Proving that if $AB=A$ and $BA=B$, then both matrices are idempotent Let $A, B$ be two matrices such that $AB=A$ and $BA=B$, how do I show that $A\cdot A=A$ and $B\cdot B=B$?

Steps I took:


Let $A= \left[\begin{array}{rr} 
    a &amp; b \\
    c &amp; d \\
   \end{array}\right]$ and let $B= \left[\begin{array}{rr} 
    w &amp; x \\
    y &amp; z \\
   \end{array}\right]$ 
$\left[\begin{array}{rr} 
    a &amp; b \\
    c &amp; d \\
   \end{array}\right] \cdot \left[\begin{array}{rr} 
    w &amp; x \\
    y &amp; z \\
   \end{array}\right] = \left[\begin{array}{rr} 
    aw+by &amp; ax+bz \\
    cw+dy &amp; cx+dz \\
   \end{array}\right] $ (Which is also equal to A)
$\left[\begin{array}{rr} 
    w &amp; x \\
    y &amp; z \\ 
   \end{array}\right] \cdot \left[\begin{array}{rr} 
    a &amp; b \\
    c &amp; d \\
   \end{array}\right] = \left[\begin{array}{rr} 
    wa+xc &amp; wb+xd \\
    ya+zc &amp; yb+zd \\
   \end{array}\right] $ (Which is also equal to B)


At this point I am stuck. I don't know how to proceed and I imagine that I started off on the wrong track to begin with. I'd like a hint to guide me in the right direction.



My proof after consultation with answerers below:

Proof:

1) Since $AB=A$, we can say that: $(AB)A=AA$ which is equal to $A^2$

2) Then, (by associativity of matrix multiplication), we can say that $A(BA) = AB$ (since $BA=B$)

3) Then, $AB=A$ (since $AB=A$ was given)

4) Therefore, $AA$ is equal to $A$

5) Since $BA=B$, we can also say that: $(BA)B=BB$ which is equal to $B^2$

6) Then, (by associativity of matrix multiplication) we can say that $B(AB)=BA$ (since $AB=A$)

7) Then, $BA=B$

8) Therefore, $BB$ is equal to $B$

9) Thus, $AA$ is equal to $A$ and $BB$ is equal to $B$

Q.E.D.
Notice that $AA = (AB)(AB) = A(BA)B = ABB = AB = A$.
Since $AB=A$ and $BA=B$, to show that $A$ is an idempotent we have to show that $A=A^2$
Therefore $A^2=(AB)^2 = A^2B^{ } = AB = A(BA) = AA = A^2 = A$. Proven