [ x 1 ; :::; x n ] and R [[ x 1 ; :::; x n ]] are No etherian. (This is the famous Hilb ert Basis The or em .) Theorem If R is No etherian and I R is a prop er ideal, then R =I is No etherian. (This follo ws immediately from the denition. This and the previous theorem sho w that No etherian is a ubiquitous prop ert y in ring theory .) Domains With Nonunique F actorizations Next are presen ted t w o of the standard examples of No etherian domains that are not unique factorization domains. Exercise Let R = Z ( p 5) = f n + m p 5 : n; m 2 Z g Sho w that R is a subring of R whic h is not a UFD. In particular 2 2 = (1 p 5) ( 1 p 5 ) are t w o distinct irreducible factorizations of 4. Sho w R is isomorphic to Z [ x ] = ( x 2 5), where ( x 2 5) represen ts the ideal ( x 2 5) Z [ x ], and R = (2) is isomorphic to Z 2 [ x ] = ( x 2 [5]) = Z 2 [ x ] = ( x 2 + [1]) ; whic h is not a domain. PAGE 122 114 App endix Chapter 6 Exercise Let R = R [ x; y ; z ] = ( x 2 y z ). Sho w x 2 y z is irreducible and th us prime in R [ x; y ; z ]. If u 2 R [ x; y ; z ], let u 2 R b e the coset con taining u Sho w R is not a UFD. In particular x x = y z are t w o distinct irreducible factorizations of x 2 : Sho w R = ( x ) is isomorphic to R [ y ; z ] = ( y z ) ; whic h is not a domain. An easier approac h is to let f : R [ x; y ; z ] R [ x; y ] b e the ring homomorphism dened b y f ( x ) = xy ; f ( y ) = x 2 and f ( z ) = y 2 Then S = R [ xy ; x 2 ; y 2 ] is the image of f and S is isomorphic to R : Note that xy ; x 2 ; and y 2 are irreducible in S and ( xy )( xy ) = ( x 2 )( y 2 ) are t w o distinct irreducible factorizations of ( xy ) 2 in S Exercise In Group Theory If G is an additiv e ab elian group, a subgroup H of G is said to b e maximal if H 6 = G and there are no subgroups prop erly b et w een H and G Sho w that H is maximal i G=H Z p for some prime p F or simplicit y consider the case G = Q : Whic h one of the follo wing is true? 1) If a 2 Q then there is a maximal subgroup H of Q whic h con tains a 2) Q con tains no maximal subgroups. Splitting Short Exact Sequences Supp ose B is an R mo dule and K is a submo dule of B As dened in the c hapter on linear algebra, K is a summand of B pro vided 9 a submo dule L of B with K + L = B and K \ L = 0 In this case w e write K L = B When is K a summand of B ? It turns out that K is a summand of B i there is a splitting map from B =K to B In particular, if B =K is free, K m ust b e a summand of B This is used b elo w to sho w that if R is a PID, then ev ery submo dule of R n is free. Theorem 1 Supp ose R is a ring, B and C are R mo dules, and g : B C is a surjectiv e homomorphism with k ernel K Then the follo wing are equiv alen t. 1) K is a summand of B 2) g has a righ t in v erse, i.e., 9 a homomorphism h : C B with g h = I : C C ( h is called a splitting map .) Pro of Supp ose 1) is true, i.e., supp ose 9 a submo dule L of B with K L = B Then ( g j L ) : L C is an isomorphism. If i : L B is inclusion, then h dened b y h = i ( g j L ) 1 is a righ t in v erse of g No w supp ose 2) is true and h : C B is a righ t in v erse of g Then h is injectiv e, K + h ( C ) = B and K \ h ( C ) = 0 Th us K h ( C ) = B PAGE 123 Chapter 6 App endix 115 Denition Supp ose f : A B and g : B C are R mo dule homomorphisms. The statemen t that 0 A f B g C 0 is a short exact se quenc e (s.e.s) means f is injectiv e, g is surjectiv e and f ( A ) = k er ( g ). The canonical split s.e.s. is A A C C where f = i 1 and g = 2 : A short exact sequence is said to split if 9 an isomorphism B A C suc h that the follo wing diagram comm utes. 0 A B C 0 A C f g i 1 2 Z Z Z Z Z Z ~ > ? W e no w restate the previous theorem in this terminology Theorem 1.1 A short exact sequence 0 A B C 0 splits i f ( A ) is a summand of B i B C has a splitting map. If C is a free R mo dule, there is a splitting map and th us the sequence splits. Pro of W e kno w from the previous theorem f ( A ) is a summand of B i B C has a splitting map. Sho wing these prop erties are equiv alen t to the splitting of the sequence is a go o d exercise in the art of diagram c hasing. No w supp ose C has a free basis T C and g : B C is surjectiv e. There exists a function h : T B suc h that g h ( c ) = c for eac h c 2 T The function h extends to a homomorphism from C to B whic h is a righ t in v erse of g Theorem 2 If R is a domain, then the follo wing are equiv alen t. 1) R is a PID. 2) Ev ery submo dule of R R is a free R mo dule of dimension 1. This theorem restates the ring prop ert y of PID as a mo dule prop ert y Although this theorem is transparen t, 1) ) 2) is a precursor to the follo wing classical result. Theorem 3 If R is a PID and A R n is a submo dule, then A is a free R mo dule of dimension n: Th us subgroups of Z n are free Z mo dules of dimension n Pro of F rom the previous theorem w e kno w this is true for n = 1. Supp ose n > 1 and the theorem is true for submo dules of R n 1 Supp ose A R n is a submo dule. PAGE 124 116 App endix Chapter 6 Consider the follo wing short exact sequences, where f : R n 1 R n 1 R is inclusion and g = : R n 1 R R is the pro jection. 