On Linear Independence of a Set 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.    
 A: 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 $span\{T^{n-m}(v),\ldots,T^{n}(v)\}\cap W=\{0\}$.
