The Moore-Penrose inverse of a factorization
Let \(A\) be a von Neumann regular matrix (i.e., \(AXA= A\) is solvable), and let \(P\), \(Q\) be given. Assume that there exist \(P'\) and \(Q'\) such that \(P'PA= A= AQQ'\). Necessary and sufficient conditions are given in order to \(PAQ\) be Moore-Penrose invertible, generalizing previous results.
scientific article; zbMATH DE number 1970621
scientific article