$V$ 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.
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.