Show that this orthogonal diagonalization is a singular value decomposition.

[s_j_sawyer]

Homework Statement

Prove that if A is an nxn positive definite symmetric matrix, then an orthogonal diagonalization A = PDP' is a singular value decomposition. (where P' = transpose(P))2. The attempt at a solution.

I really don't know how to start this problem off. I know that the singular value decomposition is of the form A = UEV' where E will be an nxn matrix containing the singular values of A, and in this case the eigenvalues of A as well. But that's about it. Any help would be greatly appreciated!

 

[lanedance]

is A a real matrix? if A is symmetric, how are the eigenvectors of A related to that of A^T

 

[s_j_sawyer]

is A a real matrix? if A is symmetric, how are the eigenvectors of A related to that of A^T

Well, A has an orthonormal set of n eigenvectors, which would therefore be the same as A^T, but I don't know how to use this in the proof.