Let $A\in \mathbb{C}^{n\times n}$ be a nilpotent matrix. I want to find a nonsingular matrix $P$, s.t. $P^{-1}AP$ is in Jordan canonical form. Denote $F_i=N(A^i).$ (where $N(A)$ is the null space of $A$).
$\textbf{Case 1}$: $\mathrm{dim}F_1=1$.
Statement 1: if $\mathrm{dim}F_i=i$, then $\mathrm{dim}F_i\leq i+1$.
Statement 2: $\mathrm{dim}F_i=i$.
Note: Statement 1 can be proved by contradiction (suppose $\mathrm{dim}F_{i+1}>i+1$). Statement 2 follows from Statement 1 and the fact that $\mathrm{dim}F_1=1$ and $\mathrm{dim}F_n=n$.
Now we know that $\{\mathrm{dim}F_1,\mathrm{dim}F_2,...,\mathrm{dim}F_n\}=\{1,2,...,n\}$. Thus there exist $x\in F_n$ but $x\notin F_{n-1}$. Then $A^ix\in F_{n-i}$. Thus $A^{n-1}x,...,Ax,x$ form a basis of $\mathbb{C}^n$.
Let $P=[A^{n-1}x\quad...\quad Ax\quad x]$, then $P^{-1}AP=\begin{bmatrix}0&1 & && \\&0& \ddots&& \\&&\ddots &\ddots&\\ &&&0&1\\&&&&0 \end{bmatrix}$.
$\textbf{Case 1}$: $n>\mathrm{dim}F_1>1$. (Define $\mathrm{dim}F_i=m_i$)
In this case there exists $1<j<n$, s.t. $A^j=0$, and $A^{j-1}\neq0$.
Statement 1: matrix $P^{-1}AP$ will contain $m_1$ Jordan blocks inside.
Statement 2: $2\mathrm{dim}F_i\geq\mathrm{dim}F_{i+1}$ i.e. $2m_i\geq m_{i+1}$.
$F_1=\mathrm{span}\{b_1,b_2,...,b_{m_1}\}$
$F_2=\mathrm{span}\{b_1,b_2,...,b_{m_1},b_{m_1+1},...,b_{m_2}\}$
$\vdots$
$F_{j-1}=\mathrm{span}\{b_1,b_2,...,b_{m_1},b_{m_1+1},...,b_{m_2},...,b_{m_{j-2}+1},...,b_{m_{j-1}}\}$.
$F_{j}=\mathrm{span}\{b_1,b_2,...,b_{m_1},b_{m_1+1},...,b_{m_2},...,b_{m_{j-2}+1},...,b_{m_{j-1}},b_{m_{j-1}+1},...,b_n\}$.
I believe that the last vectors $b_{m_{j-1}+1},...,b_n$ must be used same as $x$ in Case 1. Please help me to complete the construction of $P$.