Let G be finite group. Let V be irreducible representation of G over $\mathbb{C}$. For $f\in \mathbb{C}[G]$, consider $\phi_f$ on V, defined by $\phi_f(v) = f.v$ (scaling by f).
I know, $\phi_f$ is $\mathbb{C}$-linear. Does the converse true?
If $\phi$ is $\mathbb{C}$- linear on V, does there exist $f\in \mathbb{C}[G]$ such that $\phi=\phi_f$
Please help me with this.
We want to show for any irrep $V_i$ that the map $\rho:\mathbb{C}[G]\rightarrow End_{\mathbb{C}}(V_i)$ is surjective. We will prove this using some character theory, solets look at the (algebra) product of these maps, over a chosen set of irreducibles $V_i$.
\begin{equation} \mathbb{C}[G]\rightarrow \prod_{V_i} End_{\mathbb{C}}(V_i)\end{equation}
Comparing dimensions, we see that both sides have dimension $|G|$, by the sum of squares of dimensions formula. (This is secretly equivalent to the statement we want to prove, but it definitely seems friendlier).
So to show surjectivity, we just need to show injectivity, since these spaces have the same dimension. So if we had an element $x\in \mathbb{C}[G]$ that acted as zero in every irreducible representation, by linearity, it would act as zero on every representation.
But this cannot occur, as it would contradict the fact that $x$ acts as a nonzero linear transformation on $\mathbb{C}[G]$, viewed as a representation via left multiplication. Thus, the map is injective, hence surjective, giving your result.