For a matrix $A$, let $σ_{min}(A)$ denote the smallest $\textbf{non-zero}$ singular value of A.
I saw some materials (e.g. https://pdfs.semanticscholar.org/e74f/89ac85239811b295787619e73c99fc867724.pdf)
says that
$$σ_{i}(AB)≥σ_{i}(A)σ_{n}(B),~~~ i=1,\cdots,n$$
Here $A$ and $B$ are rectangle matrix with dimension, say, $p\times n$ and $n\times m$. This holds for even zero singular value. So if $B$ is not full rank, then $\sigma_n(B)$ is $0$ and the above bound is trivial.
My question is if the following inequality holds:
$$σ_{min}(AB)≥σ_{min}(A)σ_{min}(B)$$
Thanks.
Best Answer
In general it isn't true. E.g. when $A$ is a nonzero row vector and $B$ is a nonzero column vector such that $AB=0$, both $\sigma_\min(A)$ and $\sigma_\min(B)$ are positive but $AB$ has not any nonzero singular value.
The inequality holds, however, when $A$ has full column rank and $B\ne0$ or when $A\ne0$ and $B$ has full row rank. Since the latter case reduces to the former when we consider $(AB)^T$, we shall only deal with the first case. In this case, $Bx=0$ or $ABx=0$ if and only if $x\in\operatorname{range}(B^\ast)^\perp$, and whenever $ABx\ne0$, we have $$ \|ABx\| =\left\|A\frac{Bx}{\|Bx\|}\right\|\|Bx\| =\|Ay\|\|Bx\| $$ for some unit vector $y$. Therefore, by Courant-Fischer minimax principle,
$$ \sigma_\min(AB) =\min_{\|x\|=1,\,x\in\operatorname{range}(B^\ast)}\|ABx\| \ge \min_{\|y\|=1} \left\|Ay\right\| \min_{\|x\|=1,\,x\in\operatorname{range}(B^\ast)}\|Bx\| = \sigma_\min(A)\sigma_\min(B). $$