most recent 30 from http://mathoverflow.net 2013-05-19T18:03:32Z http://mathoverflow.net/feeds/question/84796 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/84796/1a-not-invertible-implies-1an-not-invertible Alex A 2012-01-03T10:40:18Z 2012-01-03T12:27:50Z

<p>Let $A(z)$ be a compact operator on a Hilbert space, depending on a complex parameter $z$. I want to count the number of points where $1+A(z)$ is not invertible and therefore I want to count zeros of $\det (1+A(z))$. Unfortunately $A(z)$ is not of trace class so the determinant does not make sense. However, for some positive integer power $n$ the operator $A(z)^n$ is of trace class. Can I argue that the points $z$ where $1+A(z)^n$ is not invertible include those where $1+A(z)$ is not invertible? </p> <p>EDIT: I mean $1+(-1)^{n-1}A(z)^n$ rather than $1+A(z)^n$. </p>