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Let $\rho$ afford the character $\chi$, and let $m$ be the order of a group element $g$. Then $\rho (g{)}^m=I$, so its minimal polynomial divides $T^m-1$. Over the complex numbers this polynomial has distinct polynomial roots, since its formal derivative $m{T}^{m-1}$ does not vanish at any of its polynomial roots. Thus $\rho (g)$ is diagonalizable, with eigenvalues ${\lambda }_1,\dots ,{\lambda }_d$ that are roots of unity.

Each ${\lambda }_j$ has absolute value one, so ${\lambda }_j^{-1}=\overline{{\lambda }_j}$. The eigenvalues of $\rho (g^{-1})=\rho (g{)}^{-1}$ are ${\lambda }_j^{-1}$, givingThis is the character inversion identity.

For a finite group, the number of irreducible characters which are real-valued characters equals the number of real conjugacy classes. On the complex vector space of class functions, the linear operator $Tf(g)=f(g^{-1})$ permutes the basis of class indicators by inversion and the basis of irreducible characters by pointwise complex conjugation, using the character inversion identity. The trace of a basis permutation is its number of fixed members. Equality of the trace in the two bases proves the theorem. The linear operator $T$ is complex-linear; pointwise complex conjugation on arbitrary class functions is not.

A character is real-valued when all its values are real numbers, equivalently $\chi =\overline{\chi }$. By the character inversion identity, this is equivalent to $\chi (g)=\chi (g^{-1})$ for every $g$. This condition alone does not assert that the underlying complex representation can be realized by real matrices.