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Math 220 Exam 3 Solutions Spring ’99 3 1 1. Let A = . 0 3 a) What is the characteristic polynomial chA (λ)? λ − 3 −1 = (λ − 3)2 . Solution: chA (λ) = 0 λ − 3 b) For each eigenvalue of A, find a basis for the associated eigenspace. Solution: The only eigenvalue is 3. The eigenspace for 3 is 0 1 the nullspace of A − 3I = , which is already reduced. 0 0 There is one free variable: x1 , so we set x1 = t. The first row of A − 3I gives x2 = 0. Thus, a vector x is in the eigenspace of 3 if t 1 x= =t , 0 0 1 so the eigenspace is one-dimensional with basis . 0 c) Is A diagonalizable? If so, find a matrix P such that P −1 AP is diagonal, and display the diagonal matrix P −1 AP . Solution: The geometric multiplicity of 3 is 1 and the algebraic multiplicity is 2. Since these are not equal, A is not diagonalizable. 2 Exam 3 Solutions 7 0 −10 2. Let A = 5 2 −10. Then the characteristic polynomial of A 5 0 −8 2 is (λ − 2) (λ + 3). a) For each eigenvalue of A, find a basis for the associated eigenspace. Solution: The eigenspace of 2 is the nullspace of 5 0 −10 A − 2I = 5 0 −10 . 5 0 −10 1 0 −2 This reduces to 0 0 0 . There are two free variables: x2 0 0 0 and x3 . We set x2 = s and x3 = t. The first row of the reduction gives x1 = 2x3 = 2t, so x is in the eigenspace if 2t 2 0 x = s = t 0 + s 1 . t 1 0 2 0 Thus, a basis for the eigenspace of 2 is given by 0 , 1. 1 0 10 0 −10 The eigenspace for −3 is the nullspace of A+3I = 5 5 −10. 5 0 −5 1 0 −1 This reduces to 0 1 −1, which has one free variable: x3 . 0 0 0 We set x3 = t. The first row then gives x1 = x3 = t, and the second row gives x2 = x3 = t, so x is in the eigenspace if t 1 x = t = t 1 . t 1 1 Thus, the eigenspace is one-dimensional with basis 1. 1 Exam 3 Solutions 3 b) Is A diagonalizable? If so, find a matrix P such that P −1 AP is diagonal, and display the diagonal matrix P −1 AP . Solution: The algebraic and geometric multiplicities are equal for each of the two eigenvalues, so A is diagonalizable. basis A 2 0 1 of R3 consisting of eigenvectors is given by 0, 1, 1. 1 0 1 We make these the columns of the matrix P : 2 0 1 P = 0 1 1 . 1 0 1 2 0 0 Then P −1 AP = 0 2 0 , the diagonal matrix whose diag0 0 −3 onal entries are the eigenvalues corresponding to the eigenvectors in the same-numbered columns of P . 3. Let T : R2 → R2 be the linear transformation x1 2x1 − x2 T = . x2 x 1 + x2 a) What is the matrix for T with respect to the standard basis B = {e1 , e2 } of R2 ? Solution: The matrix of T with respect to the basis standard 2 −1 . is the standard matrix of T : [T (e1 )|T (e2 )] = 1 1 3 2 b) Let B = {v2 , v2 } be the basis given by v1 = and v2 = . 2 1 What is the matrix [T ]B0 of T with respect to B0 ? (Use the transition matrix P = [I]B,B0 from B 0 to B to find it.) 0 Solution: We have P = [I]B,B0 = [ [v1 ]B | [v2 ]B ] = [v1 | v2 ], 3 2 because B is the standard basis. Thus P = . Since 2 1 4 Exam 3 Solutions 1 −2 −1 2 1 −1 det P = −1, P = −1 = . We get −2 3 2 −3 [T ]B0 = P −1 [T ]B P −1 2 2 −1 3 2 = 2 −3 1 1 2 1 6 3 = −7 −3 4. Let T : P2 → P2 be given by T (p) = p(3x + 2). a) What is the matrix of T with respect to the standard basis B = {1, x, x2 }? Solution: [T ]B = [[T (1)]B | [T (x)]B | [T (x2 )]B ] = [[1]B | [3x + 2]B | [(3x + 2)2 ]B ] 1 2 4 = 0 3 12 0 0 9 b) What is det(T )? Solution: det T = det[T ]B . Since [T ]B is upper triangular, its determinant is the product of its diagonal entries. Thus, det T = 27. c) What are the rank and nullity of T ? Solution: The rank and nullity of T are the rank and nullity of [T ]B . Because det[T ]B 6= 0, [T ]B is invertible, so its rank is 3 and its nullity is 0. d) What is the characteristic polynomial chT (λ)? λ − 1 −2 −4 λ − 3 −12 . Again, Solution: chT (λ) = ch[T ]B (λ) = 0 0 0 λ − 9 the matrix is upper triangular, so chT (λ) = (λ−1)(λ−3)(λ−9). Exam 3 Solutions 5 e) What are the eigenvalues of T ? Solution: The eigenvalues are 1, 3, 9. f) Is T diagonalizable? Solution: Yes: Each eigenvalue has algebraic multiplicity 1. Eigenvalues always have geometric multiplicity at least 1, so the two multiplicities must be equal for each eigenvalue.