I tried to find a Jordan Basis for
$$\begin{pmatrix}3&-1&1&-7\\9&-3&-7&-1\\0&0&4&-8\\0&0&2&-4\end{pmatrix}.$$
I know that Jordan form and eigenvalues are 0 and 0 with eigenvectors being {5,0,6,3} and {1,3,0,0} but I don't know what am I supposed to do next. What am i supposed to do?
By Jordan theorem you know that the Jordan normal form is:
$$J=\begin{pmatrix}0&1&0&0\\0&0&0&0\\0&0&0&1\\0&0&0&0\end{pmatrix}$$
https://en.wikipedia.org/wiki/Jordan_normal_form
Indeed note that since $A^2=0$,the number of Jordan blocks of size 2 is;
$$2\cdot \dim(\ker(A^2)-\dim(\ker(A^3)-\dim(\ker(A))=8-4-2=2$$
By Jordan theorem we know that a matrix $P$ exists such that $$P^{-1}AP=J$$
let $$P=[v_1,v_2,v_3,v_4]$$
then P has to satisfy the following system: $$AP=PJ$$ that is in your case $$Av_1=0$$ $$Av_2=v_1$$ $$Av_3=0$$ $$Av_4=v_3$$ you have already found $v_1$ and $v_3$.