abstract algebra - Proof of Artin's Theorem (linearly independent functions)

Recently I have come across one of Artin's theorems and I have not been able to crack it quite yet. The theorem is stated as follows:

Let $G$ be a group. and let $f_1,\dots, f_n\colon G\to K^*$ be distinct homomorphisms of $G$ into the multiplicative group of a field. Prove that these functions are linearly independent over $K$.

Would anyone know a (if possible quite simple) proof of this Theorem. This proof came up in a chapter regarding eigenvectors and eigenvalues, so I presume it has something to do with that?

Answer

Suppose there are nontrivial linear relations between the maps $f_1,\dots,f_n$ seen as elements of the vector space $K^G$; among them choose one with the minimum number of nonzero coefficients. Upon a reordering, we can assume it is
$$
\alpha_1f_1+\dots+\alpha_kf_k=0
$$
with all $\alpha_i\ne0$. This means that, for every $x\in G$,
$$
\alpha_1f_1(x)+\dots+\alpha_kf_k(x)=0

$$
Note that $k>1$ or we have a contradiction.

Fix $y\in G$; then also
$$
\alpha_1f(yx)+\dots+\alpha_kf_k(yx)=0
$$
and, since the maps are homomorphisms,
$$
\alpha_1f_1(y)f_1(x)+\dots+\alpha_kf_k(y)f_k(x)=0\tag{1}

$$
for every $x\in G$ and
$$
\alpha_1f_1(y)f_1(x)+\dots+\alpha_kf_1(y)f_k(x)=0\tag{2}
$$
By subtracting $(2)$ from $(1)$ we get
$$
\alpha_2(f_2(y)-f_1(y))f_2(x)+\dots+\alpha_k(f_k(y)-f_1(y))f_k(x)=0
$$
for all $x$, hence

$$
\alpha_2(f_2(y)-f_1(y))f_2+\dots+\alpha_k(f_k(y)-f_1(y))f_k=0
$$
which would be a shorter linear relation, so we conclude that
$$
f_2(y)=f_1(y),\quad
\dots,\quad
f_k(y)=f_1(y)
$$
Now, choose $y$ such that $f_1(y)\ne f_2(y)$ and you have your contradiction.