fa.functional analysis - spectra of sums and products in (Banach) algebras [was: Spectrum in Banach Algebra]

In general Banach algebras the spectral radius is neither subadditive nor submultiplicative; in particular, neither of the two properties you mention holds.

$2times 2$ (real or complex) matrices should suffice to give examples, so this is not to do with any subtleties of infinite-dimensional algebras.

For example, take $ a= left(begin{matrix} 0 & 1 \\ 0 & 0 end{matrix} right) $.
Note that this is nilpotent, so the only point in the spectrum is zero, and hence $sigma(a)sigma(b)= {0}$ for any other matrix $b$. On the other hand, we can find $b$ for which $ab$ is not nilpotent, so that $sigma(ab)notsubseteq {0}$. A simple choice which works is $b=left(begin{matrix} 0 & 0 \\ 1 & 0 end{matrix} right)$, since then $ab$ is a non-zero projection (=idempotent) and so contains $1$ in its spectrum.

The same pair also works as a counter-example for the "additive question". For since $sigma(a)=sigma(b)={0}$, we have $sigma(a)+sigma(b)={0}$. On the other hand, $a+b$ is a reflection and hence its spectrum is ${-1,1}$

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(edited 11-02-10 for a couple of typos/omissions)

Jonas Meyer points out in comments that one can pose the following converse question: let $A$ be a Banach algebra with identity, and suppose that we have

($*$) $sigma(a+b)subseteqsigma(a)+sigma(b)$ and $sigma(a)sigma(b)subseteqsigma(ab)$ for all $a,bin A$.

Must $A$ be commutative?

As Jonas also pointed out in comments, the answer is in general `no': the algebra of strictly upper-triangular matrices (or, more precisely, the algebra of scalar+strictly upper triangular $mtimes m$ matrices, for some fixed $m$) gives a counterexample, at least when $mgeq 3$.

A more careful version of this argument shows, I think, that if $A$ is a finite-dimensional algebra with identity, such that $A/{rm Rad}(A)$ is commutative, then $A$ will satisfy condition $(*)$. I suspect that the same might be true for any unital Banach algebra that is commutative modulo its radical, i.e. that the finite-dimensional hypothesis is unnecessary; but at present I'm a bit too tired to check this properly.

We could therefore modify the question yet further, and ask if a Banach algebra that satisfies condition $(*)$ must be commutative modulo its radical. The answer turns out to be yes, after a wander down memory lane and a forage on MathSciNet:

MR0461139 (57 #1124)
J. Zemánek. Spectral radius characterizations of commutativity in Banach algebras.
Studia Math. 61 (1977), no. 3, 257--268.

The MR is short and informative enough to give in full, for those without access:

It is standard that the spectral radius is subadditive and submultiplicative on any commutative complex Banach algebra. The author proves that, for a complex Banach algebra $A$, the following three conditions are equivalent: (1) the spectral radius is sub-additive on $A$, (2) the spectral radius is submultiplicative on $A$, (3) $A$ is commutative modulo its radical. Some applications of this result to other problems in Banach algebras are given, along with references to a number of related papers.