I am looking for a proof for the Jordan chain $\{L^ix, L^{i-1}x, \dots, x\}$ being independent, where $L$ is a nilpotent matrix of index $k$, $i<k$ and $x \neq 0$. I have tried the following. Suppose the set is not linearly independent, then $\exists$ at least one $m$, $0\leq m \leq i$, such that: \begin{align} L^mx &= \sum_{j=0, j\neq m}^{i} \alpha_jL^jx \hspace{5mm} \text{multiplying by $L^{k-m}$} \\ 0 &= \sum_{j = 0}^{m-1} \alpha_jL^{k+m-j}x. \end{align} I am stuck here trying to prove a contradiction. Any help would be appreciated.
I am referring to the article here.
I assume that the Jordan chain is "complete" for the given vector $x$ such that $L^i x\neq 0$ but $L^{i+1}x=0.$ Now let $$ \sum_{j=0}^{i}\alpha_j L^j x = 0 $$ Then we multiply with $L^i$ from the left: $$ L^i\sum_{j=0}^{i}\alpha_j L^j x = \sum_{j=0}^{i}\alpha_j L^{i+j} x = \alpha_0 L^i x $$ As $L^ix\neq 0,$ we can conclude that $\alpha_0=0.$ Now $$ L^{i-1}\sum_{j=0}^{i}\alpha_j L^j x = \sum_{j=0}^{i}\alpha_j L^{i+j-1} x = \alpha_0 L^{i-1} x + \alpha_1 L^i x = \alpha_1 L^i x $$ which means that $\alpha_1 =0$ and so on and so forth, such that we get $$ L^{i-m}\sum_{j=0}^{i}\alpha_j L^j x = \sum_{j=0}^{i}\alpha_j L^{i+j-m} x = \alpha_0 L^{i-m} x + \ldots + \alpha_m L^i x = \alpha_m L^i x $$ for $0\leq m\leq i$ and we can conclude $\alpha_0 = \ldots = \alpha_i = 0.$