0 R n 1 f R n 1 R R 0 0 A \ R n 1 A ( A ) 0 By induction, A \ R n 1 is free of dimension n 1 : If ( A ) = 0 ; then A R n 1 If ( A ) 6 = 0 ; it is free of dimension 1 and th us the sequence splits b y Theorem 1.1. In either case, A is a free submo dule of dimension n Exercise Let A Z 2 b e the subgroup generated b y f (6 ; 24) ; (16 ; 64) g Sho w A is a free Z mo dule of dimension 1. Also sho w the s.e.s. Z 4 3 Z 12 Z 3 splits but Z 2 Z Z 2 and Z 2 2 Z 4 Z 2 do not (see top of page 78). Euclidean Domains The ring Z p ossesses the Euclidean algorithm and the p olynomial ring F [ x ] has the division algorithm (pages 14 and 45). The concept of Euclide an domain is an abstraction of these prop erties, and the eciency of this abstraction is displa y ed in this section. F urthermore the rst axiom, ( a ) ( ab ), is used only in Theorem 2, and is sometimes omitted from the denition. An yw a y it is p ossible to just pla y around with matrices and get some deep results. If R is a Euclidean domain and M is a nitely generated R mo dule, then M is the sum of cyclic mo dules. This is one of the great classical theorems of abstract algebra, and y ou don't ha v e to w orry ab out it b ecoming obsolete. Here N will denote the set of all nonnegativ e in tegers, not just the set of p ositiv e in tegers. Denition A domain R is a Euclide an domain pro vided 9 : ( R 0 ) N suc h that if a; b 2 ( R 0 ) ; then 1) ( a ) ( ab ). 2) 9 q ; r 2 R suc h that a = bq + r with r = 0 or ( r ) < ( b ). Examples of Euclidean Domains Z with ( n ) = j n j A eld F with ( a ) = 1 8 a 6 = 0 or with ( a ) = 0 8 a 6 = 0 F [ x ] where F is a eld with ( f = a 0 + a 1 x + + a n x n ) = deg ( f ). Z [ i ] = f a + bi : a; b 2 Z g = Gaussian in tegers with ( a + bi ) = a 2 + b 2 PAGE 125 Chapter 6 App endix 117 Theorem 1 If R is a Euclidean domain, then R is a PID and th us a UFD. Pro of If I is a nonzero ideal, then 9 b 2 I 0 satisfying ( b ) ( a ) 8 a 2 I 0 Then b generates I b ecause if a 2 I 0 9 q ; r with a = bq + r No w r 2 I and r 6 = 0 ) ( r ) < ( b ) whic h is imp ossible. Th us r = 0 and a 2 bR so I = bR Theorem 2 If R is a Euclidean domain and a; b 2 R 0 then (1 ) is the smallest in teger in the image of a is a unit in R i ( a ) = (1 ). a and b are asso ciates ) ( a ) = ( b ). Pro of This is a go o d exercise. Ho w ev er it is unnecessary for Theorem 3 b elo w. The follo wing remark able theorem is the foundation for the results of this section. Theorem 3 If R is a Euclidean domain and ( a i;j ) 2 R n;t is a nonzero matrix, then b y elemen tary ro w and column op erations ( a i;j ) can b e transformed to 0BBBBBBBBBBB@ d 1 0 0 0 d 2 ... d m 0 0 0 1CCCCCCCCCCCA where eac h d i 6 = 0 ; and d i j d i +1 for 1 i < m Also d 1 generates the ideal of R generated b y the en tries of ( a i;j ). Pro of Let I R b e the ideal generated b y the elemen ts of the matrix A = ( a i;j ). If E 2 R n then the ideal J generated b y the elemen ts of E A has J I If E is in v ertible, then J = I In the same manner, if E 2 R t is in v ertible and J is the ideal generated b y the elemen ts of AE then J = I This means that ro w and column op erations on A do not c hange the ideal I Since R is a PID, there is an elemen t d 1 with I = d 1 R and this will turn out to b e the d 1 displa y ed in the theorem. The matrix ( a i;j ) has at least one nonzero elemen t d with ( d ) a mimin um. Ho w ev er, ro w and column op erations on ( a i;j ) ma y pro duce elemen ts with smaller PAGE 126 118 App endix Chapter 6 v alues. T o consolidate this approac h, consider matrices obtained from ( a i;j ) b y a nite n um b er of ro w and column op erations. Among these, let ( b i;j ) b e one whic h has an en try d 1 6 = 0 with ( d 1 ) a minim um. By elemen tary op erations of t yp e 2, the en try d 1 ma y b e mo v ed to the (1 ; 1) place in the matrix. Then d 1 will divide the other en tries in the rst ro w, else w e could obtain an en try with a smaller v alue. Th us b y column op erations of t yp e 3, the other en tries of the rst ro w ma y b e made zero. In a similar manner, b y ro w op erations of t yp e 3, the matrix ma y b e c hanged to the follo wing form. 0BBBBBBBBB@ d 1 0 0 0 ... c ij 0 1CCCCCCCCCA Note that d 1 divides eac h c i;j and th us I = d 1 R The pro of no w follo ws b y induction on the size of the matrix. This is an example of a theorem that is easy to pro v e pla ying around at the blac kb oard. Y et it m ust b e a deep theorem b ecause the next t w o theorems are easy consequences.Theorem 4 Supp ose R is a Euclidean domain, B is a nitely generated free R mo dule and A B is a nonzero submo dule. Then 9 free bases f a 1 ; a 2 ; :::; a t g for A and f b 1 ; b 2 ; :::; b n g for B with t n and suc h that eac h a i = d i b i where eac h d i 6 = 0 and d i j d i +1 for 1 i < t: Th us B = A R =d 1 R =d 2 R =d t R n t Pro of By Theorem 3 in the section Splitting Short Exact Sequences, A has a free basis f v 1 ; v 2 ; :::; v t g Let f w 1 ; w 2 ; :::; w n g b e a free basis for B where n t The comp osition R t A B R n e i v i w i e i is represen ted b y a matrix ( a i;j ) 2 R n;t where v i = a 1 ;i w 1 + a 2 ;i w 2 + + a n;i w n By the previous theorem, 9 in v ertible matrixes U 2 R n and V 2 R t suc h that PAGE 127 Chapter 6 App endix 119 U ( a i;j ) V = 0BBBBBBBBB@ d 1 0 0 0 d 2 0 ... 0 d t 0 0 1CCCCCCCCCA with d i j d i +1 Since c hanging the isomorphisms R t A and B R n corresp onds to c hanging the bases f v 1 ; v 2 ; :::; v t g and f w 1 ; w 2 ; :::; w n g ; the theorem follo ws. Theorem 5 If R is a Euclidean domain and M is a nitely generated R mo dule, then M R =d 1 R =d 2 R =d t R m where eac h d i 6 = 0 and d i j d i +1 for 1 i < t Pro of By h yp othesis 9 a nitely generated free mo dule B and a surjectiv e homomorphism B M 0. Let A b e the k ernel, so 0 A B M 0 is a s.e.s. and B = A M : The result no w follo ws from the previous theorem. The w a y Theorem 5 is stated, some or all of the elemen ts d i ma y b e units, and for suc h d i ; R =d i = 0 If w e assume that no d i is a unit, then the elemen ts d 1 ; d 2 ; :::; d t are called invariant factors They are unique up to asso ciates, but w e do not b other with that here. If R = Z and w e select the d i to b e p ositiv e, they are unique. If R = F [ x ] and w e select the d i to b e monic, then they are unique. The splitting in Theorem 5 is not the ultimate b ecause the mo dules R =d i ma y split in to the sum of other cyclic mo dules. T o pro v e this w e need the follo wing Lemma. Lemma Supp ose