Let $A\in\mathbb{R}^{n\times m}$ and $B\in\mathbb{R}^{m\times k}$ be two random matrices (each element is drawn iid from a normal distribution). Also $n<m<k$.
Let $\sigma_{min}(A)$ be the smallest singular value of $A$.
Is the following true (with prob. 1)? $$\sigma_{min}(AB) \geq \sigma_{min}(A)\sigma_{min}(B)$$ If so, why? If not, can we say something about the probability of that?
I ran a lot of simulations, and in all of them I got that the statement is true, but then again, it is only simulations.
Also, I can prove it using square matrices (using the fact that $\sigma_{min}(A)=(\sigma_{max}(A^{-1}))^{-1}$ and $(AB)^{-1}=B^{-1}A^{-1}$).
Thanks!
Both sides of the inequality are zero if $B$ has deficient column rank. When $B$ has full column rank (so that $Bx\ne0$ when $x\ne0$), $$ \sigma_\min(AB) =\min_{x\ne0}\frac{\|ABx\|_2}{\|x\|_2} =\min_{x\ne0}\frac{\|ABx\|_2}{\|Bx\|_2}\frac{\|Bx\|_2}{\|x\|_2} \ge\min_{y\ne0}\frac{\|Ay\|_2}{\|y\|_2}\min_{x\ne0}\frac{\|Bx\|_2}{\|x\|_2} =\sigma_\min(A)\sigma_\min(B). $$