Find Jordan basis for the following nilpotent matrix I need to find the Jordan basis and Jordan form for this nilpotent matrix:
$$\begin{pmatrix}
-2i & 1 & 10+5i \\
4 & 2i & -10+19i \\
0 & 0 & 0 \\
\end{pmatrix}$$
The matrix is nilpotent so all eigenvalues are 0, therefore $Av_1=0$ and from that I got that $v_1=\begin{pmatrix}
1 \\ 2i \\ 0 \\ \end{pmatrix}$
Now $Av_2=v_1$ so $v_2=\begin{pmatrix} 0 \\ 1 \\ 0 \\ \end{pmatrix}$. This is where I'm stuck there seems to be no $v_3$ such that $Av_3=v_2$. How do I continue or what am I doing wrong?

You have it a bit backwards. We have $A^3 = 0$ but $A^2 \neq 0.$ It appears you want the columns of the change of basis matrix, call that $V,$ as  you want column vectors $v_1, v_2,v_3.$ 
Well,
$$
A^2 =
\left(
\begin{array}{rrr}
0 & 0 & -i \\
0 & 0 & 2 \\
0 & 0 & 0
\end{array}
\right)
$$
In order to arrange $A^2 v_3 \neq 0,$ take
$$
v_3 =
\left(
\begin{array}{rr}
0  \\
0  \\
1
\end{array}
\right)
$$
Next
$$
v_2 =A v_3 =
\left(
\begin{array}{rr}
10 + 5i  \\
-10 + 19i  \\
0
\end{array}
\right)
$$
$$
v_1 = Av_2 =A^2 v_3 =
\left(
\begin{array}{rr}
-i  \\
2  \\
0
\end{array}
\right)
$$
we have set up $A v_1 = A^3 v_3 = 0 v_3 = 0.$
Put the columns in order, we have
$$
V =
\left(
\begin{array}{rrr}
-i & 10+5i & 0 \\
2 & -10 + 19i & 0 \\
0 & 0 & 1
\end{array}
\right)
$$
Note $\det V = -1,$ so finding $V^{-1}$ is not that bad.
Then 
$$
J = V^{-1} AV =
\left(
\begin{array}{rrr}
0 & 1 & 0 \\
0 & 0 & 1 \\
0 & 0 & 0
\end{array}
\right)
$$
