Solving a simple matrix polynomial Does there exist a $2\times 2$ Matrix $A$ such that

$A-A^2=\begin{bmatrix} 3 &amp; 1\\1 &amp; 4\end{bmatrix}$ ?
Let $A = \begin{bmatrix} a &amp; b\\ c &amp; d\end{bmatrix}$, then $A^2 = \begin{bmatrix} a &amp; b\\ c &amp; d\end{bmatrix}\cdot \begin{bmatrix} a &amp; b\\ c &amp; d\end{bmatrix} = \begin{bmatrix} a^2 + bc &amp; ab + bd\\ ac + cd &amp; bc + d^2\end{bmatrix}$. Thus:

$A - A^2 = \begin{bmatrix} a-a^2-bc &amp; b-ab-bd\\ c-ac-cd &amp; d-bc-d^2\end{bmatrix} = \begin{bmatrix} 3 &amp; 1\\1 &amp; 4\end{bmatrix}$. Thus:

$a-a^2-bc = 3$

$d-bc-d^2 = 4$

$b-ab-bd = 1$

$c-ac-cd = 1$.

Subtract the last $2$ equations to each other:

$b-c - a(b-c) - d(b-c) = 0 \to (b-c)(1-a-d) = 0$.

And subtract the first $2$ equations to obtain:

$a-d -(a^2-d^2) = -1 \to (a-d)(1-a-d) = -1$. Thus: $b-c = 0$, and $b = c$. Let's rewrite the system:

$a-a^2-b^2 = 3$

$d-b^2-d^2 = 4$

$b-ab-bd = 1$. Using Wolframalpha:

$a \approx 0.5 - 0.479i$

$b \approx 1.588i$

$d \approx 0.5 + 1.109i$.

Thus all roots are imaginary numbers, which means there is no matrix $A$ with real entries for the equation.
Assume such an $A$ exists. Let $\lambda_1$ and $\lambda_2$ be the eigenvalues of $A$. Then, the eigenvalues of $A-A^2$ are $\lambda_1-\lambda_1^2$ and $\lambda_2-\lambda_2^2$. 

The eigenvalues of $\begin{bmatrix} 3 &amp; 1 \\ 1 &amp; 4 \end{bmatrix}$ are $\dfrac{7 \pm \sqrt{5}}{2}$. Hence $\lambda_1-\lambda_1^2 = \dfrac{7 - \sqrt{5}}{2}$ and $\lambda_2-\lambda_2^2 = \dfrac{7 + \sqrt{5}}{2}$. 

Solving gives us two possibilities for each eigenvalue: 

$\lambda_1 = \dfrac{1}{2} \pm i\dfrac{1}{2}\sqrt{13-2\sqrt{5}}$ and $\lambda_2 = \dfrac{1}{2} \pm i\dfrac{1}{2}\sqrt{13+2\sqrt{5}}$. 

If the entries of $A$ are all real, then either $\lambda_1, \lambda_2 \in \mathbb{R}$ or $\lambda_1 = \overline{\lambda_2}$, which is not the case here. 

Hence, there is no real $2 \times 2$ matrix $A$ such that $A-A^2 = \begin{bmatrix} 3 &amp; 1 \\ 1 &amp; 4 \end{bmatrix}$.