$A$ a $n\times n$ matrix with real entries such that $A^3 = I$ but $A \ne I$ Let $A$ a $n \times n$ matrix with real entries  such that $A^3 = I$ but $A \ne I$.
a) Give an example that satisfies this conditions. 
b) what are the eigenvalues ​​of $A$?
Well for $a)$ i construct this matrix , let $A=\begin{bmatrix} 1 & -\frac{3}{2} & \frac{\sqrt{3}}{2} \\ 0 & -\frac{1}{2} & -\frac{\sqrt{3}}{2} \\ 0 & \frac{\sqrt{3}}{2} & -\frac{1}{2} \end{bmatrix}$. Then $A^3=I$,  and $A \ne I$ also $A$ is not orthogonally.
For $b)$ I can't found the eigenvalues for $A$ I suppose I need to find the characteristic polynomial and then find it but in general  for a matrix $n \times n$ I can't find this eigenvalues, some help please.
 A: Given

$$
\mathbf{A}_n^3 = \mathbf{I}_n
\hspace{1em} \textrm{and} \hspace{1em}
\mathbf{A}_n \ne \mathbf{I}_n. \tag {OP}
$$

Eigenvalues
Let $\lambda$ be an eigenvalue of $\mathbf{A}_n$.
Therefore
$$
\mathbf{A}_n \vec{x}_\lambda = \lambda \vec{x}_\lambda.
$$
Whence
$$
\mathbf{A}_n^k \vec{x}_\lambda = \lambda^k \vec{x}_\lambda.
$$
From (OP) follows
$$
\lambda^3 = 1.
$$

It is also clear that
$$
\lambda^3 = 1 \Longrightarrow \big( \lambda^k \big)^3 = 1.
$$
Therefore we obtain


$$
\lambda \ne 1,
p + q > 0,
p+q+r = n :
\mathbf{A}^\flat_n =
\left[
\begin{array}{ccc}
\lambda \mathbf{I}_p & 0 & 0\\
0 & \lambda^2 \mathbf{I}_q & 0\\
0 & 0 & \mathbf{I}_r
\end{array}
\right]. \tag {A1}
$$

Property of eigenvalues
Let us define
$$
z = 1 + \lambda + \lambda^2.
$$
We have
$$
\lambda z = z \Rightarrow \big( \lambda - 1 \big) z = 0 \Rightarrow
\lambda = 1 \vee z = 0.
$$
As $\lambda = 1$ excludes (OP), we have $z=0$, thus

$$
\lambda + \lambda^2 = -1. \tag{p1}
$$

Conjugation
Let $\mathbf{B}_n$ be any invertible $n \times n$ matrix.
We obtain
$$
\mathbf{I}_n
= \mathbf{B}_n \mathbf{I}_n \mathbf{B}^{-1}_n
= \mathbf{B}_n \mathbf{A}^3_n \mathbf{B}^{-1}_n
= \left( \mathbf{B}_n \mathbf{A}_n \mathbf{B}^{-1}_n \right)^3.
$$
From (A) follows that $\mathbf{A}^\flat_n$ is (not yet) a real matrix.
Let us define
$$
\mathbf{A}^\sharp_n =
\mathbf{B}^\flat_n \mathbf{A}^\flat_n \big( \mathbf{B}^\flat_n \big)^{-1}.
$$
If $\mathbf{A}^\sharp_n$ is a real $n \times n$ matrix, then at least we have that BOTH the trace $(\chi)$ and the determinant $(\Delta)$ are real.
Thus
$$
\chi\big(\mathbf{A}^\sharp_n\big) \in \mathbb{R},\\
\Delta\big(\mathbf{A}^\sharp_n\big) \in \mathbb{R}.
$$
As
$$
\chi\big(\mathbf{P}\mathbf{Q}\mathbf{P}^{-1}\big) = \chi(\mathbf{Q}),\\
\Delta\big(\mathbf{P}\mathbf{Q}\mathbf{P}^{-1}\big) = \Delta(\mathbf{Q}),
$$
we obtain
$$
p \lambda + q \lambda^2 + r \in \mathbb{R}, \tag 1
$$
$$
\lambda^{p+2q} \in \mathbb{R}. \tag 2
$$
Using (p1) we can write (1) as
$$
(p-q) \lambda + (r-q) \in \mathbb{R}.
$$
As $\lambda \ne \mathbb{R}$, we obtain
$$
p=q. \tag {c1}
$$
And (2) becomes
$$
\lambda^{p+2q} = \lambda^{3p} = 1^p = 1 \in \mathbb{R}.
$$
Therefore we obtain

$$
\lambda \ne 1,
p > 0,
2p+r = n :
\mathbf{A}^\flat_n =
\left[
\begin{array}{cc}
\left[
\begin{array}{cc}
\lambda & 0 \\
0 & \lambda^2
\end{array}
\right] \mathbf{I}_p & 0\\
0 & \mathbf{I}_r
\end{array}
\right]. \tag {A2}
$$

Real solution
As
$$
\left[
\begin{array}{cc}
\lambda & 0 \\
0 & \lambda^2
\end{array}
\right]^3 = \mathbf{I}_2,
$$
we can consider the conjugation
$$
\left[
\begin{array}{cc}
-\frac{1}{2} & \pm \frac{\sqrt{3}}{2} \\
\mp \frac{\sqrt{3}}{2} & -\frac{1}{2}
\end{array}
\right]^3 = \mathbf{I}_2.
$$
The general solution can be written as

$$
p > 0,
2p+r = n,
\mathbf{B}_n \in \mathbb{R}^n,
\exists \mathbf{B}^{-1}_n \in \mathbb{R}^n:
$$
  $$
\mathbf{A}_n =
\mathbf{B}_n
\left[
\begin{array}{cc}
\left[
\begin{array}{cc}
-\frac{1}{2} & \pm \frac{\sqrt{3}}{2} \\
\mp \frac{\sqrt{3}}{2} & -\frac{1}{2}
\end{array}
\right] \mathbf{I}_p & 0\\
0 & \mathbf{I}_r
\end{array}
\right]
\mathbf{B}^{-1}_n.
$$

A: if $\lambda$ is an eigenvalue with eigenvector $v$, $A v = \lambda v$. Then, $A^2 v = A (Av) = A (\lambda v) = \lambda A v = \lambda \lambda v = \lambda^2 v$. Finally, $v = I v = A^3 v = A (A^2 v) = A (\lambda^2 v) = \lambda^2 A v = \lambda^2 \lambda v = \lambda^3 v$. 
So, $\lambda^3 = 1$. 
A: consider the $n \times n$ permutation matrix $\sigma.I$ corresponding to a three cycle $\sigma$ in $S_n$ for $n \ge 3$. This matrix satisfies the required property. These matrices are orthogonal and you can easily find the eigen values.  Let $\sigma$ be a k-cycle in $S_n$ where $k \le n$. Then the permutation matrix $\sigma.I$ satisfies $A^k = I$.
A: Hint: If we know that $A\vec v = \lambda \vec v$, what else can we deduce about $\vec v$? We don't know anything about $\vec v$ other than that it's an eigenvector, so what else can we manipulate?
A: Geometrically, consider a matrix describing a rotation by $2\pi/3$, so that $A^3$ is a rotation by $2\pi=Id$. Then the eigenvalues of a rotation are known --and, since $3$ is odd, you have at least one Real eigenvalue; there are just three possible choices for them:
http://mathworld.wolfram.com/RotationMatrix.html
A: Note that it's minimal polynomial is $x^3-1=0$. Now use the fact that characteristic and  minimal polynomial have the same roots, except for multiplicities.
