Question: We know about a matrix $A \in \mathbb{R}^{n\times n}$, that $A^2 = I\_n$. Does the matrix $A$ have a basis of eigenvectors?

Below are my thoughts about the problem.

Some of the roots are diagonal matrices. For such cases the basis exists.

But other square roots of $I\_n$ can be antidiagonal, e.g. for $n=2$:
$$\left( \begin{array}{ccc}
0 & 1 \\
1 & 0 \\
\end{array} \right)$$

Here we have to manually check, that algebraic multiplicity of an eigenvalue equals its geometric multiplicity.

How can we do it for an arbitrary $n \in \mathbb{N}$?

Answer: 
The general fact is:

> 
> A matrix is diagonalizable if and only if its minimal polynomial splits into distinct linear factors.
> 
> 
> 

Therefore, the answer is *yes*, because $X^2-1=(X-1)(X+1)$. Therefore, the minimal polynomial of $A$ is one of $X^2-1$, $X-1$, $X+1$, all of which split into distinct linear factors.
Here is a concrete answer.

Let $v \in V = \mathbb{R}^{n}$. Then $v=u+w$, where $u=(1/2)(v-Av)$ and $w=(1/2)(v+Av)$.

Now $u \in U = V\_{-1}$ and $w \in W = V\_{1}$, where $V\_\lambda = \{ v \in V : Av = \lambda v \}$, a subspace of $V$.

Therefore, $V = U + W$. Since $U\cap W=0$, we have $V = U \oplus W$.

This means that $V$ has a basis of eigenvectors of $A$: join bases for $U$ and $W$.