If $\lambda = i$ is an eigenvalue of $A \in \mathbb R^{n \times n}$, explain why the rank of $A^3 + A$ is less than $n$ 
$A$ is a $n \times n$ matrix with real entries and $\lambda = i$ is an eigenvalue of $A$. Explain why the rank of $A^3 + A$ is less than $n$.

I understand completely how to get eigenvalues, as well as information regarding the dimension of null space, rank, etc., but have a really hard time understanding how I am supposed to relate the two, which I can only assume I have to do to answer this question. 

Let $\vec{x}$ be the eigenvector corresponding to the eigenvalue $\lambda=i$. By definition, this means that $A\vec{x}=\lambda\vec{x}=i\vec{x}$.
Consider the product $A^3\vec{x}$ and note that we can move the scalars $i$ from each product to the front, yielding $A^3\vec{x}=A(A(A\vec{x}))=iA(A\vec{x})=i^2A\vec{x}=i^3\vec{x}=-i\vec{x}$
So now $(A^3+A)\vec{x}=A^3\vec{x}+A\vec{x}=-i\vec{x}+i\vec{x}=\vec{0}$. Eigenvectors are nonzero by definition, so $\vec{x}$ is apprently in the nullspace of $A^3+A$. But since $A^3+A$ has a nontrivial nullspace, it must have rank strictly less than $n$.
