Projectors solution of a matrix equation

Question

If I have two $n \times n$ complex matrices $A$ and $B$, where $A$ and $B$ are both projectors, i.e., $A^2=A$ and $B^2=B$.

If $A B = A$ and $A \neq I$, clearly if $B=A$, or $B=I$ then the equality holds, can $B$ have other solutions?

Answer 1

Yes there can be other solutions. Consider the matrices $A = \begin{bmatrix}1&0&0\\0&0&0\\0&0&0\end{bmatrix}$ and $B = \begin{bmatrix}1&0&0\\0&1&0\\0&0&0\end{bmatrix}$.

Clearly, $A$ and $B$ are projection matrices, and $AB = A$, but $B \neq A$, $B \neq I$, and $A \neq I$.

Answer 2

We want to solve the system $A^2=A,B^2=B,AB=A$ where $A,B\in M_n$. Let $rank(A)=p$. Since $AB=A$, $rank(B)\geq rank(A)$, that is, $rank(B)=p+q$. Let $r=n-p-q$ ($p,q,r$ are non negative integers).

Proposition. There is $P\in GL_n$ and $R\in M_{r,p}$ s.t. $A=Pdiag(I_p,0_q,0_r)P^{-1},B=P\begin{pmatrix}I_p&0&0\\0&I_q&0\\R&0&0_r\end{pmatrix}P^{-1}$.