Finite multiplicative group of matrices with sum of traces equal to zero. Let  $G=\{M_1,M_2,...,M_k\}$ be a finite set such that $ M_i\in M_n(\mathbb R)$ and $(G,\;\cdot\:)$ is a group with matrix multiplication. If $\sum _{i=1}^k \operatorname{tr} (M_i)=0$ then how to prove $\sum _{i=1}^k M_i=0$ ?
 A: Let $M:=\sum_{i=1}^kM_i$. For each $i$, $M\longmapsto M_iM$ induces a bijection of $G$ onto itself. So
$$
M^2=\sum_{i=1}^k\sum_{j=1}^kM_iM_j=\sum_{i=1}^k\sum_{j=1}^kM_j=\sum_{i=1}^kM=kM.
$$
Therefore $E=\frac{1}{k}M$ is idempotent, i.e. $E^2=E$. As is well-known, in characteristic $0$, the rank of an idempotent equals its trace. Just diagonalize to check this fact. So
$$
\mbox{rank}E=\mbox{trace}\frac{1}{k}M=\frac{1}{k}\mbox{trace}M=\frac{1}{k}\sum_{i=1}^k\mbox{trace}M_i=0.
$$
Thus $E=0$, whence $M=0$.
(Not exactly relevant) note: this is a standard trick in group algebras, in particular. See these notes by Alain Valette, p.11 et seq. for related results. In particular, note that Higson and Kasparov proved that the complex group algebra $\mathbb{C}\Gamma$ has no nontrivial idempotents when $\Gamma$ is torsion-free and a-T-menable (i.e. has the Haagerup property), as an application of their work on the Baum-Connes conjecture. Apparently, this is still not accessible by algebraic tools. It is conjectured that this is true for any torsion-free group. That's called Kaplansky's conjecture. 
A: The group acts in a natural way on $\mathbb R^n$, and the hypothesis is that if $\chi$ is character afforded by that representation, then $\chi$ is orthogonal to the character of the trivial module. It follows from standard results on characters that the representation on $\mathbb R^n$ does not have any trivial subrepresentation.
Since the group acts trivially on the image of the map $\sum_iM_i:\mathbb R^n\to\mathbb R^n$, this map must be zero. This is precisely the conclusion you seek.
