Let $A$ be a unital complex Banach algebra and $a,b\in A$. Define

$r(a) = \sup_{\lambda \in \sigma(a)} |\lambda|$

where $\sigma(a)$ denotes the spectrum of $a$. Note that $\sigma (ab) \setminus \{0\} = \sigma (ba) \setminus \{0\}$. Hence if both $\sigma (ab)$ and $\sigma (ba)$ contain non-zero elements then $r(ab) = r(ba)$.

Is it possible to find an example of $A$ such that there exist $a,b \in A$ with $\sigma(ab) = \{0\}$ and $\sigma (ba) \neq \{0\}$? That is, $ab - \lambda$ invertible for all $\lambda \neq 0$ but $ba -\lambda$ not invertible for at least one $\lambda \neq 0$?

  • $\begingroup$ You just said this can't happen: if $\sigma(ab) = \{0\}$ then $\sigma(ba) \backslash \{0\} = \sigma(ab) \backslash \{0\} = \emptyset$. $\endgroup$ Feb 24, 2014 at 16:08

1 Answer 1


No. If $\sigma(ab)=\{ 0\}$, then $\sigma(ba)\subset\{ 0\}$ since, as you mentioned it, $\sigma(ba)\setminus\{ 0\}=\sigma(ab)\setminus\{0\}=\emptyset$. But $\sigma(ba)$ is nonempty, so you must have $\sigma(ba)=\{ 0\}$.

What is possible is that $\sigma(ab)$ contains $0$ and $\sigma(ba)$ does not. For example, take $A=\mathcal L(\ell^2)$, the algebra of all bounded operators on $\ell^2=\ell^2(\mathbb N)$. Let $S$ be the "forward shift" on $\ell^2$, $$S(x_1,x_2, x_3,\dots )=(0,x_1,x_2, x_3,\dots )$$ and let $B$ be the "backward shift", $$B(x_1,x_2,x_3,\dots )=(x_2,x_3,\dots )\, . $$ Then $BS=Id$, so $0\not\in\sigma(BS)$. But $$SB(x_1,x_2,x_3\dots )=(0,x_2,x_3,\dots )\, ,$$ so $SB$ is not one-to-one (and hence $0\in\sigma(SB)$ since $\ker(SB)$ contains the vector $e_1:=(1,0,0,0,\dots )$.

  • $\begingroup$ Thanks, what is $\ell^2(\mathbb N N)$? I have not seen this notation before. I suppose it's the same as $\ell^2 (\mathbb N)$? $\endgroup$
    – Student
    Feb 24, 2014 at 18:40
  • $\begingroup$ Yes, this was a typo... $\endgroup$
    – Etienne
    Feb 24, 2014 at 20:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.