I have a question which I didn't manage to solve.
Assume that there is $A \in \mathbb{F}^{nxn}$ niloptent matrix, which means that there is $k \in \mathbb{N}$ such that $A^k = 0$. I need to prove that $A^n = 0$.
They told us that we need to do it by proving the following:
If $A^k = 0$ then of course $A^{k-1} \ne 0$ so there is $v \in \mathbb{F}^n$ such that $vA^{k-1} \ne 0$. Prove that the set $ \{v,Av,A^2v,\dots,vA^{k-1}\}$ is linearly independent.
From what I understood it will help to show that the case where $k > n$ is not possible (cause we found that we have n+1 linearly independent vectors in vectorial space from size n) and then the only cases that actually can happen are when
$n = k$: $A^n = A^k = 0$ or
$n > k$: $A^n = A^k * A^{n-k} = 0 * A^{n-k} = 0$
they told me that for $k > n$ we need to do it by using inductive reasoning and the above statement but I'm not sure how to do it.
Hint: To prove the linear independence of $\{v, \dots, A^{k-1}v\}$, all you need to do is show that $\alpha_0 v + \cdots + \alpha_{k-1}A^{k-1}v = 0$ implies that $\alpha_0 = \cdots = \alpha_{k-1} = 0$. Multiplying the equation $\alpha_0 v + \cdots + \alpha_{k-1}A^{k-1}v = 0$ by $A^{k-1}$ will imply that $\alpha_0A^{k-1}v = 0$. From here, you can continue inductively to get linear independence.