[proof] **Step 1: Orthogonality of matrix coefficients.** Fix irreducible representations $\rho : G \to \operatorname{GL}(V)$ and $\rho' : G \to \operatorname{GL}(V')$. For any linear map $T : V' \to V$, define the averaging operator \begin{align*} \tilde{T} = \frac{1}{|G|} \sum_{g \in G} \rho(g) \circ T \circ \rho'(g)^{-1}. \end{align*} Then $\tilde{T}$ intertwines $\rho$ and $\rho'$: a direct computation shows $\rho(h) \circ \tilde{T} = \tilde{T} \circ \rho'(h)$ for all $h \in G$. By Schur's lemma, either $\tilde{T} = 0$ (when $\rho \not\cong \rho'$) or $\tilde{T} = \lambda \, \mathrm{id}_V$ (when $\rho \cong \rho'$, and $V = V'$) for some scalar $\lambda$. Taking $T = E_{pq}$ (the matrix unit with a $1$ in position $(p,q)$ and zeros elsewhere), and computing the $(i,j)$ matrix coefficient of $\tilde{T}$ gives the Schur orthogonality relation: \begin{align*} \frac{1}{|G|} \sum_{g \in G} \overline{\rho_{ip}(g)}\, \rho'_{jq}(g) = \begin{cases} \frac{1}{\dim V}\, \delta_{ij}\, \delta_{pq} & \text{if } \rho \cong \rho', \\ 0 & \text{if } \rho \not\cong \rho'. \end{cases} \end{align*} **Step 2: Summing diagonal entries gives character orthogonality.** The character $\chi = \sum_i \rho_{ii}$ is the sum of diagonal matrix coefficients. Setting $p = i$ and $q = j$ and summing over the diagonal ($i = j$, $p = q$) in the Schur relation gives \begin{align*} \langle \chi, \chi' \rangle = \frac{1}{|G|} \sum_{g \in G} \overline{\chi(g)}\, \chi'(g) = \begin{cases} 1 & \text{if } \rho \cong \rho', \\ 0 & \text{otherwise,} \end{cases} \end{align*} establishing Part 1. **Step 3: Spanning via the regular representation.** For the spanning claim, suppose $f \in \mathcal{C}(G)$ satisfies $\langle f, \chi_j \rangle = 0$ for all irreducible characters $\chi_j$. Consider the operator $T_f = \sum_{g \in G} f(g) \rho_j(g)$ for each irreducible $\rho_j$. Since $f$ is a class function, $T_f$ commutes with all $\rho_j(h)$, so by Schur's lemma $T_f = \lambda_j \, \mathrm{id}$. Taking the trace gives $\lambda_j \dim V_j = \sum_g f(g) \chi_j(g) = |G| \langle \bar{f}, \chi_j \rangle = 0$, hence $T_f = 0$ for every irreducible $\rho_j$. The regular representation $\mathbb{C}G$ decomposes as a direct sum of all irreducibles (each appearing with multiplicity equal to its dimension), and $T_f = 0$ on every summand implies $f = 0$ as a function on $G$ (evaluate at $1$). Thus the irreducible characters span $\mathcal{C}(G)$. [/proof]