Let $T^{m-1}v\neq 0$ but $T^mv=0$. Show that $v,Tv,\ldots,T^{m-1}v$ are linearly independent 
Let $T\in\mathcal L(V)$ and $T^{m-1}v\neq 0$ but $T^mv=0$ for some positive integer $m$ and some $v\in V$. Show that $v,Tv,\ldots,T^{m-1}v$ are linearly independent.

I had written a proof but Im not sure if it is correct. And in the case it would be correct I dont know how to write it better and clearly. So I have two questions:


*

*It is the proof below correct?

*If so, how I can write it better using the same ideas?

The attempted proof:
1) If $T^{m-1} v$ would be linearly dependent of $T^{m-2} v$ then exists some $\lambda\neq 0$ such that
$$T^{m-1}v=\lambda T^{m-2}v\implies T^mv=\lambda T(T^{m-2}v)=\lambda T^{m-1}v=0\implies \lambda=0$$
Then $T^{m-2}v$ is linearly independent of $T^{m-1}v$. 
2) Now observe that
$$\lambda_1v+\lambda_2Tv+\lambda_3T^2v=0\implies T^{m-2}(\lambda_1v+\lambda_2Tv+\lambda_3T^2v)=\lambda_1T^{m-2}v+\lambda_2T^{m-1}v=0$$
so $\lambda_1,\lambda_2=0$ as we had shown previously, so the original equation reduces to
$$\lambda_3T^2v=0\implies \lambda_3=0$$
thus $v,Tv,T^2v$ are linearly independent.
3) Repeating recursively the analysis in 2) for longer lists of vectors of the form $v,Tv,\ldots,T^kv$ for $k< m$ we can show that the list $v,Tv,\ldots,T^{m-1}v$ is linearly independent.
 A: Hint
$$a_0v+a_1Tv+\ldots+a_{m-1}T^{m-1}v=0 \quad(1)$$
multiply by $T^{m-1}$, then
$$a_0T^{m-1}v=0\to a_0=0$$
so, from $(1)$,
$$a_1Tv+\ldots+a_{m-1}T^{m-1}v=0$$
multiply by $T^{m-2}$, then
$$a_1T^{m-1}v=0\to a_1=0$$
can you finish?
A: See below for a simple proof.
Suppose that they are not linear independent; that is for some choice of $\alpha_{k}$,
$T^{m-1}v = \sum_{k=0}^{m-2}\alpha_{k}T^{k}v$.
Then $0 =T^{m}v = T\cdot T^{m-1}v =  \sum_{k=0}^{m-2}\alpha_{k}T^{k+1}v$,
so $Tv,...,T^{m-1}v$ are linearly dependent.  
But that means that $T^{m-1}v = \sum_{k=1}^{m-2}\beta_{k}T^{k}v$.
Inductively repeat the argument to eventually conclude that for some $\gamma \neq 0$, $T^{m-1}v = \gamma T^{m-2} v$, but then we would have to say that $0 = T T^{m-1} v  = \gamma T^{m-1} v \neq 0$, a contradiction.
A: More simply put: if you had a linear combination
$c_0 v + c_1 T v + \ldots c_{m-1} T^{m-1} v = 0$
with $c_j$ not all zero, let $c_i$ be the first nonzero coefficient.
Then $$T^i v = - \sum_{j=i+1}^{m-1} (c_j/c_i) T^j v$$  Applying $T^{m-i-1}$ to both sides, 
$$T^{m-1} v = - \sum_{j=i+1}^{m-1} (c_j/c_i) T^{m+j-i-1} v = 0$$
But all terms on the right are $0$, so $T^{m-1} v = 0$, contradiction!
