Dimension of a vector space of functions from a finite field to itself Let $F$ be a finite field with $q$ elements. Let $V = P(F)$ and let $W = Func(F, F)$ be the vector space of all functions from $F$ to $F$.
I am struggling with two things:


*

*Compute $\dim W$.


2.Let $T : V \to W$ be the transformation that sends a polynomial $f(x)$ to the function it represents. Prove that $T$ is a linear transformation.
I know that all functions in $W$ are polynomials, if that's true then $W$ is not finite-dimensional, right? As for $T$, what does the linear transformation do exactly? 
 A: Let's look at an example to understand what $T$ does.  Take the case of $q = 3$.  The space $V$ is infinite dimensional, and consists of all "formal sums" of the form
$$
p(x) = a_0 + a_1 x + a_2x^2 + \cdots + a_n x^n
$$
where each $a_i$ is taken from $\Bbb F_3$ (whose elements are $0,1,2$).
Now, the transformation $T$ produces the "function" associated with $p$.  For example: if $p(x) = 2x^3$, then $Tp$ is the function $f_p:\Bbb F_3 \to \Bbb F_3$ defined by
$$
f_p(0) = p(0) = 2(0)^3 =0, \quad f_p(1) = 2(1)^3 = 2, \quad f_p(2) = 2(2)^3 = 1
$$
Notice that the polynomial $q(x) = 2x$ produces exactly the same function.  That is, $p(x) = q(x)$ for every $x \in \Bbb F_3$.  However, $p(x)$ and $q(x)$ are distinct as polynomials.  In brief: we have $p \neq q$, but $f_p = f_q$. That is, $p \neq q$, but $Tp = Tq$.

Note that $Func(\Bbb F,\Bbb F)$ has a natural choice of basis.  Namely, for $k \in \{0,1,\dots,q-1\}$, define $f_k : \Bbb F_p \to \Bbb F_p$ by
$$
f_k(j) = \begin{cases}
1 & j = k\\
0 & j \neq k
\end{cases}
$$
why is this set a basis?
