# Is the minimum singular value (amplitude) of $\boldsymbol{AB}$ bounded by the counterparts of $\boldsymbol{A}$ and $\boldsymbol{B}$?

Consider two complex matrices $$\boldsymbol{A} \in \mathbb{C}^{M\times N}$$ and $$\boldsymbol{B} \in \mathbb{C}^{N\times Q}$$. Is the minimum singular value's amplitude of $$\boldsymbol{AB}$$ bounded by the minimum singular value amplitudes of $$\boldsymbol{A}$$ and $$\boldsymbol{B}$$? That is,

does $$|\sigma_{min}(\boldsymbol{AB})| \leq \min \{|\sigma_{min}(\boldsymbol{A})|,|\sigma_{min}(\boldsymbol{B})|\}$$ hold?

• Try with $A= \operatorname{diag}(1, {1 \over 2}), B= \operatorname{diag}({1 \over 2},1)$. Dec 31 '20 at 6:39
• @copper.hat Wow thanks, then what if $\min\{|\sigma_{min}(\boldsymbol{A})|,|\sigma_{min}(\boldsymbol{B})|\}$? Is $|\sigma_{min}(\boldsymbol{AB})|$ bounded by this? Dec 31 '20 at 6:57
• @copper.hat Can I intuitively believe that, since the rank of $\boldsymbol{AB}$ will be lower than $\boldsymbol{A}$ and $\boldsymbol{B}$, then $|\sigma_{min}(\boldsymbol{AB})|$ can not be greater than $|\sigma_{min}(\boldsymbol{A})|$ and $|\sigma_{min}(\boldsymbol{B})|$? Dec 31 '20 at 7:11

The inequality you do have is $$\sigma_{\min}(A B) \ge \sigma_{\min}(A) \cdot \sigma_{\min}(B)$$
• What if when both $|\sigma_{min}(\boldsymbol{A})|$ and $|\sigma_{min}(\boldsymbol{B})|$ lower than 1? Is the original proposition valid at this time? And can you give me some hint to prove your inequality? Jan 2 '21 at 9:19