MATH 2050 Assignmen 9 Winer 206 Do no need o hand in Noe ha he final exam also covers maerial afer HW8, including, for insance, calculaing deerminan by row operaions, eigenvalues and eigenvecors, similariy and diagonalizaion ec Alhough HW9 is no graded, you should work on i as serious as oher homework assignmens in order o be familiar wih he above menioned maerial ) Find he deerminan by reducing o riangular form for he following marices 0 2 (a) A = 2 4 5 ANS: We perform he Gaussian Eliminaion on A by he following process: 0 2 A = 2 4 5 5 R R 3 2 4 0 2 5 R 2 R 2 2R 0 3 6 0 2 5 R 2 R 3 0 2 0 3 6 5 R 3 R 3 +3R 2 0 2 = U 0 0 0 Noe ha marix U is a riangular marix and hence deerminan of U is 0, he produc of is diagonal elemens Also noe ha inerchange rows only effec he deerminan by a sign (ie, muliply by ), and operaion R R cr does no change he deerminan, and R cr effec he deerminan by a muliplicaion of c Hence, de(a) = de(u) = 0 (b) A = 2 5 3 0 2 3 2 0
ANS: We perform he Gaussian Eliminaion on A by he following process: A = 2 5 3 0 2 R 2 R 2 2R 0 7 5 R 3 R 3 +R,R 4 R 4 3R 0 4 3 3 2 0 0 4 4 6 R 2 R 3 0 4 3 0 7 5 0 4 4 6 R 4 R 4 +4R 0 4 3 R 3 R 3 +7R 0 0 27 6 R 4 R 4 /2 0 4 3 0 0 27 6 0 0 2 6 0 0 /2 R 4 R 3 0 4 3 0 0 /2 R 4 R 4 27R 3 0 4 3 0 0 /2 = U 0 0 27 6 0 0 0 5/2 Noe ha marix U is a riangular marix and hence deerminan of U is 5/2, he produc of is diagonal elemens Also noe ha inerchange rows only effec he deerminan by a sign (ie, muliply by ), and operaion R R cr does no change he deerminan, and R cr effec he deerminan by a muliplicaion of c Hence, de(u) = de(a) ( ) 2 /2 = 5/2 Hence, de(a) = 30 2 Find he characerisic polynomial, he real eigenvalues and he corresponding eigenspaces of each of he following marices: [ 3 (a): A = 2 0 ANS: Consider A λi 2 = [ λ 3 2 λ wih de(a λi 2 ) = λ(λ + ) 6 = λ 2 + λ 6 = 0 So he characerisic polynomial is λ 2 + λ 6 Such an equaion has wo soluion λ = 2 and λ 2 = 3, which are eigenvalues of A For eigenvalue λ = 2, one needs o find he eigenvecors for i, which is o solve he following homogeneous equaion (A 2I 2 )x = 0 We can use Gauss-Eliminaion mehod o solve his as follows [ [ 3 3 0 0 [A 2I 2 0 = 2 2 0 0 0 0,
[ Hence, here is one free variable x 2 = and x = x 2 = So he soluion is x =, [ and he eigenspace of A wih respec o eigenvalue λ = 2 is for all real number For eigenvalue λ 2 = 3, one needs o find he eigenvecors for i, which is o solve he following homogeneous equaion (A + 3I 2 )x = 0 We can use Gauss-Eliminaion mehod o solve his as follows [ [ 2 3 0 2 3 0 [A + 3I 2 0 = 2 3 0 0 0 0 Hence, here [ is one free variable x 2 = and x = 3x 2 /2 = 3/2 So he soluion 3/2 is x =, and he eigenspace of A wih respec o eigenvalue λ = 3 is [ 3/2 for all real number 2 3 (b): A = 2 6 6 2 ANS: Firs, le us calculae he deerminan of A λi 3 as follows λ 2 3 A λi 3 = 2 6 λ 6 2 λ 2 λ R R 3 2 6 λ 6 λ 2 3 2 λ R 2 R 2 2R R 3 R 3 +(λ )R 0 2 λ 2λ 4 0 2λ 4 4 λ 2 R 3 R 3 +2R 2 Hence, deerminan of U is (λ 2) 3 2 λ 0 2 λ 2λ 4 0 0 (λ 2) 2 = U Also noe ha inerchange rows only effec he deerminan by a sign (ie, muliply by ), and operaion R R cr does no change he deerminan, and R cr effec he deerminan by a muliplicaion of c Hence, de(u) = de(a λi 3 ) and hen de(a λi 3 ) = (λ 2) 3, which is he characerisic polynomial of A There is only one soluion for de(a λi 3 ) = (λ 2) 3 = 0, which is λ = 2 Then λ = 2 is he only eigenvalue of A