R is a PID and b and c are nonzero nonunit elemen ts of R Supp ose b and c are relativ ely prime, i.e., there is no prime common to their prime factorizations. Then bR and cR are comaximal ideals. (See p 108 for comaximal.) Pro of There exists an a 2 R with aR = bR + cR : Since a j b and a j c a is a unit, so R = bR + cR Theorem 6 Supp ose R is a PID and d is a nonzero nonunit elemen t of R Assume d = p s 1 1 p s 2 2 p s t t is the prime factorization of d (see b ottom of p 110). Then the natural map R =d R =p s 1 1 R =p s t t is an isomorphism of R mo dules. (The elemen ts p s i i are called elementary divisors of R =d .) Pro of If i 6 = j p s i i and p s j j are relativ ely prime. By the Lemma ab o v e, they are PAGE 128 120 App endix Chapter 6 comaximal and th us b y the Chinese Remainder Theorem, the natural map is a ring isomorphism (page 108). Since the natural map is also an R mo dule homomorphism, it is an R mo dule isomorphism. This theorem carries the splitting as far as it can go, as seen b y the next exercise. Exercise Supp ose R is a PID, p 2 R is a prime elemen t, and s 1. Then the R mo dule R =p s has no prop er submo dule whic h is a summand. T orsion Submo dules This will giv e a little more p ersp ectiv e to this section. Denition Supp ose M is a mo dule o v er a domain R An elemen t m 2 M is said to b e a torsion element if 9 r 2 R with r 6 = 0 and mr = 0 This is the same as sa ying m is dep enden t. If R = Z it is the same as sa ying m has nite order. Denote b y T ( M ) the set of all torsion elemen ts of M If T ( M ) = 0 w e sa y that M is torsion free.Theorem 7 Supp ose M is a mo dule o v er a domain R Then T ( M ) is a submo dule of M and M =T ( M ) is torsion free. Pro of This is a simple exercise. Theorem 8 Supp ose R is a Euclidean domain and M is a nitely generated R mo dule whic h is torsion free. Then M is a free R mo dule, i.e., M R m Pro of This follo ws immediately from Theorem 5. Theorem 9 Supp ose R is a Euclidean domain and M is a nitely generated R mo dule. Then the follo wing s.e.s. splits. 0 T ( M ) M M =T ( M ) 0 Pro of By Theorem 7, M =T ( M ) is torsion free. By Theorem 8, M =T ( M ) is a free R mo dule, and th us there is a splitting map. Of course this theorem is transparen t an yw a y b ecause Theorem 5 giv es a splitting of M in to a torsion part and a free part. PAGE 129 Chapter 6 App endix 121 Note It follo ws from Theorem 9 that 9 a free submo dule V of M suc h that T ( M ) V = M The rst summand T ( M ) is unique, but the complemen tary summand V is not unique. V dep ends up on the splitting map and is unique only up to isomorphism. T o complete this section, here are t w o more theorems that follo w from the w ork w e ha v e done. Theorem 10 Supp ose T is a domain and T is the m ultiplicativ e group of units of T If G is a nite subgroup of T then G is a cyclic group. Th us if F is a nite eld, the m ultiplicativ e group F is cyclic. Th us if p is a prime, ( Z p ) is cyclic. Pro of This is a corollary to Theorem 5 with R = Z The m ultiplicativ e group G is isomorphic to an additiv e group Z =d 1 Z =d 2 Z =d t where eac h d i > 1 and d i j d i +1 for 1 i < t Ev ery u in the additiv e group has the prop ert y that ud t = 0 So ev ery g 2 G is a solution to x d t 1 = 0 If t > 1, the equation will ha v e degree less than the n um b er of ro ots, whic h is imp ossible. Th us t = 1 and so G is cyclic. Exercise F or whic h primes p and q is the group of units ( Z p Z q ) a cyclic group? W e kno w from Exercise 2) on page 59 that an in v ertible matrix o v er a eld is the pro duct of elemen tary matrices. This result also holds for an y in v ertible matrix o v er a Euclidean domain. Theorem 11 Supp ose R is a Euclidean domain and A 2 R n is a matrix with nonzero determinan t. Then b y elemen tary ro w and column op erations, A ma y b e transformed to a diagonal matrix 0BBBB@ d 1 0 d 2 0 d n 1CCCCA where eac h d i 6 = 0 and d i j d i +1 for 1 i < n Also d 1 generates the ideal generated b y the en tries of A F urthermore A is in v ertible i eac h d i is a unit. Th us if A is in v ertible, A is the pro duct of elemen tary matrices. PAGE 130 122 App endix Chapter 6 Pro of It follo ws from Theorem 3 that A ma y b e transformed to a diagonal matrix with d i j d i +1 Since the determinan t of A is not zero, it follo ws that eac h d i 6 = 0 F urthermore, the matrix A is in v ertible i the diagonal matrix is in v ertible, whic h is true i eac h d i is a unit. If eac h d i is a unit, then the diagonal matrix is the pro duct of elemen tary matrices of t yp e 1. Therefore if A is in v ertible, it is the pro duct of elemen tary matrices. Exercise Let R = Z A = 3 11 0 4 and D = 3 11 1 4 P erform elemen tary op erations on A and D to obtain diagonal matrices where the rst diagonal elemen t divides the second diagonal elemen t. W rite D as the pro duct of elemen tary matrices. Find the c haracteristic p olynomials of A and D : Find an elemen tary matrix B o v er Z suc h that B 1 AB is diagonal. Find an in v ertible matrix C in R 2 suc h that C 1 D C is diagonal. Sho w C cannot b e selected in Q 2 Jordan Blo c ks In this section, w e dene the t w o sp ecial t yp es of square matrices used in the Rational and Jordan canonical forms. Note that the Jordan blo c k B ( q ) is the sum of a scalar matrix and a nilp oten t matrix. A Jordan blo c k displa ys its eigen v alue on the diagonal, and is more in teresting than the companion matrix C ( q ). But as w e shall see later, the Rational canonical form will alw a ys exist, while the Jordan canonical form will exist i the c haracteristic p olynomial factors as the pro duct of linear p olynomials. Supp ose R is a comm utativ e ring, q = a 0 + a 1 x + + a n 1 x n 1 + x n 2 R [ x ] is a monic p olynomial of degree n 1, and V is the R [ x ]mo dule