$A^n=0$ for $A$ nilpotent without the use of minimal polynomial

Question

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?

I edited my initial post. Sorry for not mentioning it in the first place. — perlman, Oct 08 '17 at 21:10
Can you show that the order of nilpotency $k$ of the matrix $A$ is $\leq$ to the order of $A$, i.e., $n$ ? Then, the claim follows trivially. — Prasun Biswas, Oct 08 '17 at 21:11
@Omnomnomnom, that would defeat the purpose of not relying on the knowledge of minimal polynomials. — Prasun Biswas, Oct 08 '17 at 21:13
We haven't mentioned Cayley Hamilton. That's why I find it difficult to prove this. I don't have almost anything in my arsenal. — perlman, Oct 08 '17 at 21:13
@PrasunBiswas well Cayley-Hamilton is about characteristic polynomials — Ben Grossmann, Oct 08 '17 at 21:13
@Omnomnomnom, IIRC, the usual statement is that the minimal polynomial divides the characteristic polynomial. — Prasun Biswas, Oct 08 '17 at 21:14
@Prasun Biswas Actually I don't know how to prove this elementary. Any suggestion? — perlman, Oct 08 '17 at 21:14
@PrasunBiswas that's not the usual statement (although perhaps that's equivalent) — Ben Grossmann, Oct 08 '17 at 21:15

score 3 · Answer 1 · answered Oct 08 '17 at 21:18

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 $n$.

score 3 · Answer 2 · answered Oct 08 '17 at 21:18

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.

score 1 · Answer 3 · answered Oct 08 '17 at 22:01

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.

$A^n=0$ for $A$ nilpotent without the use of minimal polynomial

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