Prove rank($A$)=rank($B$) if $A+B=AB,\ and\ A,B \in M_n(F)$. Let $A,B$ be matrices $n\times n$ $(A, B \in M_{n\times n}(F))$,and
$A+B=AB$.
How can we prove that
"rank($A$)=rank($B$)".
I have proved that $AB=BA$ by $(A-E)(B-E)=E\ $, but it seems to make no sense. 
 A: We know that for an $n\times n$ matrix $M$ \begin{equation}rank(M)=n-{\text nullity}(M),\end{equation} where nullity is dimension of the solution space of $M$.
We have \begin{equation}A+B=AB \ \Rightarrow \ A^{T}+B^{T}=B^{T}A^T \end{equation}
Hence for $x$  in the solution space of $B$ ($Bx=0$), the first equation gives that $x$ is also in the solution space of $A$. Hence nullity of $A$ is greater or equal to nullity of $B$. From the equation regarding rank above, it follows that \begin{equation}rank(A)\leq rank(B).\end{equation} From $A^{T}+B^{T}=B^{T}A^T$, we get $B^T=(I-B^T)A^T$, so if $z$ is in the solution space of $A^T$, then it is in the solution space of $B^T$. (The columns of $A^T,B^T$ are rows of $A$ and $B$ respectively, and $z\neq 0$ is in the solution space of $A^T$ means that there is a non trivial combination of its column vetoes adding to the zero vector.) Hence, $B$ has at least as many linearly dependent rows as $A$. Hence, \begin{equation}rank(B)\leq rank(A).\end{equation}
Combining the two inequalities gives the result.
