Take the 2-minute tour ×

A similar question was already asked in question titled "Spectra of sums and products in (Banach) algebras [was: Spectrum in Banach Algebra]".

Answer there led me to the following question.

If for elements $a,b$ in Banach algebra $A$ hold that $\operatorname{spec}(\lambda a+\mu b)\subseteq \lambda \operatorname{spec}(a)+\mu \operatorname{spec}(b)$ for every $\lambda, \mu\in \mathbb{C}$, can we say something about the subalgebra $B$ of $A$ generated by $a$ and $b$. Might we conclude something about $B/\operatorname{rad}(B)$?

Where to look for (possible) counterexample of $B/\operatorname{rad}(B)$ being commutative?

Thank you for finding a counterexample. Now I would like to add some additional assumptions. Suppose that for a selfadjoint element $a$ and a unitary element $u$ of $C^∗$-algebra holds that $\operatorname{spec}(a+\lambda a^2)=\operatorname{spec}(a+\lambda ua^2u^∗)$ for every $λ∈\mathbb{C}$. Moreover, if $\lambda \in \mathbb{R}$ elements $a+\lambda a^2$ and $a+\lambda ua^2u^∗$ are unitarily equivalent.

share|improve this question
Since you have already accepted an answer to your original question, perhaps you should post the new version as a separate question (with the c-star algebras tag)? –  Yemon Choi May 31 '11 at 23:39

1 Answer 1

up vote 4 down vote accepted

Here is a counterexample to $B/rad(B)$ being commutative. Let $S\colon l^{2}(\Bbb N)\to l^{2}(\Bbb N)$ be the unilateral shift. Then the spectrum of $S$ is $\overline{\Bbb D}=\{z\in \Bbb C:|z|\leq 1\}.$ Now $S^{*}$ has the same spectrum. Consider the algebra they generate inside $B(l^{2}(\Bbb N)),$ it is a C$^{*}$-algebra, hence its radical is zero. For any $\lambda,\mu \in \Bbb C$ we have $\|\lambda S+\mu S^{*}\|\leq |\lambda|+|\mu|.$ Thus the spectrum of $\lambda S+\mu S^{*}$ is contained in $(|\lambda|+|\mu|)\overline{\Bbb D}.$ It is straigthfoward to show that $(|\lambda|+|\mu|)\overline{\Bbb D}=\lambda\overline{\Bbb D}+\mu \overline{\Bbb D},$ so this does it.

In the case of a $C^{*}$-algebra $A$ one can at say that the closed convex hull of the spectrum. of $a\in A$ is the set of $\phi(a),$ where $\phi$ is a state. Thus you always get that the closed convex hull of $a+b$ is contained in the sum of the closed convex hull of $a$ and $b.$ I don't know if this is true in arbitrary Banach algebras, but maybe there is a similar statement involving polynomial convex hulls.

EDIT: Okay this last statement about states may not be true, as I think you need $a$ to be normal for this statement about the closed convex hull of the spectrum of $a$ to be true.

share|improve this answer
Your edit is correct (in saying that your earlier statement doesn't quite work) - just consider any quasinilpotent. The closed convex hull of the spectrum is always contained in the numerical range (this holds for all unital Banach algebras if one sets up the definitions correctly) but containment may be proper. –  Yemon Choi May 8 '11 at 1:50
Okay, thanks for your correction and clarification. Can one use this to get some sort of statement about the spectrum of $a+b.$ (I'm not familiar with the definition of numerical range as in an arbitrary unital Banach algebra). –  Benjamin Hayes May 8 '11 at 7:31

Your Answer


