1+A not invertible implies 1+A^n not invertible? - MathOverflow [closed] 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 1+A not invertible implies 1+A^n 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>