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Unformatted text preview: Purdue University MA 353: Linear Algebra II with Applications Homework 9, due Mar. 26, Some Solutions (I) Sec. 5.1:#17 Let T be the linear operator on M n × n ( R ) defined by T ( A ) = A t . (a) Show that ± 1 is the only eigenvalues of T . Proof. Let λ be an eigenvalue of T . Then T ( A ) = λA for some nonzero A . Hence A t = λA. Now since A is nonzero, there is an entry A ij 6 = 0 of A for some i and j . (Recall that A ij is the ( i,j ) entry of A .) Then A t = λA implies A ji = λA ij and A ij = λA ji . This gives A ij = λ 2 A ij . Since A ij 6 = 0, this implies λ 2 = 1, i.e. λ = ± 1. (b) Describe the eigenvectors corresponding to each eigenvalue of A . Answer. Assume λ = 1. Then if A is an eigenvector, then T ( A ) = A , i.e. A t = A . So the set of the eigenvectors for λ = 1 is precisely the set of nonzero symmetric matrices. Assume λ = 1. Then if A is an eigenvector, then T ( A ) = A , i.e. A t = A . So the set of the eigenvectors for λ = 1 is precisely the set of nonzero skew symmetric matrices. (c) Find an ordered basis β for M 2 × 2 ( R ) such that [ T ] β is a diagonal matrix. Answer. Note that a basis for the eigenspace E 1 for λ = 1 is 1 0 0 0 , 0 1 1 0 , 0 0 0 1 , and a basis for the eigenspace E 1 for λ = 1 is 1 1 0 . Hence we may choose β = 1 0 0 0 , 0 1 1 0 , 0 0 0 1 , 1 1 0 1 (c) Find an ordered basis β for M n × n ( R ) such that [ T ] β is a diagonal matrix for n > 2. Answer. Just generalize the above problem. For the eigenspace E 1 , we may choose a basis consisting of elements of the form { ( A ij ) : A ij = A ji = 1 for one pair of i,j ,A ij = 0 for any other i,j } ....
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 Spring '08
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 Linear Algebra, Algebra, Vector Space, aij, linear operator, finite dimensional vector

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