# On the nonemptiness of the spectrum

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 $M \in \mathbb{C}^{n \times n}$ be an $n \times n$ complex matrix. It is a well known fact that the spectrum of $M$ defined as

$\sigma(M) = \{ \lambda \in \mathbb{C}: M - \lambda I \text{ is not invertible}\},$

is a nonempty set. However, if we look at the $M$ as a linear transformation rather than a matrix, we can see quickly that the spectrum does not depend on $\mathbb{C}^n$ (on which $M$ acts), we may as well take operators acting on an arbitrary Hilbert space $H$. Can it happen that the spectrum of $M: H \mapsto H$ 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 $n \times n$ matrices. A complex number $\lambda$ is called an eigenvalue for $M$ if there is a vector $x \in \mathbb{C}^{n \times n}$ such that $Mx = \lambda x$. $x$ is called an eigenvector of $M$ for $\lambda$.

Theorem 1. For every $M \in \mathbb{C}^{n \times n}$ be an $n \times n$ complex matrix the spectrum $\sigma(M)$ is nonempty. Moreover, every $\lambda \in \sigma(M)$ is an eigenvalue, that is there is a vector $x$ for which $Mx = \lambda x$.

Proof. It is known that $M - \lambda I$ is not invertible if and only if $\det(M - \lambda I) = 0$. Expanding this we can see that it is a polynomial of degree $n$ in $\lambda$, which, according to the fundamental theorem of algebra, has a root, therefore the spectrum is nonempty.
If $\lambda \in \sigma(M)$, then since $M - \lambda I$ is not invertible, it has a nontrivial kernel, that is there is a vector $x$ for which $(M - \lambda I)x = 0$, which says exactly that $x$ is an eigenvector for $\lambda$. $\Box$

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 $M: \mathbb{C}^n \to \mathbb{C}^m$ where $m > n$.) 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 $A$ be a complex algebra on which there is a norm $| \cdot |$. If $A$ is a Banach space with this norm and

$| xy | \leq |x||y|$

is satisfied for all $x, y \in A$, then we call $A$ a Banach algebra. If there is an element $e \in A$ such that $xe = ex = x$ holds for all $x \in A$, then $e$ is called an unit and we say that $A$ is a Banach algebra with unit.

Examples. 1. Let $X$ be a compact Hausdorff topological space. The set of continuous functions $C(X)$ 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 $H$ be a Hilbert space. The set of its continuous linear operators $B(H)$ is a Banach algebra with the operator composition and operator norm, moreover the identity operator is a unit.
3. $L^1(\mathbb{R})$ is also a Banach algebra with the convolution product and the usual $L^1$ norm. This Banach algebra does not have an unit. (It has something called approximative unit, but we will not talk about this here.)
4. $\mathbb{C}$ is also a Banach algebra with the usual operations and absolute value as norm.

Of course, if we denote the unit with $e$, we say that $x$ is invertible if there exists an $x^{-1} \in A$ such that $x x^{-1} = x^{-1} x = e$.

The spectrum of $x \in A$ is defined in the usual way, that is

$\sigma(x) = {\lambda \in \mathbb{C}: x - \lambda e \text{ is not invertible}}.$

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

Theorem 2. (C. Neumann) Let $A$ be a Banach algebra and let $x \in A$. If $| x | < 1$, then $e - x$ is invertible and

$(e-x)^{-1} = \sum_{k=0}^{\infty} x^k$.

Proof. Define $s_n := \sum_{k=0}^{n} x^k$. Then for all $n < m$ we have

$| s_n - s_m | \leq \sum_{k=n}^{m} | x |^k < \epsilon$

for arbitrary $\epsilon > 0$ if $n$ and $m$ large enough (since $\sum_{k=0}^{\infty} | x |^k$ is a convergent geometric series), which shows that $s_n$ is a Cauchy sequence. Since $A$ is a Banach space, it implies that there is an $s \in A$ such that $\lim_{n \to \infty} s_n = s$. Moreover,

$s_n (e - x) = (e - x )s_n = e - x^{n+1}$,

which implies, since $\lim_{n \to \infty} x^{n+1} = 0$, that $s(e - x) = (e - x)s = e$, which is what we had to show. $\Box$

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. ($\mathbb{C}$ is a very special Banach algebra.) The usual definition of differentiability carries through, that is  if $f: \mathbb{C} \to A$ is a function taking values in a Banach algebra $A$, then we say that $f$ is differentiable at $z_0$ if the limit

$\lim_{z \to z_0} \frac{f(z) - f(z_0)}{z - z_0}$

exists. $f$ is analytic in the open set $D$, if it is differentiable at every point of $D$. 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 $f: \mathbb{C} \to A$ be a Banach algebra valued function and assume that it is analytic on $\mathbb{C}$. If it is bounded, that is

$\sup{ |f(z)|: z \in \mathbb{C} } < \infty$,

then it is constant.

Proof. Let $\phi : A \to \mathbb{C}$ be a continuous linear functional. (In other words, let $\phi \in A^*$, where $A^*$ denotes the dual space of $A$.) Since $\phi(f(z))$ is differentiable and bounded in the usual sense, it is constant according to the classic Liouville theorem, that is

$\varphi(f(z)) = \varphi(f(0))$

holds for all $z \in \mathbb{C}$. Since the linear functionals in $A^*$ separates the points of $A$ (an easy consequence of the Hahn-Banach theorem), it follows that $f(z) = f(0)$. $\Box$

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

$R_{x}(z): \mathbb{C} \setminus \sigma(x) \to A, z \mapsto (x - z e)^{-1}$.

$R_{x}(z)$ is analytic, since

$R_{x}(z) - R_{x}(z_0) = (x - ze)^{-1}(z_0 - z)(x - z_0 e)^{-1}$,

and this implies that

$\lim_{z \to z_0} \frac{R_{x}(z) - R_{x}(z_0)}{z - z_0} = -(x - z_0 e)^{-2}$,

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

Theorem 4. Let $A$ be a Banach algebra with unit and let $x \in A$ be arbitrary. Then the spectrum $\sigma(x)$ of $x$ is nonempty.

Proof. We shall prove the theorem indirectly. Assume that $\sigma(x)$ is empty. This means that the resolvent function $R_x(z)$ is holomorphic on the entire complex plane. If $z$ is so large such that $| x | < z$, then

$R_x(z) = z^{-1} (x/z - e)^{-1} = - \frac{1}{z} \sum_{k=0}^{\infty} \frac{x^k}{z^k}$,

which implies that

$\lim_{z \to \infty} | R_x(z) | \leq \lim_{z \to \infty} \frac{1}{|z|} \frac{1}{1-|x|/|z|} = 0$.

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. $\Box$

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 $(C([0,1]), | \cdot |_\infty)$ of continuous functions with the supremum norm. For each $f \in C([0,1])$, we can define the multiplication operator $M_f : C([0,1]) \to C([0,1])$ with $g \mapsto fg$ The spectrum of $M_f$ is the range $f([0,1])$, and it is easy to see that if $\lambda \in \sigma(M_f)$ is an element of the spectrum such that $A_\lambda = \{ x \in [0,1]: f(x) = \lambda \}$ contains no open set (i.e. the only continuous function with support in $A_\lambda$ is the constant zero function), then $\lambda$ is not an eigenvalue for $M_f$.