$A,B$ matrices , prove $Bv = \Lambda v$

Question

$A,B$ are $n \times n$ matrices and $AB = BA$

Also, there is an eigenvalue $\Lambda$ in $A$ which its geometric complexity is $1$.

Also there is $ v \ne 0 $ that $v$ is an eigenvector of $A$.

Prove that $v$ is an eigenvector of $B$

Answer 1

Suppose $\lambda$ is a eigenvalue of geometric multiplicity $1$ of $A$ and $v \neq 0$ is an associated eigenvector. If $Bv = 0$, then $v$ is an eigenvector of $B$ associated to the eigenvalue $0$. If $Bv \neq 0$, note that $$\lambda Bv = B(\lambda v) = BAv= A(Bv).$$ This shows that $Bv$ is an eigenvector of $A$ associated to the eigenvalue $\lambda$. Since the geometric multiplicity of $\lambda$ is $1$, the eigenspace $V_\lambda$ is of dimension $1$ and thus we must have $Bv = \alpha v$ for some coefficient $\alpha$ (because $Bv \in V_\lambda = \text{span} \{v\}$). This shows that $v$ is an eigenvector of $B$ associated to the eigenvalue $\alpha$.