Why is the space of linear operators on $n$-dimensional complex vector space $V$ finite dimensional I'm reading Axler Sheldon's paper Down with Determinants! and on page 7 he states the following: 

4. The Minimal Polynomial
Because the space of linear operator on $V$ is finite dimensional,
  there is a smallest positive integer $k$ such that 
$$I, T, T^2, ..., T^k$$
are not linearly independent.

In above $T$ is a linear operator on $V$ and $V$ is a $n$-dimensional complex vector space. The author doesn't seem to explain why the space of linear operators on $V$ is finite. He simply seems to states this.
Question: Why is it finite dimensional? Is this a consequence from something? 
Subquestion: The author also uses on some places of the paper the notation $T|_{V_2}$ where $V_2$ is a subspace of V. What does this notation mean? Is it the restriction of $T$ to $V_2$?
 A: Choose a basis of $V$, say $e_1, \dots, e_n$. Any linear operator $T$ on $V$ is determined uniquely by its action on the basis, i.e. by the values
$$T(e_i) = a_{i1} e_1 + \dots + a_{in} e_n, \quad 1 \le i \le n,$$
for some $a_{ij} \in \mathbb{C}$. In other words, $T$ is uniquely determined by the $n^2$ complex numbers $a_{ij}$. This implies that the space of operators has dimension $n^2$, and so is finite-dimensional. Using your basis to view the operators as matrices is another easy way to see the space is of dimension $n \times n = n^2$.
A: That is because, over any field $K$, in the finite dimensional vector space $V$ choose a basis $\mathcal B=(e_1,\dots,e_n)$ of $V$.
 Then, for any vector space $W$, we have a (non-canonical) isomorphism:
$$\operatorname{Hom}_K(V,W)\cong W^{\lvert\mathcal B\rvert}=W^n. $$
In particular
$$\operatorname{End_K}(V)=\operatorname{Hom_K}(V,V)\cong V^n.$$
Added:
Actually, this theorem is more generally true for any commutative ring $R$ and any finitely generated free $R$-module. 
