This time we shall explore a topic which really fits the name of the blog. 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
.
Small typo in proof of Theorem 1: “It is known that $M – \lambda I$ is invertible…” should be “not invertible”.
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Corrected, thank you!
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These are awesome posts, keep it up ^_^
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