Le us now calculae he eigenspace of A wih respec o 2, ha is all he soluions of (A 2I 3 )x = 0 We solve i by Gaussian Eliminaion mehod as follows 2 3 0 2 3 0 [A 2I 3 0 = 2 4 6 0 R 3 R 3 +R R 2 R 2 +2R 0 0 0 0 2 3 0 0 0 0 0 Thisequaion has wo free variables x 2 = s and x 3 =, which leads x = 3 2s Hence 3 2s 3 2 x = s = 0 + s 0 for all real numbers, s is he eigenspace of A wih respec o λ = 2 3 Are he following marices A and B similar o each oher? [ [ 2 4 2 4 (a): A =, B = 3 3 ANS: One can calculae he eigenvalues for A by solving de(a λi 2 ) = (λ 3)(λ+2) = 0 Tha is λ = 3 and λ 2 = 2 are he eigenvalues of A Similarly, one can calculae he eigenvalues for B by solving de(b λi 2 ) = (λ 6)(λ+ ) = 0 Tha is λ 3 = 6 and λ 4 = are he eigenvalues of B Noe ha wo similar marices mus have same eigenvalues By he above informaion, A and B have differen eigenvalues and hence hey are no similar o each oher [ [ 4 4 (b): A =, B = 2 3 3 2 ANS: One can calculae he eigenvalues for A by solving de(a λi 2 ) = (λ 5)(λ+) = 0 Tha is λ = 5 and λ 2 = are he eigenvalues of A Similarly, one can calculae he eigenvalues for B by solving de(b λi 2 ) = (λ 5)(λ ) = 0 Tha is λ 3 = 5 and λ 4 = are he eigenvalues of B Noe ha wo similar marices mus have same eigenvalues By he above informaion, A and B have differen eigenvalues and hence hey are no similar o each oher 4 Find a marix P such ha P AP is a diagonal marix [ 2 4 (a): A = ANS: One can calculae he eigenvalues for A by solving de(a λi 2 ) = (λ 3)(λ+2) = 0 Tha is λ = 3 and λ 2 = 2 are he eigenvalues of A Now le us calculae he eigenvecors of A as follows For eigenvalue λ = 3, one needs o find he eigenvecors for i, which is o solve he following homogeneous equaion (A 3I 2 )x = 0
We can use Gauss-Eliminaion mehod o solve his as follows [ [ 4 0 4 0 [A 3I 2 0 = 4 0 0 0 0 Hence, [ here is one free variable x 2 = and x = 4x 2 = 4 So [ he soluion is 4 4 x = Le = and one ges a special eigenvecor o be u = For eigenvalue λ 2 = 2, one needs o find he eigenvalues for i, which is o solve he following homogeneous equaion (A + 2I 2 )x = 0 We can use Gauss-Eliminaion mehod o solve his as follows [ [ 4 4 0 0 [A + 2I 2 0 = 0 0 0 0 Hence, [ here is one free variable x 2 = and x = x 2 = So [ he soluion is x = Take = and one ges a special eigenvecor o be u 2 = [ 4 Le marix P = [ 3 0 P AP = 0 2 [ 2 2 (b): A = 3 [ wih marix P = /5 4, a diagonal marix One can check ha ANS: One can calculae he eigenvalues for A by solving de(a λi 2 ) = (λ 4)(λ ) = 0 Tha is λ = 4 and λ 2 = are he eigenvalues of A Now le us calculae he eigenvecors of A as follows For eigenvalue λ = 4, one needs o find he eigenvecors for i, which is o solve he following homogeneous equaion (A 4I 2 )x = 0 We can use Gauss-Eliminaion mehod o solve his as follows [ [ 2 2 0 0 [A 4I 2 0 = 0 0 0 0 Hence, here is one free variable x 2 = and x = x 2 = So he soluion is x = [ Le = and one ges a special eigenvecor o be u = [
For eigenvalue λ 2 =, one needs o find he eigenvalues for i, which is o solve he following homogeneous equaion (A I 2 )x = 0 We can use Gauss-Eliminaion mehod o solve his as follows [ [ 2 0 2 0 [A I 2 0 = 2 0 0 0 0 Hence, [ here is one free variable x 2 = and x = 2x 2 = 2 So he [ soluion is 2 2 x = Take = and one ges a special eigenvecor o be u 2 = [ 2 Le marix P = [ 4 0 P AP = 0 5 A = D 6 [ [ 2 wih marix P = /3, a diagonal marix One can check ha Find a marix P such ha P AP = D is a diagonal marix Find soluion: de(a λi 2 ) = 0 gives λ = + i and λ 2 = i as he eigenvalues of A [ When λ = + i, an eigenvecor is X = i [ When λ = i, an eigenvecor is X 2 = i [ Le marix P = i i [ ( + i) D 6 6 0 = 0 ( i) 6 I gives P AP = D wih D = [ + i 0 0 i To simply, ( + i) 6 = ( 2e iπ/4 ) 6 = 2 8 e i4π = 2 8 and ( i) 6 = ( 2e iπ/4 ) 6 = 2 8 e i4π = 2 8, since e i is a periodic funcion wih period 2π