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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 $a^{-1}$?


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Perhaps it would be worthwhile to think about the positive case, then use the identity $a^{-1}=(a^*a)^{-1}a^*$. – Jonas Meyer Dec 13 '12 at 4:28
up vote 2 down vote accepted

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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Sorry for the delay in replying. I got your idea, but in my opinion the second polynomial is more canonical since it only involves $t$ and hence would work in any algebra containing $(a-\lambda)$, even when the algebra does not contain $1$. Although monic polynomials are nice in general, I do not really think it is a big deal to have non-monic polynomials here since most of the time the we are talking algebras over a field. – Hui Yu Jan 18 '13 at 16:20
But to even write "$a-\lambda$" you need the identity. So if you consider $a-\lambda$ in an algebra without identity, that algebra would also fail to contain $a$; not to mention that $(a-\lambda)^{-1}$ will not make sense. – Martin Argerami Jan 18 '13 at 16:25
I did not make my point clear enough. It is okay to have $1$ in a larger algebra but I think it would be nicer if we could express the inverse in the smaller algebra. – Hui Yu Jan 18 '13 at 16:38
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). – Martin Argerami Jan 18 '13 at 16:57

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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Why does the series converge? – Hui Yu Dec 8 '12 at 2:14
This answer is incorrect. If $a$ is invertible in a unital $C^*$-algebra $A$, then it is true that $a$ is also invertible in the unital $C^*$ subalgebra of $A$ generated by $\{a,a^*\}$, but $a$ need not be invertible in the unital Banach subsubalgebra genrated by $a$. Such examples rule out the possibility of such a power series working in principle. One such example is the function $z\mapsto z$ in $C(\mathbb T)$. A more direct problem with the answer is that that series diverges when $|\lambda|$ is less than the spectral radius of $a$. – Jonas Meyer Dec 13 '12 at 4:22
Of course, one has a lot of freedom to make selections so that you get a series that does converge. – Hurkyl Dec 16 '12 at 18:11
The critique is correct; I was sloppy. What I should have wrote was: if $|\lambda| > \|a\|$, then $\lambda - a$ is invertible and... – Travis Feb 16 '13 at 17:49

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