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Let $G$ be a finite group, $V$ a $\mathbb{C}G$-module with character $\chi$. Let $z:=\frac{1}{|G|}\sum\limits_{g\in G}\chi(g^{-1})g\in\mathbb{C}G$. Let $U$ be an irreducible $\mathbb{C}G$-module with characer $\psi$. We define $\zeta:U\to U$ by $\zeta(u)=zu$. It is easy to show that $\zeta$ is $\mathbb{C}G$-homomorphism. By Schur's lemma, there exists $\lambda\in\mathbb{C}$ such that $\zeta(u)=\lambda u$. Then Tr$_U(\zeta)=\lambda\dim(U)$. Now my question asks to calculate Tr$_U(\zeta)$ in terms of $\psi $ and $\chi$. How can I do this?

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By definition the trace of $g$ acting on $U$ is $\psi(g)$. So the trace of $\frac1{|G|}\sum_{g\in G}\chi(g^{-1})g$ acting on $U$ is $$\frac1{|G|}\sum_{g\in G}\chi(g^{-1})\psi(g).$$ This is an integer, the number of copies of the irreducible module $U$ within the module $V$.

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