rank(AB)=rank(BA) for non square matrix

Question

Let A $\in(a_{ij})_{m\times n}$ and B$\in(b_{ij})_{n\times m}$ where entries are real numbers, if $rank(AB)=m$, then $rank(BA)=m$

Here is some of my approach

First, we know $m=rank(AB)$ $\le$ $min$ {$rankA,rankB$} and as $rankA$ $\le$ m, so we get rankA=m, so n$\le$m and A be full rank, and we know that AB be invertible matrix,how can we also show that rank(BA)=m?

$m=\text{rank}\big((AB)^2\big)\leq \text{rank}\big((BA)\big)\leq m$ — user8675309, Aug 09 '22 at 04:27
thank,I get it. rank(ABAB) $\le$ rank(ABA) $\le$ rank(BA) since A be full rank first hold, since dim(T(V)) $\le$ dim(V) second hold. — lee, Aug 09 '22 at 04:46
It is not true even for $2\times 1$ and $1\times 2$ matrices. For example $A=[1,0],$ $B=[0,1]^t.$ — Ryszard Szwarc, Aug 09 '22 at 13:46

score 0 · Answer 1 · answered Aug 10 '22 at 14:41

First we prove that $r(B)=m$. If there exists an $x \neq\,0$ such that $Bx=0$ then $ABx=0$ and hence $N(AB)\neq${$0$}, which implies that $r(AB)\neq\,m$, contradiction. Therefore $N(B)$={$0$} which gives $r(B)=m$.

Also $m=r(AB)\leq\,min${$r(A),m$} and $r(A)\leq\,m$. But if $r(A)<m$ then $r(A)\geq\,m$ which implies $r(A)=m$. If on the other hand $r(A)\geq\,m$ then $m\leq\,r(A)\,\leq m$ which gives $r(A)=m$. In every case $r(A)=m$.

Now we apply the Sylvester's inequality for $B$ an $nxm$ matrix and $A$ an $mxn$ matrix. Then we get:

$r(B)+r(A)-m$$\leq$$ r(BA)$ which gives $r(BA)\geq\,m$.

I believe this is all we can get, because there are counterexamples mentioned in above comments which prove that equality is not true!

rank(AB)=rank(BA) for non square matrix

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