Rudimentry Mtrix Alger Mrk Sullivn Decemer 4, 217 i
Contents 1 Preliminries 1 1.1 Why does this document exist?.................... 1 1.2 Why does nyone cre out mtrices?................ 1 1.3 Wht is mtrix?........................... 2 2 Mtrix opertions 4 2.1 Addition of mtrices......................... 4 2.2 Multipliction of mtrices....................... 5 2.2.1 Sclr multipliction..................... 5 2.2.2 Multipliction of mtrices y mtrices............ 6 2.3 Division of mtrices?......................... 8 3 Mtrix equtions 1 3.1 Elementry row opertions...................... 13 3.2 Guss-Jordn elimintion....................... 16 3.3 Computing inverse mtrices..................... 22 4 Determinnts 23 4.1 Tricks for computing determinnts.................. 26 4.2 The mening of the determinnt................... 28 5 Eigenvlues, eigenvectors, generlized eigenvectors 3 5.1 Distinct, rel eigenvlues....................... 33 5.2 Repeted, rel eigenvlues...................... 36 5.3 Complex eigenvlues......................... 42 ii
1 Preliminries In this document, we introduce the sic principles of using mtrices in mthemtics nd pplied mthemtics to solve systems of equtions. In order to do this, we ll need to explin how to do lger on mtrices, just s one would do lger on the numers. The prctice of redefining nd generlizing opertions on numers to work for different mthemticl contexts is field of mthemtics known s strct lger. The prticulr cse of vectors nd mtrices is the su-field known s liner lger. These two fields re primry reserch interest for mthemticins of the pst, present nd future. There re entire enormous ooks devoted solely to these sujects, nd the uthor dmits to proclivity to rmle for hours on end out them. With such n undnce of literture in mind, you my e sking: 1.1 Why does this document exist? Unfortuntely, mny students do not tke course in strct or liner lger until lte in their undergrdute eduction if ever. This leves them mostly in the drk concerning mtrices. I wrote this document for the ske of those who wnt to lern how to use mtrices despite not hving tht ckground. So: 1.2 Why does nyone cre out mtrices? Mtrix lger is skill tht cn significntly reduce the effort required to do computtion. As visul strtegy, it cn highlight wys of reducing the mount of steps in computtion. Additionlly, computers think of systems of equtions primrily in terms of mtrices. Beyond this, mtrices cn provide link etween solving systems of equtions nd deeper concepts in liner lger. In generl, just out every person ever doing mthemticl work cn enefit from lerning out mtrices. 1
1.3 Wht is mtrix? Definition 1.1 An m n mtrix pronounced m y n mtrix is n ssignment of some mthemticl ojects to ech ordered pir i, j, where 1 i m nd 1 j n. We sy tht n m n mtrix hs m rows nd hs n columns. Mtrices re generlly visulized s rectngles mde up of squres, ech one leled y i, j, with lrger vlues for i elow smller ones, nd lrger vlues for j to the right of smller ones. In ech squre, the vlue corresponding to i, j is plced. In other words, the vlue ssigned to i, j is plced in the cell contined y the ith row nd the jth column. For exmple, 3 4 mtrix of numers mking the following ssignments: 1, 1 1 1, 2 5 1, 3 4 3 1, 4 4 2, 1 2, 2 4 2, 3 2 2, 4 3 3, 1 2 3, 2 6 3, 3 e 2 3, 4 π 1 would e visulized like this: 1 5 4 3 4 4 2 3 2 6 e 2 π. 2 However, we need not other ourselves with drwing lines dividing the cells. In- 2
sted, we ll just write the ssignments like this: 1 5 4 4 1 5 4 4 3 3 4 2 3 or 4 2 3. 3 2 6 e 2 π 2 6 e 2 π We define two mtrices s eing equl if they re the sme size nd for ech ordered pir i, j, the two mtrices ssocite i, j to equl mthemticl ojects. Thus, two mtrices re equl only if ll of their entries gree: 1 9 1 9. 4 6 7 6 6 There re plenty generliztions of mtrices. For one thing, we ve only delt with mtrices tht hve two coordintes for ech cell, creting 2-dimensionl digrm. Wht if we hd triplets, i, j, k, creting 3-dimensionl digrm? Wht if we hd qudruplets, i, j, k, l, creting 4-dimensionl digrm? Wht if we hd more thn tht? These generliztions re known in mthemtics s tensors, lthough some pplied mthemticins refer to them s n-dimensionl mtrices. For our purposes, though, we will only e need with 2-dimensionl mtrices, such s those digrmmed ove. 3