V = R [ x ] =q V is a torsion mo dule o v er the ring R [ x ], but as an R mo dule, V has a free basis f 1 ; x; x 2 ; : : : ; x n 1 g (See the last part of the last theorem on page 46.) Multiplication b y x denes an R mo dule endomorphism on V and C ( q ) will b e the matrix of this endomorphism with resp ect to this basis. Let T : V V b e dened b y T ( v ) = v x: If h ( x ) 2 R [ x ], h ( T ) is the R mo dule homomorphism giv en b y m ultiplication b y h ( x ). The homomorphism from R [ x ] =q to R [ x ] =q giv en b y m ultiplication b y h ( x ), is zero i h ( x ) 2 q R [ x ]. That is to sa y q ( T ) = a 0 I + a 1 T + + T n is the zero homomorphism, and h ( T ) is the zero homomorphism i h ( x ) 2 q R [ x ] : All of this is supp osed to mak e the next theorem transparen t. Theorem Let V ha v e the free basis f 1 ; x; x 2 ; :::; x n 1 g The companion matrix PAGE 131 Chapter 6 App endix 123 represen ting T is C ( q ) = 0BBBBBBB@ 0 : : : : : : 0 a 0 1 0 : : : 0 a 1 0 1 0 a 2 ... ... 0 : : : : : : 1 a n 1 1CCCCCCCA The c haracteristic p olynomial of C ( q ) is q ; and j C ( q ) j = ( 1) n a 0 Finally if h ( x ) 2 R [ x ], h ( C ( q )) is zero i h ( x ) 2 q R [ x ]. Theorem Supp ose 2 R and q ( x ) = ( x ) n Let V ha v e the free basis f 1 ; ( x ) ; ( x ) 2 ; : : : ; ( x ) n 1 g : Then the matrix represen ting T is B ( q ) = 0BBBBBBB@ 0 : : : : : : 0 1 0 : : : 0 0 1 ... ... ... 0 : : : : : : 1 1CCCCCCCA The c haracteristic p olynomial of B ( q ) is q ; and j B ( q ) j = n = ( 1) n a 0 Finally if h ( x ) 2 R [ x ], h ( B ( q )) is zero i h ( x ) 2 q R [ x ]. Note F or n = 1, C ( a 0 + x ) = B ( a 0 + x ) = ( a 0 ). This is the only case where a blo c k matrix ma y b e the zero matrix. Note In B ( q ), if y ou wish to ha v e the 1 s ab o v e the diagonal, rev erse the order of the basis for V Jordan Canonical F orm W e are nally ready to pro v e the Rational and Jordan forms. Using the previous sections, all that's left to do is to put the pieces together. (F or an o v erview of Jordan form, read rst the section in Chapter 5, page 96.) PAGE 132 124 App endix Chapter 6 Supp ose R is a comm utativ e ring, V is an R mo dule, and T : V V is an R mo dule homomorphism. Dene a scalar m ultiplication V R [ x ] V b y v ( a 0 + a 1 x + + a r x r ) = v a 0 + T ( v ) a 1 + + T r ( v ) a r Theorem 1 Under this scalar m ultiplication, V is an R [ x ]mo dule. This is just an observ ation, but it is one of the great tric ks in mathematics. Questions ab out the transformation T are transferred to questions ab out the mo dule V o v er the ring R [ x ]. And in the case R is a eld, R [ x ] is a Euclidean domain and so w e kno w almost ev erything ab out V as an R [ x ]mo dule. No w in this section, w e supp ose R is a eld F V is a nitely generated F mo dule, T : V V is a linear transformation and V is an F [ x ]mo dule with v x = T ( v ). Our goal is to select a basis for V suc h that the matrix represen ting T is in some simple form. A submo dule of V F [ x ] is a submo dule of V F whic h is in v arian t under T W e kno w V F [ x ] is the sum of cyclic mo dules from Theorems 5 and 6 in the section on Euclidean Domains. Since V is nitely generated as an F mo dule, the free part of this decomp osition will b e zero. In the section on Jordan Blo c ks, a basis is selected for these cyclic mo dules and the matrix represen ting T is describ ed. This giv es the Rational Canonical F orm and that is all there is to it. If all the eigen v alues for T are in F w e pic k another basis for eac h of the cyclic mo dules (see the second theorem in the section on Jordan Blo c ks). Then the matrix represen ting T is called the Jordan Canonical F orm. No w w e sa y all this again with a little more detail. F rom Theorem 5 in the section on Euclidean Domains, it follo ws that V F [ x ] F [ x ] =d 1 F [ x ] =d 2 F [ x ] =d t where eac h d i is a monic p olynomial of degree 1, and d i j d i +1 Pic k f 1 ; x; x 2 ; : : : ; x m 1 g as the F basis for F [ x ] =d i where m is the degree of the p olynomial d i Theorem 2 With resp ect to this basis, the matrix represen ting T is 0BBBBBBBBB@ C ( d 1 ) C ( d 2 ) C ( d t ) 1CCCCCCCCCA PAGE 133 Chapter 6 App endix 125 The c haracteristic p olynomial of T is p = d 1 d 2 d t and p ( T ) = 0 This is a t yp e of canonical form but it do es not seem to ha v e a name. No w w e apply Theorem 6 to eac h F [ x ] =d i This giv es V F [ x ] F [ x ] =p s 1 1 F [ x ] =p s r r where the p i are irreducible monic p olynomials of degree at least 1. The p i need not b e distinct. Pic k an F basis for eac h F [ x ] =p s i i as b efore. Theorem 3 With resp ect to this basis, the matrix represen ting T is 0BBBBBBB@ C ( p s 1 1 ) C ( p s 2 2 ) 0 0 C ( p s r r ) 1CCCCCCCA The c haracteristic p olynomial of T is p = p s 1 1 p s r r and p ( T ) = 0 This is called the R ational c anonic al form for T No w supp ose the c haracteristic p olynomial of T factors in F [ x ] as the pro duct of linear p olynomials. Th us in the Theorem ab o v e, p i = x i and V F [ x ] F [ x ] = ( x 1 ) s 1 F [ x ] = ( x r ) s r is an isomorphism of F [ x ]mo dules. Pic k f 1 ; ( x i ) ; ( x i ) 2 ; : : : ; ( x i ) m 1 g as the F basis for F [ x ] = ( x i ) s i where m is s i Theorem 4 With resp ect to this basis, the matrix represen ting T is 0BBBBBBBBBBB@ B (( x 1 ) s 1 ) 0 B (( x 2 ) s 2 ) 0 B (( x r ) s r ) 1CCCCCCCCCCCA PAGE 134 126 App endix Chapter 6 The c haracteristic p olynomial of T is p = ( x 1 ) s 1 ( x r ) s r and p ( T ) = 0 This is called the Jor dan c anonic al form for T : Note that the i need not b e distinct. Note A diagonal matrix is in Rational canonical form and in Jordan canonical form. This is the case where eac h blo c k is one b y one. Of course a diagonal matrix is ab out as canonical as y ou can get. Note also that if a matrix is in