Proof that a matrix $A$ is diagonalisable 
Let a vector $v \in \mathbb C^{3}\setminus\left\{ 0\right\} $ and $X(v)= \left\{ A \in \mathbb C^{3\times3}: \exists \lambda \in \mathbb C\mid v \in \operatorname{ker}(A-\lambda I_{3})\right\}$. Let us assume that vectors $u,v,w \in \mathbb C^{3}$ are linearly independent and $A \in X(u) \cap X(v) \cap X(w)$. Prove that a matrix $A$ is diagonalisable.

I know that matrix is diagonalisable when the entries outside the main diagonal are all zero. However I don't know how do this task because I think this information is inadequate.
Can you talk me through what I should know to do this task?
 A: Suppose that
$$u=(u_1,u_2,u_3)^{t}$$
$$v=(v_1,v_2,v_3)^{t}$$
$$w=(w_1,w_2,w_3)^{t}$$
and let $P$ be the matrix whose columns are $u$, $v$ and $w$. 
$$A\in X(u)\ \Rightarrow Au=\lambda_{1}u,\ \text{for some}\ \lambda_{1}\in\mathbb{C}$$
$$A\in X(v)\ \Rightarrow Av=\lambda_{2}v,\ \text{for some}\ \lambda_{2}\in\mathbb{C}$$
$$A\in X(w)\ \Rightarrow Aw=\lambda_{3}w,\ \text{for some}\ \lambda_{3}\in\mathbb{C}$$
So, $P$ is invertible because $u$, $v$ and $w$ are 3 linearly independent vectors in $\mathbb{C}^{3}$: 
$$ 
P = \left(
\begin{array}{ccc}
\left. u\ \middle\vert\ v\ \middle\vert w\ \right.\\
\end{array} 
\right)
=\left(
\begin{array}{ccc}
u_1 & v_1 & w_1 \\
u_2 & v_2 & w_2\\
u_3 & v_3 & w_3
\end{array} 
\right)
$$
then compute the matrix $AP$
$$ 
AP = A\left(
\begin{array}{ccc}
\left. u\ \middle\vert\ v\ \middle\vert w\ \right.\\
\end{array} 
\right)
=\left(
\begin{array}{ccc}
\left. Au\ \middle\vert\ Av\ \middle\vert Aw\ \right.
\end{array}
\right)
=
\left(
\begin{array}{ccc}
\left.
\lambda_{1}u\
\middle\vert
\lambda_{2}v\
\middle\vert
\lambda_{3}w\
\right.
\end{array}
\right)
=
\left(
\begin{array}{ccc}
\left. u\ \middle\vert\ v\ \middle\vert w\ \right.\\
\end{array} 
\right)
\left(\begin{array}{ccc}
\lambda_1&0&0\\
0&\lambda_2&0\\
0&0&\lambda_3
\end{array}\right)
=P\left(\begin{array}{ccc}
\lambda_1&0&0\\
0&\lambda_2&0\\
0&0&\lambda_3
\end{array}\right).
$$ 
this implies
$$
AP=P\left(\begin{array}{ccc}
\lambda_1&0&0\\
0&\lambda_2&0\\
0&0&\lambda_3
\end{array}\right)
$$
Finallly multiply by $P^{-1}$ on the left to get:
$$P^{-1}AP=\left(\begin{array}{ccc}
\lambda_1&0&0\\
0&\lambda_2&0\\
0&0&\lambda_3
\end{array}\right).$$
