I am trying to show that for a nilpotent matrix $A\in M_n(F)$, there exists a natural number $k\leq n$ such that $A^k=0$.
Using the Cayley-Hamilton theorem, it is straightforward since a nilpotent matrix has a characteristic polynomial $p_A(\lambda)=\lambda^n$ so apply Cayley-Hamilton to deduce $A^n=0$.
I'm asking for a solution without using the Theorem.
Thank you!
Recall or show:
If $\ker A^i = \ker A^{i+1}$ for some $i$, then $\ker A^{i} = \ker A^m$ for all $m \ge i$.
Thus, until reaching $A^k = 0$, you need strict inclusion in $(\ker A^i)_{0 \le i \le k}$, which limits the size of $k$ considering dimensions.
Denote $k$ the nilpotent index so there is some $x$ such that $A^{k-1}x\ne0$. Then the vectors $(x_0,Ax_0,\ldots,A^{k-1}x)$ are linearly independent. In fact, by contradiction let $\alpha_0,\ldots,\alpha_{k-1}$ not all $0$ such that $$\sum_i\alpha_i A^{i}x=0\quad (*)$$ and let $$p=\min\bigg\{i\in\{0,\ldots,k-1\}\mid \alpha_i\ne0\bigg\}$$ so we apply $A^{k-p-1}$ to $(*)$ we find $\alpha_{p}=0$ which is a contradiction. Now recall that the number of linearly independent vector is less that the dimension of the vectors space so $k\le n$.
Convince yourself that if $$A=\begin{bmatrix}0&a_{12}&\cdots&a_{1n}\\ &0&\cdots&a_{2n}\\ &&\ddots&\vdots\\ &&&0\end{bmatrix}$$ is a strictly upper-triangular $n\times n$ matrix, then $A^n=0$ (notice that the zeros "move up" the superdiagonals after each multiplication).
Recall that every square matrix is unitarily equivalent to one that is upper-triangular and whose eigenvalues appear on the diagonal (Jordan canonical form will also do the trick here). Since your matrix is nilpotent, all of its eigenvalues are $0$ and hence it is unitarily equivalent to a strictly upper-triangular matrix. Since these matrices satisfy $A^n=0$ (by the above), so too must your nilpotent.
