True or False - Matrix Equation $$A^2 - AB - BA + B^2 = 0 \implies A = B $$
I know this is false but how can I go about proving it. I got to:
$$A^2 - AB - BA + B^2$$
$$ = (A - B)(A-B) = 0$$
I know that if A and B are matrices such that AB = 0 then AB need not be 0 but how can I show this equation wise without resorting to a counter examples which would a bit annoying in this case since it is matrix multiplication?
 A: Hint: False,choose $A$ to be zero matrix and $B$ such that $B^2=0$.For instance:
$ B=\begin{bmatrix}0 & 1\\0 & 0\end{bmatrix}$ and $A=\begin{bmatrix}0 & 0\\0 & 0\end{bmatrix}$
A: HINT:
$$A(A-B) -B(A-B)=0$$
$$(A-B)(A-B)=0$$
But from this does not follow that $A=B$, because $A-B$ is not necessary $0$.
Example:
$$M=
  \left( {\begin{array}{cc}
   0 & 1\\  
   0 & 0      
\end{array} } 
\right)
$$
$$M^2=0$$
Every two matrices $A, B$ such that $A-B=M$ give a counter example !
A: Suppose that $det(A-B)\neq0$, then there exists inverse matrix of $A-B$. In the equation: $$(A-B)(A-B)=0$$
multiplicate by the $(A-B)^{-1}$, then we get:
$$A-B=0, $$
from there we get that the $A=B$, but it's impossible, because $det(A-B)\neq0$. So contradiction. Hence $det(A-B)=0$. So we don't necessary need A to be equal to B, just the determinant to be zero.  
A: The most general answer is that matrices do not form a division algebra. A division algebra is basically an algebra in which multiplication has an inverse. There are only four division algebras: real numbers, complex numbers, quanternions and octonions. Square matrices are neither.
A specific consequence (as someone mentioned in an answer comment) is that there exist nilpotent matrices, specifically nonzero square matrices $X$ for which $X^2=0$. So given an arbitrary square matrix $A$ and a degree-2 nilpotent matrix $X$ of the same order, you can make $B=A+X$ and thus, $(A-B)\cdot(A-B)=X^2=0$ even though $A- B=X\ne 0$ so $A\ne B$.