2 Mtrix opertions It will e convenient to e le to dd, sutrct, multiply nd sort of divide mtrices, in the sme wy tht one would dd, sutrct, multiply nd sometimes divide rel numers. So first, we ll hve to explin wht these opertions men. 2.1 Addition of mtrices Addition of mtrices is unnervingly simple. Given two m n mtrices, one dds them y simply dding the entries in corresponding positions: 11 12... 1n 11 12... 1n 11 + 11 12 + 12... 1n + 1n 21 22... 2n...... + 21 22... 2n...... = 21 + 21 22 + 22... 2n + 2n....... m1 m2... mn m1 m2... mn m1 + m1 m2 + m2... mn + mn 5 For exmple: 1 4 9 + 4 1 + 9 4 1 + = = 1 2 1 8 1 + 1 2 + 8 1 12 7 4 6 12 + 4 7 + 6 8 13 + = = 6 2 1 5 6 + 1 2 + 5 5 7 1 1 2 2 2 1 + 2 + 2 1 + 2 3 2 3 1 1 + 2 2 = 1 + 1 + 2 + 2 = 1 3 2 1 1 2 + 1 + 2 1 + 1 1 6 1 8 9 1 4 2 1 1 4 + = 1 2 2 1 4 6 12 3 5 8 14 2 1 1 4 9 13 2 + 1 = 3 2 2 6 1 5 6. 4
The sum of two m n mtrices will lwys e n m n mtrix. Addition of mtrices mkes sense only when the mtrices hve the sme numers of rows nd columns; one cnnot dd 2 3 mtrix to 4 3 mtrix: 5 1 3 1 7 + 1 1 1 2 8 1 12 is meningless 7 8 9 1 2.2 Multipliction of mtrices There re two sorts of multiplictions for mtrices, depending on wht kind of mthemticl oject you wnt to multiply y mtrix. 2.2.1 Sclr multipliction Sclr multipliction is the multipliction of sclr, tht is, numer y mtrix. This, too, is very simple; just multiply ech entry y the sclr: 11 12... 1n k 11 k 12... k 1n k 21 22... 2n...... = k 21 k 22... k 2n....... 8 m1 m2... mn k m1 k m2... k mn For exmple: 1 5 12 6 12 = 3 2 1 36 24 12 1 2 1 5 2 = 1 1 9 5
1 2 8 = 2 8 1 2 3 4 6 12 18 24 6 5 6 7 8 9 1 11 12 = 3 36 42 48 54 6 66 72 13 14 15 16 78 84 9 96 The product of sclr nd mtrix will lwys e mtrix of the sme size. 1 2.2.2 Multipliction of mtrices y mtrices Multiplying two mtrices together cn e significntly more complicted. 11 12... 1n 11 12... n 1p i=1 n 1i i1 i=1 1i i2... n i=1 1i ip 21 22... 2n 21 22... 2p............ = n i=1 n 2i i2 i=1 2i i2... n i=1 2i ip....... m1 m2... mn n1 n2... n np i=1 n mi i1 i=1 mi i2... n i=1 mi ip 11 To summrize, the i, jth entry of the product of two mtrices will e the sum of the products of the entries in the ith row of the first fctor nd the jth column of the second fctor. For exmple: 1 3 2 1 1 3 2 1 = 3 2 1 2 6 4 2 4 12 8 4 3 1 2 = 11 4 1 2 2 1 1 2 4 1 2 1 2 = 1 1 1 1 1 1 1 12 6
1 1 1 2 1 1 2 1 2 = 4 1 1 4 2 1 1 1 2 1 1 1 3 2 1 2 1 7 3 2 8 2 1 1 2 = 1 1 5 1 4 3 2 1 2. 13 1 1 1 1 1 1 1 3 3 1 1 1 1 = 2 2 1 1 1 1 1 3 3 1 1 Mtrix multipliction mkes sense only when the numer of columns in the first fctor is equl to the numer of rows in the second fctor. Id est, given mtrices A nd B, in order to mke sense of the product AB, the numer of columns of A must equl the numer of rows of B. If this condition is not stisfied, then the two mtrices simply cnnot e multiplied. Additionlly, the product of n m n mtrix nd n n p mtrix will lwys e n m p mtrix. An interesting feture of mtrix multipliction is tht it is not commuttive; this mens tht, given mtrices A nd B, AB might not e the sme s BA. First of ll, it s possile tht one of those multiplictions mkes sense, ut not the other. For exmple, if A is 2 3 mtrix nd B is 3 5 mtrix, then AB is 2 5 mtrix, ut BA is not defined. Besides, even if the dimensions do mtch in oth cses, the two products might still e different, s in the following exmple: 1 1 1 1 2 = 1 1 1 1 1. 14 1 1 1 1 = 1 1 1 1 2 Thnkfully, however, mtrix multipliction is ssocitive which mens tht, for mtrices A, B nd C, ABC = ABC nd distriutive which mens tht, for mtrices A, B nd C, AB + C = AB + AC nd A + BC = AC + BC. 7
2.3 Division of mtrices? Given two numers, nd, wht exctly do we men y? We ll need to introduce some terminology first. Definition 2.1 Let e nonzero rel or complex numer. The multiplictive inverse of is numer x such tht x = 1 nd x = 1. We typiclly denote the multiplictive inverse of y 1, or 1. When we refer to, wht we relly men is 1. Therefore, s long s one cn define multiplictive inverse, one cn define division. In order to explin wht we men y division of mtrices, we ll need to do similr investigtion. First, it s necessry to come up with mtrix tht cn serve the sme purpose s 1 does for the rel nd complex numers: Definition 2.2 The n n identity mtrix is mtrix I n such tht if A is ny m n mtrix, then AI n = A, nd if B is ny n m mtrix, then I n B = B. The n n identity mtrix is lwys just mtrix of 1 s long the digonl nd s elsewhere: For exmple, 1... 1... I n = 1.... 15.......... 1 1 1 1 I 2 =, I 3 = 1, I 4 = 1 1 1. 16 1 1 Now we re in position to define multiplictive inverse of mtrix. 8
