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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?

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  • $\begingroup$ Try with $A= \operatorname{diag}(1, {1 \over 2}), B= \operatorname{diag}({1 \over 2},1)$. $\endgroup$
    – copper.hat
    Dec 31 '20 at 6:39
  • $\begingroup$ @copper.hat Wow thanks, then what if $\min\{|\sigma_{min}(\boldsymbol{A})|,|\sigma_{min}(\boldsymbol{B})|\}$? Is $|\sigma_{min}(\boldsymbol{AB})|$ bounded by this? $\endgroup$
    – McZhang
    Dec 31 '20 at 6:57
  • $\begingroup$ @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 $|\sigma_{min}(\boldsymbol{B})|$? $\endgroup$
    – McZhang
    Dec 31 '20 at 7:11
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No, it doesn't even work for scalar matrices.

An inequality of this type should be homogenous, of the correct degree.

The inequality you do have is $$\sigma_{\min}(A B) \ge \sigma_{\min}(A) \cdot \sigma_{\min}(B)$$

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  • $\begingroup$ 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? $\endgroup$
    – McZhang
    Jan 2 '21 at 9:19

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