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For a unital $C^*$-algebra $\mathcal{A}$ the spectral permanence gives \begin{equation} \sigma_{\mathcal{B}}(a)=\sigma_{\mathcal{A}}(a) \end{equation} for any unital $C^*$-subalgebra $\mathcal{B}$.

It is natural to look at the smallest such subalgebras, namely, the $C^*$-subalgebra generated by $1,a$ and $a^*$. Then the permanence says if $\lambda-a$ is invertible, then $\lambda-a$ is in the closed linear span of products of $1,a$ and $a^*$ (although order of multiplications matters here and it is not actually a polynomial).

I am wondering whether there is some canonical way to construct these 'polynomials'. That is, given $a\in\mathcal{A}$ invertible, how can one find explicitly the linear span of products of $1,a$ and $a^*$ that converges to $a^{-1}$?

Thanks!

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    Perhaps it would be worthwhile to think about the positive case, then use the identity $a^{-1}=(a^*a)^{-1}a^*$.2012-12-13

2 Answers 2

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My feeling (not formally justified) is that that there is no canonical choice. Take a look at the simplest example: let $ a=\begin{bmatrix}2&0\\0&3\end{bmatrix}, $ $\lambda=1$. So $a$ is selfadjoint, and of course $ (a-\lambda)^{-1}=\begin{bmatrix}1&0\\0&1/2\end{bmatrix}=p(a-\lambda) $ for an appropriate polynomial. Now, what is the canonical polynomial that takes $1,2$ to $1,1/2$? Let us assume we want the minimum degree possible (i.e. $2$); this is already arbitrary. You could ask that the polynomial be monic, and in this case $ p(t)=t^2-\frac72\,t+\frac72; $ or you could want $p(0)=0$, in which case $ p(t)=-\frac34\,t^2+\frac74\,t. $ I don't really see a reason that makes one more canonical than the other.

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    I don't get it. If you have an operator ($a-\lambda$ in this case) that is invertible in some C$^*$-algebra, then it is invertible in any C$^*$-subalgebra that contains it (because you can obtain $a-\lambda$ from $(a-\lambda)^{-1}$ via functional calculus).2013-01-18
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To answer the question in your title:

If $\lambda -a$ is invertible you can write its inverse as a power series in $a$:

$\frac{1}{\lambda - a} = \frac{1}{\lambda(1 - \frac{a}{\lambda})} = \sum_{n = 0}^\infty \frac{a^n}{\lambda^{n+1}}.$

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    The critique is correct; I was sloppy. What I should have wrote was: if |\lambda| > \|a\|, then $\lambda - a$ is invertible and...2013-02-16