quiz3solutions - Quiz III Math 403 15 September 2010 Name...

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Quiz III, Math 403, 15 September, 2010 Name: Solutions 1. (30 points:) Find the eigenvalues and eigenvectors of the matrix A = ± 1 - 2 - 2 1 ² . Solution: Consider A - λI = ± 1 - λ - 2 - 2 1 - λ ² . Then the eigenvalues of A are the roots of characteristic polynomial: p ( λ ) = det( A - λI ) = (1 - λ )(1 - λ ) - ( - 2)( - 2) = ( λ - 1) 2 - 4 = λ 2 - 2 λ + 1 - 4 = λ 2 - 2 λ - 3 = ( λ - 3)( λ + 1) . The eigenvalues are λ 1 = - 1 and λ 1 = 3. Eigenvector associated to λ 1 = - 1: A - λ 1 I = ± 2 - 2 - 2 2 ² . Solving the homogeneous system ( A - λ 1 I ) v 1 = 0 , we find ± 2 - 2 0 - 2 2 0 ² 1 2 R 1 R 1 --------→ ± 1 - 1 0 - 2 2 0 ² 2 R 1 + R 2 R 2 ± 1 - 1 0 0 0 0 ² . Then, since it is a free variable, v 2 = t , for any t R . It follows that v 1 = t as well. Hence v 1 = t ± 1 1 ² , 1
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for any t R . Picking t = 1, we find v 1 = ± 1 1 ² . Eigenvector associated to λ 2 = 3: A - λ 2
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quiz3solutions - Quiz III Math 403 15 September 2010 Name...

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