0.1 Diagonalization and its Applications
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1 .. DIAGONALIZATION AND ITS APPLICATIONS. Diagonalization and its Applications Exercise.. Let A be a -matrix whose characteristic polynomial is A (t) = t t + t +. a) Express A in term of powers A, A, and A of A. b) Express A 5 in term of powers A, A, and A of A. Solution: a) By the Cayley Hamilton theorem, A A + A + I =. Hence I = A + A A = A ( I + A ) A. Thus A = I + A A. b) Since A = A A I, we have A 5 = A A = A (A A I) = A A A A = = A(A A I) (A A I) A = 4A 6A 4A 6A +9A+6I A = = 4A 4A + 5A + 6I = 4(A A I) 4A + 5A + 6I = = 8A A 8I 4A + 5A + 6I = I 7A 6A. Exercise.. Let A = a) Find an invertible matrix P such that P AP = b) Find the matrix P A P for the matrix P which is found in a). 8 c) Find an invertible matrix Q such that Q A Q = 7
2 Solution: a) We have λ =, λ =, λ =, where λ is from (A λi)x =. / λ = ; y is free, hence P = λ = λ = 4 ; z is free, hence P = b) P A P = (P AP ) = c) Q A Q = λ = λ = λ = ( ) P = / / ; x is free, hence P = = = D for D = 5 / / ; x is free, hence Q = ; z is free, hence Q = Q = Hence Q AQ = ; y is free, hence Q = / / /
3 .. DIAGONALIZATION AND ITS APPLICATIONS Exercise.. Let B = x What must be value of x so that B is diagonalizable? Solution: B (t) = det(b λi) = (λ )(λ ) (λ ); λ = λ =. B is diagonalizable iff there are two linearly independent eigenvectors corresponding to λ =. λ =, (B λi)x =, x x In order to have two fundamental solutions, x must be zero. Exercise..4 Find all eigenvalues of the matrix A = [ Solution: λi A = λ λ 4 = (λ 5)(λ 4) = λ 9λ+ = λ 9λ. λ, = 9 ± =,. Exercise..5 The eigenvalues of the matrix B = λ =, λ = 5. Show that B is not diagonalizable. Solution: ) λ = λ = λ =, (λi B)X =, are λ = x y z =., the variable z is free.
4 4 So there is only one fundamental solution F = ) λ = λ = 5, (λi B)X =, x y z 8 So there is only one fundamental solution F = 8 =., the variable z is free. Since we have only two linearly independent eigenvectors correspondent to λ, λ, and λ, the matrix B is not diagonalizable. Exercise..6 The eigenvalues of the real symmetric matrix C = are λ = λ =, λ =. Diagonalize C by means of an orthogonal matrix. Solution: λ = λ = λ =, λi C =, the variables y and z are free. P = λ = λ =, λi C =, P =, the variable z is free. P = Use the Gram Schmidt orthogonalization. R +R R +R / / / / Q = P = Q = P (P Q ) (Q Q ) Q = = / /
5 .. DIAGONALIZATION AND ITS APPLICATIONS 5 Q = P (P Q ) (Q Q ) Q (P Q ) (Q Q ) Q = P Q Q = Normalizing {Q, Q, Q } one gets / Q = /, Q = Q = [ Q, Q, Q = / 6 / 6 / 6, Q = Hence the orthogonal matrix which diagonalizes C is / / / Exercise..7 Determine whether or not the following matrix is diagonalizable, and if it is, find a diagonalizing matrix P and a diagonal matrix D such that P AP = D. 4 a) A = b) B = Solution: a) λi A = λ λ λ + ) (λ + ) =. The eigenvalues are λ, = and λ =. λ =, (I A)X = with X = [x, y, z T. I A =, the variables x and z are free. P = λ =, ( I A)X =. I A = variable z is free. P = / = (λ )(λ 4) = (λ 4, P =, the
