# Is sum of matrices of a faithful representation degenerate?

I noticed that for faithful representations of some groups sum of corresponding matrices is degenerate. E.g. for a representation of $$S_2$$ which permutes basis vectors we have $$\begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix} + \begin{pmatrix} 0 & 1 \\ 1 & 0 \\ \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 1 & 1 \\ \end{pmatrix}$$ Analogously for other permutation groups.

For the representation of cyclic groups by rotations we have sum of matrices to be zero. The same is for the quaternionic group,for the $$M16$$ group and standard representations of permutation groups.

Is it just a coincidence or not?

I know that any finite dimensional representation of a finite group decomposes uniquely to a direct sum of irreducible ones. So the problem can be reduced to irreducible representations only. Also I know that representations of finite groups are equivalent to unitary ones. But all that doesn't help.

Let $$\rho : G \to GL_n(\mathbb{C})$$ be a complex representation of a nontrivial finite group $$G$$, and put \begin{align*} \pi = \frac{1}{\#G}\sum_{g\in G} \rho(g)\in M_n(\mathbb{C}). \end{align*} Then \begin{align*} \gamma \pi(x) = \frac{1}{\#G} \sum_{g\in G} \rho(\gamma g) x = \frac{1}{\#G} \sum_{g\in G} \rho(g) x = \pi(x) \end{align*} for any $$\gamma\in G$$. Thus $$\pi$$ is the projection onto the subspace $$(\mathbb{C}^n)^G\subset \mathbb{C}^n$$ fixed pointwise by $$G$$ (acting by $$\rho$$). If $$G$$ does not act trivially on $$\mathbb{C}^n$$ (e.g., $$\rho$$ is faithful), that subspace is a proper one, and so $$\pi$$ is degenerate.