Let $A\in M_3(\mathbb{R})$ be a non-null matrix such as $A^3=-A$. Show that $A$ is similar to the matrix $$P = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & 0 \\ \end{bmatrix}$$.
Hey, I hope you are doing well. I'm trying to do this:
If $A^3=-A$, then $A^3+A=0$, so the characteristic polynomial of $A$ is $p_A(x)=x^3+x=x(x^2+1)$,
from this we can conclude that the minimal polynomial is $m_A(x)=x(x^2+1)=x^3+x$.
We also have that the companion matrix for $m_A$ is $$B=\begin{bmatrix} 0 & 0 & 0 \\ 1 & 0 & -1 \\ 0 & 1 & -1 \\ \end{bmatrix}$$.
Then, Can I only show that $B$ is similar to $P$ and I am done?
The roots of the characteristic polynomial are always roots of the minimal polynomial. Therefore, the minimal polynomial is $x^3+x$.
Since $0$ is a root of the characteristic polynomial with multiplicity $1$, $\dim\ker A=1$. Take $v\in\ker A$ and let $R$ be the range of $A$. By the rank-nullity theorem, $\dim R=2$. For each $w\in R$, $w=A.v$ for some $v\in\Bbb R^3$, and therefore\begin{align}(A^2+\operatorname{Id}).w&=(A^2+\operatorname{Id}).(A.v)\\&=(A^3+A).v\\&=0.\end{align}Take $w_1\in R\setminus\{0\}$ and let $w_2=A.w_1$. Then$$A.w_2=A^2.w_1=-w_1.$$It is not hard to prove that $\{v,w_1,w_2\}$ is a basis of $\Bbb R^3$ and, since $A.v=0$, $A.w_1=w_2$, and $A.w_2=-w_1$, $A$ is indeed similar to$$\begin{bmatrix}0&0&0\\0&0&-1\\0&1&0\end{bmatrix}.$$