I have a problem about linear independence of sets.
Let $V$ be a vector space and let $T:V\rightarrow V$ be a linear operator. Suppose that $n$ and $k$ are positive integers.
(i) If $w\in V$ s.t. $T^k(w)\neq0$ and $T^{k+1}(w)=0$, must $\{w, T(w),...,T^k(w)\}$ linearly independent?
(ii) Let $W$ be the subspace of $V$ spanned by $\{w, T(w),...,T^k(w)\}$ as in part (i). If $v\in V$ s.t. $T^n(v)\notin W$ and $T^{n+1}(v)\in W$ must $\{w, T(w),...,T^k(w),v,T(v),\ldots,T^n(v)\}$ be linearly independent?
Based on definition of linear independent set and the assumption of the problem in both parts, I think in both parts the sets are L.I. That is kind of you friends to help in answering this problem.
Part(i) can be proved similarly by induction, so I'll just assume part(i) holds. Then part(ii) is equivalent to:
Let $W$ be a subspace of $V$, $v\in V $ but $v\notin W$. If $n+1$ is the first index such that $T^{n+1}(v)\in W$, then $span\{T^{0}(v),\ldots,T^{n}(v)\}\cap W=\{0\}$.
Put differently, we can prove $\forall m \in \{ 0,1,\ldots,n\}$, $span\{T^{n-m}(v),\ldots,T^{n}(v)\}\cap W=\{0\}$. Note that due to part(i) $\{T^{n-m}(v),\ldots,T^{n}(v)\}$ are L.I., and W is invariant under T, which means $\forall w \in W, T(w) \in W$. If you express $w$ with basis and apply $T$ to it you'll easily see it.
Induction on $m$:
$m=0$ holds since $T^n(v) \notin W$.
$m\rightarrow m+1$:
Suppose $\exists a_{n-m},\ldots,a_{n}$ such that $\sum_{i=n-m}^{n}a_i T^i(v)\in W$, then we have $T(\sum_{i=n-m}^{n}a_i T^i(v))=\sum_{i=n-m+1}^{n}a_{i-1} T^i(v) \in W$ since $T^{n+1}(v)\in W$.
By induction hypothesis, $a_{n-m},\ldots, a_{n-1}$ are all zero.
Since $T^n(v)$ is nonzero, we can conclude $span\{T^{n-m}(v),\ldots,T^{n}(v)\}\cap W=\{0\}$.