Definition 2.3 Let A e n m n mtrix. An inverse mtrix of A is n n m mtrix B such tht AB = I m nd BA = I n. We often denote the inverse mtrix of A y A 1. No one ever uses the nottion 1 A. Of prticulr interest re the squre mtrices: Definition 2.4 A squre mtrix is n m n mtrix such tht m = n. In other words, squre mtrix hs s mny rows s columns. As it turns out, only squre mtrices cn hve inverse mtrices. This fct is not esy to prove, ut using it, we cn restte the definition: Definition 2.5 Let A e n n n mtrix. An inverse mtrix of A is n n n mtrix B such tht AB = I n = BA. Do ll squre mtrices hve inverses? No, in the sme wy tht some numers nmely, do not hve multiplictive inverses. The following re some exmples of squre mtrices tht do not hve inverses: 1 1 1 1 1 1. 17 1 There re infinitely mny for ech size n n. How would one compute the inverse mtrix of given mtrix? In order to nswer this, we ll need to understnd the procedure of row reduction of mtrices, which is the topic of the next section. 9
3 Mtrix equtions A mtrix eqution is exctly wht the nme implies; n eqution with mtrices in it. In prticulr, we ll e interested in equtions of the form Ax = B, where A is n m n mtrix, x is n n 1 mtrix, nd B is n m 1 mtrix: 11 12... 1n x 1 1 21 22... 2n x 2..... = 2. 18... x n } m1 m2... {{ mn }}{{} A x m }{{} B The gol here will e to solve for the n 1 mtrix x. Here re some exmples, nd their solutions in the following, c, c 1, c 2 nd c 3 refer to ritrry sclrs: x 1 1 1 2 x 2 = 1 x 3 1 x 1 1 1 = 1 1 x 2 x 3 1 1 1 1 1 1 x 1 x 2 x 3 x 4 2 1 = Solution: Solution: Solution: x 1 x 2 x 3 x 1 x 2 x 3 x 4 = x 1 x 2 x 3 c 2 c 1 = 1 2 c 1 c 2 = 1 c 1 c 2 19 x 5 4 2 1 1 1 x 1 3 1 = 3 x 2 5 4 1 7 x 1 y = z x 5 c 3 x 1 2 Solution: = x 2 1 x Solution: y = z c 1 c 2 1
1 1 1 x 1 1 1 1 x 2 = x 3 1 x 1 1 1 1 x 2 x 3 = 1 1 1 1 x 4 1 1 2 x 1 1 1 = 2 x 2 2 1 1 2 1 1 1 1 x 1 1 = 1 x 2 1 1 1 3 9 x 1 1 2 6 1 = 3 3 9 4 x 2 x 3 7 Solution: Solution: x 1 x 2 x 3 x 1 x 2 x 3 x 4 No solution No solution No solution = = c c 2 As with ny other kind of eqution, some mtrix equtions hve one solution, some hve no solution, nd some hve multiple solutions. A theorem from liner lger indictes tht mtrix eqution tht hs multiple solutions will lwys hve infinitely mny solutions. This is why some of the systems ove involve ritrry sclrs in their solutions; y choosing ny yes, ny vlue for the constnts, one cn crete new vector x tht is solution of the system. A system of liner equtions in n vriles x 1, x 2,..., x n corresponds exctly to mtrix eqution. To see this, consider the following exmple: 1 2 3 x 1 x 2 4 = 5 21 11
By computing the mtrix multipliction, we get: which exctly mens tht 1x 1 + 2x 2 4 =, 22 x 1 + 3x 2 5 1x 1 + 2x 2 = 4 x 1 + 3x 2 = 5. 23 On the other hnd, y just tking the coefficients on ech vrile, we cn revert ck to the originl mtrix eqution: 1 x 1 + 2 x 2 = -4 x 1 + 3 x 2 = 5 1 2 3 x 1 x 2 4 =. 24 5 The result is tht mtrix equtions cn e used s shorthnd for liner systems. In fct, we cn go even further. If we know tht we re deling in the vriles x 1 nd x 2 which, y the wy, re just nmes; clling them something different would hve chnged nothing, then we cn write the mtrix eqution ove in n even simpler form: 1 2 3 This nottion is known s n ugmented mtrix. x 1 x 2 4 1 2 4 =. 25 5 3 5 Since systems of liner equtions nd mtrix equtions correspond exctly to ugmented mtrices, we cn solve them nd/or mtrix equtions y deling only with ugmented mtrices. This process is clled Guss-Jordn elimintion. 12
3.1 Elementry row opertions We d like to discuss procedure of solving mtrix equtions commonly referred to s row reduction. First, we ll hve to define the moves tht re legl in the procedure. Definition 3.1 Let 11 12... 1n A = 21 22... 2n...... m1 m2... mn 26 e n m n mtrix. The following re the elementry row opertions of A. i Multipliction of ll of the entries in single row sy, the ith row y nonzero constnt k: 11 12... 1n 21 22... 2n...... i1 i2... in...... m1 m2... mn R i kr i ii Swpping two rows sy, the ith row nd the jth row: 11 12... 1n 21 22... 2n...... k i1 k i2... k. 27 in...... m1 m2... mn 11 12... 1n 11 12... 1n 21 22... 2n 21 22... 2n............ i1 i2... in R i R j j1 j2... jn........ 28..... j1 j2... jn i1 i2... in............ m1 m2... mn m1 m2... mn iii Adding nonzero constnt k times ll of the entries in row sy, the ith row 13
