[proofplan] The proof uses only the homomorphism property of the representation and the invariance of trace under conjugation. First we rewrite $\rho(hgh^{-1})$ as a conjugate of $\rho(g)$ by the invertible [linear map](/page/Linear%20Map) $\rho(h)$. Then we prove directly, in a basis, that conjugate linear maps have the same trace. [/proofplan]

[step:Rewrite the represented conjugate as a conjugate linear map] Let $g,h \in G$ be arbitrary. Since $\rho: G \to GL(V)$ is a [group homomorphism](/page/Group%20Homomorphism), we have \begin{align*} \rho(hgh^{-1}) &= \rho(h)\rho(g)\rho(h^{-1}) \\ &= \rho(h)\rho(g)\rho(h)^{-1}. \end{align*} The second equality follows because $\rho(h^{-1})=\rho(h)^{-1}$ for a group homomorphism into $GL(V)$. Therefore \begin{align*} \chi_\rho(hgh^{-1}) = \operatorname{tr}\!\left(\rho(h)\rho(g)\rho(h)^{-1}\right). \end{align*} [/step]

[step:Show that conjugate linear maps have the same trace]Let $n := \dim_{\mathbb{C}} V$, and choose a basis of $V$. Let $A \in M_n(\mathbb{C})$ be the matrix of $\rho(g)$ in this basis, and let $P \in GL_n(\mathbb{C})$ be the matrix of $\rho(h)$ in the same basis. Then $PAP^{-1}$ is the matrix of $\rho(h)\rho(g)\rho(h)^{-1}$. We prove that $\operatorname{tr}(PAP^{-1})=\operatorname{tr}(A)$. Write $Q := P^{-1}$, so $PQ=QP=I_n$. Using the definition of matrix multiplication and the trace, \begin{align*} \operatorname{tr}(P A Q) &= \sum_{i=1}^{n} (P A Q)_{ii} \\ &= \sum_{i=1}^{n}\sum_{j=1}^{n}\sum_{k=1}^{n} P_{ij}A_{jk}Q_{ki} \\ &= \sum_{j=1}^{n}\sum_{k=1}^{n} A_{jk}\left(\sum_{i=1}^{n} Q_{ki}P_{ij}\right) \\ &= \sum_{j=1}^{n}\sum_{k=1}^{n} A_{jk}(QP)_{kj} \\ &= \sum_{j=1}^{n}\sum_{k=1}^{n} A_{jk}(I_n)_{kj} \\ &= \sum_{j=1}^{n} A_{jj} \\ &= \operatorname{tr}(A). \end{align*} Hence \begin{align*} \operatorname{tr}\!\left(\rho(h)\rho(g)\rho(h)^{-1}\right) = \operatorname{tr}(\rho(g)). \end{align*}[/step]

[guided]We need to justify the only linear-algebra fact used in the proof: conjugating a linear map does not change its trace. Choose a basis of the finite-dimensional complex [vector space](/page/Vector%20Space) $V$, and let $n := \dim_{\mathbb{C}} V$. In that basis, let $A \in M_n(\mathbb{C})$ denote the matrix of $\rho(g)$, and let $P \in GL_n(\mathbb{C})$ denote the matrix of $\rho(h)$. Since $\rho(h)$ is invertible, $P$ is invertible; define $Q := P^{-1}$. The matrix of the conjugated linear map $\rho(h)\rho(g)\rho(h)^{-1}$ is then $P A Q$. We compute its trace directly from the entries. By definition of trace and matrix multiplication, \begin{align*} \operatorname{tr}(P A Q) &= \sum_{i=1}^{n} (P A Q)_{ii} \\ &= \sum_{i=1}^{n}\sum_{j=1}^{n}\sum_{k=1}^{n} P_{ij}A_{jk}Q_{ki}. \end{align*} Now regroup the finite sum by collecting the factors involving $P$ and $Q$: \begin{align*} \operatorname{tr}(P A Q) &= \sum_{j=1}^{n}\sum_{k=1}^{n} A_{jk}\left(\sum_{i=1}^{n} Q_{ki}P_{ij}\right). \end{align*} The inner sum is the $(k,j)$-entry of the product $QP$. Since $Q=P^{-1}$, we have $QP=I_n$, and therefore \begin{align*} \operatorname{tr}(P A Q) &= \sum_{j=1}^{n}\sum_{k=1}^{n} A_{jk}(I_n)_{kj}. \end{align*} The identity matrix satisfies $(I_n)_{kj}=0$ when $k \ne j$ and $(I_n)_{jj}=1$, so the double sum collapses to the diagonal entries of $A$: \begin{align*} \operatorname{tr}(P A Q) &= \sum_{j=1}^{n} A_{jj} = \operatorname{tr}(A). \end{align*} Thus conjugate linear maps have the same trace, so \begin{align*} \operatorname{tr}\!\left(\rho(h)\rho(g)\rho(h)^{-1}\right) = \operatorname{tr}(\rho(g)). \end{align*}[/guided]

[step:Conclude that the character is constant on conjugacy classes] Combining the previous two steps gives \begin{align*} \chi_\rho(hgh^{-1}) &= \operatorname{tr}\!\left(\rho(h)\rho(g)\rho(h)^{-1}\right) \\ &= \operatorname{tr}(\rho(g)) \\ &= \chi_\rho(g). \end{align*} Since $g,h \in G$ were arbitrary, $\chi_\rho$ is constant on every [conjugacy class](/page/Conjugacy%20Class) of $G$. Hence $\chi_\rho$ is a class function. [/step]