Calculate the Jordan matrix of a matrix $C$ where $C$ exists of the unity matrix and a matrix $A$ which is diagonalisable. Let $A \in M^{3 \times 3}(\mathbb{C})$ be diagonalisable matrix with 3 different eigenvalues $\lambda_1, \lambda_2, \lambda_3$ and their corresponding eigenvalues $v_1, v_2, v_3$. Consider the matrix $C \in M^{6 \times 6} (\mathbb{C})$ with
$C = \left( \begin{matrix} A & I \\ 0 & A \end{matrix} \right) $.
Calculate the Jordan Form/Matrix of $C$ and find a matrix P such that $P^{-1}CP = J$.

I know that the characteristic polynomial will be $f_C = (X- \lambda_1)^2(X- \lambda_2)^2(X- \lambda_3)^2$.
To be honest, I don't really know what else I can deduct from the information I have.
I know that in $A$ for each $\lambda_i$ it holds that $\text{dim(Ker}(A-\lambda_i)) = 1$ but I don't know how this translates to $C$.
Can somebody help me?
 A: Let $V$ be the $3\times 3$ matrix with columns $v_1, v_2, v_3$.
Let $L$ be the diagonal matrix with diagonal entries $\lambda_1, \lambda_2, \lambda_3$.
Then, using block matrix computations,
$$
\begin{aligned}
AV
&=A[v_1\ v_2\ v_3]\\
&=[Av_1\ Av_2\ Av_3]\\
&=[v_1\ v_2\ v_3]L\\
&=VL\ .
\end{aligned}
$$
So $V^{-1}AV$ is the diagonal matrix $L$. Similarly:
$$
\begin{aligned}
C
\begin{bmatrix}V & \\ & V\end{bmatrix}
&=
\begin{bmatrix}A & I\\ & A\end{bmatrix}
\begin{bmatrix}V & \\ & V\end{bmatrix}
\\
&=
\begin{bmatrix}AV & V\\ & AV\end{bmatrix}
\\
&=
\begin{bmatrix}VL & V\\ & VL\end{bmatrix}
\\
&=
\begin{bmatrix}V & \\ & V\end{bmatrix}
\begin{bmatrix}L & I\\ & L\end{bmatrix}
\text{ so}\\
\begin{bmatrix}V & \\ & V\end{bmatrix}^{-1}
C
\begin{bmatrix}V & \\ & V\end{bmatrix}
&=
\begin{bmatrix}L & I\\ & L\end{bmatrix}
\ .
\end{aligned}
$$
Using a permutation matrix conjugation, the last $(3+3)\times (3+3)$ matrix is written as a $(2+2+2)\times(2+2+2)$ matrix with diagonal blocks
$$
\begin{bmatrix}\lambda_j & 1\\ & \lambda_j\end{bmatrix}\ ,\qquad j=1,2,3
\ .
$$
