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quiz3_1806_sol_s05

# quiz3_1806_sol_s05 - Exam 3 Friday May 4th 2005 Solutions...

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Exam 3, Friday May 4th, 2005 Solutions Question 1. (a) The characteristic polynomial of the matrix A is 2 3 9 1 1 3 + 2 + = ( 1) 2 16 16 4 and thus the eigenvalues of A are 1 with multiplicity one and 1 with multiplicity two. Clearly 4 the eigenvectors with eigenvalue 1 are the non-zero multiples of the vector (1 , 1 , 1) T . The remaining eigenvectors are all non-zero vector of the orthogonal complement of (1 , 1 , 1) T : they are the vectors a b , with ( a, b ) = (0 , 0) a b and an orthogonal basis for this vector space is � ⎞ 1 1 1 , 1 0 2 (b) For the matrix S we may choose the orthogonal matrix 1 1 1 S = 3 2 6 1 1 1 3 2 6 1 2 0 3 6 1 and the matrix is then the diagonal matrix with entries 1, 4 , and 1 along the diagonal. 4 1 0 0 We have lim A k = S lim k S 1 = S 0 0 0 S T k �� k �� 0 0 0 and therefore 1 0 0

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