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Mathematics 7 Online
OpenStudy (anonymous):

More Linear Algebra to prove. Fun! Suppose A is an n x n symmetric matrix with n distinct eigenvalues, λ1, ...,λn. If B is the n x n matrix with Bi, the ith column of B, being a (normalized) eigenvector of A corresponding to the eigenvalue λi of A (i = 1,...,n), then A = BLB' where L is the diagonal matrix with the eigenvalues of A on its diagonal.

OpenStudy (anonymous):

Anyone?

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