Math 725 Homework on Singular Value Decomposition Due November 16, 2010

Proposition 1. M is injective if and only if its singular value decomposition M = UDV H has a V that is square and invertible. In this case, MM is invertible and M = (MHM)−1MH . Proof. Let M be an r × c matrix. Suppose that M is injective, so that rank(M) = c because the kernal is zero. Then D is a c × c matrix and so V H is also c× c. V H must already be injective (lest M not be injective), and since it is also onto as part of a singular value decomposition, it is bijective thus invertible (also true because it’s an operator). V is of course also square and has an inverse. Now suppose that V is square an has an inverse. Then V H is also square and has an inverse, and is c × c. Thus D is also c × c and so the rank of M is c. The dimension of the image is the same as the domain, so M must be injective. Thus in this case MM = (UDV )UDV H = V DUUDV H = V DDV H is really the product of four invertible square matricies and so is invertible itself. We see that M = V D−1UH = (V D−1(DH)−1V )(V DU) = (V DDV H)−1MH = (MHM)−1MH