An Exploratory Approach to Kaplansky's Lemma Leads to a Generalized Resultant

Abstract

This result actually holds in arbitrary dimensions and the extension from finite to infinite dimensions is fairly straightforward [6, p. 64]. Aupetit [1] gives an elegant proof when F= C: (valid for any dimensions) using Liouville's theorem from complex analysis. His proof is reproduced, slightly polished, in Prasolov's recent intriguing survey of linear algebra emphasizing modern proofs of classical results [5, p. 46]. Henceforth, we assume a finite-dimensional space, so T has a minimal polynomial of degree n, say. The maximum k can be is n-1 and so we may as well assume k = n-1. Prasolov then gives a slick proof, but only for infinite fields, using minimal vector-annihilating polynomials and their subspaces [5, 13.1.2, p. 73]. It relies on the result: a vector space cannot be expressed as a finite set-theoretic union of proper subspaces, and this of course requires the underlying field to be infinite. For arbitrary fields, one can readily construct a suitable vector v using the theory of the rational canonical form [4, p. 198]. This says that T acts as a block diagonal matrix, the diagonal blocks being the companion matrices of the invariant factors d1(x) l d2(x)l . . . Ids(x) of T. The vector v consisting of O's except for a 1 in the first position corresponding to the ds(x) block has minimal annihilating polynomial ds(x). This vector v fits the bill since ds(x) is also the minimal polynomial of T. But suppose out of curiosity we look for a suitable v using the version of the rational canonical form involving elementary divisors [3, p. 262]. Now T acts as a block diagonal matrix whose diagonal blocks are the companion matrices of the elementary divisors {pi(X)e'l} of T. Here the polynomials pi(x) are irreducible and distinct. Let fi(x) = pi(X)r' 1 < i < m, be the elementary divisors of highest degree, so that flim= 1 ti(x) is the minimal polynomial of T with degree n = Lim=ldeg fi. In fact, it is sufficient to assume these are the only elementary divisors; after dealing with this case, v can be augmented with zeros to cover the general case. Thus we assume T is the n-square matrix diag(A1, A2, . . ., Am) where Ai is the companion matrix of fi, 1 < i < m. Let (vlT v2T vmT)T be the corresponding partition of our desired (column) vector v. For vi, let's try the simplest possible nonzero vector: just one nonzero entry a 1, say which should be the first entry