# On the spectral radius $r(ab)$

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 $\lambda \neq 0$?

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

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 $e_1:=(1,0,0,0,\dots )$.
• Thanks, what is $\ell^2(\mathbb N N)$? I have not seen this notation before. I suppose it's the same as $\ell^2 (\mathbb N)$? Feb 24, 2014 at 18:40