We prove this using Schur's Lemma and properties of the trace.

Step 1: Construct an intertwining map

Let $T: V \to W$ be any linear map. Define: $S = \sum_{g \in G} \rho_W(g) \circ T \circ \rho_V(g^{-1})$

We claim $S$ is a $G$-homomorphism. For any $h \in G$: $\rho_W(h) \circ S = \sum_{g \in G} \rho_W(h) \circ \rho_W(g) \circ T \circ \rho_V(g^{-1})$ $= \sum_{g \in G} \rho_W(hg) \circ T \circ \rho_V((hg)^{-1}) \circ \rho_V(h)$

Since summing over $g \in G$ is the same as summing over $hg \in G$: $= \left(\sum_{g \in G} \rho_W(g) \circ T \circ \rho_V(g^{-1})\right) \circ \rho_V(h) = S \circ \rho_V(h)$

Thus $S \in \text{Hom}_G(V, W)$.

Step 2: Apply Schur's Lemma

By Schur's Lemma:

• If $V \not\cong W$, then $\text{Hom}_G(V, W) = \{0\}$, so $S = 0$
• If $V \cong W$, then $S = \lambda \cdot \text{id}_V$ for some scalar $\lambda$

Step 3: Compute the trace

Taking traces in the defining equation for $S$: $\text{tr}(S) = \sum_{g \in G} \text{tr}(\rho_W(g) \circ T \circ \rho_V(g^{-1}))$

Using the cyclic property of trace: $= \sum_{g \in G} \text{tr}(T \circ \rho_V(g^{-1}) \circ \rho_W(g))$

Step 4: Case $V \not\cong W$

If $V \not\cong W$, then $S = 0$, so $\text{tr}(S) = 0$ for all choices of $T$. In particular, choose $T$ with matrix entries $T_{ij} = \delta_{ik}\delta_{jl}$ (matrix with single 1 in position $(i,j)$). Then: $0 = \sum_{g \in G} \text{tr}(T \circ \rho_V(g^{-1}) \circ \rho_W(g))$

Summing over all such $T$ and using linearity gives: $\sum_{g \in G} \overline{\chi_V(g)} \chi_W(g) = 0$

Therefore $\langle \chi_V, \chi_W \rangle = 0$.

Step 5: Case $V \cong W$

If $V = W$, then $S = \lambda \cdot \text{id}_V$, so $\text{tr}(S) = \lambda \dim V$. We need to find $\lambda$.

Taking $T = \text{id}_V$: $\lambda \dim V = \text{tr}(S) = \sum_{g \in G} \text{tr}(\rho_V(g) \circ \rho_V(g^{-1})) = \sum_{g \in G} \text{tr}(\text{id}_V) = |G| \dim V$

Thus $\lambda = |G|$.

Now compute $\langle \chi_V, \chi_V \rangle$ by choosing appropriate $T$: $\sum_{g \in G} |\chi_V(g)|^2 = \sum_{g \in G} \overline{\chi_V(g)} \chi_V(g)$

By the argument above (using $T = \text{id}_V$ and relating to $\text{tr}(S)$): $\sum_{g \in G} |\chi_V(g)|^2 = |G|$

Therefore $\langle \chi_V, \chi_V \rangle = \frac{1}{|G|} \sum_{g \in G} |\chi_V(g)|^2 = 1$.