Finding a diagonalizing matrix associated with Jordan normal form Find the Jordan normal form $J$ of the upper triangular matrix
$$A = \begin{pmatrix}2 & 0 & 1 & 2 \\ 0 & 2 & 2 & 1 \\ 0 & 0 & 2 & 1\\ 0 &0 & 0& 3 \end{pmatrix}$$
and find a matrix $M$ such that $M^{-1}AM = J$.
Note that the characteristic polynomial is $\det(A- \lambda I) = (2- \lambda)^{3}(3-\lambda)$.  We have
$$(A - 2I)(A-3I) = \begin{pmatrix}0 & 0 & -1 & 1 \\ 0 & 0 & 0 & 2 \\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{pmatrix} \neq 0_{4 \times 4}$$
and $(A - 2I) \neq 0_{4 \times 4}$ and $(A - 3I) \neq 0_{4 \times 4}$, but $(A-2I)^{2}(A-3I) = 0_{4 \times 4}$. Hence, the minimal polynomial is: $m_{A}(x) = (x-2)^{2}(x -3)$.
It follows that the other invariant factor must be $(x-2)$. This gives us the Jordan form as
$$J = \begin{pmatrix}2 & 0 & 0 & 0 \\ 0 & 2 & 1 & 0 \\ 0 & 0 & 2 & 0\\ 0 &0 & 0& 3 \end{pmatrix}.$$
Now, $v_{4} = \begin{pmatrix} 3 \\ 3 \\ 1 \\ 1 \end{pmatrix}$ is an eigenvector associated with $\lambda = 3$.
We note that for $k \geq 2$, we have
$$B_{k} = (A - 2I)^{k} = \begin{pmatrix}0 & 0 & 0 & 3 \\ 0 & 0 & 0 & 3 \\ 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 1 \end{pmatrix}.$$
The null space of $B_{k}$ is three-dimensional with the standard basis vectors as a basis for instance. My difficulty stems from not being able to find a seed vector $v_{i}$ in this basis such that I am able to generate the three generalized eigenvectors in the usual manner: $v_{i-1} = (A - 2I)^{i}v_{i}$. I will appreciate a fix to this of course in a generalizable way though.

You have found that there are two Jordan blocks for the eigenvalue $2$, so $\dim\ker(A-2I)=2$. There is a $1\times1$-block and a $2\times2$-block, which means that $\dim\ker(A-2I)^2=3$. Now pick a vector $v_3\in\ker(A-2I)^2$ that is not containd in $\ker(A-2I)$. Then $v_2=(A-2I)v_3$ is an eigenvector with eigenvalue $2$, and you can pick another $v_1\in\ker(A-2I)$ to complete your basis.
