Linear Algebra: describing all possible matrices $A$ Suppose that $A$ is an $n\times n$ matrix such that $A^2+2A-8I=0$. 
Solve the equation in $A$, namely describe all possible matrices $A$. 

So, I found the spectrum of $A$, namely $\{-4,2\}$, so do I describe all possible matrices by putting those eigenvalues on the diagonal of the matrix for all upper triangular combinations that are possible? 
 A: If we have that $A^2+2A-8I=0$ we know that the polynomial $x^2+2x-8 \in I(A)$ where $I(A)$ is the ideal of the endomorphism we are considering. So this is clearly a multiple of the minimal polynomial of the matrix (which we know generates the ideal). If we decompose this polynomial we have $x^2+2x-8=(x+4)(x-2)$ from which we know the possible eigenvalues, or the spectrum: $sp(f)=\{-4,2\}$.
The minimal polynomial of this matrix so can be only $(x-4), (x-2)$ or $(x-4)(x-2)$.
By using the Jordan Normal form theorem, we can conclude that all the matrices which respect this identity are diagonal: they can have all twos, all fours or both.
Example of all the possible matrices:
$$
  D_1 =
  \begin{bmatrix}
    -4 & & \\
    & \ddots & \\
    & & -4
  \end{bmatrix}
$$
$$
  D_2 =
  \begin{bmatrix}
    -2 & & \\
    & \ddots & \\
    & & -2
  \end{bmatrix}
$$
$$
  D_3 =
  \begin{bmatrix}
    -4 & & \\
    & \ddots & \\
    & & -4 && \\ &&& -2 && \\
    &&&& \ddots & \\ &&&&&-2
  \end{bmatrix}
$$
Hope you're satisfied!
A: Since this polynomial has only simple roots and is a multiple of the minimal polynomial, $A$ can be written in Jordan Normal Form using blocks of size $1$: i.e. it is diagonalisable. Hence such matrices are precisely the diagonalisable ones with eigenvalues from $\{-4,2\}$.
