Proof that $[v, Tv, T²v, ... , T^n v]$ is a basis for $V$ ($\dim(V)=n$)? Let $T:V\to V$ be a linear map from a finite dimensional vector space over a field $F$ to itself. Assume $[v,Tv,T²v,...]$ spans $V$ for some $v \in V$.
Don't know at all how to prove that $[v, Tv, T²v, ... , T^n v]$ is a basis for $V$. Can you help me? I've tried to work with the size of the set being the dimension of the vector space $V$ but I think I'm missing a theorem or something...
Thank you!
 A: I think you mean that $[v,Tv,T^2v,\dotsc,T^{\color{red}{n-1}}v]$ is a basis.  (This sequence has $n$ elements, while the one in the question has $n+1$.)
Let $j$ be the smallest nonnegative integer such that $v,Tv,T^2v,\dotsc,T^jv$ are linearly dependent.  Then
$$ T^j v \in \operatorname{span}\ \{v,Tv,T^2v,\dotsc,T^{j-1}v\} $$
Show by induction that
$$ \forall m\ge j : T^m v \in \operatorname{span}\ \{v,Tv,T^2v,\dotsc,T^{j-1}v\} $$
and conclude that
$$ V = \operatorname{span}\ \{v,Tv,T^2v,\dotsc\} = \operatorname{span}\ \{v,Tv,T^2v,\dotsc,T^{j-1}v\} $$
and so
$$ n = \dim V = \dim \operatorname{span}\ \{v,Tv,T^2v,\dotsc,T^{j-1}v\} = j $$
The definition of $j$ then yields that $v,Tv,T^2v,\dotsc,T^{n-1}v$ are linearly independent; since there are $n$ of them, this shows that they are a basis.
A: Let's assume $\dim V>0$ or the problem is trivial (zero elements from the set span $V$). In particular $v\ne0$, so we can consider the greatest $k\ge0$ such that
$$
\{v,Tv,\dots,T^kv\}
$$
is linearly independent. Such a $k$ surely exists, because $v=T^0v$ forms a linearly independent set and the size of linearly independent sets is bounded by $\dim V$.
If $\{v,Tv\dots,T^kv\}$ doesn't span $V$, we can say that $T^rv$ does not belong to the span of $\{v,Tv,\dots,T^kv\}$ for some $r>k$ (because the set of all $T^jv$ spans $V$). We can assume $r$ is the minimum such. If $r\ne k+1$, then $T^jv$ is in the span of $\{v,Tv\dots,T^kv\}$, for $j=k+1,\dots,r-1$. In particular
$$
T^{r-1}v=\alpha_0v+\alpha_1Tv+\dots+\alpha_kT^kv
$$
so
\begin{align}
T^rv=T(T^{r-1}v)&=T(\alpha_0v+\alpha_1Tv+\dots+\alpha_kT^kv)\\
&=\alpha_0Tv+\alpha_1T^2v+\dots+\alpha_jT^{k+1}v
\end{align}
which is a contradiction, because $T^{k+1}v$ belongs to the span of $\{v,Tv,\dots,T^kv\}$. The contradiction implies that $r=k+1$, against maximality of $k$.
