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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?

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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 $A$.

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    $\begingroup$ Simple and concise. Thank you! $\endgroup$
    – Um Shmum
    Dec 20, 2018 at 19:20

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