Eigenvalue and proper subespace.

Question

I have the follow problem:

Suppose that $A,B\in{\cal M}_n(\mathbb{R})$ such that $AB = BA.$ Show that if $v$ is an eigenvector of $A$ associated to the eigenvalue $\lambda$, with $Bv\neq 0$ and dim$(S_\lambda)=1$, then $v$ is also an eigenvector of B.

Thanks in advance!

Answer 1

$ABv = BAv = B\lambda v$ = $\lambda Bv$

So $Bv$ is an eigenvector of $A$ with eigenvalue $\lambda$

Since $dim(S_\lambda) = 1$ we have that $v \tilde{} Bv$. In other words: $v$ is an eigenvector of $B$.

Answer 2

Hint: If $\lambda \neq 0$, then $Bv = B\left( \frac 1\lambda Av\right)$.

Or, think of it this way: $(A - \lambda I)Bv = B(A - \lambda I) v = 0$