Schur's Lemma and Character Orthogonality

Step 1 (Schur's Lemma, kernel and image). Let $f: V \to W$ be $G$-equivariant, i.e. $f(\rho_V(g)v) = \rho_W(g)f(v)$ for all $g,v$. Then $\ker f \subseteq V$ is a $G$-invariant subspace (if $f(v)=0$ then $f(\rho_V(g)v) = \rho_W(g)f(v) = 0$), and since $V$ is irreducible, $\ker f$ is either $0$ or all of $V$.

Step 2 (conclude $f=0$ or $f$ injective, then surjective). If $\ker f = V$, then $f = 0$. Otherwise $\ker f = 0$, so $f$ is injective; the image $f(V) \subseteq W$ is also $G$-invariant (by equivariance), and since $W$ is irreducible and $f(V) \ne 0$ (as $f\ne0$), $f(V) = W$, making $f$ bijective, hence an isomorphism.

Step 3 ($V=W$ forces $f=\lambda I$). Since $\mathbb{C}$ is algebraically closed, $f$ has an eigenvalue $\lambda$ with eigenspace $E_\lambda = \ker(f-\lambda I) \ne 0$. Because $f$ is $G$-equivariant, so is $f - \lambda I$, so $E_\lambda$ is a nonzero $G$-invariant subspace of the irreducible $V$, forcing $E_\lambda = V$; hence $f = \lambda I$.

Step 4 (orthogonality from averaging). Fix bases and let $f: V_j \to V_i$ be an arbitrary linear map; form the equivariant map $\bar f = \dfrac{1}{|G|}\sum_{g} \rho_i(g) f \rho_j(g)^{-1}$ exactly as in Maschke's theorem. By Steps 1–3, $\bar f = 0$ if $V_i \not\cong V_j$, and $\bar f = \lambda I$ (with $\lambda$ computable via trace) if $V_i = V_j$. Choosing $f$ to be elementary matrix units and comparing matrix entries of $\bar f$ on both sides of this identity yields, after summing over the diagonal to extract traces, exactly $\dfrac{1}{|G|}\sum_{g \in G} \chi_i(g)\overline{\chi_j(g)} = \delta_{ij}$.