On the Product of Two Alternating Matrices

Abstract

article may be also of interest to the readers who are not familiar with the purely algebraic method of specialization which can often be used as a substitute for the usual topological arguments in the setting of finite dimensional vector spaces over R or C. THEOREM. Let A = BC where B and C are alternating n by n matrices over R, let f(A) be the characteristic polynomial of A, and assume that n is even, say n = 2m. Then there exists a monic polynomial p(A) over R of degree m such that f(A) =p(A)2, p(A) = 0. (1) REMARK. When R is a field of characteristic 7 2 it was shown in an Elementary Problem in [2] that A has no simple eigenvalues. In fact the solution of that problem given by 0. P, Lossers shows that the geometric multiplicity of each eigenvalue of A is at least 2. Both of these results are obvious consequences of our theorem. We recall some properties of pfaffians and refer the reader to Lang's book [1] for more details. Let X be an alternating n by n matrix over R. If n is odd then det(X) = 0. If n = 2m is even then det(X) = pf(X)2 where Pf(X), the pfaffian of X, is a homogeneous polynomial of degree m in the entries of X. Consequently we have Pf(X') = Pf(-X) = (- )mPf(X). If we assume in addition that X is invertible then X1 is also alternating. Indeed XX-1 = I,, implies that (X-1)'X'