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Let $A$ be a Banach algebra, and suppose that $a,b\in A$ have spectra that satisfy: $\sigma(a) \subset U$, and $\sigma(b)\subset U$, where $U$ is the open right half-plane of complex numbers with positive real part.

Is is true that $\sigma(ab)$ does not contain any element of the form $-r$ for $r\geq 0$ ?

That's obviously the case when $a$ and $b$ commute, but I can't find a counterexample in the noncommutative case.

Michael
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  • Here's something related though it does not have an answer to your question: http://math.stackexchange.com/questions/19576/spectrum-of-a-product-of-operators-on-a-banach-space And another: http://mathoverflow.net/questions/14246/spectra-of-sums-and-products-in-banach-algebras-was-spectrum-in-banach-algeb – Cameron Williams Jul 22 '14 at 02:13

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Found a counterexample (finally!). Let $A = \left(\begin{array}{rr} 1 & 0 \\ 4 & 1\end{array}\right)$ and $B = \left(\begin{array}{rr} 1 & -1 \\ 0 & 1\end{array}\right)$, then $AB = \left(\begin{array}{rr} 1 & -1 \\ 4 & -3\end{array}\right)$. Then we have that $\sigma(A) = \{1\}$, $\sigma(B) = \{1\}$ but $\sigma(AB) = \{-1\}$.

This was done by trial and error so there is not much rhyme nor reason to my choices for coefficients beyond the fact that I wanted $A$ and $B$ to be triangular (so that I could easily control their eigenvalues).

Cameron Williams
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    You made it look easy; these counterexamples never are. Well done! – Ben Grossmann Jul 22 '14 at 02:58
  • @Omnomnomnom Thanks a lot! Means a bunch coming from you. You should have seen the example I had before. I had something like $A = \left(\begin{array}{rr} 1 & 0 \ 20 & 1\end{array}\right)$ and $B = \left(\begin{array}{rr} 1 & -10 \ 0 & 4\end{array}\right)$. Then I tried to make it nicer by finding better numbers to work with. – Cameron Williams Jul 22 '14 at 02:59
  • Well, whatever works! But to be sure, a real mathematician never uses numbers higher than $8$, as you've exemplified :P – Ben Grossmann Jul 22 '14 at 04:03