Linear Independence for point-wise addition

Question

Let X be any set, and let $V = \{f: X \to F\}$ be the vector space of functions on $X$ with pointwise addition, i.e. $$(f + g)(x) = f(x) + g(x),\quad f, g \in V.$$

1. For each $x \in X$, let $fx \in V$ be the vector defined by $$fx(y)=\begin{cases}1, & \text{if } y=x \\ 0, & \text{if } y\neq x \end{cases}$$ For any finite subset $\{x_1, x_2, \ldots , x_n\} \subset X$, show that $\{fx_1,fx_2, \ldots ,fx_n\}$ is linearly independent.

Please help! I have no clue how to even begin solving this question....

Answer 1

Let $\alpha_1 ... \alpha_n \in K$, where K is the underlying field.
Lets assume there exsits those coefficients such that $$ \sum_{i=1}^n \alpha_i f_{x_i}(y)=0 $$ Here the $0$ on the right-hand side is the 0 function $f(x)=0$. I will also suppose that $x_i \neq x_j$, otherwise you can bundle the coefficients to $(\alpha_j+\alpha_i)f_{x_i}(y)$. If we evalue the sum at an arbitrary $x_k \in \{x_1,....x_n \} $ we get that $$ \sum_{i=1}^n \alpha_i f_{x_i}(x_k)=\alpha_k $$ Therefore $\alpha_k=0$, since our sum has to be zero for all points. Since $x_k$ was arbitrary, we deduce that $\alpha_i=0$ for all $i \leq n$. Can you see now why there is linear indepence?