Bourbaki

Exercise 24, § 8

Algebra, III, 643

Let XiX_i (1≤i≤m1 \leq i \leq m) be square matrices of the same order nn over the commutative ring A. Show that if mm is even then (Exercise 23)

Tr⁡[X1X2…Xm]=0 \operatorname{Tr}[X_1X_2 \ldots X_m] = 0

and if mm is odd

Tr⁡[X1X2…Xm]=m⋅Tr⁡(Xm[X1X2…Xm−1]). \operatorname{Tr}[X_1 X_2 \ldots X_m] = m \cdot \operatorname{Tr}(X_m[X_1 X_2 \ldots X_{m-1}]) .

(If λ\lambda is the cyclic permutation (1,2,…,m)(1, 2, \ldots, m), CC the cyclic subgroup of SmS_m generated by λ\lambda and HH the subgroup of SmS_m consisting of the permutations leaving mm invariant, note that every permutation in SmS_m can be written uniquely as στ\sigma \tau with σ∈H\sigma \in H and τ∈C\tau \in C; then use the fact that the trace of a product of matrices is invariant under cyclic permutation of the factors.)

Solution

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Transcribed from the printed volume. Not translated.