Let $A$ and $B$ be matrixes. If $BA=B$ and $Rank\space A = Rank\space B$, prove $A^2=A$

Ok, so I can see that:





but I don't know how to keep following. Any hint? Also how would I apply the Rank thing?

EDIT: I messed up ABA implies B is regular which I don't know.

  • $\begingroup$ Is $B$ a square matrix as well? Otherwise $ABA$ might not not be defined. $\endgroup$ – cthl Dec 13 '18 at 17:15
  • $\begingroup$ Ok we don't know that good point I didn't see $\endgroup$ – iggykimi Dec 13 '18 at 17:19

Let $x$ in $ker(A), A(x)=0$ implies that $BA(x)=B(x)=0$ implies $x$ in $Ker (B)$ since $rank A=rank B$ we deduce that $dimker(A)=dimker B$ and $ker A=ker B$ since the previous argument shows that $ker A$ is contained in $kerB$. $BA(x)=B(x)$ implies that $B(A(x)-x)=0, A(x)-x$ is in $Ker(B)=ker(A)$ implies $A(A(x)-x)=0$. Which implies that $A^2(x)=A(x)$.

  • $\begingroup$ I think after you showed that $\text{ker } A=\text{ker B}$ you need to say that now you take an arbitrary $x$ since you want to have $(A^2-A)x=0$ for every $x$ to conclude that $A^2=A$. $\endgroup$ – user9077 Dec 13 '18 at 17:28

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