Does there exists a non-diagonal $2 \times 2$ matrix $A$, such that $A^3 =I$ Does there exist a $2 \times 2$ non-diagonal matrix $A$ such that $A^3 = I$ ?
Suppose it exists then $\lambda^3 = 1$ where $\lambda$ is the eigenvalue of $A$
Now a $2 \times 2$ matrix can have $2$ eigenvalues.
Those look like cube roots of unity
$1,\omega , \omega ^ 2$.
Now, can I generate a $2 \times 2$ matrix whose eigen values are 1,1 and it's non diagonal?
I think yes
$A = \begin{bmatrix} 1 & 1\\ 0 & 1\\ \end{bmatrix}$
But I see that cubing this does not give me $I$, even though it satisfies the determinant and trace conditions.
Is this approach correct?
Any interesting problem can be formulated from this?
 A: Note that $(x-\omega)(x-\omega^2)=x^2+x+1$ so look for a matrix with
characteristic function $x^2+x+1$, that is with trace $-1$ and determinant $1$, for instance
$$\pmatrix{-1&1\\-1&0}.$$
A: Of course, a rotation by $2\pi/3$ would have the property that $A^3=I$. The matrix for that is:
$$A=\begin{pmatrix}
-1/2 & -\sqrt{3}/2 \\
\sqrt{3/2} & -1/2
\end{pmatrix}$$
As for eigenvalues this one has only complex eigenvalues that happen to cube up to $1$.
A: Note that $x^3-1=(x-1)(x^2+x+1)=(x-1)(x-j)(x-j^2)$ where $j=\exp(2i\pi/3)$; since $A$ is real, $spectrum(A)=\{1,1\}$ or $spectrum(A)=\{j,j^2\}$. Since $A$ is not diagonal, the first case does not work because $A$ should be similar to $B=\begin{pmatrix}1&1\\0&1\end{pmatrix}$ which does not satisfy $B^3=I$.
Thus $(*)$ $spectrum(A)=\{j,j^2\}$ and, consequently $A^2+A+I=0$ -that implies $A^3=I$-. We conclude with the following gneralization.
EDIT 2. Considering the @Lord Shark the Unknown's comment , I rewrite my two propositions into one.
Proposition. Let $P(x)$ be a REAL polynomial of degree $p\geq 2$ and $q$ be an even integer $\geq 2$ (resp. of odd degree $p\geq 2$ and $q$ be an odd integer $\geq 2$); then there is a non-diagonal real $q\times q$ matrix $A$ s.t. $P(A)=0$.
Proof. We may write $P(x)=(x^2+ax+b)Q(x)$ where $x^2+ax+b$ and $Q$ are real polynomials. Let $U\in M_2(\mathbb{R})$ be the companion matrix of $x^2+ax+b$.
Case 1. $q=2r$, we choose $A=diag(A_1,\cdots,A_r)$ where $A_i=U$. 
Note that when $P(x)=x^2+1$ and $q$ is odd, such a matrix $A\in M_q$ does not exist.
Case 2. $p$ and $q=2r+1$ are odd. Then $P$ admits a real root $u$ and we choose  $A=diag(U_1,\cdots,U_r,u)$.
