<s>[INST] finding the jordan canonical with one eigenvalue Let $$A= \left[
  \begin{array}{ c c }
     -2 & -3 & 1 \\
     0 & -2 & 0  \\
     0 & 0 & -2
  \end{array} \right]
$$
Determine the Jordan canonical form of A.
The only eigenvalue I found was $\lambda=-2$
So finding the eigenvector of this is
$$ \left[
  \begin{array}{ c c }
     0 & -3 & 1 \\
     0 & 0 & 0  \\
     0 & 0 & 0
  \end{array} \right]. \left[
  \begin{array}{ c c }
     x \\
     y \\
     z
  \end{array} \right]=0\implies -3x+z=0\iff z=3x
$$
So let $x$ and $y$ be free variables so $x=a$ and $y=b$ with $a,b\in\mathbb{F}$. So $$\left[
  \begin{array}{ c c }
     a \\
     b \\
     3b
  \end{array} \right]=a\left[
  \begin{array}{ c c }
     1 \\
     0 \\
     0
  \end{array} \right]+b\left[
  \begin{array}{ c c }
     0 \\
     1 \\
     3
  \end{array} \right]$$
So we have two eigenvectors that are LI: $$v_1=\left[
  \begin{array}{ c c }
     1 \\
     0 \\
     0
  \end{array} \right], v_2=\left[
  \begin{array}{ c c }
     0 \\
     1 \\
     3
  \end{array} \right]$$
But what would $v_3$ be? Since correct me if I am wrong but our Jordan canonical matrix $J=P^{-1}AP$ where $P$ is the matrix made up of $v_1,v_2,v_3$. 
 [/INST] Hint: You can find a third generalized and linearly independent eigenvector by solving (use Gaussian Elimination):
$$[A-\lambda I]v_3 = [A + 2I]v_3 = v_1$$
You should get:
$$v_3 = \left(0, -\dfrac 13, 0 \right)$$
You can then use the eigenvectors to find (this uses your current eigenvectors) and $v_3$ as:
$$J = P^{-1} A P = \begin{bmatrix}-2 &0&0 \\0 & -2 & 1 \\ 0 & 0 & -2\end{bmatrix}$$
Where $P$ is made up of your column eigenvectors.
</s>