Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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

Let $\vert \vert . \vert \vert$ be the 2-norm. Since this norm is submultiplicative, we know that for any two square matrices $A, B \in \mathbb{R}^{n \times n}$,

$$ \vert \vert A B \vert \vert \leq \vert\vert A \vert \vert \vert \vert B \vert \vert \leq \sigma_{\textrm{max}}(A) \vert \vert B \vert \vert.$$

What I am looking for is an inequality of the form

$$ \sigma_{\textrm{min}}(A) \vert \vert B \vert \vert \leq \vert \vert A B \vert \vert. $$

The first inequality is true because this norm simply satisfies the submultiplicative property. But what about the second inequality? Is it true? And if not, is it only true for special type of matrices?

share|cite|improve this question
up vote 6 down vote accepted

The inequality is true. It is obvious when $A$ is singular. When $A$ is invertible, for any unit vector $x$, we have $\|x^TA\|\ge\sigma_\min(A)$. Therefore \begin{align} \|AB\| &= \max_{\|x\|=1} \|x^TAB\|\\ &= \max_{\|x\|=1} \|x^TA\|\left\|\frac{x^TA}{\|x^TA\|}B\right\|\tag{1}\\ &\ge \max_{\|x\|=1} \sigma_\min(A)\left\|\frac{x^TA}{\|x^TA\|}B\right\|\\ &= \max_{\|y\|=1} \sigma_\min(A)\left\|y^TB\right\|\tag{2}\\ &= \sigma_\min(A)\|B\|, \end{align} where we have used the assumption that $A$ is invertible in $(1)$ (so that we can divide by $\|x^TA\|\neq0$) and $(2)$ (so that every $x$ corresponds to a unique $y$ and vice versa).

share|cite|improve this answer

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.