$a_0 + a_1T + \cdots + a_k T^k$ is nilpotent if and only if $a_0 = 0$, where $T$ is nilpotent

Question

Question:

Let $T: V \rightarrow V$ be a nilpotent linear transformation of an F-vector space. Show that $H := \sum^k_{i=0} a_i T^i$ with $a_0, \ldots, a_k \in F$ is nilpotent if and only if $a_0 = 0$

My attempt:

I have so far shown the backwards direction.

If $p$ is the smallest positive integer such that $T^p = T_0$, then I assume that $k < p$. Otherwise, I can just find maximal $n < p$ and then $H = \sum^n_{i=0} a_i T^i$ (everything greater than n "dies"), and I can apply the same argument below.

Then, if $a_0 = 0$, then the smallest power of $T$ in $H^p = (a_1T + \cdots + a_kT^k)^p$ is $T^p$. And so every term in the expansion of $H^p$ is $T_0$, so $H^p = T_0$. Hence, $H$ is nilpotent.

But I'm having trouble showing the forwards direction. My first thought was via a contradiction. So first I suppose $a_0 \neq 0$. Then I said that $H^m = a_0^m + b_1T + \cdots +b_{km}T^{km}$, for any positive integer $m$, where $b_i \in F$

But looking at the above, it seems possible that when I apply this to some $v \in V$, then the $a_0^mv$ and $b_1T(v), \ldots, b_{p-1}T^{p-1}(v)$ will somehow cancel out, and still leave $H$ being nilpotent, which is not what I want as I want to show some contradiction.

I would appreciate any feedback on my approach so far, and how one would show the forwards direction. Thank you!

Answer 1

First, you need at least one $a_i\neq 0$, otherwise the statement is false. Suppose $H$ is Nilpotent. Then, there is $p>0$ s.t. $H^p=0$. Moreover, $$0=H^p=a_0^p+\sum_{i=1}^{kp}\beta _iT^i,$$ for some $\beta _i\in \mathbb R$. Notice that at least one $\beta _j\neq 0$ (since $a_i\neq 0$ for some $i$). Let $J=\min\{j\mid \beta _j\neq 0\}$. Then $$0=a_0^p+T^J\underbrace{\left(\beta _j+\sum_{s=j+1}^{kp}\beta _jT^{s-j}\right)}_{=:q(T)\neq 0}=a_0^p+T^Jq(T).$$

Therefore $a_0=0$ and $T^J=0$ as wished.

Answer 2

Lemma: Let $R$ be any ring, $a,r\in R$, where $a$ is a unit and $r$ is nilpotent. Furthermore suppose that $a^{-1}$ commutes with $r$. Then $a+r$ is a unit.
Proof: Let $s=-a^{-1}r$ and $k\in \Bbb N$ such that $r^k=0$. As $a^{-1}$ and $r$ commute we also have $s^k=0$. Then $$(1-s)b=1-s^{k}=1=b(1-s)$$ where $b=1+s+\dots+s^{k-1}$. Hence $1-s$ and therefore also $a+r=a(1-s)$ is a unit.

Now assume that $a:=a_0\ne0$. By what you already showed $r:=a_1T+\dots+a_kT$ is nilpotent. Then (letting $R=\operatorname{End}(V)$ or $F[T]$) the hypotheses of the lemma are satisfied, so we get that $H=a+r$ is a unit. In particular it cannot be nilpotent.