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.
To prove.

Question

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.
To prove.

anonymous · Answer

Ugh I am so glad I finished linear algebra, however I have used eigenvalues several times since then.