Jordan form, its trace is the sum of the eigen v alues and its determinan t is the pro duct of the eigen v alues. Finally this section is lo osely written, so it is imp ortan t to use the transp ose principle to write three other v ersions of the last t w o theorems. Exercise Supp ose F is a eld of c haracteristic 0 and T 2 F n has trace( T i ) = 0 for 0 < i n Sho w T is nilp oten t. Let p 2 F [ x ] b e the c haracteristic p olynomial of T The p olynomial p ma y not factor in to linears in F [ x ], and th us T ma y ha v e no conjugate in F n whic h is in Jordan form. Ho w ev er this exercise can still b e w ork ed using Jordan form. This is based on the fact that there exists a eld F con taining F as a subeld, suc h that p factors in to linears in F [ x ]. This fact is not pro v ed in this b o ok, but it is assumed for this exercise. So 9 an in v ertible matrix U 2 F n so that U 1 T U is in Jordan form, and of course, T is nilp oten t i U 1 T U is nilp oten t. The p oin t is that it sucies to consider the case where T is in Jordan form, and to sho w the diagonal elemen ts are all zero. So supp ose T is in Jordan form and trace ( T i ) = 0 for 1 i n Th us trace ( p ( T )) = a 0 n where a 0 is the constan t term of p ( x ). W e kno w p ( T ) = 0 and th us trace ( p ( T )) = 0 and th us a 0 n = 0 Since the eld has c haracteristic 0, a 0 = 0 and so 0 is an eigen v alue of T This means that one blo c k of T is a strictly lo w er triangular matrix. Remo ving this blo c k lea v es a smaller matrix whic h still satises the h yp othesis, and the result follo ws b y induction on the size of T This exercise illustrates the p o w er and facilit y of Jordan form. It also has a cute corollary Corollary Supp ose F is a eld of c haracteristic 0, n 1, and ( 1 ; 2 ; ::; n ) 2 F n satises i1 + i2 + + in = 0 for eac h 1 i n: Then i = 0 for 1 i n Minimal p olynomials T o conclude this section here are a few commen ts on the minimal p olynomial of a linear transformation. This part should b e studied only if y ou need it. Supp ose V is an n dimensional v ector space o v er a eld F and T : V V is a linear transformation. As b efore w e mak e V a mo dule o v er F [ x ] with T ( v ) = v x PAGE 135 Chapter 6 App endix 127 Denition A nn ( V F [ x ] ) is the set of all h 2 F [ x ] whic h annihilate V i.e., whic h satisfy V h = 0 This is a nonzero ideal of F [ x ] and is th us generated b y a unique monic p olynomial u ( x ) 2 F ( x ) ; A nn ( V F [ x ] ) = uF [ x ]. The p olynomial u is called the minimal p olynomial of T : Note that u ( T ) = 0 and if h ( x ) 2 F [ x ], h ( T ) = 0 i h is a m ultiple of u in F [ x ] : If p ( x ) 2 F [ x ] is the c haracteristic p olynomial of T p ( T ) = 0 and th us p is a m ultiple of u No w w e state this again in terms of matrices. Supp ose A 2 F n is a matrix represen ting T Then u ( A ) = 0 and if h ( x ) 2 F [ x ], h ( A ) = 0 i h is a m ultiple of u in F [ x ]. If p ( x ) 2 F [ x ] is the c haracteristic p olynomial of A then p ( A ) = 0 and th us p is a m ultiple of u The p olynomial u is also called the minimal p olynomial of A Note that these prop erties hold for an y matrix represen ting T and th us similar matrices ha v e the same minimal p olynomial. If A is giv en to start with, use the linear transformation T : F n F n determined b y A to dene the p olynomial u No w supp ose q 2 F [ x ] is a monic p olynomial and C ( q ) 2 F n is the companion matrix dened in the section Jordan Blo c ks. Whenev er q ( x ) = ( x ) n let B ( q ) 2 F n b e the Jordan blo c k matrix also dened in that section. Recall that q is the c haracteristic p olynomial and the minimal p olynomial of eac h of these matrices. This together with the rational form and the Jordan form will allo w us to understand the relation of the minimal p olynomial to the c haracteristic p olynomial. Exercise Supp ose A i 2 F n i has q i as its c haracteristic p olynomial and its minimal p olynomial, and A = 0BBBB@ A 1 0 A 2 0 A r 1CCCCA : Find the c haracteristic p olynomial and the minimal p olynomial of A Exercise Supp ose A 2 F n 1) Supp ose A is the matrix displa y ed in Theorem 2 ab o v e. Find the c haracteristic and minimal p olynomials of A 2) Supp ose A is the matrix displa y ed in Theorem 3 ab o v e. Find the c haracteristic and minimal p olynomials of A 3) Supp ose A is the matrix displa y ed in Theorem 4 ab o v e. Find the c haracteristic and minimal p olynomials of A PAGE 136 128 App endix Chapter 6 4) Supp ose 2 F Sho w is a ro ot of the c haracteristic p olynomial of A i is a ro ot of the minimal p olynomial of A Sho w that if is a ro ot, its order in the c haracteristic p olynomial is at least as large as its order in the minimal p olynomial. 5) Supp ose F is a eld con taining F as a subeld. Sho w that the minimal p olynomial of A 2 F n is the same as the minimal p olynomial of A considered as a matrix in F n (This funn y lo oking exercise is a little delicate.) 6) Let F = R and A = 0B@ 5 1 3 0 2 0 3 1 1 1CA Find the c haracteristic and minimal p olynomials of A Determinan ts In the c hapter on matrices, it is stated without pro of that the determinan t of the pro duct is the pro duct of the determinan ts (see page 63). The purp ose of this section is to giv e a pro of of this. W e supp ose R is a comm utativ e ring, C is an R mo dule, n 2, and B 1 ; B 2 ; : : : ; B n is a sequence of R mo dules. Denition A map f : B 1 B 2 B n C is R multiline ar means that if 1 i n and b j 2 B j for j 6 = i then f j ( b 1 ; b 2 ; : : : ; B i ; : : : ; b n ) denes an R linear map from B i to C Theorem The set of all R m ultilinear maps is an R mo dule. Pro of F rom the rst exercise in Chapter 5, the set of all functions from B 1 B 2 B n to C is an R mo dule (see page 69). It m ust b e seen that the R m ultilinear maps form a submo