$K^S$ := { $f | f: S \rightarrow K$ is a function} is a vector space. Determine a basis for $K^S$ when $S$ = {1, 2, ... , n}.
It is assumed that $K$ is either the real numbers or the complex numbers.
So far, I have determined that the basis must have the form:
$B$ = { $f_{1}$, $f_{2}$, ... , $f_{n}$ } where $f_{i} \in K^S$.
I have tried defining $f_{i}(s) = s^i +1$ for $s \in S$.
Checking for linear independence:
$\sum_{i=1}^{n} a_i f_i(s)$ = $\sum_{i=1}^{n} a_i (s^i +1)$ = $\sum_{i=1}^{n} a_i s^i + a_i = 0$
So, $\sum_{i=1}^{n} a_i s^i + \sum_{i=1}^{n} a_i = 0$
Since, $\sum_{i=1}^{n} a_i s^i = - \sum_{i=1}^{n} a_i$
Meaning that either: $s^i = -1$ or $a_i = 0$. Since, $s \in S$, $s^i ≥ 1$ So, $a_i = 0$. Therefore, $f_1 , ... , f_n$ are linearly independent.
I have tried to show that $B \cup f_{n+1}$ is linearly dependent with no success. Is this because my choice for $f_i$ is not a basis of $K^S$?
$f_i(s) = \begin{cases} 1&s = i\\0&\ s\ne i\end{cases}$