Solving a simple matrix polynomial Does there exist a $2\times 2$ Matrix $A$ such that
$A-A^2=\begin{bmatrix} 3 & 1\\1 & 4\end{bmatrix}$ ?

Yes, there are complex solutions, but no real solutions.  Using the brute force method also mentioned by JimmyK, you can see that $A$ must be symmetric.  The top left element then becomes $a-a^2-b^2$ which must equal $3$, which can obviously only happen if $A$ has complex elements.
