If $A \in M_n \mathbb{(R)}$, $A^2=E$, then $\mbox{Rank} (A+E) + \mbox{Rank} (A-E) = n$ How can I prove this:
If $A \in M_n \mathbb{(R)}$, $A^2=E$, then $\mbox{Rank} (A+E) +  \mbox{Rank} (A-E) = n$.
I've tried (without success) the rank nullity approach so any hint would be appreciate.
 A: If $E$ is simply $A^2$ and $A$ is arbitrary, the result is false. Let 
$$
A=\begin{bmatrix}1&0&0&\cdots&0\\ 0&0&0&\cdots&0\\\vdots&\vdots&\vdots&\ddots&\vdots\\0&0&0&\cdots&0
\end{bmatrix}
$$
Then $A^2=A$, $A+A^2=2A$, $A-A^2=0$. So the rank of $A+A^2$ is $2$, and then rank of $A-A^2$ is zero. 
If we assume that $E$ is the identity $I$, the result is true. 
Since $A^2=I$, the only possible eigenvalues of $A$ are $1$ and $-1$. Looking at the Jordan blocks, they all have to be $1\times1$, or we wouldn't have $A^2=I$. So $A$ is diagonalizable, $A=SDS^{-1}$ with $D$ diagonal with the diagonal containing only $1$ and $-1$. 
We have $A+I=S(D+I)S^{-1}$ and $A-I=S(D-I)S^{-1}$, so 
$$\text{Rank}(A+I)+\text{Rank}(A-I)=\text{Rank}(D+I)+\text{Rank}(D-I).$$
Since $D$ is diagonal with $1$ and $-1$ in the diagonal, the rank of $D+I$ is the number of $1$ in the diagonal, and the rank of $D-I$ is the number of $-1$ in the diagonal. Then 
$$
\text{Rank}(A+I)+\text{Rank}(A-I)=\text{Rank}(D+I)+\text{Rank}(D-I)=n.$$
