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Suppose we are given two arbitrary $m \times n$ matrices, $A$, $B$, where we know $B$ has full column rank. Let $m>>n$. Can we always find a square $m \times m $ matrix $X$, such that $A=XB$? I do not care if $X$ is unique or not, as long as one exists.

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  • $\begingroup$ In optimization, can we optimize wrt $X$ instead of $A$? $\endgroup$ – user25004 Oct 18 '12 at 5:58
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Yes. The rank of $B$ is $n$, so $B$ has a trivial nullspace. Take $X = AB^{+}$, where $B^{+}$ is the Moore-Penrose pseudoinverse of $B$. Then $$XB = AB^{+}B = AI_n = A.$$

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