6. EIGENVALUES AND EIGENVECTORS 3 = 3 2


 Kelly Holly Griffin
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1 EIGENVALUES AND EIGENVECTORS The Characterstc Polynomal If A s a square matrx and v s a nonzero vector such that Av v we say that v s an egenvector of A and s the correspondng egenvalue Av v Example : Let A and v Then Av So s an egenvalue of A and s a correspondng egenvector The name egen comes from German and means characterstc The egenvalues are mportant characterstcs of the matrx n so far as they contan valuable nformaton about the matrx Egenvalues are sometmes called characterstc roots or latent roots The word latent means hdden The latent roots, or egenvalues, are bured somewhat and need a certan amount of work for them to be revealed The tradtonal symbol for egenvalues s the Greek letter lambda () (correspondng to the Englsh letter lower case L), probably because of the word latent The characterstc polynomal of a square matrx A s I A Ths s a polynomal n We denote the characterstc polynomal of A by χ A () or, f the matrx A s understood, just χ() The word characterstc n Englsh s derved from the Greek χαρακτηριστικ meanng the same thng In turn ths comes from the Greek word χαρακτηρ meanng a stampng tool, a tool for stampng a desgn or symbol on an object Ths s why t s tradtonal to use the Greek letter χ Example : If A fnd χ A () Soluton: ( )( ) Theorem : The egenvalues of a square matrx A are the zeros of ts characterstc polynomal Proof: s an egenvalue of A f and only f Av v has a nonzero soluton for v Ths holds f and only f I A whch means that s a zero of the characterstc polynomal Example : Fnd the characterstc polynomal of the matrx A egenvalues and egenvectors Soluton: χ A () and hence fnd ts
2 ( )( ) ( )( ) Hence the egenvalues are, For each egenvalue we must fnd the nonzero solutons to the equaton x y However t s more convenent to use the equvalent equaton (A I)v, because we can smply take the orgnal matrx and subtract off the egenvalue from the dagonal components For we consder x y, that s x y Now ths has nfntely many solutons x k, y k, for any k, but we smply quote any nonzero soluton, such as For we solve x y, gvng the egenvector A magc square s an n n matrx where each row, each column and each dagonal has the same total It s generally arranged for the components to be the ntegers from to n, each once but ths s not oblgatory For example s a magc square A very famous magc square appears n a woodcut by Albrecht Dürer [ ] It s: For a magc square, wth row/column/dagonal total T, we can easly see that T s an egenvalue snce s an egenvector correspondng to T For the Dürer square T The other egenvalues are and ± Fndng the characterstc polynomal and solvng t s the normal way to fnd egenvalues And once we fnd the egenvalues we can fnd the correspondng egenvectors But, n the case of magc squares, we can fnd an egenvector frst and then fnd ts egenvalue Often we can guess all the egenvectors and so get all the egenvalues, and then the characterstc polynomal the exact reverse of the usual procedure Consder the followng example
3 Example : Fnd the egenvalues and egenvectors of A Soluton: It would take a lot of work to follow the tradtonal path of computng the characterstc polynomal frst It turns out to be and so there would reman the dffcult job of fndng ts zeros But note that the sum of every row s the same Ths means that s an egenvector, for s just another way of sayng that every row adds to So n ths case we ve found the egenvector frst, and then the egenvalue What about the other egenvalues? Note that each row s the same as the one above but rotated one place to the left, wth the component that falls off the lefthand end gong down to the rghthand end Ths pattern can be k encapsulated by the equaton ( k k k k ) f k k k k k k So for every th root of, s an egenvector and k k k s the correspondng k k egenvalue Puttng k,, and we conclude that the egenvalues of A are, ± and So χ A () ( )( )( )( ) Theorem : Let A be an n n matrx wth egenvalues,,, n (Here multple zeros are counted accordng to ther multplctes) () tr(a) n (trace s the sum of ts egenvalues); () A n (determnant s the product of the egenvalues) Proof: () Let A (a j ) be a square matrx and let ts characterstc polynomal be n c n n c c Then the sum of the egenvalues s c n But the only term of degree n n I A comes from ( a )( a ) ( a nn ) and ths has coeffcent (a a a nn ) tr(a) So c n tr(a) and hence the sum of the egenvalues s tr(a) () The product of the egenvalues s () n c Now the constant term s the value of the characterstc polynomal when A () n A (Remember that takng out the factor of from each row changes the sgn of the determnant) Hence A n
4 Theorem : The egenvalues of an uppertrangular matrx, or a lowertrangular matrx, are the dagonal components a a a Proof: n a an a nn ( a )( a ) ( a nn ) Smlar Matrces Two n n matrces A, B are sad to be smlar f B S AS for some nvertble matrx S In ordnary algebra, the algebra of a feld, the equaton B S AS would gve us B A by cancellng the S and the S But whle such remote cancellng s permssble n a feld, t s not permtted for a noncommutatve system such as the system of n n matrces We can only cancel f the element and ts nverse are adjacent In a feld t doesn t matter f they re not because we can rearrange terms untl they are For matrces we just can t do that because matrces n general don t commute Theorem : Smlarty s an equvalence relaton Proof: Reflexve: Suppose A s an n n matrx Then A I AI so A s smlar to A Symmetrc: Suppose A s smlar to B Then A S BS for some nvertble S Hence B SAS (S ) A(S ) and so B s smlar to A Transtve: Suppose A s smlar to B and B s smlar to C Then A S BS and B T CT for some nvertble S, T A S (T CT)S (TS) C(TS) so A s smlar to C Smlar matrces are ndeed smlar n the nontechncal sense n that they have many common propertes Theorem : Smlar matrces have the same characterstc polynomal Proof: Suppose that B S AS for some nvertble matrx S The characterstc polynomal for B s I B I S AB S IS S AS S (I A)S S I A S I A Corollary: Smlar matrces have the same determnant and the same trace Propertes of Egenvalues If p(x) s a polynomal and A s an n n matrx (wth the coeffcents of p(x) and the components of A comng from the same feld) we defne p(a) to be the n n matrx obtaned by substtutng A for x and replacng the constant term, a, by a I Example : If p(x) x x and A then p(a)
5 Theorem : If p(x) s a polynomal the egenvectors of p(a) are the same as for A If the egenvalues of A are,,, n, the egenvalues of p(a) are p( ), p( ),, p( n ) Proof: Suppose p(x) a n x n a n x n a x a and suppose that Av v Then p(a)v (a n A n a n A n a A a I)v (a n n a n n a a )v Theorem : If A s nvertble the egenvectors of A are the same as for A If the egenvalues of A are,,, n, the egenvalues of A are,,, n Proof: Suppose Av v Then multplyng both sdes on the left by A, v A v so A v v Theorem : The egenvalues of A T are the same as those for A (The egenvectors wll generally be dfferent) Proof: I A T (I A) T I A Theorem : If p(a) then every egenvalue of A s a zero of p(x) Proof: Suppose Av v Then A r r v for all r and so p()v p(a)v Snce v, p() Example : () A nlpotent matrx s one for whch A m for some m The egenvalues of a nlpotent matrx satsfy m and so are all zero () An dempotent matrx s one where A A The egenvalues of an dempotent matrx satsfy and so are all s and s Cayley Hamlton Theorem The Cayley Hamlton Theorem states that a square matrx satsfes ts characterstc polynomal That s, χ A (A) Example : Suppose A, Then χ A () A A I Snce χ A () I A t s temptng to attempt to prove ths by substtutng A The problem s that n dong ths we would have the determnant of a matrx where the components are a mxture of scalars and matrces a b a b For example, f A then I A If we substtute A for we get c d c d A a b whch doesn t make sense We could replace the scalars a, b, c, d by ai, bi, ci, di c A d A ai bi respectvely, but we would get and there s no way ths s the determnant of the ci A di zero matrx the way AI A would be The followng proof s due to Yshao Zhou
