This time we shall explore a topic which really fits the name of the blog. (**Note:** This was posted on my blog Nonempty Spaces, from which I migrated the old posts here.) Let be an complex matrix. It is a well known fact that the spectrum of defined as

is a nonempty set. However, if we look at the as a linear transformation rather than a matrix, we can see quickly that the spectrum does not depend on (on which acts), we may as well take operators acting on an arbitrary Hilbert space . Can it happen that the spectrum of is empty?

This post is devoted to exploring this question. The answer will be surprising and the methods developed are very interesting and beautiful.

To start off, let’s see the classical proof of the nonemptiness of spectrum for matrices. A complex number is called an *eigenvalue for * if there is a vector such that . is called an *eigenvector of for *.

**Theorem 1.** For every be an complex matrix the spectrum is nonempty. Moreover, every is an eigenvalue, that is there is a vector for which .

*Proof.* It is known that is not invertible if and only if . Expanding this we can see that it is a polynomial of degree in , which, according to the fundamental theorem of algebra, has a root, therefore the spectrum is nonempty.

If , then since is not invertible, it has a nontrivial kernel, that is there is a vector for which , which says exactly that is an eigenvector for .

One can easily see that nothing in this proof works for operators acting on infinite dimensional cases. First, there is no determinant. Second, in infinite dimensional vector spaces noninvertibility does not imply nontrivial kernel. (We do not need infinite dimensions for this to fail, it is enough to take a linear transformation where .) To top this, it is not even true that every element in the spectrum is an eigenvalue, but more on this later. We shall develop the theory not only for operators, but rather for elements of a so-called Banach algebra, for which one well known example is the algebra of operators acting on a Hilbert space.

**Definition.** Let be a complex algebra on which there is a norm . If is a Banach space with this norm and

is satisfied for all , then we call a *Banach* algebra. If there is an element such that holds for all , then is called an unit and we say that is a *Banach algebra with unit.*

**Examples. 1.** Let be a compact Hausdorff topological space. The set of continuous functions with the pointwise multiplication and supremum norm is a Banach algebra. It is easy to see that the constant function is an unit for this algebra.

**2.** Let be a Hilbert space. The set of its continuous linear operators is a Banach algebra with the operator composition and operator norm, moreover the identity operator is a unit.

**3. ** is also a Banach algebra with the convolution product and the usual norm. This Banach algebra does not have an unit. (It has something called approximative unit, but we will not talk about this here.)

**4.** is also a Banach algebra with the usual operations and absolute value as norm.

Of course, if we denote the unit with , we say that is invertible if there exists an such that .

The spectrum of is defined in the usual way, that is

The following theorem plays a central role in the theory of Banach algebras.

**Theorem 2.** (C. Neumann) Let be a Banach algebra and let . If , then is invertible and

.

*Proof.* Define . Then for all we have

for arbitrary if and large enough (since is a convergent geometric series), which shows that is a Cauchy sequence. Since is a Banach space, it implies that there is an such that . Moreover,

,

which implies, since , that , which is what we had to show.

Note that the previous theorem was named after Carl Neumann, not John von Neumann.

It is also very natural to study functions which takes its values in Banach algebras. ( is a very special Banach algebra.) The usual definition of differentiability carries through, that is if is a function taking values in a Banach algebra , then we say that is *differentiable* at if the limit

exists. is analytic in the open set , if it is differentiable at every point of . Most famous theorems about complex analytic functions carries through in a heartbeat, for example the Liouville theorem.

**Theorem 3.** (Liouville theorem for Banach algebra valued functions) Let be a Banach algebra valued function and assume that it is analytic on . If it is bounded, that is

,

then it is constant.

*Proof.* Let be a continuous linear functional. (In other words, let , where denotes the dual space of .) Since is differentiable and bounded in the usual sense, it is constant according to the classic Liouville theorem, that is

holds for all . Since the linear functionals in separates the points of (an easy consequence of the Hahn-Banach theorem), it follows that .

One important function is the so-called resolvent function defined as

.

is analytic, since

,

and this implies that

,

which is not that surprising, because it is the same as usual. Now we are ready to prove the main theorem.

**Theorem 4.** Let be a Banach algebra with unit and let be arbitrary. Then the spectrum of is nonempty.

*Proof.* We shall prove the theorem indirectly. Assume that is empty. This means that the resolvent function is holomorphic on the entire complex plane. If is so large such that , then

,

which implies that

.

This means that the resolvent function is not only holomorphic, it is bounded in the entire complex plane. Therefore the generalized Liouville’s theorem implies that it is constant, moreover it is zero everywhere. Since zero is not invertible, it is clearly impossible, therefore the spectrum cannot be empty.

It is not an accident that the sentence *“Moreover, every element of a spectrum is an eigenvalue”* is omitted from the theorem. First of all, eigenvalues are defined for operators and not for elements of a Banach algebra, but this is the least of the problems. Consider the Banach space of continuous functions with the supremum norm. For each , we can define the multiplication operator with The spectrum of is the range , and it is easy to see that if is an element of the spectrum such that contains no open set (i.e. the only continuous function with support in is the constant zero function), then is not an eigenvalue for .