to ll of the corresponding entries in nother row sy, the jth row: 11 12... 1n 11 12... 1n 21 22... 2n 21 22... 2n............ i1 i2... in R j R j +kr i i1 i2... in............. j1 j2... jn. j1 + k i1 j2 + k i2... jn + k in........... m1 m2... mn m1 m2... mn 29 For exmple: 1 5 R 2 3R 2 1 5 1 2 3 6 1 5 R 1 R 2 1 2. 3 1 2 1 5 1 5 R 1 R 1 +2R 2 1 9 1 2 1 2 We hve nme for the reltionship etween two mtrices when one cn e gotten from the other y doing only elementry row opertions. Definition 3.2 Let A nd B e m n mtrices. We sy tht A nd B re row equivlent mtrices provided tht there exists sequence A = M 1 M 2 M 3... M k 1 M k = B 31 of m n mtrices M 1, M 2,..., M k such tht M 1 = A, M k = B nd for ech i where 1 i < k, M i+1 cn e produced from n elementry row opertion on M i. In other words, two mtrices re equivlent if you cn get one from the other y 14
sequence of elementry row opertions. For exmple, A = 1 2 9 18 1 2 nd B = 32 re row equivlent mtrices, ecuse you cn produce B y dding 9 times the first row of A to the second row of A: 1 2 R 2 R 2 9R 1 1 2. 33 9 18 As nother exmple, 1 1 1 A = 2 2 nd 4 1 1 1 34 re row equivlent mtrices, ecuse 1 1 1 1 1 1 2 2 R 3 R 3 4R 1 2 2 R 3 R 3 +2R 2 4 4 4 1 1 1 1 1 1 1 2 2 R 2 1 2 R 2 1 1 R 1 R 1 R 2 1 1. 35 Note tht row equivlent mtrices re not necessrily equl unless, of course, ll of the elementry row opertions done to get from one to the other cncel ech other out. As we mentioned in the first section, two mtrices re equl only if ll of their corresponding entries re equl. Wht, then, is so gret out this reltionship of row equivlence? The following sections nswer this question. 15
3.2 Guss-Jordn elimintion Here is the min interest of row equivlence: Theorem 3.3 If two liner systems hve row equivlent ugmented mtrices, then the two liner systems hve the sme solutions. This mens tht for ny ugmented mtrix, we cn do elementry row opertions without worrying out whether the solutions will e ltered. Let s look t the following exmple: 1x 1 + x 2 + 1x 3 = 1 1x 1 + 1x 2 + 2x 3 = 1. 36 1x 1 + x 2 + 2x 3 = 1 1x 1 1x 2 + 2x 3 = Even when we look t the ugmented mtrix for this, it s not relly cler how to proceed: 1 1 1 1 1 2 1 1 2 1. 37 1 1 2 However, this ugmented mtrix is row equivlent to: 1 1 1 1. 38 1 Bsed on the theorem, this ugmented mtrix cn then e thought of s resttement of the originl system: 1x 1 + x 2 + x 3 = 1 x 1 + 1x 2 + x 3 =. 39 x 1 + x 2 + 1x 3 = x 1 + x 2 + x 3 = 1 16
It s now cler wht the solutions of this system re. In this cse, there ren t ny, since x 1 +x 2 +x 3 cnnot e nything other thn, ut the lst eqution requires tht it e 1. This exmple illustrtes tht if we cn use elementry row opertions to find sufficiently simple ugmented mtrix, then solving the mtrix eqution will e esy. But how simple is sufficiently simple? The following definitions will formlize this concept. Definition 3.4 Let A e n m n mtrix. A row echelon form of A is mtrix B tht is row equivlent to A such tht: i The left-most nonzero entry of ech row of B is 1. ii The left-most nonzero entry of ech row of B contins only zeros elow it in its column. iii If i, j nd k, l re two positions of left-most nonzero entries in B nd i < k, then j < l. iv Any rows tht do not contin nonzero entries re t the ottom of the mtrix. Here re some exmples of mtrices tht meet the criteri of row echelon form, nd some tht don t: 1 3 2 1 is row echelon. 1 1 1 is row echelon. 1 3 2 1 is not; it doesn t stisfy criterion i. 1 1 1 is not; it doesn t stisfy criterion ii. 1 1 2 1 is row echelon. is not; it doesn t stisfy criterion iii. 1 1 2 1 1 1 1 1 is row echelon. is not; it doesn t stisfy criterion iv. 1 4 17
To summrize: row echelon form mtrices hve only 1 s s the left-most nonzero entries of ech row, elow which re only s; further, the left-most nonzero 1 s must move to the right s one moves down the rows. Row echelon form is simple, ut it cn get even simpler: Definition 3.5 Let A e n m n mtrix. A reduced row echelon form of A is mtrix B tht is row equivlent to A such tht: i The left-most nonzero entry of ech row of B is 1. ii The left-most nonzero entry of ech row of B contins only zeros ove nd elow it in its column. iii If i, j nd k, l re two positions of left-most nonzero entries in B nd i < k, then j < l. iv Any rows tht do not contin nonzero entries re t the ottom of the mtrix. To summrize: reduced row echelon mtrices re row echelon mtrices with the dditionl condition tht ech left-most nonzero entry in ech row is the only nonzero entry in its entire column. Here re some exmples of reduced row echelon mtrices, nd some exmples which re row echelon ut not reduced row echelon: 1 1 1 1 is reduced row echelon. 1 is not. 1 1 1 1 3 is reduced row echelon. is not. 1 1 1 1 1 1 1 41 is reduced row echelon. is not. 1 4 1 4 1 1 is reduced row echelon. 1 is not. 1 1 As long s we cn find reduced row echelon form of n ugmented mtrix, we 18
