$A^3+A=0$ We need to show $\mathrm{rank}(A)=2$ 
Let $A\ne 0$ be a $3\times 3$ matrix with real entries such that $A^3+A=0$. We need to show $\mathrm{rank}(A)=2$.

$\det A(A^2+I)=0\Rightarrow\det A=0\Rightarrow \mathrm{rank}(A)<3$,  Suppose $\mathrm{rank}(A)=1$, Then I showed one matrix with rank $1$ which do not satisfies the given relation, is my answer is ok? Thank you for help and discussion 
 A: The "standard" way to solve this kind of problems is to examine the minimal polynomial of $A$. Yet, for this particular problem, there are other approaches as well. Here is one of them:


*

*By considering the determinants of both sides of $A^3 = -A$, argue that $\operatorname{rank}(A)\ne3$.

*If $\operatorname{rank}(A)=1$, then $A=uv^T$ for some nonzero vectors $u,v\in\mathbb{R}^3$. Show that $A^3+A$ is a nonzero scalar multiple of $A$ and hence it cannot be zero.

*As $\operatorname{rank}(A)\ne3,1$ and $A\ne0$ by assumption, the only possibility is that $\operatorname{rank}(A)=2$.

A: By the Cayley Hamilton theorem, the minimal polynomial of a square matrix divides its characteristic polynomial. Therefore, since $A \neq 0$ by assumption, $(x^2 + 1) | f_A(x)$ which means that $A$ has at least the eigenvalues $\pm i$. Since these are distinct eigenvalues, the corresponding eigenvectors are linearly independent. Since they lie in the column space of $A$, the rank of $A$ is at least $2$. As you mentioned, a determinant argument can be used to show that it is less than $3$.
