Show that the set has a linearly independent subset which is a basis for V Let T : V → V be a linear transformation of a finite dimensional vector
space over a field F to itself. Assume that {$v, T v, T^2v, . . . $} span V for some v ∈ V . 
Show that
(i) there exists a k such that $v, T v, . . . , T^{k-1}v$ are linearly independent and
for some $α_i$ ∈ F
$T^kv = α0v + α1T v + · · · + αk−1T^{k−1}v$
Hence show {$v, T v, T^2v, ...,T^{k-1}v $} is a basis for V.
 A: No knowledge of the Cayley-Hamilton theorem is necessary.
The result is not true for $V=\{0\}$, in the form it is formulated, so I assume $\dim V>0$. In particular $v\ne0$ or the given set cannot span $V$.


*

*There exists $k\ge0$ such that $\{v=T^0v,Tv,T^2v,\dots,T^{k-1}v\}$ is linearly independent, for instance $k=1$, because $v\ne0$.

*There exists $k\ge0$ such that $\{v=T^0v,Tv,T^2v,\dots,T^{k-1}v\}$ is linearly dependent, because $V$ is finite dimensional, so $k=1+\dim V$ suffices.

*There exists a maximum $k\ge0$ such that $\{v=T^0v,Tv,T^2v,\dots,T^{k-1}v\}$ is linearly independent, because of the two facts above.
In particular $T^{k}v$ is a linear combination of $\{v=T^0v,Tv,T^2v,\dots,T^{k-1}v\}$, so
$$
T^kv=\alpha_0v+\alpha_1Tv+\dots+\alpha_{k-1}T^{k-1}v
$$
Now, by induction on $n$, we prove that $T^{k+n}v$ is a linear combination of  $\{v=T^0v,Tv,T^2v,\dots,T^{k-1}v\}$.
The base case has already been established. So, assume
$$
T^{k+n}v=\beta_0v+\beta_1Tv+\dots+\beta_{k-1}T^{k-1}v
$$
Then
\begin{align}
T^{k+n+1}v
&=T(\beta_0v+\beta_1Tv+\dots+\beta_{k-1}T^{k-1}v)\\
&=\beta_0Tv+\beta_1T^2v+\dots+\beta_{k-1}T^{k}v\\
&=\beta_0Tv+\beta_1T^2v+\dots+
\beta_{k-1}(\alpha_0v+\alpha_1Tv+\dots+\alpha_{k-1}T^{k-1}v)
\end{align}
and the statement is proved.
Since any vector $u\in V$ can be written as
$$
v=\sum_{i=0}^m\gamma_iT^iv
$$
we have proved that $\{v=T^0v,Tv,T^2v,\dots,T^{k-1}v\}$ is a basis of $V$.
A: I'll give you a sketch of the proof so you can fill in the gaps. 
Let $v \in V$ be non zero.  Consider the space
$$B= \{g(T) (v) : g(x) \in F[x]\} $$
This space $B$ is finite linear combinations of the elements $\{v, T(v), \cdots, \}$ over $F$.
Prove this is indeed a vector space if you want.  But this is the space spanned by $ \{v, T(v), T^2(v), \cdots,\}$ 
Consider the set 
$$A = \{g(x) \in F[x] : g(T)(v) = 0\} $$
How do we know this set isn't just zero?
$T: V \to V$ is linear and $dim(V) < \infty$ so that the minimal polynomial $p_T(x)$ exist and is automatically in $A$.  I mean $p_T(T)(w) = 0$ for every $w \in V$ since $p_T(T) = 0$.  
$A$ is generated by some unique monic polynomial $g(x) \in F[x]$.
(This comes from abstract algebra, F[x] is euclidean domain and hence a principle ideal domain.  $A$ is indeed an ideal so it has a generator.) 
We prove the following. 
Let $g$ be the minimal monic polynomial for $A$.  let $k$ be the degree of $g$.  Then $v, T(v), \cdots, T^{k-1}(v)$ is a basis for $B$.  
For any polynomial $f \in F[x]$, since $F$ is a field we can the division algorithm.
$$ f = q g + r$$
for some $q, r \in F[x]$
where $r = 0$ or $deg(r) < deg(g) = k$.
Then since $g(x) \in A$, evaluation at $v \in V$,
\begin{align*}
f(T) v &= (qg)(T)v + r(T) v \\
&= r(T)v
\end{align*}
Either $r = 0$ or $deg(r) < k$.  In other words, $v, T(v), \cdots, T^{k-1}(v)$ spans $B$.  
Note that $g$ is of minimal degree.  If there is a non trivial linear combination such that 
$$c_1 v + \cdots + c_{k-1}T^{k-1}(v) = 0 $$ 
the polynomial $c_1 + \cdots + c_{k-1} x^{k-1}$ clearly belongs in $A$ and has degree less than $g$, a contradiction. So $v, T(v), \cdots, T^{k-1}(v)$ is linearly independent and spans $B$.
By your assumption, $B = V$.  
A: Suppose the characteristic polynomial of $T$ is $p_T(x)=x^n+a_{n-1}x^{n-1}+...+a_1x+a_0$.
Then by Cayley-Hamilton Theorem: $p_T(T)=T^n+a_{n-1}T^{n-1}+...+a_1T+a_0I=0$.
This means that $T^{n}=-a_{n-1}T^{n-1}-...-a_1T-a_0I$.
And subsequently, if $k>n$, then $T^k=T^{k-n}T^n=T^{k-n}(-a_{n-1}T^{n-1}-...-a_1T-a_0I)$. If $k-n<n$, then we are done. If not, then continue in the same fashion until all the terms are below $n$.
In a nutshell, this proves that for any $k \geq n$, $T^k$ can be represented as a linear combination of $I,T,T^2,...,T^{n-1}$. This is a direct result of Cayley-Hamilton.
Clarification: $T: V \to V$ and $Dim(V)=n$.