6 Theorem (CAYLEYHAMILTON & YISHAO ZHOU): If A s an n n matrx then χ A (A) That s, every square matrx satsfes ts characterstc polynomal Proof: Let χ A () n a n n a Let B() adj(i A) and wrte the components of B() as b j () Snce these are cofactors of the n n matrx adj(i A), they wll each be a polynomal of degree at most n Let b j () b j (n) n b j () b j () Here the (r) are superscrpts, not powers Defne B (k) (b j (k) ) So the components of B (k) are the coeffcents of k n the components of B() Hence adj(i A) B() n B (n) B () B () Remember that Madj(M) M I for all square matrces M (ths s what gves the cofactor formula for M when M ) So (I A)adj(I A) I A I Hence (I A)[ n B (n) B () B () ] I A I χ A ()I n I a n n I a I Equatng coeffcents of the varous powers of we get B (n) I, whence A n B (n) A n ; B (n) AB (n) a n I, whence A n B (n) A n B (n) a n A n ; B (n) AB (n) a n I, whence A n A n B (n) a n A n ; B () AB () a I, whence AB () A B () a A; and fnally AB () a I Addng these equatons we get A n a n A n a n A n a A a I You may fnd all ths confusng What a pty we can t just put A n I A To help you, here s an example that you can follow n parallel wth the proof Example : Let A Then χ() ( ) ( )[( )( ) ] [ ] [ ] ( )[ ] [ ] [ ] [ ] [ ] [ ] I A ( )( ) ( ) ( ) B() adj(i A) ( ) ( )( ) ( ) ( ) ( ) ( )( ) T
7 so B () ; B () ; B () A B () A ; B () AB () I So A (B () AB () A B () AB () I So A(B () AB () ) I AB () So A A A A B () [A B () A B () ] [AB () A B () ] AB ()
8 Calculatng the Characterstc Polynomal Fndng the characterstc polynomal s dffcult, and error prone, when dong t by hand And, although there are algorthms for evaluatng a determnant they are dffcult to mplement on a computer f the determnant nvolves a varable What we need s an algorthm that calculates the coeffcents of the characterstc polynomal and so only works wth numbers For matrces bgger than the followng algorthm s preferable If k, the kth trace, tr k (A), of a square matrx A s the sum of all the k k subdetermnants that can be obtaned from A by deletng correspondng rows and columns (so that the dagonal of each subdetermnant concdes wth the dagonal of A) We defne tr (A) Example : If A, tr (A), tr (A) tr (A) j: () () () () () () tr (A) j: Fnally tr (A) A
9 Lemma (COOPER): Suppose A s an n n matrx and D D(,, n ) a a a n a a a n s a polynomal n a n a n n a nn n commutng varables,, n For every subset S {s,, s k } of {,,, n} wth s < s < < s k the coeffcent of s sk n D s () nk tmes the subdetermnant got by deletng all rows and columns except those correspondng to the elements of S Proof: We prove ths by nducton on k If k, S and the requred coeffcent s the constant term of D whch s A () n A Suppose k Expandng D along row s the only term n s st arses from ( s a s s ) tmes the determnant D got from D by deletng row s and column s Ths s the coeffcent of s sk n D whch, by nducton, s () (n)(k) tmes the subdetermnant D got from D by deletng the rows and columns correspondng to s,, s k But ths s () nk tmes the subdetermnant got from D by deletng the rows and columns correspondng to the elements of S Theorem (COOPER): The characterstc polynomal of the n n matrx A s χ A () n tr(a) n tr (A) n () nk tr k (A) k () n A Proof: Wth D(,, n ) defned above χ A (A) D(,, ) Example : Fnd the characterstc polynomal of A Soluton: tr (A), tr (A), tr (A) and so the characterstc polynomal s Example : Let A tr(a), tr (A) A So χ A () Compare the amount of work n computng χ A (A) to that n Example
