<s>[INST] Minimal polynomial of a matrix satisfying $A^t=A^2$ 
Let $A\in M_n(\mathbb R)$ be a non-zero non-identity matrix such that $A^t=A^2$. Then is it possible to find the minimal polynomial of $A$?

My  try is: the given condition implies that $A^4=A$, hence the minimal polynomial $p(t)$ of $A$ divides $t^4-t=t(t-1)(t^2+t+1)$. So $p(t)$ can be some factor of $t(t-1)(t^2+t+1)$.
Also if $\lambda$ is an eigenvalue of $A$ and $x$ is the corresponding eigenvector (considered as a column vector), then $Ax=\lambda x \implies \lambda x^t=x^t A^t\implies \lambda x^t=x^tA^2\implies \lambda(x^t x)=x^t(A^2x)\implies \lambda(x^t x)=\lambda^2(x^tx)\implies (\lambda^2-\lambda)(x^tx)=0\implies\lambda^2-\lambda=0,\text{ as } x^tx \text{ is non-zero}.$
Hence $0$ and $1$ are the only eigenvalues of $A$. So the minimal polynomial is of the form $p(t)=t(t-1).$ Is this correct or did I misunderstood something?
P.S. $A^t$ denotes the transpose of the matrix $A$.
 [/INST] Since
$$
A^4=\left(A^2\right)^2=\left(A^T\right)^2=\left(A^2\right)^T=\left(A^T\right)^T=A
$$
the minimal polynomial must divide $x^4-x=x(x-1)(x^2+x+1)$.
For example,
$$
A=\begin{bmatrix}
-\frac12&\frac{\sqrt3}2\\
-\frac{\sqrt3}2&-\frac12
\end{bmatrix}
$$
has minimal polynomial $x^2+x+1$ and $A^2=A^T$.
$$
B=\begin{bmatrix}
-\frac12&\frac{\sqrt3}2&0\\
-\frac{\sqrt3}2&-\frac12&0\\
0&0&1
\end{bmatrix}
$$
has minimal polynomial $x^3-1$ and $B^2=B^T$.
$$
C=\begin{bmatrix}
-\frac12&\frac{\sqrt3}2&0\\
-\frac{\sqrt3}2&-\frac12&0\\
0&0&0
\end{bmatrix}
$$
has minimal polynomial $x^3+x^2+x$ and $C^2=C^T$.
$$
D=\begin{bmatrix}
-\frac12&\frac{\sqrt3}2&0&0\\
-\frac{\sqrt3}2&-\frac12&0&0\\
0&0&1&0\\
0&0&0&0
\end{bmatrix}
$$
has minimal polynomial $x^4-x$ and $D^2=D^T$.
</s>