If $\vec{v}$ is an eigenvector of $A$, then also $B\vec{v}$, when $AB = BA$

Question

I have the following problem:

Let $V$ be a finite dimensional vector space. Let $A$, $B$ be linear maps of $V$ into itself. Assume that $AB = BA$. Show that if $\vec{v}$ is an eigenvector of $A$, with eigenvalue $\lambda$, then $B\vec{v}$ is an eigenvector of $A$ with eigenvalue $\lambda$ also if $B\vec{v} \neq \vec{0}$.

Now, I have to show that if $\vec{v}$ is an eigenvector of $A$ with eigenvalue $\lambda$, then also $B\vec{v}$ with the same eigenvalue.

So I started by assuming $$A\vec{v} = \lambda \vec{v}$$ Then I multiplied both sides by $B$:

$$BA\vec{v} = B \lambda \vec{v}$$ Since $AB = BA$, then $$A(B\vec{v}) = \lambda (B \vec{v})$$ Which should prove what I needed to prove.

Is my proof correct?

Answer 1

Suppose $Av=\lambda v$. Compute: $$ A(Bv) = (AB)v = (BA)v = B(Av) = B(\lambda v) = \lambda(Bv), $$ so $Bv$ is an eigenvector of $A$ if $Bv \neq 0$.