I have my matrix A, which is
$$A=\begin{bmatrix} 3 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 3 \\ \end{bmatrix}$$
and I have been instructed to find a chain basis for it and then put it into Jordan-Canonical form.
I can see that it is upper triangular so finding the eigenvalues isn't hard, you just take the determinant of A - $\lambda$I and set it equal to zero. For that I have:
$det(A - \lambda I) = p(\lambda) = (\lambda - 2)(\lambda - 3)^2$
and then by setting $p(\lambda) = 0$ I find that $\lambda \in \{ 2,3 \}$
so now we find the eigenspaces for each corresponding eigenvalue.
When $\lambda = 2$, the eigenspace is
$$E_{\lambda = 2}=\left\{\begin{bmatrix}x_1 \\x_2 \\x_3\end{bmatrix}=r\begin{bmatrix}-1\\1\\0\end{bmatrix} : r \in \mathbb{R} \right\}$$
and for $\lambda = 3$, we find the eigenspace to be (I think)
$$E_{\lambda = 3}=\left\{\begin{bmatrix}x_1 \\x_2 \\x_3\end{bmatrix}=s\begin{bmatrix}1\\0\\0\end{bmatrix} : s \in \mathbb{R} \right\}$$
And now this is where I get lost. I know I need a $\lambda$-chain for $\lambda = 2$ and for $\lambda = 3$, but I don't know how to figure out how long the chains are supposed to be and how many $\lambda$-chains I need for $\lambda = 2$ and $\lambda = 3$ so I can't even proceed to put it in JCF because I'm not sure how to find my chain basis.
Thanks in advance for all the help!