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 $n$

Question

$A$ 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.

Answer 1

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 $n$.

Answer 2

Let $\textbf{x}$ be an eigenvector of $A$ with eigenvalue $i$. $(A^3+A)\textbf{x}=A^3\textbf{x}+A\textbf{x}=i^3\textbf{x}+i\textbf{x}=-i\textbf{x}+i\textbf{x}=\textbf{0}$. Note that the real part of a real matrix times a complex vector is just the matrix times the real part of the vector. So we have $(A^3+A)\Re(\textbf{x})=\textbf{0}$. Now all that's left is to show there is a real part of $\textbf{x}$. If not, then $\textbf{x}=i\textbf{x}'$ for some real $\textbf{x}'$, so $A\textbf{x}=iA\textbf{x}'$, which is a purely imaginary (nonzero, by definition of eigenvector) vector but $A\textbf{x}=i\textbf{x}=i*i*\textbf{x}'=-\textbf{x}'$, a purely real vector. So the real part is nonzero. So we have a nonzero real vector such that $(A^3+A)\Re(\textbf{x})=0$, so the rank of $A^3+A$ is not full.

Answer 3

An eigenvalue of $A$ is $i$.

An eigenvalue of $A^3$ is $i^3$

An Eigenvalue of $A^3+A$ is $i+i^3=0$ which implies $A$ cannot have full rank.