$I + A + A^2 +A^3 = 0$ prove $A^2$ similar to a real diagonalized matrix. The first part was to prove that A is invertible which is done like that:
$$
I = A(-I-A-A^2)
$$
I need to prove that $A^2$  is similar to some diagonalized matrice and to calculate $A^{2000}$.
also after some algebraic manipulation I was able to prove that $A=-A^{-1}$ I don't know if that's the wanted direction but it might help.
EDIT: proof:
I know now that A is invertible therefore there exists $A^{-1}$ and we know what it is.
$$
A^2 = -I -A -A^3
$$
multiply by $A^{-1}$
$$
A= -A^{-1} -I -A^2
$$
$$
A^2 +A = -A^{-1} -I
$$
$$
A(A+I)= -A^{-1}(I + A)
$$
$$
A=-A^{-1}
$$
Any help is appreciated!
 A: $Q(A^2) = 0$ with $Q(X) = X^2 - 1$ has real simple roots. This proves that $A^2$ is similar to some real diagonal matrix.
Then, as $A^4 = I$, $A^{2000} = (A^4)^{500} = I$.
A: Given
$$
\textbf{A}^3 + \textbf{A}^2 + \textbf{A} + \textbf{I} = 0
$$
This can be written as
$$
\Big( \textbf{A} + \textbf{I} \Big) \Big( \textbf{A}^2 + \textbf{I} \Big ) = 0
$$
So the eigenvalue of $\textbf{A}^2$ is given by $-1$, i.e. similar to a real diagonalized matrix.
We also have
$$
\textbf{A}^4 + \textbf{A}^3 + \textbf{A}^2 + \textbf{A} = 0
$$
So
$$
\Big(\textbf{A}^4 + \textbf{A}^3 + \textbf{A}^2 + \textbf{A} \Big) -
\Big(\textbf{A}^3 + \textbf{A}^2 + \textbf{A} + \textbf{I} \Big) = \textbf{A}^4 - \textbf{I} = 0
$$
thus
$$
\textbf{A}^4 = \textbf{I}
$$
and
$$
\textbf{A}^{2000} = \textbf{I}
$$
A: Just to be clear we work over the reals. A matrix is diagonalizable if and only if its minimal polynomial is a product of distinct linear factors. Thus we will show that the minimal polynomial of $A^2$ is (a divisor) of $(x-1)(x+1)=x^2-1$. This amounts to proving that $A^4-I=0$  
If we multiply our equation by $A$ we have 
$$A^4+A^3+A^2+A=0$$
or 
$$A^4-I=-(A^3+A^2+A+I)=0$$
