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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?

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up vote 3 down vote accepted

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 $\mu_i(AB) \le \|B\| \mu_i(A)$.

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Thanks for this fast answer. As far as I understand the same reasoning doesn't work if $A,B$ are symmetric and if the $\mu_i$'s denote eigenvalues instead of singular values. Am I wrong? Do in that case the inequalities not hold? –  Hans Aug 17 '12 at 20:38
    
Why $A^T A$ and $A A^T$ have the same eigenvalues? –  Hans Aug 17 '12 at 21:01
    
ok, I got it: if $u$ is an eigenvector of $A^TA$ then $Au$ is an eigenvector of $AA^T$ with same eigenvalue, and viceversa. –  Hans Aug 17 '12 at 21:06
    
@Filippo: if $A$ and $B$ are symmetric, $AB$ is generally not symmetric, and its eigenvalues may not even be real. –  Robert Israel Aug 17 '12 at 21:17
    
For example, try $A = \pmatrix{2 & 1\cr 1 & 0\cr}$, $B = \pmatrix{1 & -1\cr -1 & 0\cr}$. Their product has eigenvalues $\pm i$. –  Robert Israel Aug 17 '12 at 21:29
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