10 Matrces and Graphs A drected graph s a set of vertces, G, together wth drected edges from some vertces to others A drected edge s smply an ordered par of vertces Hence we can thnk of a drected graph as beng a set G together wth a subset of G G Example : The above pcture represents a graph wth vertces and edges (The doubleheaded arrow represents two edges) As a set of ordered pars the graph s {(, ), (, ), (, ), (, ), (, ), (, ), (, ), (, )} A path n an ordered graph s a sequence of edges of the form (x, x ), (x, x ), (x, x ),, (x n, x n ) The length of a path s the number of edges, n ths case, n A path of length s smply an edge Example : In the above example there s a path (, ), (, ), (, ) from to We can wrte ths as There s a path of length from to (represented by the crcular arrow), a path of length () and a path of length () There s even a path of length, namely There are at least four paths of length from to : and many more How can we count the number of paths of length n from vertex to vertex j? The adjacency matrx of a drected graph on vertces,,, n s the matrx A (a j ) where a j f there s an edge from to j n the graph, and otherwse Example : The matrx of the drected graph n example s A Theorem : If A s the matrx of a drected graph, the number of paths of length N from vertex to vertex j s the j component of A N Proof: It s true for N Suppose that t s true for N A path of length N from to j s a path from to some vertex k followed by an edge from k to j By nducton the number of paths from to k s a (N) k, the k component of A N Snce the number of paths from k to j s a kj the total number of paths of length N from to j s ( N ) a k akj whch s the j component of A N A A N Hence t s true for all N k
11 Example : The number of paths n the drected graph n example, of length between varous vertces are gven by the components of A, where A s the adjacency matrx of that graph, as gven n example A A A So there are paths of length from vertex to vertex These are the paths of length from vertex to vertex, followed by an edge from vertex to vertex plus the edges of length from vertex to vertex followed by the edge from vertex to vertex Ths gves us,,, EXERCISES FOR CHAPTER Exercse : If A fnd χ A () Exercse : If A fnd χ A () Exercse : If A fnd χ A ()
12 Exercse : Fnd the egenvalues and egenvectors of A Exercse : Fnd the egenvalues and egenvectors of A Exercse : Fnd the egenvalues and egenvectors of A Exercse : Fnd the egenvalues of the n n matrx A Exercse : If v a n a a Fnd the egenvalues of the n n matrx A vv T Exercse : Prove that t s mpossble for a magc square to have all ts egenvalues real and postve Example : Fnd the egenvalues of A SOLUTIONS FOR CHAPTER Exercse : tr(a), A So χ A () Exercse : tr(a), tr (A) A () () So χ A () Exercse : tr(a), tr (A)
13 tr (A) () A () So χ A () Exercse : tr(a) tr (A) A () () χ A () ( ) Hence the egenvalues are ± : A So s an egenvector : A I R R R (), R R R ( )R, R R R () R () So s an egenvector
14 : Smlarly s an egenvector Exercse : tr(a), tr (A), A χ A () ( ) Hence the egenvalues are (twce) and : A so and are ndependent egenvectors : s an egenvector Exercse : tr(a), tr (A) ( ) ( ) ( ) A ( ) ( ) ( ) () () () χ A () ( )( ) So the egenvalues are and (twce) : A I R R R R, R R R () R R Hence s an egenvector : A I R R, R
15 Hence R R, R R and are ndependent egenvectors Exercse : tr(a) n For r >, tr r (A) because the r r sub determnants are all zero Hence χ A () n n n n ( n) Hence the egenvalues are n and (n tmes) Exercse : The j component of A s a a j Hence tr(a) Σa Let ths be N For r >, tr r (A) because the r r sub determnants are all zero Hence χ A () n N n n ( N) n Hence the egenvalues are a and (n tmes) Exercse : Let T be the row/column/dagonal total Snce the row totals are all equal to T then s an egenvector for the egenvalue T Snce the trace s equal to T the sum of the remanng egenvalues must be zero Therefore they cannot all be postve Example : Fnd the egenvalues of A Soluton: tr(a) tr (A) tr (A) () () () () () () Rather than work out A n the standard way we can be a lttle smart Note that C C C C Ths means that s an egenvector correspondng to the egenvalue of Hence A Hence χ A () ( ) The egenvalues are (twce) plus the two zeros of the quadratc, that s,,, ±
16
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