Finding a Jordan basis for a $4\times 4$ matrix and corresponding canonical form.

Find a Jordan basis for

$$X=\begin{bmatrix} 2 & 0 & -1 & 3 \\ 0 & 2 & 2 & -1 \\ 0 & 0 & 2 & -1 \\ 0 & 0 & 0 & 2 \end{bmatrix}$$

and write down the corresponding Jordan canonical form.

I know the characteristic polynomial ($2-\lambda^4$) and such $\lambda=2$.

This gives the eigenvectors $$v_1 = \begin{bmatrix} 1\\0\\0\\0 \end{bmatrix} \quad\text{and}\quad v_2 = \begin{bmatrix} 0\\1\\0\\0 \end{bmatrix}$$

Where do I go from here to complete the question?

• In fact, we get the two separate eigenvectors $(1,0,0,0),(0,1,0,0)$ – Omnomnomnom Jan 13 '17 at 0:21
• I've edited the question with correct eigenvectors. – B.K97 Jan 13 '17 at 0:38

The standard method consists in examining the dimensions $r_i$ of $\ker (A-2I)^i$ for $i=1, 2,\dots$. The case $i=1$ corresponds to the eigenspace $E_2$.
The fundamental results is that $r_i-r_{i-1}$ is equal to the number of Jordan blocks of size $\ge i$. Conventionally, $r_0=0$, so that the dimension of the eigenspace is the total number of Jordan blocks ($2$ in the present case. Further, one obtains $r_2=3$, $r_3=4$. We conclude there is a Jordan block of size $1$ and another one of size $3$, and the Jordan normal form will be: $$\begin{bmatrix}2&0&0&0\\0&2&1&0\\0&0&2&1\\ 0&0&0&2\end{bmatrix}$$ Now to have a Jordan basis, take a non-zero vector $e_4=\left[\begin{smallmatrix}x\\y\\z\\t\end{smallmatrix}\right]\in\ker(A-2I)^3\smallsetminus\ker(A-2I)^2$. This condition means $t\ne 0$.
Then set $u_3=(A-2I)u_4$; this vector $u_3\in\ker(A-2I)^2\smallsetminus\ker(A-2I)$. $(A-2I)u_3=u_2$ is an eigenvector. Complete the basis othe eigenspace with another, linearly independent vector, and you have your Jordan basis.
Here one obtains $$u_4=\begin{bmatrix}0\\0\\0\\1\end{bmatrix},\; u_3=\begin{bmatrix}3\\-1\\-1 \\0\end{bmatrix}, \;u_2=\begin{bmatrix}1\\-2\\0\\0\end{bmatrix},$$ which one can complete with, say, $$u_1=\begin{bmatrix}1\\0\\0\\0\end{bmatrix}.$$