$A,B$ same eigenvalue $\lambda$ 
Let $V$ be a finite dimensional vector space over the complex numbers. Let $A,B$ be two linear maps of $V$ into itself. Show that $A,B$ have a common eigenvector.[Hint: If $\lambda$ is an eigenvalue of $A$, consider the space $V_\lambda$ consisting of all vectors $v$ such that $Av=\lambda v$, and show that $B$ maps this space into itself. Then proceed on your own]Linear 
  Algebra,Serge Lang.

I think of $Av=\lambda v $ and it is the only eigenvalues possible it means $\dim V_\lambda=1$. Therefore $Bv$ must be equal to $\lambda v$. However I do not know if this is right. I have no other idea on  how to prove this statement.
Questions:
How do I prove the statement? Can someone provide me a proof?
Thanks in advance
 A: You are missing an important hypothesis - you need to require that $AB = BA$. Without this hypothesis, the result is not true as the example
$$ A = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix},\quad B = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}$$
demonstrates. Assuming that $AB = BA$, choose an eigenvalue $\lambda$ of $A$ and consider the space $V_{\lambda} = \ker(A - \lambda I)$. Let us show that $B(V_{\lambda}) \subseteq V_{\lambda}$ (we say that $V_{\lambda}$ is $B$-invariant). If $v \in V_{\lambda}$ then $$Av = \lambda v \quad\text{and}\quad A(Bv) = B(Av) = B(\lambda v) = \lambda (Bv),$$ so $Bv \in V_{\lambda}$. 
This means we can restrict $B$ to $V_{\lambda}$ and obtain an operator $B|_{V_{\lambda}}$ from $V_{\lambda}$ to itself. Since we are working over the complex numbers, $B|_{V_{\lambda}}$ has an eigenvalue $\mu$ and if $w \in V_{\lambda}$ is an associated eigenvector then $Bw = \mu w$ and $Aw = \lambda w$.
A: The statement is false, as the matrices
$$A=\begin{bmatrix}1&0\\0&2\end{bmatrix}\\
B=\begin{bmatrix}3&-1\\-1&3\end{bmatrix}$$
demonstrate.
Each eigenvector of $A$ is of the form $\begin{bmatrix}z\\0\end{bmatrix}$
or of the form $\begin{bmatrix}0\\z\end{bmatrix}$
while each eigenvector of $B$  is of the form $\begin{bmatrix}z\\z\end{bmatrix}$
or of the form $\begin{bmatrix}-z\\z\end{bmatrix}$
for some $z\in\mathbb C$.
