A question on finite matrix groups

December 29th, 2007 ~ Posted in: Mathematics

I came across this problem from a Putnam exam: Let G = \{M_i\}_{i=1}^r be a finite group of matrices in M_n(\R) such that \sum_{i=1}^r \text{tr}(M_i) = 0. Show that \sum_{i=1}^r M_i = 0 .

I haven’t been able to solve it in three days of sporadic efforts, but I’m slowly getting a handle on this: I’ve convinced myself that the M_i are diagonalizable with only \pm 1 in their spectrum.

Solution Sweet! Lieven from Neverendingbooks came up with this neat proof: Note that P = \frac{1}{\# G} \sum_i M_i is a projection onto its eigenspace– this follows from the basic group theoretical fact that M_k \sum_i M_i = \sum_i M_i. We know that the dimension of the range space of a projection operator is the trace of the operator, so voila!

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