$A^n=0$ for $A$ nilpotent without the use of minimal polynomial I want to use the fact that for a $(n \times n)$ nilpotent matrix $A$, we have that $A^n=0$, but we haven't yet introduced the minimal polynomials -if we had, I know how to prove this.
The definition for a nilpotent matrix is that there exists some $k\in \mathbb{N}$ such that $A^k=0$.
Any ideas?
 A: Let $T: V \to V$ be any linear transformation. Then the following facts are true:


*

*For all $k \in \mathbb{N}$, $\operatorname{ker}(T^k) \subseteq \operatorname{ker}(T^{k+1})$.

*If $\operatorname{ker}(T^k) = \operatorname{ker}(T^{k+1})$, then $\operatorname{ker}(T^k) = \operatorname{ker}(T^{k+m})$ for all $m \in \mathbb{N}$.


From this, you should be able to see that the nilpotency degree is at most $n$.
A: Note that if $\operatorname{rank}(A^k) = \operatorname{rank}(A^{k+1})$, then $\operatorname{rank}(A^j) = \operatorname{rank}(A^k)$ for all $j \geq k$.  To see that this is the case, note if $\operatorname{rank}(A^k) = \operatorname{rank}(A^{k+1})$, then the restriction of $A$ to the image of $A^{k}$ is an invertible map.
Thus, if $A$ is nilpotent and if $A^{n-1} \neq 0$, we must have
$$
n > \operatorname{rank}(A) > \operatorname{rank}(A^2) > \cdots > \operatorname{rank}(A^n)
$$
and of course, the rank of any matrix is an integer.
A: Let $k$ be the smallest positive integer such that $A^k=0$.  Then we are done if $k \leq n$, since then $A^n=A^kA^{n-k}=0$. Suppose $k>n$. We look at $A$ as a linear map $\mathbb{R}^n \rightarrow \mathbb{R}^n$. Notice that we have the following sequence of nested subspaces of $\mathbb{R}^n$: 
$$\mathbb{R}^n \supset A(\mathbb{R}^n) \supset \cdots \supset A^k(\mathbb{R}^n)=0. $$ 
Moreover, all above inclusions are strict, since if $A^j(\mathbb{R}^n) =A^{j+1}(\mathbb{R}^n)$ for some $j \in \{0, ..., k-1 \}$, then $A^{k-(j+1)}(A^j(\mathbb{R}^n))=A^{k-(j+1)}(A^{j+1}(\mathbb{R}^n))$, so $A^{k-1}(\mathbb{R}^n)=0$, a contradiction with the minimality of $k$. Now we take dimensions to see 
$$n > \dim A(\mathbb{R}^n) > \cdots > \dim A^k(\mathbb{R}^n)=0,$$
obtaining more than $n$ nonnegative integers strictly smaller than $n$, a contradiction. 
