# Singular Values of Symmetric Matrix

I saw the following claim in this thread:

How to compute the SVD of a symmetric matrix?

Claim: The singular values of a symmetric matrix $$A$$ are the absolute values of its eigenvalues.

I understand why is it true for positive definite symmetric matrices as their be orthogonal diagonalization and SVD are the same.

But how can I prove it for a symmetric matrix which is not necessarily positive definite?

Let $$A=UDU^*$$ be the orthogonal diagonalization, where $$D = \mathrm{diag}(s_1,\dots,s_k,s_{k+1},\dots,s_n)$$ with $$s_1,\dots,s_k\geq 0$$ and $$s_{k+1},\dots,s_n<0$$.
Let $$V$$ be the matrix with the same firs $$k$$ columns as $$U$$ and the last $$n-k$$ columns which are the opposite as those of $$U$$: $$V=(u_1,\dots,u_k,-u_{k+1},\dots,-u_n),$$ where $$U=(u_1,\dots,u_n)$$. Moreover, let $$\Sigma = \mathrm{diag}(s_1,\dots,s_k,-s_{k+1},\dots,-s_n).$$ Then $$V$$ is also orthogonal and $$A=U\Sigma V^*$$ is the SVD of $$A$$.