HW14MATH110S17 - HW 14 Solutions MATH 110 with Professor...

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HW 14 Solutions, MATH 110 with Professor Stankova 5.2 #1a-g (a) False. Take the identity matrix. (b) False. They can be scalar multiples of each other. (c) True by definition. (d) True. If v had two eigenvalues, λ and μ , then Av = λv = μv , so ( λ - μ ) v = 0, so λ - μ = 0. (e) True. (f) True. It means there is a basis of eigenvectors. (g) True. If A = PDP - 1 , then Pe 1 is an eigenvector with eigenvalue the first diagonal entry of D . 5.2 #2bdf (b) The characteristic polynomial is (1 - t ) 2 - 9 = t 2 - 2 t - 8 = ( t - 4)( t + 2). The matrix is diagonalizable. The eigenvalues are 4 , - 2 with eigenvectors (1 , 1) t , (1 , - 1) t respectively. Q is the matrix whose columns are the eigenvectors, so Q = 1 1 1 - 1 . (d) The characteristic polynomial is (7 - t )( - 5 - t )(3 - t )+4(8(3 - t )) = (3 - t )((7 - t )( - 5 - t )+32) = (3 - t )( t 2 - 2 t +3). The polynomial t 2 - 2 t +3 has no real roots, so the matrix is not diagonalizble. (f) The matrix is not diagonalizable since for the eigenvalue 1, which has multiplicity two, A - I = 0 1 0 0 0 2 0 0 0 , and the nullspace of this matrix has dimension 1. 5.2 #3abe (a) In the standard basis, this linear transformation is strictly upper triangular with zeroes on the diagonal (since derivatives lower the degree of polynomials). Thus the only eigenvalue

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