For a complex character, $\chi (1)=\dim V$, $\chi$ is a class function, and $\mathrm{\mid }\chi (g)\mathrm{\mid }\le \chi (1)$ with equality exactly at scalars

In any ordered basis of $V$ the matrix of $\rho (1)={\mathrm{id}}_V$ is the identity matrix, whose trace is the number of basis vectors. Hence $\chi (1)=\mathrm{tr}({\mathrm{id}}_V)=\dim V$, which is claim 1.

For the same reason $\rho (g)$ and $\rho (h)$ compose as in the group, so $\chi (gh{g}^{-1})=\mathrm{tr}(\rho (g)\rho (h)\rho (g{)}^{-1})=\mathrm{tr}(\rho (h)\rho (g{)}^{-1}\rho (g))=\mathrm{tr}\rho (h)=\chi (h)$, the middle step being [A3] applied to $A=\rho (g)$ and $B=\rho (h)\rho (g{)}^{-1}$.

By [F2], $g$ has finite order $n$ with $n\mid \mathrm{\mid }G\mathrm{\mid }$, so $\rho (g{)}^n=\rho (g^n)={\mathrm{id}}_V$. Since $\mathbb{C}$ is algebraically closed of characteristic zero, [A1] gives a basis of $V$ in which $\rho (g)$ is diagonal with diagonal entries ${\lambda }_1,\dots ,{\lambda }_d$, where $d=\dim V$; each ${\lambda }_i^n=1$, so each ${\lambda }_i$ is a root of unity.

By step 1.2 the value of $\chi$ does not change under conjugation, so $\chi$ is constant on each conjugacy class and hence is a class function in the sense of [A6], which is claim 2.

In that basis the characteristic polynomial of $\rho (g)$ is ${\prod }_{i=1}^d(x-{\lambda }_i)$, so [A2] gives $\chi (g)=\mathrm{tr}\rho (g)={\sum }_{i=1}^d{\lambda }_i$, a sum of $d=\dim V$ roots of unity, which is claim 3.

Conversely, if $\rho (g)=\lambda {\mathrm{id}}_V$ for a scalar $\lambda$, first consider the degenerate case $V=0$. Then $\chi (g)=0=\chi (1)$, so the equality clause of claim 4 holds. If $V\mathrm{\ne }0$, then the identity $\rho (g{)}^n={\mathrm{id}}_V$ of step 1.3 reads ${\lambda }^n{\mathrm{id}}_V={\mathrm{id}}_V$, and evaluating it on a nonzero vector gives ${\lambda }^n=1$. Thus $\lambda$ is a root of unity and $\mathrm{\mid }\lambda \mathrm{\mid }=1$; then $\chi (g)=\lambda \dim V$ and $\mathrm{\mid }\chi (g)\mathrm{\mid }=\dim V=\chi (1)$. This closes the biconditional in claim 4.

The inverse operator has the inverse eigenvalues, and a root of unity $\lambda$ satisfies ${\lambda }^{-1}=\bar{\lambda }$ because $\mathrm{\mid }\lambda \mathrm{\mid }=1=\lambda \bar{\lambda }$.

Applying [A5] to the eigenvalues of step 1.3 gives $\mathrm{\mid }\chi (g)\mathrm{\mid }=\mathrm{\mid }{\sum }_i{\lambda }_i\mathrm{\mid }\le {\sum }_i\mathrm{\mid }{\lambda }_i\mathrm{\mid }=d=\dim V=\chi (1)$, which is the inequality in claim 4.

If $V=0$, its only endomorphism is scalar, so the equality case is immediate. If $V\mathrm{\ne }0$, equality holds in step 3.1 exactly when the equality clause of [A5] applies: since every $\mathrm{\mid }{\lambda }_i\mathrm{\mid }=1$, all the eigenvalues share one argument, so all the ${\lambda }_i$ are equal to one root of unity $\lambda$. The diagonal form of step 1.3 then shows $\rho (g)=\lambda {\mathrm{id}}_V$, a scalar operator.

Hence, using [A2] for $\rho (g^{-1})=\rho (g{)}^{-1}$ and the additivity of conjugation from [A4], $\chi (g^{-1})={\sum }_i{\lambda }_i^{-1}={\sum }_i\overline{{\lambda }_i}=\overline{{\sum }_i{\lambda }_i}=\overline{\chi (g)}$, which is claim 5.