Eigenvalue and proper subespace. 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!

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