Section Let A =. Then A has characteristic equation λ. 2 4λ + 3 = 0 or (λ 3)(λ 1) = 0. Hence the eigenvalues of A are λ 1 = 3 and λ 2 = 1.

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Prudence Marsh
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1 Sectio Let A The A has characteristic equatio λ 2 4λ + 3 or (λ 3)(λ ) Hece the eigevalues of A are λ 3 ad λ 2 λ 3 The correspodig eigevectors satisfy (A λ I 2 )X, or 3 3 or equivaletly x 3y Hece x 3y y y ad we take X 3, 3 y Similarly for λ 2 we fid the eigevector X 2 3 Hece if P X X 2 P AP, the P is o sigular ad 3 Hece A P 3 P ad cosequetly A 3 P P A I

2 3/5 4/5 2 Let A 2/5 /5 λ 2 /5, with correspodig eigevectors 2 X ad X 2 The if P X X 2, P is o sigular ad P AP ad A P /5 The we fid that the eigevalues are λ ad /5 P Hece A P ( /5) P P P /3 2/3 3 /3 /3 2 3 The give system of differetial equatios is equivalet to Ẋ AX, where 3 2 x A ad X 5 4 y 2 The matrix P is a o-sigular matrix of eigevectors correspodig to eigevalues λ 2 ad λ 2 The 5 P 2 AP The substitutio X P Y, where Y x, y t, gives Ẏ 2 Y, 66

3 or equivaletly x 2x ad y y Hece x x ()e 2t ad y y ()e t To determie x () ad y (), we ote that x () P x() 3 3 y () y() Hece x 3e 2t ad y 7e t Cosequetly x 2x + y 6e 2t + 7e t ad y 5x + y 5e 2t + 7e t 4 Itroducig the vector X x y, the system of recurrece relatios x + 3x y y + x + 3y, 3 becomes X + AX, where A Hece X 3 A X, where X 2 To fid A we ca use the eigevalue method We get A Hece X (2 4 ) (2 + 4 ) (3 2 4 )/2 ( )/2 Hece x 2 (3 2 4 ) ad y 2 ( ) a b 5 Let A be a real or complex matrix with distict eigevalues c d λ, λ 2 ad correspodig eigevectors X, X 2 Also let P X X 2 (a) The system of recurrece relatios x + ax + by y + cx + dy 67

4 has the solutio x y ( ) A x λ P P x y λ 2 y λ P λ P x 2 y λ X X 2 α λ 2 β λ X X 2 α λ 2 β λ αx + λ 2 βx 2, where α β P x y (b) I matrix form, the system is Ẋ AX, where X x y X P Y, where Y x, y t The We substitute Ẋ P Ẏ AX A(P Y ), so Ẏ (P λ AP )Y λ 2 Hece x λ x ad y λ 2 y The x y x x ()e λ t ad y y ()e λ 2t But so x () y () x() y() P x () y () P x() y(), α β Cosequetly x () α ad y () β ad x x αe λ t P X y X 2 βe λ 2t y αe λ t X + βe λ 2t X 2 68

5 a b 6 Let A be a real matrix with o real eigevalues λ a + ib c d ad λ a ib, with correspodig eigevectors X U +iv ad X U iv, where U ad V are real vectors Also let P be the real matrix defied by P U V Fially let a + ib re iθ, where r > ad θ is real (a) As X is a eigevector correspodig to the eigevalue λ, we have AX λx ad hece A(U + iv ) (a + ib)(u + iv ) AU + iav au bv + i(bu + av ) Equatig real ad imagiary parts the gives AU au bv AV bu + av (b) a AP AU V AU AV au bv bu+av U V b b a P a b b a Hece, as P ca be show to be o sigular, P a b AP b a (The fact that P is o sigular is easily proved by showig the colums of P are liearly idepedet: Assume xu + yv, where x ad y are real The we fid (x + iy)(u iv ) + (x iy)(u + iv ) Cosequetly x+iy as U iv ad U +iv are eigevectors correspodig to distict eigevalues a ib ad a + ib ad are hece liearly idepedet Hece x ad y ) (c) The system of recurrece relatios x + ax + by y + cx + dy 69

