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Here is the question:

Let $F$ be an infinite field and let c$_0$, c$_1$, ..., c$_n$ $\in$ $F$ be distinct. Then for $i$ = 0 ,1, ..., n, define

$f_i(x) = \prod^n_{k=0,k\neq i}\frac{x-c_k}{c_i-c_k}$

Is {$f_0, f_1,...,f_n$} then a linearly independent set? Why?

So I tried to find a linear combination of {$f_0, f_1,...,f_n$}, namely $\exists$ $a_0, a_1, ..., a_n \in F$ and $a_0f_0+a_1f_1+...+a_nf_n=0$

I got the part that $\forall$ $i$, $f_i(c_i) = 1$ and $\forall$ $j\neq i, f_i(c_j)=0$

But then I figured that for $f_i(x)$, $x$ is arbitrary, then how is that if $x\neq c_i$?

Then take the linear combination, suppose $g=\sum^n_{i=0}a_if_i=0$ for some scalars $a_0, a_1,..., a_n$

Then, $g=\sum^n_{i=0}a_if_i(c_j)=0$ for $j=0, 1,..., n$,

and by the fact that $g=\sum^n_{i=0}a_if_i(c_j)=a_j$ for each $j$.

Then $a_j=0$ for all $j$.

Comes to the conclusion that {$f_0, f_1, ..., f_n$} is linearly independent.

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    Take a linear combination of them, say $g$. What happen when you plug $g(c_i)$ for $i=1,\ldots n$? Also try to explain what you tried so far. – 2017-02-14

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First show that $f_i(c_j) = \delta_{ij}$, the kronecker delta. Then, upon assumption that $g = \alpha_0 f_0 + \cdots + \alpha_n f_n$ is the zero function, it will follow that $0 = g(c_j)=\sum_{i=0}^n \alpha_if_i(c_j) = \sum_{i=0}^n \alpha_i \delta_{ij} = \alpha_j$ for each $j$.

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    I think I have forgotten the fact that $f_i(c_j)=1$ if $i = j$. – 2017-02-14