Solve the equation $X^2+X=\text{a given matrix}$ I want to solve the quadratic matrix equation 
$$X^2+X=\begin{pmatrix}1&1\\1&1\end{pmatrix}$$
If I put $X$ in the form 
$$X=\begin{pmatrix}a&b\\c&d\end{pmatrix}$$
then I find complicated equations. Is there a simple way to tackle the problem without using diagonalization?

There is a matrix $S$ such that
$$
J=S\begin{bmatrix} 2 & 0 \\ 0 & 0 \end{bmatrix}S^{-1}
$$
If you set $Y=S^{-1}XS$, then $X=SYS^{-1}$ and the equation becomes
$$
S(Y^2+Y)S^{-1}=S\begin{bmatrix} 2 & 0 \\ 0 & 0 \end{bmatrix}S^{-1}
$$
that simplifies to
$$
Y^2+Y=\begin{bmatrix} 2 & 0 \\ 0 & 0 \end{bmatrix}
$$
We can try and complete the square:
$$
4Y^2+4Y=\begin{bmatrix} 8 & 0 \\ 0 & 0 \end{bmatrix}
$$
so the equation becomes
$$
(2Y+I)^2=\begin{bmatrix} 9 & 0 \\ 0 & 1 \end{bmatrix}
$$
Now the problem is simpler; if $2Y+I=\left[\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right]$, then the equation becomes
$$
\begin{bmatrix}
a^2 + bc & ab + bd \\
ac + cd & bc + d^2
\end{bmatrix}=
\begin{bmatrix} 9 & 0 \\ 0 & 1 \end{bmatrix}
$$
Case 1: $b=0$; then $a=\pm3$, $d=\pm1$ and $c=0$
Case 2: $a+d=0$: then $a^2+bc=9$ and $a^2+bc=1$: a contradiction.
Thus we have four solutions and
$$
Y=\frac{1}{2}\left(\begin{bmatrix} \pm3 & 0 \\ 0 & \pm1 \end{bmatrix}-I\right)
$$
so
$$
X=\frac{1}{2}S\left(\begin{bmatrix} \pm3 & 0 \\ 0 & \pm1 \end{bmatrix}-I\right)S^{-1}
$$
How do you determine the matrix $S$? Just determine an eigenvector relative to $2$ and one relative to $0$:
$$
S=\begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix}
$$