6 has solutio x y A x y a b P P x b a y r cos θ r si θ α P r si θ r cos θ β P r cos θ si θ α si θ cos θ β r cos θ si θ α U V si θ cos θ β r α cos θ + β si θ U V α si θ + β cos θ r {(α cos θ + β si θ)u + ( α si θ + β cos θ)v } r {(cos θ)(αu + βv ) + (si θ)(βu αv )} (d) The system of differetial equatios dx dt dy dt ax + by cx + dy is attacked usig the substitutio X P Y, where Y x, y t The Ẏ (P AP )Y, so x y a b b a x y Equatig compoets gives Now let z x + iy The x ax + by y bx + ay ż x + iy (ax + by ) + i( bx + ay ) (a ib)(x + iy ) (a ib)z 7

7 Hece z z()e (a ib)t x + iy (x () + iy ())e at (cos bt i si bt) Equatig real ad imagiary parts gives Now if we defie α ad β by α β x e at {x () cos bt + y () si bt} y e at {y () cos bt x () si bt} P x() y() we see that α x () ad β y () The x x P y y e U V at (α cos bt + β si bt) e at (β cos bt α si bt), e at {(α cos bt + β si bt)u + (β cos bt α si bt)v } e at {cos bt(αu + βv ) + si bt(βu αv )} a b 7 (The case of repeated eigevalues) Let A ad suppose that c d the characteristic polyomial of A, λ 2 (a + d)λ + (ad bc), has a repeated root α Also assume that A αi 2 (i) λ 2 (a + d)λ + (ad bc) (λ α) 2 Hece a + d 2α ad ad bc α 2 ad λ 2 2αλ + α 2 (a + d) 2 4(ad bc), a 2 + 2ad + d 2 4ad 4bc, a 2 2ad + d 2 + 4bc, (a d) 2 + 4bc 7

8 (ii) Let B A αi 2 The B 2 (A αi 2 ) 2 A 2 2αA + α 2 I 2 A 2 (a + d)a + (ad bc)i 2, But by problem 3, chapter 24, A 2 (a + d)a + (ad bc)i 2, so B 2 (iii) Now suppose that B The BE or BE 2, as BE i is the i th colum of B Hece BX 2, where X 2 E or X 2 E 2 (iv) Let X BX 2 ad P X X 2 We prove P is o sigular by demostratig that X ad X 2 are liearly idepedet Assume xx + yx 2 The xbx 2 + yx 2 B(xBX 2 + yx 2 ) B xb 2 X 2 + ybx 2 xx 2 + ybx 2 ybx 2 Hece y as BX 2 Hece xbx 2 ad so x Fially, BX B(BX 2 ) B 2 X 2, so (A αi 2 )X ad Also Hece AX αx (2) X BX 2 (A αi 2 )X 2 AX 2 αx 2 The, usig (2) ad (3), we have AX 2 X + αx 2 (3) Hece ad hece AP AX X 2 AX AX 2 AP P αx X + αx 2 α X X 2 α α α 72

9 P α AP α 8 The system of differetial equatios is equivalet to the sigle matrix 4 equatio Ẋ AX, where A 4 8 The characteristic polyomial of A is λ 2 2λ + 36 (λ 6) 2, so we ca use the previous questio with α 6 Let 2 B A 6I The BX 2 4 P X X 2, we have, if X 2 P AP 6 6 Now make the chage of variables X P Y, where Y Ẏ (P AP )Y Also let X BX 2 The if 6 6 Y, or equivaletly x 6x + y ad y 6y Solvig for y gives y y ()e 6t Cosequetly x 6x + y ()e 6t Multiplyig both side of this equatio by e 6t gives x d dt (e 6t x ) e 6t x 6e 6t x y () e 6t x y ()t + c, where c is a costat Substitutig t gives c x () Hece e 6t x y ()t + x () y The ad hece x e 6t (y ()t + x ()) 73

