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Let $A \in Mat_{mxm}$ and $B \in Mat_{nxn}$ be invertible matrices and $X \in Mat_{mxn}$ a matrix.

Prove $\text{rank}(AXB)=\text{rank}(X)$

So I proved $\text{rank}(AX)=\text{rank}(X)$ using the nullity spaces of $AX$ and $X$, but got stuck.

Any help or hints appreciated.

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    Do you know the fact that $rank(XB)=rank(X)$? – 2017-02-26
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    @Giulio Why is that true? – 2017-02-26
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    I proved it in my answer – 2017-02-26

2 Answers 2

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Combining your knowledge with $$ \operatorname{rank}(X) = \operatorname{rank}(X^T) $$ and $$ \operatorname{rank}(XB) = \operatorname{rank}(B^TX^T) $$ gives the desired result.

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Let's prove that $\mathrm {rank}(BX)=\mathrm {rank}(X)$, knowing that $B$ is invertible.

Proof: Let $V$ be a matrix such that $B^TX^TV=0$. Multiply both sides by $(B^T)^{-1}\implies X^TV=0$. This tells us that $X^T$ shares the same nullspace as $B^TX^T$ which implies $\mathrm {rank}(X)=\mathrm {rank}(X^T)=\mathrm {rank}(B^TX^T)=\mathrm {rank}((XB)^T)=\mathrm {rank}(XB)$, knowing the fact that $\mathrm {rank}(C)=\mathrm {rank}(C^T)$ for a matrix $C$.

So now we know that $\mathrm {rank}(XB)=\mathrm {rank}(X)$ and $A$ is invertible, which gives us $$\mathrm {rank}(AXB)=\mathrm {rank}(A(XB))=\mathrm {rank}(XB)=\mathrm {rank}(X)$$ where the second equality comes under the hypothesis $A$ invertible.

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    Could you please explain why the second equality in the last row is right? I know you proved $\text{rank}(AX)=\text{rank}(X)$, but why can we place $X$ instead of $AX$ in the rank? – 2017-02-26
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    The passage was different, I changed it and explained. There I used the hypothesis that $B$ was invertible, hope now it's clear! – 2017-02-26
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    It is. Thanks ! – 2017-02-26