Finding Jordan normal form when $(A-\lambda I)$ is singular and nilpotent I'm having trouble finding generalised eigenvectors of the following matrix:
$$A=
\begin{bmatrix}
    7 &1 \\ -4 &3
\end{bmatrix}
$$
This matrix has a double eigenvalue $\lambda=5$, with one corresponding eigenvector $\vec{v_1}=\left[\matrix{-1\\2}\right]$. Now if I try to use the Jordan technique, I get:
$$(A-\lambda I)^2\vec{v_2}=\begin{bmatrix}
    2 &1 \\ -4 &-2
\end{bmatrix}^2\vec{v_2}=\vec{0}
$$
but since 
$$\begin{bmatrix}
    2 &1 \\ -4 &-2
\end{bmatrix}^2=0$$
this doesn't get me very far.
Now I also tried
$$(A-\lambda I)\vec{v_2}=\begin{bmatrix}
    2 &1 \\ -4 &-2
\end{bmatrix}\vec{v_2}=\vec{v_1}$$
But again, since $\begin{bmatrix}
    2 &1 \\ -4 &-2
\end{bmatrix}$ is singular, this doesn't get me anywhere either.
Now how do I proceed in finding the Jordan normal form of $A$?

As you get $(A-\lambda I)^2=0$, the equation
$$
0\cdot \vec v_2=0
$$
has lots of solutions to pick - you can take any vector you like. Then to set $\vec v_1=(A-\lambda A)\vec v_2$ and you are done.
In your second attempt: there is a solution, namely,
$$
\vec v_2=\left[\matrix{0\\-1}\right]
$$
even if the matrix is singular.