6 6 Hence P = [P P P = b) λi B = / λ 4 λ λ and D = = (λ )((λ 4)(λ ) + ) = (λ )(λ 6λ + 9) = (λ ) =. So, the only eigenvalue is λ =. λ =, (I B)X =. I B = y and z are free., the variables Thus we have only two fundamental solutions which are not enough for diagonalizing a matrix. Hence B is not diagonalizable. Exercise..8 Given a diagonal matrix D = and an orthogo- nal matrix P = P AP = D., find a real symmetric matrix A such that Solution: Since P is orthogonal, P = P T. Then A = P DP = P DP T. P D = A = P D P T = = ( ), P T = = / / / / =
7 .. DIAGONALIZATION AND ITS APPLICATIONS 7 Exercise..9 The characteristic polynomial of an invertible -matrix A is given by x x 6x + 8. a) Write A as a polynomial matrix in A. b) Write A 5 as a linear combination of I, A, A, A. Solution: a) By the Cayley-Hamilton theorem, A A 6A + 8I =. Hence A A 6I + 8A = and A = 8 A + 8 A + 4 A = 4 I + 8 A 8 A. b) From (a), A A 6I + 8A = then A 5 A 4 6A + 8A = and thus A 5 = A 4 + 6A 8A. Analogously A 4 A 6A + 8A = and thus A 4 = A + 6A 8A. So we have A 5 = (A + 6A 8A) + 6A 8A = 9A + 8A 4A + 6A 8A = Exercise.. Let A = a) Find eigenvalues of A. = 4A + A + 5A. b) Determine whether A is invertible or not. c) Find an orthogonal matrix P and a diagonal matrix D such that P T AP = D. λ Solution: a) = A λi = λ λ = ( λ) λ λ λ + λ = ( λ)(λ 4) ( λ 4) + (4 + λ) = (λ + )( λ + λ + 8) = (λ + ) (λ 4) then λ, = and λ = 4. b) The determinant A = = ( ) + = ( )( 4)+ 4 = 6. Since 6, the matrix A is invertible.
8 8 c) λ = 4, λi A = the variable z is free. y = z; x = y z = z. P = λ =, λi A = free. x = y z. The fundamental solutions are P = Use the Gram Schmidt orthogonalization. Q = P =, Q = P = since P P ,, the variables y and z are Q = P (P Q ) (Q Q ) Q (P Q ) (Q Q ) Q = P P = Normalizing {Q, Q, Q } one gets / Q = / /, Q = P = [ Q, Q, Q = Exercise.. Let A = / /, Q = D =, P = / / / 6 / 6 / 6 4 a) Determine the characteristic polynomial and all eigenvalues of the matrix A. b) Find the eigenvectors of the matrix A. c) Diagonalize the matrix A by means of an orthogonal matrix Q such that Q T AD is diagonal.
9 .. DIAGONALIZATION AND ITS APPLICATIONS 9 Solution: a) λi A = λ + λ λ = (λ + )(λ ) (λ + ) = (λ+) λ (λ ) = λ λ λ is a characteristic polynomial of A and λ =, λ =, and λ = are the eigenvalues of A. b) λ =, λi A =, the variable x is free. y = z =. P = λ =, λi A = x =, y = z. P = λ =, λi A = y = z. P =, the variable z is free., the variable z is free. x =, c) The vectors P,P, and P are orthogonal so we need only to norming them. P = P P = P = P = P P = P / = P = P P = P / = / / / /
10 So the required orthogonal matrix is Q = [ P P P = / / / / ; Q = Q T = Q. D = Q T AQ = QAQ =
11 .. LINEAR TRANSFORMATIONS. Linear Transformations Exercise.. Consider the linear operator T : R R defined by T (x, x, x ) = (x + x, x + x ). a) Is T one-to-one? Explain. b) Find a basis for the kernel (= null space) N = T () of T. c) Extend the basis which is found in b) to a basis for R. d) Find the dimension of N = T () and the dimension of T (R ). e) Find the matrix representation of T with respect to the standard bases in R and R. Solution: a) T is not one-to-one, since, for example, T (,, ) = (, ) = T (,, ). { x + x b) T (x, x, x ) = (x + x, x + x ) = (, ). Thus = x + x =. [ ; x is free. P = B = {P T } = {(,, )} is a basis for T (). c) [P e e e = = R +R R +R Hence {P, e, e } is a basis for R. is a single fundamental solution. Hence d) dim(n) = and dim(t (R )) = =. e) T (,, ) = (, ) = (, ) + (, ); T (,, ) = (, ) = (, ) + (, ); T (,, ) = (, ) = (, ) + (, ). A T = [. R +R