cn solve the system. So, which mtrices hve reduced row echelon forms? All of them: Theorem 3.6 Every m n mtrix hs unique reduced row echelon form. Thus, whtever mtrix we re given, we cn perform some sequence of elementry row opertions to rrive t reduced row echelon mtrix. The process of using elementry row opertions to find the reduced row echelon form of n ugmented mtrix is commonly clled row reduction, or Guss-Jordn elimintion. forms: Here re some exmples of ugmented mtrices, nd their reduced row echelon 1 2 2 1 5 2 1 2 4 1 1 1 1 1 1 2 2 2 3 1 4 4 2 15 6 6 24 2 1 5 2 6 2 3 1 1 3 3 1 3 5 1 1 2 4 1 1 1 3 2 3 4 1 21 6 4 6 3 27 6 6 4 2 3 28 6 2 1 28 9 3 1 4 1 RREF: 1 1 1 RREF: 1 2 1 2 1 RREF: 1 1 2 1 1 RREF: 1 3 1 4 RREF: 1 6 1 3 1 2 4 RREF: 1 6 1 1 3 1 4 RREF: 1 1 4 42 19
2 4 2 2 1 2 1 1 1 1 1 1 1 1 1 2 RREF: 1 1 1 43 1 1 3 1 When deling with liner system, it is importnt to understnd how the reduced row echelon form of the corresponding mtrix eqution should e interpreted. Let s look t the following exmple: 3w + 1x 7y 3z = 2 2w + 1x 5y + z = 5 w + 1x 1y + 1z = 2w + x + 4y + z = 4 44 First, convert this system into n ugmented mtrix: 3 1 7 3 2 2 1 5 5 1 1 1, 45 2 4 4 nd find its reduced row echelon form: 1 2 1 1 1 1 1 1. 46 Wht does this men? In order to interpret this, let s revert from the ugmented mtrix to the system nottion: 1w + x + 2y + z = 1 w + 2y = 1 w + 1x 1y + z = 1 x y = 1. 47 w + x + y + 1z = 1 z = 1 w + x + y + z = 2
At first, this does not seem helpful. It ppers tht there s still more to solve. However, tht is not the cse; these re the only restrictions on the vlues of w, x, y nd z. In other words, their vlues cn e nything, s long s they stisfy tht w + 2y = 1, x y = 1, nd z = 1. Select vlue c for y. This completely determines the vlues for w, x, y nd z: w + 2c = 1 x c = 1 y = c z = 1 w = 2c 1 x = c 1 y = c z = 1, or w 2c 1 x y = c 1 c. 48 z 1 In liner lger, we cll y free vrile, since ny chosen vlue for y will produce solution to the system. Some systems hve more thn one free vrile. For exmple, 1 1 1 1 1x 1 + x 2 + x 3 1x 4 + 1x 5 = 1 1 2 2 x 1 + 1x 2 + x 3 + 2x 4 + 2x 5 = 1 1 which gives the equtions x 1 + x 2 + 1x 3 + x 4 + x 5 = 1 x 1 x 4 + x 5 = 1 49 x 2 + 2x 4 + 2x 5 =. 5 x 3 = 1 The sitution is similr here. Agin, ny vlues for x 1, x 2, x 3, x 4 nd x 5 which stisfy the ove reltionships will e solution. By choosing vlues x 4 = c 1 nd x 5 = c 2, we cn completely determine the solution: x 1 x 2 x 3 x 4 1 + c 1 c 2 2c 1 2c 2 = 1. 51 c 1 x 5 c 2 21
3.3 Computing inverse mtrices 22
4 Determinnts A common theme in mthemtics is drwing informtion from expressions y ssociting simpler expressions to them. For exmple, given qudrtic eqution x 2 + x + c =, how cn we tell whether the solutions will e rel, non-rel, or repeted? In this cse, we look t the ssocited discriminnt, the expression 2 4c, nd study it in order to lern out the originl eqution. Just from this single numer, we cn tell whether the eqution hs two rel solutions in the cse tht the discriminnt is positive, nd whether they re rtionl if the discriminnt is nonzero perfect squre or irrtionl if the discriminnt is not perfect squre, one rel solution in the cse tht the discriminnt is zero, or two non-rel solutions in the cse tht the discriminnt is negtive. In this section, we ll see wht we cn lern out squre mtrix from n ssocited expression known s its determinnt: Definition 4.1 Let A e n n n squre mtrix. The determinnt of A is defined recursively s follows. i If n = 2, nd A = then the determinnt deta = d c. ii If n > 2 nd then the determinnt c d, 52 11 12... 1n A = 21 22... 2n......, 53 n1 n2... nn deta = n 1 i+1 1i detm 1i, 54 i=1 23
