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I have some trouble proving the following statement:

Let $A$ be a self-adjoint element of a $C^*$-algebra $\mathcal{B}$ and let $\mathcal{A}$ denote the unital subalgebra of $\mathcal{B}$ that is generated by 1 and $A$.

If $A$ is invertible in $\mathcal{B}$, then it is invertible in $\mathcal{A}$.

I got a hint to consider the subalgebra $\mathcal{A}_0$ of $\mathcal{B}$ generated by adding $A^{-1}$ to $\mathcal{A}$ and then use functional calculus with the map $z \mapsto \frac{1}{z}$ but I do not see quite how to do this.

Thanks for any help!

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    What have you learned about functional calculus? There are a number of ways to approach it, and it would help to know what background you are allowed to assume in your solution.2012-05-12
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    Another approach is to use use spectral permanence for Banach algebras: http://planetmath.org/encyclopedia/SpectralPermanenceTheorem.html2012-05-12
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    @JonasMeyer OP is last seen on the site the day he asked the question. I really have had no exposure to functional calculus, but can read on it. I would be interested in understanding the problem and the solution conceptually. Suppose a quantum-mechanical system has finite state space. Then $A$ must be a finite-dimensional Hermitian matrix. Can the statement to be proved be understood on this setting, or am I getting it all wrong? Thanks!2012-05-16
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    @Sasha: It isn't really possible to tell what the terms of the original question were to give a good answer. For your particular problem, functional calculus takes less ground work. If $A$ is a finite dimensional invertible Hermitian matrix, then $A=\sum_k \lambda_k P_k$, where $\lambda_1,\lambda_2,\ldots$ are the (nonzero) eigenvalues of $A$, and $P_1,P_2,\ldots$ are orthogonal projections onto the eigenspaces. Let $f(x)$ be a polynomial such that $f(\lambda_k)=1/\lambda_k$ for each $k$ (e.g., a Lagrange interpolating polynomial). Then $f(A)=A^{-1}$ is in the algebra generated by $A$.2012-05-17
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    @JonasMeyer Thank for for the explanation, I should have seen that. Explicit polynomial $f(A)$ follows from the [Cayley-Hamilton theorem](http://en.wikipedia.org/wiki/Cayley%E2%80%93Hamilton_theorem). If you post these comments as an answer, I will award the bounty to you.2012-05-17

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