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Let $A$ and $B$ be two (nonzero) real square matrices and suppose that $\operatorname{rank} (AB) = \operatorname{rank} (BA)$ for any $B$. Can one prove that $A$ is invertible? (The converse is true and a simple linear algebra question but I'm stuck on this one.)

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If $A$ is not invertible, there is a nonzero vector $u$ in the null space of $A$. Also, since you assume $A \neq 0$, there is a nonzero vector $v$ in the column space of $A$. Use these observations to construct a rank one matrix $B$ such that $AB = 0$ and $BA \neq 0$.

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  • $\begingroup$ Technically we should consider the row space of $A$ (the column space of $A^T$) $\endgroup$ – Omnomnomnom Aug 28 '16 at 15:43
  • $\begingroup$ @Omnomnomnom: I don't think it's obligatory to talk about the row space of $A$, I was just thinking of taking $B = u v^t$ where $u \in \mathrm{null}(A)$ and $v \in \mathrm{col}(A)$ are nonzero, but I agree one could also argue slightly differently using the row space of $A$. $\endgroup$ – Mike F Aug 28 '16 at 17:25
  • $\begingroup$ It is not generally true that $v^TA\neq 0$ (Or, for that matter, that $Av=0$) when $v$ is in col(A). $\endgroup$ – Omnomnomnom Aug 29 '16 at 2:47
  • $\begingroup$ That should be true! If $v \in \mathrm{col}(A)$ is nonzero, then take some $w$ with $Aw=v$. Note that $v^T A w = v^T v = \|v\|^2 \neq 0$, so $v^TA \neq 0$ too. $\endgroup$ – Mike F Aug 29 '16 at 3:05
  • $\begingroup$ ah, so you're right. I didn't know, and I had a different solution in mind, but that all works. $\endgroup$ – Omnomnomnom Aug 29 '16 at 3:37
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Let's construct the $B$ matrix along the lines given by Mike F. In the image of $A$ we have a nonzero vector $v$ which has a nonzero antecedent $w$ so that $v=A w$. In components we define $B_{ij}=u_i w_j$. This matrix is nonzero. Left multiplication by $A$ gives 0 because $u$ is in the kernel of $A$. If we do the right-multiplication by $A$ we find a matrix which is $u_i v_j$ which is nonzero. Strictly speaking there is a transposition when dealing with the right stuff. Great ! thanks.

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