Lower bound on norm of product of two matrices

Question

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?

Answer 1

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).

Answer 2

By use of the multiplicativity of the spectral norm: $$\|ABB^{-1}\|_2 \le \|AB\|_2\|B^{-1}\|_2,$$ which implies that $$\|AB\|_2\ge \|A\|_2 \frac{1}{\|B^{-1}\|_2}.$$