Htykuut

Suppose $T$ is in $L(V)$, $m$ is a positive integer, and $v$ in vector space $V$ is such that

$(T^{m-1})v neq 0$,

and

$(T^m)v = 0$.

Prove that $[v, Tv, T^2v, ... , T^{m-1}v]$ is linearly independent

I get that $(T^j)v neq 0 forall j < m$. Additionally, since $T$ is nilpotent, $V$ has a basis with respect to which the matrix of $T$ has $0$s on and below the diagonal.

However, I'm not sure if these can be used to show linear independence or if they're even relevant to the problem at all.

Any help is appreciated!

It is simple to do it directly:

If $a_0 v + a_1 Tv +a _2 T^2v + cdots + a_{m-1} T^{m-1}v=0$, then apply $T^{m-1}$ and get $a_0 T^{m-1}v=0$, which implies $a_0=0$. Now apply $T^{m-2}$ to get $a_1=0$. And so on.

The matrix $T$ need not be nilpotent. Let's take a simple example: $n=2$.
Say you have a vector $v$ with $Tvne0$ but $T^2v=0$. Let $w=Tv$.
We need to show that $av+bw=0$ implies $a=b=0$. But applying $T$ gives
$$0=T(av+bw)=aTv+bTw=aw.$$
As $wne0$, $a=0$. Therefore $bw=0$ and so $b=0$.

Can you do something similar for $n=3$? General $n$?