dule. It is easy to see that if f 1 and f 2 are R m ultilinear, so is f 1 + f 2 : Also if f is R m ultilinear and r 2 R then ( f r ) is R m ultilinear. F rom here on, supp ose B 1 = B 2 = = B n = B Denition 1) f is symmetric means f ( b 1 ; : : : ; b n ) = f ( b (1) ; : : : ; b ( n ) ) for all p erm utations on f 1 ; 2 ; : : : ; n g 2) f is skewsymmetric if f ( b 1 ; : : : ; b n ) = sign( ) f ( b (1) ; : : : ; b ( n ) ) for all PAGE 137 Chapter 6 App endix 129 3) f is alternating if f ( b 1 ; : : : ; b n ) = 0 whenev er some b i = b j for i 6 = j Theorem i) Eac h of these three t yp es denes a submo dule of the set of all R m ultilinear maps. ii) Alternating ) sk ewsymmetric. iii) If no elemen t of C has order 2, then alternating ( ) sk ewsymmetric. Pro of P art i) is immediate. T o pro v e ii), assume f is alternating. It sucies to sho w that f ( b 1 ; :::; b n ) = f ( b (1) ; :::; b ( n ) ) where is a transp osition. F or simplicit y assume = (1 ; 2). Then 0 = f ( b 1 + b 2 ; b 1 + b 2 ; b 3 ; :::; b n ) = f ( b 1 ; b 2 ; b 3 ; :::; b n ) + f ( b 2 ; b 1 ; b 3 ; :::; b n ) and the result follo ws. T o pro v e iii), supp ose f is sk ew symmetric and no elemen t of C has order 2, and sho w f is alternating. Supp ose for con v enience that b 1 = b 2 and sho w f ( b 1 ; b 1 ; b 3 ; : : : ; b n ) = 0 If w e let b e the transp osition (1 ; 2), w e get f ( b 1 ; b 1 ; b 3 ; : : : ; b n ) = f ( b 1 ; b 1 ; b 3 ; : : : ; b n ), and so 2 f ( b 1 ; b 1 ; b 3 ; : : : ; b n ) = 0 and the result follo ws. No w w e are ready for determinan t. Supp ose C = R In this case m ultilinear maps are usually called multiline ar forms Supp ose B is R n with the canonical basis f e 1 ; e 2 ; : : : ; e n g (W e think of a matrix A 2 R n as n column v ectors, i.e., as an elemen t of B B B .) First w e recall the denition of determinan t. Supp ose A = ( a i;j ) 2 R n Dene d : B B B R b y d ( a 1 ; 1 e 1 + a 2 ; 1 e 2 + + a n; 1 e n ; :::::; a 1 ;n e 1 + a 2 ;n e 2 + + a n;n e n ) = P all sign( )( a (1) ; 1 a (2) ; 2 a ( n ) ;n ) = j A j The next theorem follo ws from the section on determinan ts on page 61. Theorem d is an alternating m ultilinear form with d ( e 1 ; e 2 ; : : : ; e n ) = 1 If c 2 R dc is an alternating m ultilinear form, b ecause the set of alternating forms is an R mo dule. It turns out that this is all of them, as seen b y the follo wing theorem. Theorem Supp ose f : B B : : : B R is an alternating m ultilinear form. Then f = d f ( e 1 ; e 2 ; : : : ; e n ). This means f is the m ultilinear form d times the scalar f ( e 1 ; e 2 ; :::; e n ). In other w ords, if A = ( a i;j ) 2 R n then f ( a 1 ; 1 e 1 + a 2 ; 1 e 2 + + a n; 1 e n ; :::::; a 1 ;n e 2 + a 2 ;n e 2 + + a n;n e n ) = j A j f ( e 1 ; e 2 ; :::; e n ). Th us the set of alternating forms is a free R mo dule of dimension 1, and the determinan t is a generator. PAGE 138 130 App endix Chapter 6 Pro of F or n = 2, y ou can simply write it out. f ( a 1 ; 1 e 1 + a 2 ; 1 e 2 ; a 1 ; 2 e 1 + a 2 ; 2 e 2 ) = a 1 ; 1 a 1 ; 2 f ( e 1 ; e 1 ) + a 1 ; 1 a 2 ; 2 f ( e 1 ; e 2 ) + a 2 ; 1 a 1 ; 2 f ( e 2 ; e 1 ) + a 2 ; 1 a 2 ; 2 f ( e 2 ; e 2 ) = ( a 1 ; 1 a 2 ; 2 a 1 ; 2 a 2 ; 1 ) f ( e 1 ; e 2 ) = j A j f ( e 1 ; e 2 ) : F or the general case, f ( a 1 ; 1 e 1 + a 2 ; 1 e 2 + + a n; 1 e n ; :::::; a 1 ;n e 1 + a 2 ;n e 2 + + a n;n e n ) = P a i 1 ; 1 a i 2 ; 2 a i n ;n f ( e i 1 ; e i 2 ; :::; e i n ) where the sum is o v er all 1 i 1 n; 1 i 2 n; :::; 1 i n n: Ho w ev er, if an y i s = i t for s 6 = t that term is 0 b ecause f is alternating. Therefore the sum is just P all a (1) ; 1 a (2) ; 2 a ( n ) ;n f ( e (1) ; e (2) ; : : : ; e ( n ) ) = P all sign( ) a (1) ; 1 a (2) ; 2 a ( n ) ;n f ( e 1 ; e 2 ; : : : ; e n ) = j A j f ( e 1 ; e 2 ; :::; e n ). This incredible classication of these alternating forms mak es the pro of of the follo wing theorem easy (See the third theorem on page 63.) Theorem If C ; A 2 R n then j C A j = j C jj A j Pro of Supp ose C 2 R n Dene f : R n R b y f ( A ) = j C A j In the notation of the previous theorem, B = R n and R n = R n R n R n If A 2 R n ; A = ( A 1 ; A 2 ; :::; A n ) where A i 2 R n is column i of A and f : R n R n R has f ( A 1 ; A 2 ; :::; A n ) = j C A j Use the fact that C A = ( C A 1 ; C A 2 ; :::; C A n ) to sho w that f is an alternating m ultilinear form. By the previous theorem, f ( A ) = j A j f ( e 1 ; e 2 ; :::; e n ). Since f ( e 1 ; e 2 ; :::; e n ) = j C I j = j C j it follo ws that j C A j = f ( A ) = j A jj C j Dual Spaces The concept of dual mo dule is basic, not only in algebra, but also in other areas suc h as dieren tial geometry and top ology If V is a nitely generated v ector space o v er a eld F its dual V is dened as V = Hom F ( V ; F ). V is isomorphic to V but in general there is no natural isomorphism from V to V Ho w ev er there is a natural isomorphism from V to V and so V is the dual of V and V ma y b e considered to b e the dual of V This remark able fact has man y expressions in mathematics. F or example, a tangen t plane to a dieren tiable manifold is a real v ector space. The union of these spaces is the tangen t bundle, while the union of the dual spaces is the cotangen t bundle. Th us the tangen t (cotangen t) bundle ma y b e considered to b e the dual of the cotangen t (tangen t) bundle. The sections of the tangen t bundle are called v ector elds while the sections of the cotangen t bundle are called 1forms. In algebraic top ology homology groups are deriv ed from c hain complexes, while cohomology groups are deriv ed from the dual c hain complexes. The sum of the cohomology groups forms a ring, while the sum of the homology groups do es not. PAGE 139 Chapter 6 App endix 131 Th us the concept of dual mo dule has considerable p o w er. W e dev elop here the basic theory of dual mo dules. Supp