# On the spectrum of the sum of two commuting elements in a Banach algebra

Original: Soit A une algèbre de Banach unitaire et a et b deux éléments tels que $$a*b=b*a$$. Pourquoi $$σ (a+b) \subset σ(a)+σ(b)$$. Et qu’elle est la relation entre σ (a*b) et σ(a) et σ(b)?

Translation: Let $$A$$ be a unitary Banach algebra and the elements $$a$$ and $$b$$ such that $$ab = ba$$. Why is the spectrum of $$a + b$$ contained in the sum of the spectra of $$a$$ and $$b$$? What relations do we have between the spectrum of $$ab$$ and the spectra of $$a$$ and $$b$$?

• For C*-algebras one can use the functional calculus to see this, at least for normal $a,b$ ... Commented Mar 9, 2013 at 12:48
• Related, in case you thought about dropping the commutation assumption. Commented Mar 13, 2013 at 13:34
• Commented Aug 1, 2013 at 4:06

Two noncommuting quasinilpotent elements $a,b$ can give a non quasinilpotent sum $a+b$ and product $ab$. E.g., in $M_2(\mathbb{C})$: $$a=\pmatrix{0&1\\0&0}\quad b=\pmatrix{0&0\\1&0}\qquad \sigma(a)=\sigma(b)=\{0\}\quad\sigma(a+b)=\{\pm 1\}\quad \sigma(ab)=\{0,1\}.$$ But, indeed:

In a unital Banach algebra, for every commuting elements $ab=ba$, the spectra of $a+b$ and $ab$ satisfy $$\sigma(a+b)\subseteq \sigma(a)+\sigma(b)\qquad \sigma(ab)\subseteq \sigma(a)\sigma(b).$$

Sketch: 1) We do it in the commutative case 2) We restrict to the commutative case by considering the bicommutant of $\mathbb{C}[x,y]$.

Proof: assume that $A$ is commutative first. Then by Gelfand, for every $x\in A$, we have $$\sigma(x)=\{\phi(x)\;;\;\phi\in \hat{A}\}$$ where $\hat{A}$ denotes the set of characters (nonzero algebra homomorphisms from $A$ to $\mathbb{C}$).

It follows readily that for all $x,y\in A$: $$\sigma(x+y)\subseteq\sigma(x)+\sigma(y)\qquad \mbox{and}\qquad\sigma(xy)\subseteq\sigma(x)\sigma(y).$$

Now back to the general case where $A$ is not assumed to be commutative, take $x,y\in A$ such that $xy=yx$.

Let $B=\mathbb{C}[x,y]$ be the unital algebra generated by $x,y$. And denote $B'$ its commutant, that is the set of all $z$ in $A$ which commute with every element of $B$ (equivalently, with $x$ and $y$). Note that for any sets $S\subseteq T$, we have $T'\subseteq S'$. Also $S$ is commutative if and only if $S\subseteq S'$.

Since $B$ is commutative, we have: $$B\subseteq B'\qquad \Rightarrow \qquad B''\subseteq B'\qquad\Rightarrow\qquad B''\subseteq (B'')'.$$ So $B''$ is commutative. Moreover, one checks easily that it is closed (hence a unital Banach algebra) and contains $B$.

We need to compare the spectra relative to $B''$ and to $A$ for elements of $B''$. So take $z$ in $B''$. Of course, if $z$ is invertible in $B''$, it is invertible in $A$. Now conversely, assume that $z$ is invertible in $A$. Now for all $u$ in $B'$, we have $zu=uz$ so $uz^{-1}=z^{-1}u$. Hence $z^{-1}$ belongs to $B''$. Therefore $$z\;\mbox{invertible in } A\quad\Leftrightarrow\quad z\;\mbox{invertible in}\; B''$$ for all $z\in B''$. Applying this to $z=b-\lambda 1$, we see that $$\sigma_A(b)=\sigma_{B''}(b)\qquad\forall b\in B''.$$ This shows that we can restrict to the commutative unital Banach algebra $B''$ and apply the above commutative case to $x,y,x+y,xy$ which all belong to $B$, hence to $B''$. $\Box$

Note: to see why this result is natural, and why the inclusions are not equalities in general, consider the case of $A=M_n(\mathbb{C})$. If $x$ and $y$ commute, they can be triangularized simulatneously. Then the result is clear. You see also why you don't necessarily have equalities.

• Hi! Only one question: what is the definition of the unital algebra generated by $x,y$? Commented Jan 8, 2021 at 4:45
• @CarlosJiménez Smallest $C^*$ subalgebra that contains $x,y$ Commented Feb 22, 2021 at 3:40