Let $$A = \left(\begin{array}{cccc} 1&0&0&0\\3&-2&0&0\\14&0&-2&0\\8&-1&1&-2 \end{array}\right)$$
Easy to verify that $f_(x) = (x-1)(x+2)^3$. So the eigen-values are: $\lambda = 1,-2$. Some more calculations gives the result that the Jordan form is:
$$J = \left(\begin{array}{cccc} 1&0&0&0\\0&-2&0&0\\0&0&-2&0\\0&0&1&-2 \end{array}\right)$$
Now, I want to find $P$ such that $P^{-1}AP = J$. Meaning, we wish to find $AP = PJ$. More detailed. Let's denote $P=(v_1,v_2,v_3,v_4)$. Then,
$$Av_1 = v_1 \implies (A-I)v_1 = 0 \\ Av_2 = -2v_2 \implies (A+2I)v_2 = 0 \\ Av_3 = -2v_3 + v_4 \implies (A+2I)v_3 = v_4 \\ Av_4 = -2v_4 \implies (A+2I)v_4 = 0$$
Now, that means that $v_3 \in \ker (A+2I)^2$ since $$(A+2I)^3v_3 = (A+2I)(A+2I)v_3 = (A+2I)v_4 = 0$$
I evaluated that:
$$ \ker(A+2I) = \text{span}\{(0,0,0,1), (0,1,1,0)\} $$
$$ \ker(A+2I)^2 = \text{span}\{(0,0,0,1), (0,1,0,0), (0,0,1,0)\} $$
First we choose $v_4\in\ker(A+2I)$. We may select $v_4 = (0,0,0,1)$.
Now, we want $v_3$ such that $$(A+2I)v_3 = v_4$$
So I picked $v_3 = (0,1,1,0)$ which answers the criteria.
But now, I need to pick $v_2,v_1$ which linearly independent of $v_3,v_4$ but that's impossible I'm afraid.
What am I doing wrong/ understand wrong?