ose R is a comm utativ e ring and W is an R mo dule. Denition If M is an R mo dule, let H ( M ) b e the R mo dule H ( M )=Hom R ( M ; W ) : If M and N are R mo dules and g : M N is an R mo dule homomorphism, let H ( g ) : H ( N ) H ( M ) b e dened b y H ( g )( f ) = f g Note that H ( g ) is an R mo dule homomorphism. M N W f g H ( g )( f ) = f g ? Z Z Z Z Z Z Z Z ~ Theorem i) If M 1 and M 2 are R mo dules, H ( M 1 M 2 ) H ( M 1 ) H ( M 2 ). ii) If I : M M is the iden tit y then H ( I ) : H ( M ) H ( M ) is the iden tit y iii) If M 1 g M 2 h M 3 are R mo dule homomorphisms, then H ( g ) H ( h ) = H ( h g ) : If f : M 3 W is a homomorphism, then ( H ( g ) H ( h ))( f ) = H ( h g )( f ) = f h g M 1 M 2 M 3 g h f f h W f h g ? P P P P P P P P P P P P P P P q Z Z Z Z Z Z ~ Note In the language of the category theory H is a con tra v arian t functor from the category of R mo dules to itself. PAGE 140 132 App endix Chapter 6 Theorem If M and N are R mo dules and g : M N is an isomorphism, then H ( g ) : H ( N ) H ( M ) is an isomorphism with H ( g 1 ) = H ( g ) 1 Pro of I H ( N ) = H ( I N ) = H ( g g 1 ) = H ( g 1 ) H ( g ) I H ( M ) = H ( I M ) = H ( g 1 g ) = H ( g ) H ( g 1 ) Theorem i) If g : M N is a surjectiv e homomorphism, then H ( g ) : H ( N ) H ( M ) is injectiv e. ii) If g : M N is an injectiv e homomorphism and g ( M ) is a summand of N ; then H ( g ) : H ( N ) H ( M ) is surjectiv e. iii) If R is a eld and g : M N is a homomorphism, then g is surjectiv e (injectiv e) i H ( g ) is injectiv e (surjectiv e). Pro of This is a go o d exercise. F or the remainder of this section, supp ose W = R R : In this case H ( M ) = Hom R ( M ; R ) is denoted b y H ( M ) = M and H ( g ) is denoted b y H ( g ) = g Theorem Supp ose M has a nite free basis f v 1 ; :::; v n g Dene v i 2 M b y v i ( v 1 r 1 + + v n r n ) = r i Th us v i ( v j ) = i;j Then v 1 ; : : : ; v n is a free basis for M called the dual b asis Therefore M is free and is isomorphic to M Pro of First consider the case of R n = R n; 1 with basis f e 1 ; : : : ; e n g where e i = 0BBBBBB@ 0 1 i 0 1CCCCCCA W e kno w ( R n ) R 1 ;n i.e., an y homomorphism from R n to R is giv en b y a 1 n matrix. No w R 1 ;n is free with dual basis f e 1 ; : : : ; e n g where e i = (0 ; : : : ; 0 ; 1 i ; 0 ; : : : ; 0). F or the general case, let g : R n M b e giv en b y g ( e i ) = v i Then g : M ( R n ) sends v i to e i : Since g is an isomorphism, f v 1 ; : : : ; v n g is a basis for M Theorem Supp ose M is a free mo dule with a basis f v 1 ; : : : ; v m g and N is a free mo dule with a basis f w 1 ; : : : ; w n g and g : M N is the homomorphism giv en b y A = ( a i;j ) 2 R n;m This means g ( v j ) = a 1 ;j w 1 + + a n;j w n Then the matrix of g : N M with resp ect to the dual bases, is giv en b y A t PAGE 141 Chapter 6 App endix 133 Pro of Note that g ( w i ) is a homomorphism from M to R : Ev aluation on v j giv es g ( w i )( v j ) = ( w i g )( v j ) = w i ( g ( v j )) = w i ( a 1 ;j w 1 + + a n;j w n ) = a i;j Th us g ( w i ) = a i; 1 v 1 + + a i;m v m ; and th us g is represen ted b y A t Exercise If U is an R mo dule, dene U : U U R b y U ( f ; u ) = f ( u ). Sho w that U is R bilinear. Supp ose g : M N is an R mo dule homomorphism, f 2 N and v 2 M Sho w that N ( f ; g ( v )) = M ( g ( f ) ; v ). No w supp ose M = N = R n and g : R n R n is represen ted b y a matrix A 2 R n Supp ose f 2 ( R n ) and v 2 R n Use the theorem ab o v e to sho w that : ( R n ) R n R has the prop ert y ( f ; Av ) = ( A t f ; v ). This is with the elemen ts of R n and ( R n ) written as column v ectors. If the elemen ts of R n are written as column v ectors and the elemen ts of ( R n ) are written as ro w v ectors, the form ula is ( f ; Av ) = ( f A; v ). Of course this is just the matrix pro duct f Av Dual spaces are confusing, and this exercise should b e w ork ed out completely Denition \Double dual" is a \co v arian t" functor, i.e., if g : M N is a homomorphism, then g : M N F or an y mo dule M dene : M M b y ( m ) : M R is the homomorphism whic h sends f 2 M to f ( m ) 2 R i.e., ( m ) is giv en b y ev aluation at m: Note that is a homomorphism. Theorem If M and N are R mo dules and g : M N is a homomorphism, then the follo wing diagram is comm utativ e. M M N N g g ? ? Pro of On M ; is giv en b y ( v ) = M ( ; v ) : On N ; ( u ) = N ( ; u ). The pro of follo ws from the equation N ( f ; g ( v )) = M ( g ( f ) ; v ). Theorem If M is a free R mo dule with a nite basis f v 1 ; : : : ; v n g then : M M is an isomorphism. Pro of f ( v 1 ) ; : : : ; ( v n ) g is the dual basis of f v 1 ; : : : ; v n g ; i.e., ( v i ) = ( v i ) PAGE 142 134 App endix Chapter 6 Note Supp ose R is a eld and C is the category of nitely generated v ector spaces o v er R In the language of category theory is a natural equiv alence b et w een the iden tit y functor and the double dual functor. Note F or nitely generated v ector spaces, is used to iden tify V and V Under this iden tication V is the dual of V and V is the dual of V Also, if f v 1 ; : : : ; v n g is a basis for V and f v i ; : : : ; v n g its dual basis, then f v 1 ; : : : ; v n g is the dual basis for f v 1 ; : : : ; v n g In general there is no natural w a y to iden tify V and V Ho w ev er for real inner pro duct spaces there is. Theorem Let R = R and V b e an n dimensional real inner pro duct space. Then : V V giv en b y ( v ) = ( v ; ) is an isomorphism. Pro of is injectiv e and V and V ha v e the same dimension. Note If is used to iden tify V with V then V : V V R is just the dot pro duct V V R Note If f v 1 ; : : : ; v n g is an y orthonormal basis for V ; f ( v 1 ) ; : : : ; ( v n ) g is the dual basis of f v 1 ; : : : ; v n g that is ( v i ) = v i The isomorphism : V V denes an inner pro duct on V and under this structure, is an isometry If f v 1 ; : : : ; v n g is an orthonormal basis for V ; f v 1 ; : : : ; v n g is an orthonormal basis for V : Also, if U is another n dimensional IPS and f : V U is an isometry then f : U V is an isometry and the follo wing diagram comm utes. V V U U f f ? 