Partial Invertibility is Impossible: 10487

Abstract

Solution Iby the National SecurityAgency Problems Group, FortMeade, MD. For any matrix B over R we have (xB) C (x). Thus, when A is invertible, (x) = (xAA-1) C (xA) C (x), implying (x) = (xA). Since R is a commutative ring with 1, Cramer's rule implies that A is invertible if and only if its determinant is a unit in R. Thus, when A is not invertible, there is a maximal ideal M of R that contains the determinant of A. Furthermore, A induces a singular linear transformation of kn to itself, where k = R/M. Hence, there is a vector x over R that is nonzero modulo M (i.e., not all of its entries are in M) for which xA is zero modulo M. But then (xA) C M, while (x) is not a subset of M. Therefore (xA) :A (x).