## 1+A not invertible implies 1+A^n not invertible? [closed]

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?

EDIT: I mean $1+(-1)^{n-1}A(z)^n$ rather than $1+A(z)^n$.

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I don't understand your title : take $A=-1$ and $n=2$. – Laurent Berger Jan 3 2012 at 10:47
Do you want to look at all $n$ beyond a certain lower bound? – Wilberd van der Kallen Jan 3 2012 at 10:51
Sorry, it seems I should rather consider $1+(-1)^{n-1}A^n$. – Alex A Jan 3 2012 at 12:23
Well then if $1+ (-1)^{n-1} A^n$ is invertible, you can write $(1+ (-1)^{n-1} A^n)B = 1$ and then $(1+ A)(1-A+A^2-\cdots \pm A^{n-1})B = 1$ so that $1+A$ is invertible too. – Laurent Berger Jan 3 2012 at 12:28
besides, $A$ may be nilpotent. – Pietro Majer Jan 3 2012 at 16:13
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