TITLE: Linear independent vectors of nilpotent transformation
QUESTION [0 upvotes]: $V$ is a vector space.
$N$ is a nilpotent transformation $N:V\rightarrow V$ such that $N^k=0$ ($k$ is the lowest).
$v \in V$, $v \notin \text{ker}\ N^{k-1}$ (in other words: $N^{k-1}v \ne 0$).
Let $v_1,\ldots,v_k$ be:
$v_1 = v$
$v_2 = Nv_1$
...
$v_k = Nv_{k-1}$
Show that $v_1, v_2,\ldots,v_k$ are linearly independent.
Well I know that $\text{ker}\ N  \in \text{ker}\ N^2 
\in \ldots \text{ker}\ N^{k-2} \in \text{ker}\ N^{k-1}  \in V = N^k$
But how do I continue from here? Any suggestions? Thanks.

REPLY [2 votes]: Let $a_1,\ldots, a_k\in \Bbb R$ such that
$$a_1v_1+\cdots+a_k v_k=0$$
assume that the coefficients $a_i$ aren't all $0$ and let 
$$p=\min_{1\le i\le k}\{i\;|\; a_i\ne0\}$$
so we have
$$a_{p}v_p+\cdots+a_kv_k=0\tag1$$
Now apply $N^{k-p+1}$ to $(1)$ we get
$$a_p N^{k-p+1}v_p=a_pN^{k-1}v=0$$
and since $N^{k-1}v\ne0$ then $a_p=0$ which's a contradiction. Conclude.
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