10 However, sice we are assumig x() y(), we have x () P x() y () y() Hece x e 6t ( 3 2 t + 4 ) ad y 3 2 e6t Fially, solvig for x ad y, x 2 x y 4 y 2 4 e 6t ( 3 2 t + 4 ) 3 2 e6t ( 2)e 6t ( 3 2 t + 4 ) e6t 4e 6t ( 3 2 t + 4 ) e 6t ( 3t) e 6t (6t + ) Hece x e 6t ( 3t) ad y e 6t (6t + ) 9 Let A /2 /2 /4 /4 /2 /4 /4 /2 /4 3/2 (a) We first determie the characteristic polyomial ch A (λ) λ /2 /2 ch A (λ) det (λi 3 A) /4 λ /4 /2 /4 /4 λ /2 ( λ ) λ /4 /2 2 /4 λ /2 + /4 /2 2 /4 λ /2 ( λ ) {( λ ) ( λ ) } + { ( λ ) } ( λ ) ( λ 2 3λ ) λ {( λ λ ) ( λ 3 ) }

11 ( λ λ(λ ) ) λ 2 5λ ( λ 4 ) (b) Hece the characteristic polyomial has o repeated roots ad we ca use Theorem 622 to fid a o sigular matrix P such that P AP diag(,, 4 ) We take P X X 2 X 3, where X, X 2, X 3 are eigevectors correspodig to the respective eigevalues,, 4 Fidig X : We have to solve (A I 3 )X we have A I 3 /2 /2 /4 3/4 /2 /4 /4 /2 Hece the eigespace cosists of vectors X x, y, z t satisfyig x z ad y z, with z arbitrary Hece z X z z z ad we ca take X,, t Fidig X 2 : We solve AX We have /2 /2 A /4 /4 /2 /4 /4 /2 Hece the eigespace cosists of vectors X x, y, z t satisfyig x y ad z, with y arbitrary Hece y X y y ad we ca take X 2,, t Fidig X 3 : We solve (A 4 I 3)X We have A /4 /2 2 4 I 3 /4 /2 /4 /4 /4 75

12 Hece the eigespace cosists of vectors X x, y, z t satisfyig x 2z ad y z, with z arbitrary Hece 2z 2 X z z ad we ca take X 3 2,, t 2 Hece we ca take P (c) A P diag(,, 4 )P so A P diag(,, 4 )P Hece Let A A (a) We first determie the characteristic polyomial ch A (λ) ch A (λ) (λ 3) λ λ λ 5 λ λ R 3 R 3 + R 2 λ λ 5 2 λ 3 λ 3

13 C 3 C 3 C 2 (λ 3) (λ 3) λ λ 5 λ + 7 λ λ + 7 (λ 3) {(λ 5)( λ + 7) + 8} (λ 3)( λ 2 + 5λ + 7λ ) (λ 3)( λ 2 + 2λ 27) (λ 3)( )(λ 3)(λ 9) (λ 3) 2 (λ 9) We have to fid bases for each of the eigespaces N(A 9I 3 ) ad N(A 3I 3 ) First we solve (A 3I 3 )X We have A 3I Hece the eigespace cosists of vectors X x, y, z t satisfyig x y+z, with y ad z arbitrary Hece y + z X y y + z, z so X,, t ad X 2,, t form a basis for the eigespace correspodig to the eigevalue 3 Next we solve (A 9I 3 )X We have A 9I Hece the eigespace cosists of vectors X x, y, z t satisfyig x z ad y z, with z arbitrary Hece z X z z z ad we ca take X 3,, t as a basis for the eigespace correspodig to the eigevalue 9 77

14 The Theorem 623 assures us that P X X 2 X 3 is o sigular ad P AP

Eigenvalues and Eigenvectors

Eigenvalues and Eigenvectors 5 Eigevalues ad Eigevectors 5.3 DIAGONALIZATION DIAGONALIZATION Example 1: Let. Fid a formula for A k, give that P 1 1 = 1 2 ad, where Solutio: The stadard formula for the iverse of a 2 2 matrix yields