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The following inequality is known: $\parallel AB\parallel\geq\parallel A\parallel \sigma_{n}(B)$. However, it is only valid where both $A$ and $B$ are square.

Is there an analogue for rectangular matrices? Specifically, given that $A$ is of size $n \times nk$, $B$ is of size $nk \times nk$ and $C$ is of size $nk \times n$, what is the lower bound, in terms of singular values, of $\parallel ABC\parallel$?

Thank you.

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  • $\begingroup$ Sorry, I should have used the opposite sign. Indeed, I meant the smallest singular value. In case both $A$ and $B$ have the same dimension ($n$), it is clear that the n-th singular value should be taken. I ask what should be the corresponding ineqaulity in case the matrices are rectangular. $\endgroup$ – DanDan Feb 23 '13 at 19:41
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I don't think there is any close analogue, because the singular values of $A,B,C$ can be all nonzero but $ABC=0$. For example, we have $$\begin{pmatrix}1&0\end{pmatrix}\begin{pmatrix}0&1\\1&0\end{pmatrix}\begin{pmatrix}1\\0\end{pmatrix}=0.$$

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