<s>[INST] 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?
 [/INST] Your method also works, though cumbersome (not asap):
$$X^2+X=1 \Rightarrow \begin{pmatrix}a&b\\c&d\end{pmatrix}^2+\begin{pmatrix}a&b\\c&d\end{pmatrix}=\begin{pmatrix}1&1\\1&1\end{pmatrix} \Rightarrow\begin{cases}a^2+a+bc=1\\ ab+bd+b=1\\ ac+c+cd=1\\ bc+d^2+d=1\end{cases}.$$
Subtract the last from the first:
$$a^2-d^2+a-d=0 \Rightarrow (a-d)(\underbrace{a+d+1}_{\ne 0 \ \text{from (2)}})=0 \Rightarrow d=a.$$
Plug $d=a$ in the second and third equations and subtract:
$$(2ab+b)-(2ac+c)=0 \Rightarrow (b-c)(\underbrace{2a+1}_{\ne 0 \ \text{from (2)}})=0 \Rightarrow c=b.$$
Plug $c=b$ into the first and second equations and add them: 
$$a^2+a+b^2+2ab+b=2 \Rightarrow (a+b)^2+(a+b)-2=0 \Rightarrow \\
a+b=-2;1 \Rightarrow 1) \ b=-a-2; \ 2) \ b=-a+1.$$
$1$) Plug $b=-a-2$ into the first equation:
$$a^2+a+(-a-2)^2=1 \Rightarrow 2a^2+5a+3=0 \Rightarrow a=-\frac32; -1 \Rightarrow \\
X_1=\begin{pmatrix}a&b\\c&d\end{pmatrix}=\begin{pmatrix}-\frac32&-\frac12\\-\frac12&-\frac32\end{pmatrix};\\
X_2=\begin{pmatrix}a&b\\c&d\end{pmatrix}=\begin{pmatrix}-1&-1\\-1&-1\end{pmatrix}.$$
$2$) Plug $b=-a+1$ into the first equation:
$$a^2+a+(-a+1)^2=1 \Rightarrow 2a^2-a=0 \Rightarrow a=0; \frac12 \Rightarrow \\
X_3=\begin{pmatrix}a&b\\c&d\end{pmatrix}=\begin{pmatrix}0&1\\1&0\end{pmatrix};\\
X_4=\begin{pmatrix}a&b\\c&d\end{pmatrix}=\begin{pmatrix}\frac12&\frac12\\\frac12&\frac12\end{pmatrix}.$$
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