Prove: If T : V → V is a linear transformation and U is a T-cyclic subspace generated by a I must prove:

If $T : V \to V$ is a linear transformation and $U$ is a $T$-cyclic subspace
  generated by a vector $u \in V$, then for $\dim U = m$, 
  $\{u, Tu, T^2u, \dotsc, T^{m-1}u\} $ will be a basis for $U$.

I'm not sure where to even begin with this.  I am aware that in order to prove that {$u,Tu, T^2u, ..., T^{m-1}u$} is a basis for U that we need to show that the elements of the set are linearly independent and that $U=$ span({$u,Tu, T^2u, ..., T^{m-1}u$}).  And, since U is a T-cyclic subspace generated by $u$ then we can state $U=$ span({$u,Tu, T^2u, ...$})
Side note:  Since we know that dim U = m then can we just state that $U=$ span({$u,Tu, T^2u, ..., T^{m-1}u$})?  Or can we not assume that there are a finite amount of vectors in the span since U is finite dimensional?
Any guidance would be greatly appreciated!  Thanks :)
 A: $T$ restricts to an operator $T:U \to U$. By Cayley-Hamilton, such an operator necessarily satisfies a nontrivial linear dependence relation $T^m = \sum_{i=0}^{m-1}a_i T^i$. By induction, this implies that $T^{m+i}u$ is a linear combination of $\{u,Tu,\ldots,T^{m-1}u\}$ for all $i \geq 0$. Hence, the first $m$ elements in $U = \operatorname{span}_{i \geq 0}(T^i u)$ actually span the whole space.
A: Since you are given that $U$ has dimension $m$, the set
$$
\{u,Tu,T^2u,\dots,T^{m-1}u,T^mu\}
$$
is linearly dependent. Let $\alpha_0,\alpha_1,\dots,\alpha_m$ be scalars, not all zero, such that
$$
\alpha_0u+\alpha_1Tu+\alpha_2T^2u+\dots+\alpha_{m-1}T^{m-1}u+
\alpha_mT^mu=0
$$
Let $k$ be the largest index such that $\alpha_k\ne0$; note that $k\le m$. Then we get
$$
T^ku=\beta_0u+\beta_1Tu+\dots+\beta_{k-1}T^{k-1}u
$$
and therefore, by an easy induction, for $j\ge k$,
$$
T^ju\in\operatorname{span}\{u,Tu,\dots,T^{k-1}u\}
$$
It follows that
$$
U=\operatorname{span}\{u,Tu,\dots,T^{k-1}u\}
$$
Thus $k\ge m$, because $U$ has dimension $m$.
