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 $A$.

  • 1
    $\begingroup$ Simple and concise. Thank you! $\endgroup$
    – Um Shmum
    Dec 20 '18 at 19:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.