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I've wasted quite a bit of time in this proof and I'd like to know if I got the gist of it right. I'll only show one implication because the other one is true by definition.

(1) If $f$ is nilpotent, then it's not a monomorphism: if it were, then it would be an isomorphism (since it's a mapping between spaces of identical dimension). But given that isomorphisms preserve subspace structure, we have

$$\forall i \in \mathbb{N} \ \ \ \ \dim (\ker f^i)=0$$

which is inconsistent with our initial assumption.

(2) Having settled that $\dim (\ker f)\neq 0$, we try to prove that

$$\forall i \in \mathbb{N} \ \ \ \ \dim (\ker f^i) < \dim (\ker f^{i+1})$$

(unless they're both $n$). It's pretty clear that

$$\ker f^{i} \subset \ker f^{i+1} $$

(if $x \in \ker f^i, f^{i+1}(x)=f(f^i(x))=f(0)=0$). However

$$\ker f^{i+1} \subset \ker f^{i}$$

is not true. Let $x$ be such that $f^i(x)\in \ker f \setminus \{0\}$ (we know such an $x$ must exist because otherwise $f^i$ would be an isomorphism... how can I justify this better?). It follows that $x\neq0,x\not\in \ker f^i,x\in \ker f^{i+1}$. Hence, the dimensions of this sequence of kernels is strictly increasing. Since $n-1$ is an upper bound for this sequence, $f^n=0$.

Deep_Television
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  • What is your definition of a nilpotent endomorphism? – asdq Sep 27 '19 at 08:44
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    The property you are trying to prove in (2) is false, you can take for example $f = 0$, or for a less nontrivial example, $f$ is block diagonal with each block looking like $\begin{pmatrix} 0&1\0&0\end{pmatrix}$. What you need to show is that the dimension of the kernels of $f^i$ are strictly increasing until they stabilise. Then argue that they first stabilise at $f^i$ where $i \leq n$. – Joppy Sep 27 '19 at 09:11
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    If you want to try a different approach take the minimal polynomial $\mu\in K[X]$ of $f$. Compare $\deg \mu\leq n$ and $f^N=0$ for some $N\in \mathbb{N}$. – TomTom314 Sep 27 '19 at 09:29

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Heh. As noted, this is clear if we know that the minimal polynomial has degree no larger than $n$. But that's not at all trivial; here's an ad hoc version of that argument. I find myself unable to write "$f$" for a linear transformation, sorry:

If $X$ is a vector space of dimension $n\ge1$ and $T:X\to X$ is linear and nilpotent then $T^n=0$.

(The case $n=0$ is not of much interest; in that case it's harder to figure out what the statement means than to prove it...)

The proof is by induction on $n$. It's clear that if $n=1$ and $T$ is nilpotent then $T=0$.

In general, let $$Y=TX.$$If $Y=\{0\}$ then $T=0$; assume $\dim(Y)\ge1$.

Now $T$ nilpotent implies $T$ is not invertible, so $Y\ne X$. Define $$S:Y\to Y$$by $$S=T|_Y.$$Since $1\le\dim(Y)<\dim(X)$, induction on $n$ shows that $$S^{n-1}=0.$$Which is to say $$T^{n-1}y=0\quad(y\in Y),$$or in other words $$T^{n-1}Tx=0\quad(x\in X).$$

David C. Ullrich
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