Linear Algebra proof on Eigenvectors and Eigenvalues

Question

Suppose that $A$ and $B$ are $n \times n$ matrices and $\lambda$ is an eigenvalue of A. Furthermore, the vector $\vec{x}$ is the eigenvector of A corresponding to the eigenvalue $\lambda$, and $\vec{x}$ is the fixed point of B. Prove that $\lambda$ is an eigenvalue of AB and BA, and the vector $\vec{x}$ is the corresponding eigenvector of AB and BA.

So just from reading the question I get the idea that either $A=B$ or either $A$ or $B$ is an identity matrix of size $n\times n$. The part of the question that really throws me off is where it states that $\vec{x}$ is a fixed point of $B$ considering that $\vec{x}$ is a vector. Can anyone help guide me through this problem?

Answer 1

No, it's not true that $A=B$ or that either is the identity.

Hint: If $x$ is a fixed point of $B$ and an eigenvector of $A$ for eigenvalue $\lambda$, calculate $ABx$ and $BAx$.

Answer 2

It needn't be the case that $A=B$ or that $A$ or $B$ be the identity matrix. For example take $n=2$, let $A=2I$, let $B$ be reflection in the $x$-axis and let $\vec x = (1,0)$, then $\vec x$ is a fixed point of $B$ and an eigenvector of $A$.

What this question wants you to prove is that if $A \vec x = \lambda x$ and $B\vec x = \vec x$, then $\vec x$ is an eigenvector of both $AB$ and of $BA$. My advice would be to evaluate $AB\vec x$ and $BA \vec x$ and see what happens.

Answer 3

A (nonzero) fixed vector is just an eigenvector for eigenvalue$~1$. And an eigenvector of $M$ for$~\lambda$ is a nonzero vector such that on the subspace of dimension$~1$ that it spans, multiplication by$~M$ acts as multiplication by the scalar$~\lambda$.

By what is given $A$ acts on that subspace as multiplication by$~\lambda$, and $B$ as multiplication by$~1$. Then on that subspace both $AB$ and $BA$ act as multiplication by$~\lambda$ (scalar multiplications commute); that's all. You have no information whatsoever about what $A$ and $B$ do outside that subspace (nor do you need such information).