Suppose $\Psi:G\rar\UU(E)$ is irreducible and $\dim E=l$. For any $A\in\Hom(E)$ put $f(x)\colon=l\tr(A\Psi(x))$. Prove that $$ A=\frac{l}{|G|}\sum_{x\in G} f(x)\Psi(x)^* $$ and conclude that $\Psi:L_2(G)\rar\Hom(E)$ is onto, i.e. any operator $A$ is a linear combination of the unitary operators $\Psi(x)$, $x\in G$.

1. We need to check that $\Psi(f)=A$, i.e. $$ A =\frac1{|G|}\sum_{x}\tr(A\Psi(x))\Psi(x)^* =\frac1{|G|}\sum_{x}\tr(A\Psi(x)^*)\Psi(x)~. $$ Therefore we need to calculate the matrix entries of the right hand side. Since $\psi_{kj}*\psi_{rs}=\tfrac1l\d_{jr}\psi_{ks}$, we get \begin{eqnarray*} \Big(l\sum_{x}\tr(A\Psi(x))\Psi(x)^*\Big)_{rs} &=&l\sum_x\sum_{j,k}a_{jk}\psi_{kj}(x)\psi_{rs}(x^{-1})\\ &=&l\sum_{j,k}a_{jk}\psi_{kj}^*\psi_{rs}(e) =\sum_{j,k}a_{jk}\d_{jr}\psi_{ks}(e) =\sum_{j,k}a_{jk}\d_{jr}\d_{ks} =a_{rs} \end{eqnarray*} 2. Alternatively, by the great orthogonality theorem (cf. exam): $$ \frac l{|G|}\sum_{g\in G}\Psi(g)\otimes\Psi(g)^*=P_{ex}, $$ and since $\tr_1(A\otimes1)P_{ex}=A$ (cf. exam) we conclude: $$ A =\tr_1(A\otimes1)P_{ex} =\frac l{|G|}\sum_{g\in G}\tr(A\Psi(g))\Psi(g)^* $$