46th Putnam 1985

Problem B6

G is a finite group consisting of real n x n matrices with the operation of matrix multiplication. The sum of the traces of the elements of G is zero. Prove that the sum of the elements of G is the zero matrix.



Let the elements of G be A1, A2, ... , Am. Let A = ∑ Ai. Then AiA = A (since G is a group the multiplication just rearranges the order of the terms in the sum). Hence A2 = mA (*).

Now if we allow complex values, A has m eigenvalues (some possibly repeated). Similarly, if we allow complex vectors, then each eigenvalue has an eigenvector. But if Av = λv, then (*) implies that λ2 = mλ, so λ = 0 or m. But the trace of A is the sum of its eigenvalues, so m cannot be an eigenvalue. In other words, det(A - mI) is non-zero and A - mI is invertible. Suppose its inverse is B. We may write (*) as A(A - mI) = 0. Hence 0 = A(A - mI)B = A.



46th Putnam 1985

© John Scholes
7 Jan 2001