$\newcommand{\rank}{\operatorname{rank}}$Let $A$ be an $n \times n$ real matrix with $n \geq 2$ and characteristic polynomial $x^{n-2}(x^2-1)$, then
- $A^n=A^{n-2}$
- $\rank(A) \geq 2$
- $\rank(A) = 2$
- There exists non zero vectors $x,y \in \mathbb R^n$ such that $A(x+y)=x-y$
By Cayley-Hamilton theorem we can easily see the first option, also i can eliminate 3rd option by taking the matrix
$$\left(\begin{array}{cccc}0&1&0&0\\0&0&0&0\\0&0&1&0\\0&0&0&-1\end{array}\right),$$
but I couldn't clear 2nd and fourth option.
Best Answer
For #4, take $x$ and $y$ to be eigenvectors associated with $1$ and $-1$ respectively.
For #2, note that $x$ and $y$ (as above) are linearly independent vectors in the column space (image) of $A$.