12 Exercise.. The linear transformation T of R id given by T (x, y, z) = (x + y, y + z, x + z). a) Find the matrix representation of T relative to the standard basis for R. b) Find a vector of norm one which is orthogonal to the vector space T (R ) relative to the standard inner product in R. Solution: a) T (e ) = T (,, ) = (,, ); T (e ) = T (,, ) = (,, ); T (e ) = T (,, ) = (,, ). The required matrix is A T = b) (T (e ) (x, y, z)) = ((,, )) (x, y, z)) = x z = ; (T (e ) (x, y, z)) = ((,, )) (x, y, z)) = x + y = ; (T (e ) (x, y, z)) = ((,, )) (x, y, z)) = y + z =. x z = We have x + y =. y + z = [[x[y[z = R +R R +R Since z is free, we find the fundamental solution that is F = (F F ) = /. The required vector is /, F = P = ± F F = ± / / /
13 .. LINEAR TRANSFORMATIONS Exercise.. Let L : R R be the linear transformation given by L(A) = (A AT ). a) Write a basis and find the dimension of the Ker(L) = {A R L(A) = R }. b) Write a basis and find the dimension of the Im(L) = {L(A) A R }. c) Find the matrix representation of L with respect to the standard ordered basis for R which is {E, E, E, E }. [ a b Solution: a) A = c d b = c. Thus {[ a b Ker(L) = c d And the standard basis for Ker(L) is [ a =, b = d = : [. B = (is the funda- Hence dim(ker(l)) =. ([ ) a b b) L(A) = L c d x = y = b c z = c b t = mental solution). [. We have L(A) = } {[ a b b = c = b d ; a =, b =, d = : {[ = [ [, b c c b [, b c c b } a, b, d R. [, so y = z, and z is free. We have P = {[ Thus the basis for Im(L) is {[ c) B = {E, E, E, E } = }. [ x y = z t. =, hence ; a = b =, d = : }, and hence dim(im(l)) =. [, [, [, }.
14 4 = [[[ A L = [[L(E ) B [L(E ) B [L(E ) B [L(E ) B = [[ [[ / / / / B B B = [[ B = Exercise..4 Let L : V V be a linear transformation such that L(v ) = v and L(v ) = v, where v, v. a) Show that v, v are linearly independent. b) Show that there is no λ R such that L(v + v ) = λ(v + v ). c) Let B = {w, w, w, w 4 } be a basis for V such that B = {w, w } is a basis for Ker(L) = {v V L(v) = }. Show that the vectors L(w ), L(w 4 ) are linearly independent. Solution: a) Find α, β R such that αv + βv =. It means that = L(αv + βv ) = αl(v ) + βl(v ) = αv βv = αv + βv =. Then αv = that is α = (since v by the condition) and so β = α =. Thus v and v are linearly independent. b) If such a λ exists then L(v + v ) = v v = λ(v + v ) that is equivalent to ( λ)v = v + λv = (λ + )v. If λ then v = +λv λ that contradicts to a) (v and v are linearly independent). If λ = then ( + λ)v = and v = that contradicts to the condition that v, v. So there is no such a λ. c) {w, w } is a basis for Ker(L) then L(αw + βw ) = = αl(w ) + βl(w ) for all α, β R. If L(w ) and L(w 4 ) are linearly dependent then there are γ and δ (γ, δ ) such that γl(w ) + δl(w 4 ) = that is equivalent to L(γw + δw 4 ) = i.e. γw + δw 4 Ker(L). This means that there are a, b R (a, b ) such that γw + δw 4 = aw +bw that contradicts to the condition that w, w, w, and w 4 are linearly independent as basis vectors in B.
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