# let$G=\{M_1,M_2,…,M_k\}$ be a finite group if $\sum _{i=0}^k \operatorname{tr} (M_i)=0$ then how prove $\sum _{i=0}^k M_i=0$

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 group with operations of matrix multiplication

If $\sum _{i=1}^k \operatorname{tr} (M_i)=0$ then how to prove $$\sum _{i=1}^k M_i=0$$

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What have you tried? Or are you just stuck? –  Pedro Tamaroff Apr 29 '13 at 18:18
@Peter Tamaroff: i try to prove A=$\sum _{i=0}^k M_i$ is nil-potent or idempotent by $\{M_1,M_2,...,M_k\}=G=\{M_iM_1,M_iM_2,...,M_iM_k\}$ but i couldn't do it –  Maisam Hedyelloo Apr 29 '13 at 18:23
@MaisamHedyelloo I had not seen your comment. This is the right idea for an elementary argument. You were almost there! All you missed was my first remark below. For a fixed $i$, $M_iG=G$. –  1015 Apr 29 '13 at 18:48
@julien:hi i complete my idea like you but it has small different thanks so much –  Maisam Hedyelloo Apr 29 '13 at 18:53
Your idea was great, especially if you came up with it on your own. I was aware of the trick, so I have no merit. –  1015 Apr 29 '13 at 19:05

## 2 Answers

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.

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The results mentioned in your note of course do not apply to the case at hand, though, which has lots of torsion and, in fact, lots of non-trivial idempotents :-) –  Mariano Suárez-Alvarez Apr 29 '13 at 22:40
@MarianoSuárez-Alvarez Right. But I think it's in these notes that I've figured out for the first time that such an average yields an idempotent. So I thought I'd mention it. I agree it is not really relevant to this particular problem... –  1015 Apr 30 '13 at 2:30

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.

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Of course, all this is incomprehensible if one does not know the basics of character theory :-) Let it count as motivation to learn it! –  Mariano Suárez-Alvarez Apr 29 '13 at 18:27