6 Exercise Supp ose R is a comm utativ e ring, T is an innite index set, and for eac h t 2 T R t = R Sho w ( M t 2 T R t ) is isomorphic to R T = Y t 2 T R t No w let T = Z + ; R = R and M = M t 2 T R t Sho w M is not isomorphic to M PAGE 143 IndexAb elian group, 20, 71 Algebraically closed eld, 46, 97 Alternating group, 32 Ascending c hain condition, 112 Asso ciate elemen ts in a domain, 47, 109 Automorphism of groups, 29 of mo dules, 70 of rings, 43 Axiom of c hoice, 10 Basis or free basis canonical or standard for R n 72, 79 of a mo dule, 78, 83 Bijectiv e or onetoone corresp ondence,7 Binary op eration, 19 Bo olean algebras, 52 Bo olean rings, 51 Cancellation la w in a group, 20 in a ring, 39 Cartesian pro duct, 2, 11 Ca yley's theorem, 31 Ca yleyHamilton theorem, 66, 98, 125 Cen ter of group, 22 Change of basis, 83 Characteristic of a ring, 50 Characteristic p olynomial of a homomorphism, 85, 95 of a matrix, 66 Chinese remainder theorem, 50, 108 Classical adjoin t of a matrix, 63 Cofactor of a matrix, 62 Comaximal ideals, 108, 120 Comm utativ e ring, 37 Complex n um b ers, 1, 40, 46, 47, 97, 104 Conjugate, 64 Conjugation b y a unit, 44 Con tra v arian t functor, 131 Copro duct or sum of mo dules, 76 Coset, 24, 42, 74 Cycle, 32 Cyclic group, 23 mo dule, 107 Determinan t of a homomorphism, 85 of a matrix, 60, 128 Diagonal matrix, 56 Dimension of a free mo dule, 83 Division algorithm, 45 Domain euclidean, 116 in tegral domain, 39 of a function, 5 principal ideal, 46 unique factorization, 111 Dual basis, 132 Dual spaces, 130 Eigen v alues, 95 Eigen v ectors, 95 Elemen tary divisors, 119, 120 Elemen tary matrices, 58 135 PAGE 144 136 Index Elemen tary op erations, 57, 122 Endomorphism of a mo dule, 70 Equiv alence class, 4 Equiv alence relation, 4 Euclidean algorithm, 14 Euclidean domain, 116 Ev aluation map, 47, 49 Ev en p erm utation, 32 Exp onen tial of a matrix, 106 F actorization domain (FD), 111 F ermat's little theorem, 50 Field, 39 F ormal p o w er series, 113 F ourier series, 100 F ree basis, 72, 78, 79, 83 F ree R mo dule, 78 F unction or map, 6 bijectiv e, 7 injectiv e, 7 surjectiv e, 7 F unction space Y T as a group, 22, 36 as a mo dule, 69 as a ring, 44 as a set, 12 F undamen tal theorem of algebra, 46 Gauss, 113 General linear group GL n ( R ), 55 Generating sequence in a mo dule, 78 Generators of Z n 40 Geometry of determinan t, 90 GramSc hmidt orthonormalization, 100 Graph of a function, 6 Greatest common divisor, 15 Group, 19 ab elian, 20 additiv e, 20 cyclic, 23 m ultiplicativ e, 19 symmetric, 31 Hausdor maximalit y principle, 3, 87, 109 Hilb ert, 113 Homogeneous equation, 60 Homormophism of groups, 23 of rings, 42 of mo dules, 69 Homomorphism of quotien t group, 29 mo dule, 74 ring, 44 Ideal left, 41 maximal, 109 of a ring, 41 prime, 109 principal, 42, 46 righ t, 41 Idemp oten t elemen t in a ring, 49, 51 Image of a function, 7 Indep enden t sequence in a mo dule, 78 Index of a subgroup, 25 Index set, 2 Induction, 13 Injectiv e or onetoone, 7, 79 Inner pro duct spaces, 98 In tegers mo d n 27, 40 In tegers, 1, 14 In v arian t factors, 119 In v erse image, 7 In v ertible or nonsingular matrix, 55 Irreducible elemen t, 47, 110 Isometries of a square, 26, 34 Isometry 101 Isomorphism PAGE 145 Index 137 of groups, 29 of mo dules, 70 of rings, 43 Jacobian matrix, 91 Jordan blo c k, 96, 123 Jordan canonical form, 96, 123, 125 Kernel, 28, 43, 70 Least common m ultiple, 17, 18 Linear com bination, 78 Linear ordering, 3 Linear transformation, 85 Matrix elemen tary 58 in v ertible, 55 represen ting a linear transformation, 84 triangular, 56 Maximal ideal, 109 indep enden t sequence, 86, 87 monotonic sub collection, 4 subgroup, 114 Minimal p olynomial, 127 Minor of a matrix, 62 Mo dule o v er a ring, 68 Monomial, 48 Monotonic collection of sets, 4 Multilinear forms, 129 Multiplicativ e group of a nite eld, 121 Nilp oten t elemen t, 56 homomorphism, 93 No etherian ring, 112 Normal subgroup, 26 Odd p erm utation, 32 On to or surjectiv e, 7, 79 Order of an elemen t or group, 23 Orthogonal group O ( n ), 102 Orthogonal v ectors, 99 Orthonormal sequence, 99 P artial ordering, 3 P artition of a set, 5 P erm utation, 31 Pigeonhole principle, 8, 39 P olynomial ring, 45 P o w er set, 12 Prime elemen t, 110 ideal, 109 in teger, 16 Principal ideal domain (PID), 46 Principal ideal, 42 Pro duct of groups, 34, 35 of mo dules, 75 of rings, 49 of sets, 2, 11 Pro jection maps, 11 Quotien t group, 27 Quotien t mo dule, 74 Quotien t ring, 42 Range of a function, 6 Rank of a matrix, 59, 89 Rational canonical form, 107, 125 Relation, 3 Relativ ely prime in tegers, 16 elemen ts in a PID, 119 Righ t and left in v erses of functions, 10 Ring, 38 Ro ot of a p olynomial, 46 Ro w ec helon form, 59 Scalar matrix, 57 PAGE 146 138 Index Scalar m ultiplication, 21, 38, 54, 71 Self adjoin t, 103, 105 Short exact sequence, 115 Sign of a p erm utation, 60 Similar matrices, 64 Solutions of equations, 9, 59, 81 Splitting map, 114 Standard basis for R n 72, 79 Strips (horizon tal and v ertical), 8 Subgroup, 14, 21 Submo dule, 69 Subring, 41 Summand of a mo dule, 77, 115 Surjectiv e or on to, 7, 79 Symmetric groups, 31 Symmetric matrix, 103 T orsion elemen t of a mo dule, 121 T race of a homormophism, 85 of a matrix, 65 T ransp ose of a matrix, 56, 103, 132 T ransp osition, 32 Unique factorization, in principal ideal domains, 113 of in tegers, 16 Unique factorization domain (UFD), 111 Unit in a ring, 38 V ector space, 67, 85 V olume preserving homomorphism, 90 Zero divisor in a ring, 39 