Given a finite linear group of order $n$, why the sum of their matrices add up to $nI$? I'm reading Steinberg's Representation theory of finite groups. In the second and third exercise here:



I have made some experiments like thinking about the representation 
$$\Bbb{Z}/3\Bbb{Z}\to \{\begin{bmatrix}
1 &0 \\ 
 0&1 
\end{bmatrix},\begin{bmatrix}
 1&1 \\ 
 0&1 
\end{bmatrix},\begin{bmatrix}
 1&-1 \\ 
 0&1 
\end{bmatrix}  \} $$
And noticed that they indeed add up to $\begin{bmatrix}
 3&0 \\ 
 0& 3
\end{bmatrix}$
But I can't explain why. Can you give me a hint? It is perhaps something very silly I may be missing.
 A: It doesn't hold in general. For example, consider the cyclic group of order $n$, with a trivial 1D representation over $\mathbb{C}$ (by $n$-th roots of unity), or, if you prefer, with a 2D representation over $\mathbb{R}$ (by rotation matrices). [In both cases we have $V^G=\{0\}$.]
A: Put a $G$-invariant inner product on $V$ using Weyl's unitary trick. That is, pick any inner product, then "symmetrize" it, i.e. average it over its $G$-orbit, so $\langle u,v\rangle_G=\sum_{g\in G}\langle \rho(g)u,\rho(g)v\rangle$.
Let $W$ be the orthogonal complement of $V^G$. So $V=V^G\oplus W$ is a direct sum of representations. Write $\Sigma$ for the sum $\sum_{g\in G}\rho(g)$ of linear transformations $\rho(g)$ corresponding to group elements $g$.
Because this is a direct sum of representations, every $\rho(g)$ must be block-diagonal. On $V^G$ each $\rho(g)$ acts trivially, so $\sum$ must act as $nI$ where $n=|G|$. However if $w\in W$, then $\Sigma w$ is a fixed point, so it must be within $V^G$, a contradiction unless $\Sigma w=0$. Therefore, it has the block-diagonal form
$$ \Sigma = \begin{bmatrix} nI & 0 \\ 0 & 0 \end{bmatrix}. $$
