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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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(I am assuming that $\mathcal A$ is the closed unital subalgebra of $\mathcal B$ generated by $A$.)

Here I will not be worried about what background is expected in the problem or potential circular reasoning (as in, the tools used may depend somewhere on the fact to be proved).

The fundamental fact at the heart of continuous functional calculus says that there is a $*$-isomorphism $\Gamma:\mathcal A\to C(\sigma(A))$, such that $\Gamma(1)$ is the constant function $x\mapsto 1$, and $\Gamma(A)$ is the function $x\mapsto x$. (The operations on $C(\sigma(A))$ are pointwise addition, multiplication, and complex conjugation.) If $A$ is invertible, and $f:\sigma(A)\to\mathbb C$ is defined by $f(x)=1/x$, then $f\in C(\sigma(A))$, and $f=\Gamma(A)^{-1}$. Therefore $\Gamma^{-1}(f)=A^{-1}$ is in $\mathcal A$.

An alternative approach is to use spectral permanence for Banach algebras. Let $\sigma_{\mathcal A}(A)$ and $\sigma_{\mathcal B}(A)$ denote the spectra of $A$ relative to $\mathcal{A}$ and $\mathcal B$, respectively. It is clear that $\sigma_{\mathcal B}(A)\subseteq \sigma_{\mathcal A}(A)$, because invertibility in $\mathcal A$ implies invertibility in $\mathcal B$. The spectral permanence theorem implies that the boundary $\partial\sigma_{\mathcal A}(A)$ in the complex plane is a subset of $\sigma_{\mathcal B}(A)$. Since $A$ is a self-adjoint element of the C*-algebra $\mathcal A$, $\sigma_{\mathcal A}(A)$ is a closed subset of the real line, and therefore $\partial\sigma_{\mathcal A}(A)=\sigma_{\mathcal A}(A)$. Therefore, $\sigma_{\mathcal A}(A)=\sigma_{\mathcal B}(A)$. In particular, $0$ is in $\sigma_{\mathcal A}(A)$ if and only if $0$ is in $\sigma_{\mathcal B}(A)$.

In the case where $A$ is a self-adjoint finite matrix, the spectral theorem yields $A=\sum_k \lambda_k P_k$, where $\lambda_1,\lambda_2,\ldots$ are the distinct eigenvalues of $A$, and $P_k$ is the orthogonal projection onto the eigenspace for the eigenvalue $\lambda_k$. The $P_k$s are self-adjoint idempotents satisfying $P_kP_j=0$ for $j\neq k$. If $A$ is invertible, then $\lambda_k\neq 0$ for each $k$, and there is an interpolating polynomial $f$ such that $f(\lambda_k)=1/\lambda_k$ for each $k$. You can check that $f(A)=A^{-1}$, and since $f$ is a polynomial, $f(A)$ is in $\mathcal A$.