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

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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 0 0 2 (b) A 2 = 0 - 2 2 1 3 - 1 0 0 2 (c) A 3 = 1 2 0 0 1 2 0 0 1 (d) A 4 = 1 - 7 - 4 - 1 1 - 4 1 3 6 Answer: (a) The characteristic polynomial is: det( A 1 - λI ) = fl fl fl fl fl fl 4 - λ 6 4 - 2 - 3 - λ - 4 0 0 2 - λ fl fl fl fl fl fl = (2 - λ ) fl fl fl fl 4 - λ 6 - 2 - 3 - λ fl fl fl fl = (2 - λ )( λ 2 - λ ) = (2 - λ ) λ ( λ - 1) The eigenvalues are λ = 0 , λ = 1 , and λ = 2 To find the corresponding eigenvectors, we solve: 4 6 4 - 2 - 3 - 4 0 0 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 0 4 6 4 - 2 - 3 - 4 0 0 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 0 4 6 4 - 2 - 3 - 4 0 0 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 0 1 1
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(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( λI - A ) = fl fl fl fl fl fl - λ - 2 2 1 3 - λ - 1 0 0 2 - λ fl fl fl fl fl fl = (2 - λ ) fl fl fl fl - λ - 2 1 3 - λ fl fl fl fl = (2 - λ )( λ 2 - 3 λ + 2) = (2 - λ )( λ - 2)( λ - 1) The eigenvalues are λ = 1 and λ = 2 The eigenvalue λ = 2 is a multiple root of the characteristic polynomial with algebraic multiplicity 2. To find the corresponding eigenvectors, we solve: 0 - 2 2 1 3 - 1 0 0 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 0 0 - 2 2 1 3 - 1 0 0 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 0 , 1 0 1 (Any nonzero multiple of the first vector would also be correct, as would any other basis for the λ = 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 - λI ) = fl fl fl fl fl fl 1 - λ 2 0 0 1 - λ 2 0 0 1 - λ fl fl fl fl fl fl = (1 - λ ) 3 Therefore, the only eigenvalue is λ = 1 This eigenvalue is a multiple root of the characteristic polynomial, with algebraic multiplicity 3. To find the corresponding eigenvectors, we solve: 1 2 0 0 1 2 0 0 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 0 0 (Any multiple of this vector would also a basis for the eigenspace.) Since the eigenspace for λ = 1 is only one-dimensional, there is no basis for R 3 consisting of eigevectors, so the matrix A 3 is not diagonalizable . 2
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(d) The characteristic polynomial is: det( λI - A ) = fl fl fl fl fl fl 1 - λ - 7 - 4 - 1 1 - λ - 4 1 3 6 - λ fl fl fl fl fl fl = (1 - λ ) fl fl fl fl 1 - λ - 4 3 6 - λ fl fl fl fl + 7 fl fl fl fl - 1 - 4 1 6 - λ fl fl fl fl - 4 fl fl fl fl - 1 1 - λ 1 3 fl fl fl fl = (1 - λ )( λ 2 - 7 λ + 18) + 7( λ - 2) - 4( λ - 4) = - λ 3 + 8 λ 2 - 22 λ + 20
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