Clarifying how to prove a statement using induction

Question

Let $T \in \mathcal{L}(V), m \in \mathbb{N}, v \in V$ such that $T^{m-1}v\neq 0$ but $T^{m}v=0$. Prove $v, Tv, T^2v, ..., T^{m-1}v$ linearly independent.

My attempt: let $(*)$ represent the equation $a_0v+a_1Tv+...+a_{m-1}T^{m-1}v=0.$

I understand what to do: to apply $T^{m-1}$ to $(*)$ to show that $a_0=0$, to apply $T^{m-2}$ to $(*)$ to show $a_1=0$, and continue in this fashion until you reach $a_m = 0$. To show this rigourously, do I use induction? Please let me know if my inductive argument makes sense and is correct. Is this what is known as finite induction and if so, have I applied it correctly.

my proof:

inductive argument: I claim $\forall n\in S=\{x\in \mathbb{N}\cup\{0\}:x\leq m-1\}, a_0=a_1=...=a_n=0$.

base case ($n=0$). Apply $T^{m-1}$ to $(*)$, concluding that $a_0=0$.
Now suppose it is true for all $n \lt m-1$ and consider $n+1$. Since $n+1\leq m-1$, apply $T^{m-(n+2)}$ to $(*)$ to get $a_0T^{m-(n+2)}v+a_1T^{m-(n+2)+1}v+...+a_{n+1}T^{m-1}v=0$. Then by inductive hypothesis, since $a_0=a_1=...=a_n=0$ and $T^{m-1}v\neq 0$, we can conclude that $a_{n+1}=0$. This completes induction, and therefore shows that $a_0=a_1=...=a_{m-1}=0$.

Answer 1

Yes, you use, more or less implicitly, induction.

To make it fully rigourous and shorter, you may proceed as follows:

1. Prove $a_0=0$ as you did.
2. Consider the resulting equation, which becomes $\;a_1Tv+...+a_{m-1}T^{m-1}v=0$, and set $w=Tv$, so that the relation becomes $$a_1w+\dots+a_{m-1}T^{m-2}w,\quad\text{ with }\;T^{m-2}w\ne 0,\;\text{but}\;T^{m-1}w=0,$$ and apply the inductive hypothesis (stepping from $m-1$ to $m$).