3
$\begingroup$

I stumbled on the following inequality for singular values (stated without proof), and would like to understand it better:

Let $A,B$ be two $n\times n$ real matrices. Denoting by $\mu_i(C)$ the $i$-th singular value of a matrix $C$ and $by$ $\|\cdot\|$ the operator norm we have for $i=1,\dots,n$

$\mu_i(AB) \ \leq \|A\| \ \mu_i(B)$

and

$\mu_i(AB) \ \leq \|B\| \ \mu_i(A)$

Why is this true? Standard references? Is this inequality a specific property of singular values or does it work also for eigenvalues?

1 Answers 1

6

The singular values of $A$ are the square roots of the eigenvalues of $A^T A$ (or equivalently of $A A^T$). Thus the singular values of $AB$ are the square roots of the eigenvalues of $B^T A^T A B$ or $A B B^T A^T$. Now for any vector $v$, $v^T A^T A v = \|A v\|^2 \le \|A\|^2 v^T v$. Apply that to $v = B w$ and you get $w^T B^T A^T A B w \le \|A\|^2 w^T B^T B w$. By the Min-Max Theorem http://en.wikipedia.org/wiki/Min-max_theorem it follows that $\mu_i(AB) \le \|A\| \mu_i(B)$. Similarly, using $A B B^T A^T$ instead of $B^T A^T A B$ you get $\mu_i(AB) \le \|B\| \mu_i(A)$.

  • 0
    @RobertIsrael Thanks for your thorough and lucid answer, Robert. I used it for a proof in a research project, will definitely acknowledge you in the "Acknowledgement" once it got published. Thanks again!2018-09-05