where M 1i is the sumtrix defined vi 21 22... 2i 1 2i+1... 2n M 1i = 31 32... 3i 1 3i+1... 3n............ 55 n1 n2... ni 1 ni+1... nn This formul my seem somewht complicted, nd tht s ecuse it is. Thnkfully, for this document, we will only consider determinnts of 2 2 or 3 3 mtrices, since computing determinnts for lrger mtrices is esy, ut tiresome. As the definition sys, for 2 2 mtrix, the determinnt is just A = deta = c On the other hnd, for 3 3 mtrix, the determinnt is c d, 56 = d c. 57 d 11 12 13 A = 21 22 23, 58 31 32 33 11 12 13 22 23 deta = 21 22 23 = 11 31 32 33 32 33 21 23 12 31 33 + 21 22 13 31 32. 59 To summrize: one tkes ech entry long the first row, with lternting ± signs, multiplies them y the determinnt of the sumtrix otined y deleting the row nd the column of the entry in question, nd then dding it ll up. With some prctice, you ll see tht the process is esy, ut it cn get very tedious. 24
Here re some exmples: 1 2 = 15 2 = 5 5 1 1 1 1 = 1 3 1 4 = 7 1 2 3 4 = 2 1 1 1 1 1 1 1 = 1 1 1 1 1 1 + 1 1 = 1 2 1 1 1 1 = 2 1 2 2 1 2 3. 6 4 5 6 = 7 8 9 1 2 3 4 5 = 24 6 1 2 3 = 18 4 5 6 2 3 = 3 5 1 1 1 = 1 1 25
4.1 Tricks for computing determinnts It didn t tke long for people do relize tht the determinnt is useful, ut tiresome to compute. Therefore, some mthemticins strted to look for wys to simplify the process. The following theorem from liner lger doesn t provide ny new theory, nd it s not prticulrly importnt for doing this kind of work, ut it cn help to reduce the pin of figuring out the determinnt of mtrix y hnd. Theorem 4.2 The following sttements re true. i deti n = 1. ii If A nd B re n n mtrices, then detab = deta detb. iii If A is n n n mtrix, then det A = det A T. iv If A is n n n mtrix nd B is otined from A y multiplying nonzero constnt r y single row or column of A, then detb = r deta. v If A is n n n mtrix nd B is otined from A y swpping two different rows or columns of A, then detb = deta. vi If A is n n n mtrix nd B is otined from A y dding nonzero sclr multiple of one row of A to nother row of A, then detb = deta. Here A T refers to the trnspose mtrix, which is the mtrix you get y exchnging ll of the rows nd columns of A: 11 12... 1m 11 21... n1 A = 21 22... 2m...... AT = 12 22... n2....... 61 n1 n2... nm 1m 2m... nm Among the implictions of this theorem is the fct tht one cn do elementry row nd column opertions in order to simplify the clcultion of determinnt. For exmple, given the mtrix 87 37 9 A = 3 3 1, 62 19 8 2 26
it would e rther nnoying to compute the determinnt of A directly. However, using the theorem ove, 87 37 9 5 2 5 2 deta = 3 3 1 = 3 3 1 = 3 3 1 19 8 2 19 8 2 1 1 4 5 2 5 2 3 2 = 33 13 = 1 33 13 = 1 2 13 1 1 4 1 1 4 1 6 4 3 2 3 2 1 = 1 2 13 = 1 1 1 = 1 1 1 1 1 3 2 1 1 = 1 1 3 2 1 2 + 1 = 1. 63 3 Another impliction of this theorem is tht the first row is not specil; one cn compute the determinnt y using ny kth row or column, just y multiplying 1 k y ech term: 3 2 1 2 1 = 3 2 = 1 1 1 1 1 1 1 1 1 3 2 1 1 + 3. 64 1 1 All of these sorts of pproches re common: simplify the mtrix or look t prticulr row or column of it until the determinnt ecomes esy. This usully involves lot of entries eing zero. 27
4.2 The mening of the determinnt Wht is the point of ll this? The point is the following theorem, which forms the core of every first course in liner lger. Theorem 4.3 Let A e n n n mtrix. The following sttements re equivlent. i deta =. ii There exists vector v such tht A v =. iii The mtrix A is not invertile. By vector, in this context, we simply men n n 1 mtrix: 1 v = 2.. 65 n In prticulr, the zero vector is the vector whose entries re ll zero: =. 66 Let s tke moment to exmine the implictions of this theorem. Suppose tht we re doing clcultion tht involves finding ll the solutions of this system of equtions: 2x y + z = x + 1y + 1z =. 67 1x + 2y + 2z = This is the sme s the mtrix eqution 2 1 1 x 1 1 y =. 68 1 2 2 z 28
So now, in order to find the solutions, we hve no choice ut to do Guss-Jordn elimintion to find the reduced row echelon form of the ugmented mtrix, right? Wrong. Using finding the reduced row echelon form of the ugmented mtrix would work, ut lterntively, we could just find the determinnt of the mtrix: 2 1 1 1 1 1 1 = 2 1 2 2 2 2 1 1 1 2 + 1 1 = 2. 69 1 2 By the theorem, since the determinnt is nonzero, there cnnot e vector v which is not tht stisfies the eqution. As result, our only solution is the zero vector: x y =, 7 z nd so x =, y =, nd z =. This sort of use of the theorem is of fundmentl importnce to the theory of liner trnsformtions, ut tht is topic est left for course in liner lger. In the next section, we will use this theorem to determine nother importnt piece of dt concerning mtrices: eigenvlues. 29
