<s>[INST] Solving a simple matrix polynomial Does there exist a $2\times 2$ Matrix $A$ such that
$A-A^2=\begin{bmatrix} 3 & 1\\1 & 4\end{bmatrix}$ ?
 [/INST] 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 $\begin{bmatrix} 3 & 1 \\ 1 & 4 \end{bmatrix}$ 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 $A-A^2 = \begin{bmatrix} 3 & 1 \\ 1 & 4 \end{bmatrix}$. 
</s>