Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $i=1,\dots,n$

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


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

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

share|cite|improve this question
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 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 $\mu_i(AB) \le \|B\| \mu_i(A)$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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