5 Eigenvlues, eigenvectors, generlized eigenvectors An eigenvlue of squre mtrix is sclr tht cn simplify the multipliction of the mtrix y certin vectors. Definition 5.1 Let A e n n n mtrix. Given sclr λ, we sy tht λ is n eigenvlue of A provided tht there exists vector v such tht A v = λ v. Any such vector v is clled n eigenvector of A corresponding to λ. Let s discuss the process of finding eigenvlues. First, we know tht if A is ny n n mtrix nd λ is n eigenvlue of A, then there hs to e vector v such tht: A v = λ v. 71 We cn write this s: A v λ v =. 72 Now, given ny mtrix X with n rows, we know tht I n X = X, y definition of I n s the identity mtrix. Therefore, this eqution cn e written s A v λi n v =. 73 As we riefly mentioned erlier, mtrix multipliction cn distriute over ddition, so we cn fctor out v to write this s: A λi n v =. 74 By Theorem 4.3, this equtions will hve solutions with v if nd only if the determinnt det A λi n =. Therefore, sclr λ is n eigenvlue of n n n mtrix A if nd only if det A λi n =. In order to find the eigenvlues of mtrix, we ll ssume this eqution nd then solve for λ. We demonstrte the procedure in the cse of 1 3 A =. 75 3 1 3
1. Set = det A λi 2 nd solve s follows: 1 3 1 1 λ 3 = det A λi 2 = det λ = 3 1 1 3 1 λ = 1 λ 1 λ 33 = 1 2λ + λ 2 9 = λ 2 2λ 8. 76 The expression λ 2 2λ 8 is known s the chrcteristic polynomil of A, nd the eqution λ 2 2λ 8 = is known s the chrcteristic eqution of A. 2. Solve the chrcteristic eqution: = λ 2 2λ 8 = λ 4 λ + 2, 77 nd so A hs two eigenvlues: λ 1 = 4 nd λ 2 = 2. Now, how cn one find the eigenvectors ssocited to n eigenvlue? We will gin refer to the eqution A λi n v =. 78 Once the eigenvlues hve een determined, one cn sustitute them into this eqution to solve for the eigenvectors. We demonstrte the procedure for the cse of 1 3 A =. 79 3 1 1. Find the eigenvlues of A. By the exmple ove, these re λ 1 = 4 nd λ 2 = 2. 2. For ech eigenvlue, find the solutions of the eqution A λi n v =. In our cse, for the eigenvlue λ 1 = 4: A 4I 2 1 3 1 v 1 = 4 v1. 8 3 1 1 Thus, finding the eigenvectors v 1 corresponding to λ 1 = 4 is reduced to the pro- 31
lem of solving the mtrix eqution 3 3 =. 81 3 3 By using Guss-Jordn elimintion, we find tht this eqution hs the sme solutions s the eqution 1 1 =. 82 In the cse of 2 2 mtrices, such s this one, the solutions re very esy to see even without Guss-Jordn elimintion, ut for lrger mtrices, things could e more opque t first. From this, we get the reltionship =, 83 which cnnot e solved for n explicit vlue of nd. This is norml. In generl, ny nonzero sclr times n eigenvector will lso e n eigenvector corresponding to the sme eigenvlue, so we should expect to hve infinitely mny solutions. In this cse, the eigenvectors corresponding to λ 1 = 4 re v1 =, 84 where is ny vlue whtsoever except, known s free vrile. Similrly, for the eigenvlue λ 2 = 2, we hve the mtrix eqution 1 3 v2 =, 85 3 1 whose solutions re of the form 3 v2 =, 86 where is ny vlue whtsoever except. 32
5.1 Distinct, rel eigenvlues Here re some exmples of mtrices with eigenvlues tht re rel nd distinct: 1 1 3 1 2 1 1 2 2 2 1 2 1 1 3 2 2 9 1 5 2 5 7 hs eigenvlues λ 1 = 1, λ 2 = 1 with eigenvectors for λ 1, for λ 2 hs eigenvlues λ 1 = 3, λ 2 = 1 3 with eigenvectors for λ 1, for λ 2 hs eigenvlues λ 1 = 2, λ 2 = 1 with eigenvectors for λ 1, for λ 2 2 hs eigenvlues λ 1 = 3, λ 2 = 2 2 with eigenvectors for λ 1, for λ 2 2 hs eigenvlues λ 1 = 4, λ 2 = 2 5 with eigenvectors for λ 1, for λ 2 hs eigenvlues λ 1 = 5, λ 2 = 4 2 with eigenvectors for λ 1, for λ 2 3 3 hs eigenvlues λ 1 = 5, λ 2 = 2 2 with eigenvectors for λ 1, for λ 2 5 87 33
5 1 12 4 2 1 2 2 12 4 1 1 8 8 11 2 7 1 7 8 2 1 1 7 3 9 2 1 3 hs eigenvlues λ 1 = 6, λ 2 = 1 with eigenvectors for λ 1, for λ 2 4 hs eigenvlues λ 1 = 1, λ 2 = 2 with eigenvectors for λ 1, for λ 2 5 hs eigenvlues λ 1 = 2, λ 2 = 14 3 with eigenvectors for λ 1, for λ 2 hs eigenvlues λ 1 = 1, λ 2 = 9 8 with eigenvectors for λ 1, for λ 2 11 hs eigenvlues λ 1 = 7, λ 2 = 1 with eigenvectors for λ 1, for λ 2 7 hs eigenvlues λ 1 = 1, λ 2 = 9 4 with eigenvectors for λ 1, for λ 2 hs eigenvlues λ 1 = 6, λ 2 = 2 7 with eigenvectors for λ 1, for λ 2 3 hs eigenvlues λ 1 = 2, λ 2 = 3 with eigenvectors for λ 1, for λ 2 88 34
