Show that if A is an $n$ by $n$ positive definite symmetric matrix, then an orthogonal diagonalization $A = PDP^{T}$ is a singular value decomposition of A.
My attempt in showing this:
$$A=U\Sigma V^{T}$$
$$A^{T}A=(U\Sigma V^{T})^{T}U\Sigma V^{T}$$
$$A^{T}A=(V\Sigma^{T} U^{T})U\Sigma V^{T}$$
$$A^{T}A=V\Sigma^{T}\Sigma V^{T}$$
We can clearly see that $\Sigma^{T}\Sigma = D$, and since $A$ is n by n, $V$ is also n by n. In addition, since it is orthogonally diagonalizable, $P =V$ (they are both orthogonal matrices of the same same size). Hence the singular value decomposition is the same form as the orthogonal diagonalization.
Is there a better way of showing this?
Best Answer
It should be $\Sigma^\top \Sigma = D^2$, since $A^\top A = A^2 = P D^2 P^\top$.
I think there is an even more direct approach: just say $A = U \Sigma V^\top$ with $U=V=P$ and $\Sigma=D$. Since $A$ is positive definite (note that you have not used this yet in your argument) the diagonal entries of $D$ are positive, so this does not contradict the constraint that $\Sigma$ has nonnegative diagonal entries. (This argument would fail if $D$ had a negative entry.)