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I want to prove that $r(AB) = r(BA)$, where $r$ is a rank. $A$ is a square matrix and $B$ is an invertible matrix.

Can I use this?

$r(AB) \le min \{r(A), r(B)\} = r(A)$, and from the other side

$r(A) = r(BB^{-1}ABB^{-1}) \le min \{r(BB^{-1}), r(AB), r(B^{-1})\} = r(AB)$

hence $\;$ $r(AB) = r(A)$, and

$r(BA) \le min \{r(B), r(A)\} = r(A)$, and from the other side

$r(A) = r(B^{-1}BAB^{-1}B) \le min \{r(B^{-1}), r(BA), r(B^{-1}B))\} = r(BA)$

hence $\;$ $r(BA) = r(A)$,

and we proved that $r(AB)=r(BA)$.

Is this proof correct? Thank you.

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    Why not reason conceptually ? $r(AB)$ is the dimension of the image of $AB$, which is the same as that of $A$ or $BA$, as invertible maps preserve the dimension. – 2017-02-11
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    Because it is a chapter before the linear mappings, so I wanted to prove it somehow by the rank theorem, but thank you for the idea. – 2017-02-11

2 Answers 2

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Your proof is correct, assuming (of course) you know the fact that $r(AB)\leq \min\{r(A), r(B)\}$. One thing I wonder about is why you write $A=B^{-1}BAB^{-1}B$? Is it not enough to write $A=B^{-1}BA$?

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    Yes, that is pretty easier, thanks. – 2017-02-11
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Image:

$Im(AB)\subset Im(A)=Im(A(BB^{-1}))=Im(AB(B^{-1}))\subset Im(AB)$

So for $B$ invertible we have $Im(AB)=Im(A)$

Kernel:

$x\in Ker(A)\Rightarrow Ax=0\Rightarrow BAx=0\Rightarrow x\in Ker(BA)$ $x\in Ker(BA)\Rightarrow BAx=0\Rightarrow Ax=B^{-1}0=0\Rightarrow x\in Ker(A)$

So for $B$ invertible we have $Ker(BA)=Ker(A)$

Now all this being in finite dimension we can apply the rank-theorem, and for any B invertible: $$rank(A)=rank(AB)=rank(BA)$$