# HW9 Solutions Math 110 Fall 2016 - HW 9 Solutions MATH 110...

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HW 9 Solutions, MATH 110 with Professor Stankova 5.4 #2eg (e) The characteristic polynomial is - λ ( - λ (1 - λ ) + 1) + 1 = - λ 3 + λ 2 - λ + 1 = - λ ( λ 2 + 1) + ( λ 2 + 1) = (1 - λ )( λ 2 + 1). This polynomial has roots λ = 1 , ± i , so it is diagonalizable. The eigenvectors are (1 , 0 , 1) t , (1 , - i - 1 , i ) t , (1 , i - 1 , - i ) t , so Q = 0 @ 1 1 1 0 - i - 1 i - 1 1 i - i 1 A and D = 0 @ 1 0 0 0 i 0 0 0 - i 1 A . (g) The characteristic polynomial is (3 - λ )((4 - λ )(1 - λ )+2) - (2(1 - λ )+ 2)+( - 2+(4 - λ )) = (3 - λ )(4 - λ )(1 - λ )+(4 - λ ) = (4 - λ )( λ 2 - 4 λ +4) = (4 - λ )(2 - λ ) 2 , so the roots are λ = 4 , 2. However, A - 2 I = 0 @ 1 1 1 2 2 2 - 1 - 1 1 1 A is rank 2, so the 2-eigenspace is one- dimensional and we cannot have a basis of eigenvectors, so the matrix is not diagonalizable. 5.4 #3cdf (c) This transformation is not diagonalizable. Any diagonalizable transformation is diagonal- izable when restricted to an invariant subspace, and T rotates the xy -plane by - 90 degrees. (d) This is not diagonalizable. Using the standard basis the matrix is A = 0 @ 1 0 0 1 1 1 1 1 1 1 A which has characteristic polynomial (1 - t ) ( - t ); the eigenvalues t = 1 has only one eigenvector since A - I is rank 2. (f) This matrix is diagonalizable. Its eigenspaces are three 3-dimensional subspace of symmetric matrices (which has eigenvalue 1 under T since T ( A ) = A if A is sym- metric) and the one-dimensional subspace of skew-symmetric matrices (which has eigenvalue - 1). 5.4 #4 Let β = { b i } be the basis of eigenvectors. Put them into the columns of a matrix Q which is invertible since it is a basis. Then we claim that Q - 1 AQ is a diagonal matrix. To see this, note that Q - 1 AQe i = Q - 1 Ab i = Q - 1 λ i b i = λ i e i . 5.4 #5 (a) If A is diagonalizable then det( A - tI ) splits as a polynomial over t . If A is diagonalizable, say A = QDQ - 1 , then det( A - tI ) = det( QDQ - 1 - tI ) = det( Q ( D - tI ) Q - 1 ) = det( D - tI ) = Q ( λ i - t ). 5.4 #6 (a) A linear transformation is diagonalizable if and only if it has a basis of eigenvectors. Let E λ denote the λ -eigenspace; this is equivalent to asking that P λ dim( E λ ) = dim( V ). By Theorem 5.7, dim( E λ ) is the multiplicity of the root λ in the characteristic polynomial. Since eigenspaces for di erent eigenvalues have zero intersection, if any dim( E λ ) < the algebraic multiplicity, then it would be impossible for the dimension of the eigenspaces to be dim( V ). Thus, we need equalities across the board, and we need for every root of the characteristic polynomial to be in the field F . (b) Same. Find a direct formula for the terms of the sequence: a n +1 = 4 a n - 3 a n - 1 , a 0 = 0, a 1 = 1. Solution: We have a n +1 a n = 4 - 3 1 0 ◆ ✓ a n a n - 1 = 4 - 3 1 0 n a 1 a 0 Denote the matrix A . We will diagonalize A ; it has characteristic polynomial t 2 - 4 t + 3 = ( t - 3)( t - 1). Its eigenvalues are 1 , 3, with eigenvectors (1 , 1) t and (3 , 1) t respectively. So, 1
HW 9 Solutions, MATH 110 with Professor Stankova A = QDQ - 1 where Q = 1 3 1 1 and D = 1 0 0 3 . We can compute Q - 1 = 1 2 - 1 3 1 - 1 .