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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 onedimensional, there is no basis for R 3 consisting of eigevectors, so the matrix A 3 is not diagonalizable ....
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 Spring '08
 alndy
 Matrices, Complex Numbers

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