Matrix Equation Simplification Is there a way to solve this equation $A^4 − 3A^3 + A^2 − I_4 = 0$ with $A$ being a 4x4 Matrix ?
My Idea was this:
$A^4 − 3A^3 + A^2 = I_4 $
$A^2(A^2 − 3A + I) = I_4 $
But I don't know if those operations are allowed and I'm kinda blocked there.
Edit:
Could you make one example of how I could find one possible matrix that fits the polynomial?
 A: The companion matrix of the polynomial is always a solution. It always works regardless of the ground field.
A: If you are looking for an example of finding 'a' matrix $A$ satisfying this equation, here is one way to approach it: 
Consider the polynomial $p(x)=x^4-3x^3+x^2-1$. The (complex) roots of this polynomial are quite nasty but can be calculated exactly (Wolfram Mathematica does it for you). Approximately the roots are: a complex conjugate pair $\lambda_{\pm}\approx 0.44709 \pm 0.672117 i$ plus two real roots $\mu=-0.572892$ and $\nu=2.67871$.
The important point here is that this polynomial has four distinct roots. Write $p(x)=(x-\lambda_+)(x-\lambda_-)(x-\mu)(x-\nu)$. You are looking for a $4\times 4$ (let's say complex matrix) such that $p(A)=0$. Well one obvious example would be
$$
A=\begin{pmatrix}
\lambda_+ & 0 & 0 & 0\\
0& \lambda_- & 0 & 0\\
0 & 0 & \mu & 0\\
0 & 0 & 0 & \nu
\end{pmatrix}
$$
or any matrix similar to it. Why is that the case? Given any matrix $M$, one defines the characteristic polynomial as $p_M(x)=\det(xI-M)$. The solutions to $p_M(x)$, by definition, are the eigenvalues. Now Cayley-Hamilton theorem states that $p_M(M)=0$. In other words any matrix satisfies its own characteristic equation. Now, back to your question, therefore it is enough to design a matrix $A$ such that its charcteristic polynomial is $p(x)$. This is why the above example is the trivial solution.
A more general solution, would be
$$
A=\begin{pmatrix}
\lambda_+ & \alpha_1 & \alpha_2 & \alpha_3\\
0& \lambda_- & \alpha_4 & \alpha_5\\
0 & 0 & \mu & \alpha_6\\
0 & 0 & 0 & \nu
\end{pmatrix}
$$
with $\alpha_1, \cdots, \alpha_6$ arbitrary complex numbers. The transpose of the above matrix is also a solution. Any matrix similar to above is also a solution.
