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

Question

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.

Answer 1

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

Answer 2

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 $C$.

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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.

Answer 3

It is known that: $$ rank(XY) \le min(rank(X), rank(Y)) $$

Suppose that $Y=AXB$, then $X=A^{-1}YB^{-1}$.

$$ rank(Y)=rank(AXB) \le min(rank(A), rank(X), rank(B))=rank (B) $$ $$ rank(X) = rank(A^{-1} X B^{-1}) \le min(rank(A^{-1}), rank(B), rank(Y^{-1}))=rank(B) $$

$rank(Y) \le rank(X), \space rank(X) \le rank (Y) \Rightarrow rank(X)=rank(Y)$

P.S. You may notice that there is a little gap in my proof. Here I'll briefly explain what just happended: $$ rank(AXB) \le min (rank(AX), rank(B)), \\ rank(AX) \le min(rank(A), rank(X)) \\ $$ And here we got: $$ rank(AXB) \le min(rank(AX), rank(B) ) \le \\ \le min(min(rank(A), rank(X)), rank(B)) \le \\ \le min(rank(A), rank(X), rank(B)) $$