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Unformatted text preview: Math 3A (44351)
Quiz 1 Key 4 3 . Find the 12 5 eigenspace (eigenvector) corresponding to at least one of these eigenvalues. Problem 1. Compute the eigenvalues of A = Solution: det(A  I) = (4  )(5  )  36 = 2 +  56 = ( + 8)(  7), which tells us that the two eigenvalues of A are 1 = 7 and 2 = 8. For 2 = 8, the eigenspace will be the set of all solutions to Ax = 7x: 4 3 x1 4x1 + 3x2 7x1 = = 12 5 x2 12x1  5x2 7x2 So we have these two equations 4x1 +3x2 = 7x1 and 12x1 5x2 = 7x2 which both reduce to the same equation x1 = x2 , which means that the eigenspace is all vectors of the form (, ) = (1, 1), and thus the eigenvector is (1, 1). Problem 2. Prove that if a matrix A is both idempotent and nilpotent, then it must be the identity matrix, A = I. Solution: A is idempotent, which means that A2 = A. Thus, A3 = A A = AA = A2 = A, and in fact An = A for all positive integers n.
2 Since A is nilpotent, An = I for some positive integer n. Well, this A = A from the argument above, and so An = A = I.
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This note was uploaded on 05/08/2010 for the course MAT MATH taught by Professor Pantano during the Spring '10 term at UC Irvine.
 Spring '10
 PANTANO
 Math

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