Article ID | Journal | Published Year | Pages | File Type |
---|---|---|---|---|
4601600 | Linear Algebra and its Applications | 2010 | 10 Pages |
Let A be a contraction on a Hilbert space H. The defect index dA of A is, by definition, the dimension of the closure of the range of I-A∗A. We prove that (1) dAn⩽ndA for all n⩾0, (2) if, in addition, An converges to 0 in the strong operator topology and dA=1, then dAn=n for all finite n,0⩽n⩽dimH, and (3) dA=dA∗ implies dAn=dAn∗ for all n⩾0. The norm-one index kA of A is defined as sup{n⩾0:‖An‖=1}. When dimH=m<∞, a lower bound for kA was obtained before: kA⩾(m/dA)-1. We show that the equality holds if and only if either A is unitary or the eigenvalues of A are all in the open unit disc, dA divides m and dAn=ndA for all n, 1⩽n⩽m/dA. We also consider the defect index of f(A) for a finite Blaschke product f and show that df(A)=dAn, where n is the number of zeros of f.