3 1 1 3 7 5 4 7 5 7 3 8 1 1 2 1 hs eigenvlues λ 1 = 2, λ 2 = 4 with eigenvectors for λ 1, for λ 2 hs eigenvlues λ 1 = 12, λ 2 = 2 with eigenvectors for λ 1, for λ 2 hs eigenvlues λ 1 = 11, λ 2 = 1 3 with eigenvectors for λ 1, for λ 2 7 with eigenvectors hs eigenvlues λ 1 =, λ 2 = 1, λ 3 = 1 for λ 1, for λ 2, 3c 1 1 1 1 hs eigenvlues λ 1 = 1, λ 2 = 1, λ 3 = 2 1 1 with eigenvectors c c for λ 3 c 2 for λ 1, for λ 2, c for λ 3 c 1 1 2 hs eigenvlues λ 1 = 1, λ 2 = 2, λ 3 = 3 1 with eigenvectors for λ 1, 2 for λ 2, for λ 3 4 c 89 35
5.2 Repeted, rel eigenvlues The chrcteristic polynomil is, of course, polynomil. Therefore, it my sometimes hve repeted roots. Definition 5.2 Let A e n n n mtrix with chrcteristic polynomil p λ = λ n + n 1 λ n 1 +... + 1 λ +. 9 Suppose tht the chrcteristic polynomil fctors s: pλ = λ r 1 m 1 λ r 2 m 2...λ r k m k, 91 where r 1, r 2,..., r k 1 nd r k re ll distinct. For ech i, the multiplicity of the eigenvlue r i is the corresponding exponent m i. In other words, the multiplicity of n eigenvlue is the numer of times the fctor λ ppers in the chrcteristic polynomil. To hve distinct roots literlly mens tht the multiplicity of ech eigenvlue is 1. If n eigenvlue hs multiplicity greter thn 1, it is clled repeted eigenvlue. Some sources lso refer to these s multiple eigenvlues. The opinion of the uthor is tht this term is extremely confusing. 36
Here re some exmples of mtrices with repeted eigenvlues: 1 2 1 1 1 1 1 3 3 7 1 1 4 3 5 2 2 1 12 12 3 6 6 15 hs eigenvlue λ = 1 with multiplicity 2 with eigenvectors for λ hs eigenvlue λ = 1 with multiplicity 2 with eigenvectors for λ hs eigenvlue λ = 4 with multiplicity 2 with eigenvectors for λ hs eigenvlue λ = 1 with multiplicity 2 with eigenvectors for λ 2 hs eigenvlue λ = 3 with multiplicity 2 with eigenvectors for λ hs eigenvlue λ = 12 with multiplicity 2 with eigenvectors for λ hs eigenvlue λ = 9 with multiplicity 2 with eigenvectors for λ 92 37
6 4 4 2 3 6 6 15 1 1 3 2 1 hs eigenvlue λ = 2 with multiplicity 2 with eigenvectors for λ hs eigenvlue λ = 9 with multiplicity 2 with eigenvectors for λ hs eigenvlues λ 1 = 1 with multiplicity 2, λ 2 = 2 with multiplicity 1 with eigenvectors for λ 1, 2 for λ 2 2 1 1 1 1 2 hs eigenvlue λ = 1 with multiplicity 3 1 with eigenvectors for λ 9 4 6 1 hs eigenvlues λ 1 = 3 with multiplicity 2, λ 2 = 5 with multiplicity 1 6 4 3 2 c with eigenvectors 3 for λ 1, c for λ 2 c 93 38
2 1 2 1 1 1 1 1 2 1 1 1 1 1 1 1 1 2 1 1 1 1 2 1 4 4 4 4 4 4 4 4 λ 1 = 1 with multiplicity 2, hs eigenvlues λ 2 = 1 with multiplicity 1, λ 3 = 2 with mulitiplicity 1 c with eigenvectors for λ 1, c 2c for λ 2, for λ 3 + d hs eigenvlues λ 1 = 2 with multiplicity 2, λ 2 = 1 with multiplicity 2 with eigenvectors for λ 1, 3 for λ 2 hs eigenvlues λ 1 = 1 with multiplicity 3, λ 2 = 4 with multiplicity 1 with eigenvectors for λ 1, for λ 2 hs eigenvlue λ = 4 with multiplicity 4 with eigenvectors for λ 94. 39
There is strngeness to some repeted eigenvlues. Some eigenvlues hve property known s defectiveness. Definition 5.3 Let A e mtrix with n eigenvlue λ. The dimension of the eigenspce corresponding to λ is the numer of free vriles in the eigenvectors of λ. For exmple, the mtrix 5 6 3 A = 3 4 3 95 2 hs the eigenvlue λ 1 = 2, with multiplicity 2, nd the eigenvlue λ 2 = 1, with multiplicity 1. The eigenvectors corresponding to these tke the form: + 2 c for λ 1, c for λ 2, 96 where, nd c re ny rel numers. We sy tht λ 1 hs n eigenspce of dimension 2, since ny of its eigenvectors cn e completely determined y specifying the vlues of t lest 2 free vriles: nd. As for λ 3, there is only one free vrile involved in its eigenvectors: c. Therefore, λ 3 hs n eigenspce of dimension 1. Definition 5.4 Let A e mtrix with n eigenvlue λ. The defect of λ is the multiplicity of λ minus the dimension of the eigenspce corresponding to λ. In the exmple ove, λ 1 hd multiplicity 2 nd n eigenspce of dimension 2, so its defect ws. Similrly, λ 2 hd multiplicity 1 nd n eigenspce of dimension 1, so its defect ws lso. On the other hnd, the mtrix A = 1 1 1 4 97
hs the eigenvlue λ = 1 with multiplicity 2, ut the eigenvectors corresponding to it hve the form. 98 With only one free vrile tht is,, λ hs n eigenspce of dimension 1. Therefore the defect of λ is 2 1 = 1. An eigenvlue with defect of is clled complete eigenvlue, while one with defect greter thn is clled defective eigenvlue. 41
5.3 Complex eigenvlues 42