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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 λI ) = fl fl fl fl fl fl 4 λ 6 4 2 3 λ 4 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 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( λI A ) = fl fl fl fl fl fl λ 2 2 1 3 λ 1 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:  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 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 1 λ 2 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 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 λ = 1 is only onedimensional, there is no basis for R 3 consisting of eigevectors, so the matrix A 3 is not diagonalizable ....
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This homework help was uploaded on 04/01/2008 for the course MATH 601 taught by Professor Alndy during the Spring '08 term at A.T. Still University.
 Spring '08
 alndy
 Matrices, Complex Numbers

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