Given a square matrix $A$, find an invertible matrix $S$ such that $S^{-1}AS = J_A$ My matrix is 
\begin{bmatrix}7&4\\-4&-1\end{bmatrix} 
I found the characteristic polynomial to be $(t-3)^2$, so my eigenvalue is the repeated root of this polynomial $t = 3$.  I proceeded to find the corresponding eigenvector(s) by finding the nullspace of $A - 3I_2$, which ended up being $Span((-1, 1)^T)$.  Since this is a line, I know my Jordan canonical form for $A$ will be: 
\begin{bmatrix}3&1\\0&3\end{bmatrix}  
Then to find my other eigenvectors, I need to solve the following equation (given by the second column of $J_A$: 
$v_1 + 3v_2 = Av_2$
We can let $v_1$ be $(-1, 1)^T$ from above, then we have: 
$(-1, 1)^T + 3(a, b)^T = A(a, b)^T$ 
This gives me the system: 
$-1 + 3a = 7a + 4b$ and $1 + 3b = -4a - b$ 
The system reduces to $0 = 0$ which makes no sense.  
How do I find my other eigenvector?   
 A: The other eigenvector is no eigenvector in the ordinary sense, it is a vector that resides in the generalized eigenspace which is $$\det({\bf A}-3{\bf I})^k=0, k>1$$If you calculate for example $k=2$ you get the 0 matrix. Therefore the generalized eigenspace for $\lambda = 3$ is all of ${\mathbb R}^2$ in our case.
You can either solve the matrix equation:
$$\bf AS = SJ$$
Where $\bf J$ is the Jordan matrix block you describe and $\bf A$ is the matrix and first column in $\bf S$ is the eigenvector.
Or we can try minimizing:
$$\|({\bf A}-3{\bf I}){\bf v} - [-1,1]^T\|_2^2$$
Now maybe it gets easier to solve.
A: Strictly speaking, the other Jordan basis vector you’re trying to compute is a generalized eigenvector, but that’s bit of terminology isn’t the crucial thing here. There’s no such thing as the (generalized) eigenvector, since any scalar multiple of it is also a (generalized) eigenvector of the same eigenvalue.  
The fact that your system of equations “reduces to $0=0$” means that the two equations aren’t independent and have an infinite number of solutions. Solve one of the equations for $b$ and then choose a convenient value of $a$, and you’ll have your generalized eigenvector.
