The following matrix has zero eigenvalue with algebraic multiplicity four,
$$\begin{bmatrix}0 &0&0&0\\0&0&-1&1 \\-1&-1&0&1\\0&0&0&0\end{bmatrix}$$
Find P so that $P^{-1}AP=J$, where J is in Jordan normal form.
Solution: I found the three generalized eigenvectors, and 2 of the normal eigenvectors.
I love these. The characteristic polynomial is $x^6$ while the minimal polynomial is $x^3.$ Since you confirmed dimension $3$ of genuine eigenvectors ( perhaps with row echelon form) this settles the overall shape. There is a 3 by 3 Jordan block, the a 2 by 2, then a singleton. So, $A^3= 0,$ but $A^2 \neq 0.$ For the 3 by 3 block, we want a (nonzero) vector $w$ such that $A^2 w \neq 0.$ The easy way to find $w,$ which will be a column vector, is $(0,0,0,0,0,1)^T,$ the transpose of the displayed row vector. Then $v = Aw$ and $u = Av = A^2 w$ is a genuine eigenvector, as $Au = A^2v = A^3 w = 0.$ The $u,v,w$ are the three right-hand columns of $P.$
so far we have $$ P = \left( \begin{array}{ccc|ccc} ? & ? & ? &0 & 0 & 0 \\ ? & ? & ? &-1 & 1 & 0 \\ ? & ? & ? &-1 & 1 & 0 \\ ? & ? & ? &0 & 0 & 0 \\ ? & ? & ? & 1 & 0 & 0 \\ ? & ? & ? &0 & 0 & 1 \\ \end{array} \right) $$
Good start; For the 2 by 2 block, we need vector $t$ with $At \neq 0$ but $A^2 t = 0.$ Then $s=At$ is a genuine eigenvector. However, we need to ensure that $s,t$ are INDEPENDENT of $u,v,w,$ which is not guaranteed as they have the same eigenvalue.
The kernel of $A^2$ can be anything we want, as long as the sixth entry is zero. Looking at $u,v,w,$ the three right-hand columns, we also have freedom in the first and fourth entries. I took $t = (0,0,0,1,0,0)^T.$ Then $s = At = (0,0,1,0,0,0)^T$ uses just the third entry. We reach columns stuvw, or $$ P = \left( \begin{array}{ccc|ccc} ? & 0 & 0 &0 & 0 & 0 \\ ? & 0 & 0 &-1 & 1 & 0 \\ ? & 1 & 0 &-1 & 1 & 0 \\ ? & 0 & 1 &0 & 0 & 0 \\ ? & 0 & 0 & 1 & 0 & 0 \\ ? & 0 & 0 &0 & 0 & 1 \\ \end{array} \right) $$
Finally, we need a genuine eigenvector $r$ that is independent of everything so far, although just the first two eigenvectors su will be involved. Alright, drawing some pictures, we can have what we like in the first and fourth entries... indeed, we o get independence. So, $r = (1,0,0,1,0,0)^T$
$$ P = \left( \begin{array}{ccc|ccc} 1 & 0 & 0 &0 & 0 & 0 \\ 0 & 0 & 0 &-1 & 1 & 0 \\ 0 & 1 & 0 &-1 & 1 & 0 \\ \hline 1 & 0 & 1 &0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 &0 & 0 & 1 \\ \end{array} \right) $$ With so many zero elements, various ways to invert..
$$ P^{-1} = \left( \begin{array}{ccc|ccc} 1 & 0 & 0 &0 & 0 & 0 \\ 0 & -1 & 1 &0 & 0 & 0 \\ -1 & 0 & 0 &1 & 0 & 0 \\ \hline 0 & 0 & 0 &0 & 1 & 0 \\ 0 & 1 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 &0 & 0 & 1 \\ \end{array} \right) $$