Solutions6 - Math 601 Solutions to Homework 6 1. Find the...

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Unformatted text preview: Math 601 Solutions to Homework 6 1. Find the eigenvalues for each of the following matrices. For each eigen- value, find a basis of the corresponding eigenspace. Determine whether the matrix is diagonalizable over the complex numbers. (a) A 1 = 4 6 4- 2- 3- 4 2 (b) A 2 = - 2 2 1 3- 1 2 (c) A 3 = 1 2 1 2 1 (d) A 4 = 1- 7- 4- 1 1- 4 1 3 6 Answer: (a) The characteristic polynomial is: det( A 1- I ) = fl fl fl fl fl fl 4- 6 4- 2- 3- - 4 2- fl fl fl fl fl fl = (2- ) fl fl fl fl 4- 6- 2- 3- fl fl fl fl = (2- )( 2- ) = (2- ) ( - 1) The eigenvalues are = 0 , = 1 , and = 2 To find the corresponding eigenvectors, we solve: 4 6 4- 2- 3- 4 2 x 1 x 2 x 3 = 0 x 1 x 2 x 3 x 1 =- 3 2 x 2 x 2 is free x 3 = 0 - 3 2 4 6 4- 2- 3- 4 2 x 1 x 2 x 3 = 1 x 1 x 2 x 3 x 1 =- 2 x 2 x 2 is free x 3 = 0 - 2 1 4 6 4- 2- 3- 4 2 x 1 x 2 x 3 = 2 x 1 x 2 x 3 x 1 =- 2 x 3 x 2 = 0 x 3 is free - 2 1 1 (Any nonzero multiples of these vectors would also be correct answers.) Since these three vectors are a basis for R 3 , the matrix A 1 is diagonalizable . (b) The characteristic polynomial is: det( I- A ) = fl fl fl fl fl fl- - 2 2 1 3- - 1 2- fl fl fl fl fl fl = (2- ) fl fl fl fl- - 2 1 3- fl fl fl fl = (2- )( 2- 3 + 2) = (2- )( - 2)( - 1) The eigenvalues are = 1 and = 2 The eigenvalue = 2 is a multiple root of the characteristic polynomial with algebraic multiplicity 2. To find the corresponding eigenvectors, we solve: - 2 2 1 3- 1 2 x 1 x 2 x 3 = 1 x 1 x 2 x 3 x 1 =- 2 x 2 x 2 is free x 3 = 0 - 2 1 - 2 2 1 3- 1 2 x 1 x 2 x 3 = 2 x 1 x 2 x 3 x 1 =- x 2 + x 3 x 2 is free x 3 is free - 1 1 , 1 1 (Any nonzero multiple of the first vector would also be correct, as would any other basis for the = 2 eigenspace.) Since these three vectors are a basis for R 3 , the matrix A 2 is diagonalizable . (c) The characteristic polynomial is: det( A 2- I ) = fl fl fl fl fl fl 1- 2 1- 2 1- fl fl fl fl fl fl = (1- ) 3 Therefore, the only eigenvalue is = 1 This eigenvalue is a multiple root of the characteristic polynomial, with algebraic multiplicity 3. To find the corresponding eigenvectors, we solve: 1 2 1 2 1 x 1 x 2 x 3 = 1 x 1 x 2 x 3 x 1 is free x 2 = 0 x 3 = 0 1 (Any multiple of this vector would also a basis for the eigenspace.) Since the eigenspace for = 1 is only one-dimensional, there is no basis for R 3 consisting of eigevectors, so the matrix A 3 is not diagonalizable ....
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Solutions6 - Math 601 Solutions to Homework 6 1. Find the...

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