Difference between the spectrum and point spectrum of an operator.

Question

I have the following two definitions in my notes:

The spectrum of an operator:

We define $\sigma(T)$, the spectrum of T, by,

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

The point spectrum of an operator:

The set of all eigenvalues of T is called the point spectrum of T and is denoted by $\sigma_p(T):$

$$\sigma_p(T):=\{\lambda \in \mathbb C :\lambda\, \text{ is an eigenvalue of $T$}\}$$

After everything I have done with eigenvalues, I would have thought that these definitions were equivalent. The standard eigenvalue problem is to find all $\lambda \in \mathbb C$ where,

$$Tx=\lambda x$$

For an operator $T$ and some vector in the space we are working in, $x$. From this it follows that,

$$(T-\lambda I) x = 0$$

Where the operator $T-\lambda I$ must be singular (for the eigenvalue problem).

Am I right in saying that these two definitions are equivalent? If they were, I am unsure as to why they are given two seperate definitions. And, if they are not, could somebody give me an example that differentiates the two definitions?

Cheers!

Answer 1

Suppose $T$ is a bounded operator on a Banach space $X$. $\lambda\in\rho(T)$ iff $T-\lambda I$ is a linear bijection. In that case, the inverse $(T-\lambda I)^{-1}$ is automatically continuous by the closed graph theorem.

There are three basic things that can stand in the way of $T-\lambda I$ being invertible.

1. $T-\lambda I$ is not injective. Equivalently, $Tx=\lambda x$ for some $x\ne 0$, which means that $\lambda$ is an eigenvalue of $T$.

2. The range of $T-\lambda I$ is not dense in $X$. Equivalently, there is a non-zero bounded linear functional $x^{\star}\in X^{\star}$ such that $x^{\star}((T-\lambda I)y)=0$ for all $y$; this, in turn is equivalent to $(T-\lambda I)^{\star}x^{\star}=0$, which means $\overline{\lambda}$ is an eigenvalue of $T^{\star}$.

3. $T-\lambda I$ is injective and has dense, non-closed range. In this case, the inverse of $T-\lambda I$ cannot be bounded, which gives the existence of a sequence $\{ y_n \}$ of unit vectors in the range of $\mathcal{R}(T-\lambda I)$ such that $\lim_n\|(T-\lambda I)^{-1}y_n\|=\infty$. After renormalization, you obtain a sequence of unit vectors $\{ x_n \}$ such that $\lim_n\|(T-\lambda I)x_n\|=0$. So $\lambda$ is an approximate eigenvalue of $T$ in this case.

You have an example of (1); there are plenty of examples of eigenvalues.

Case (2) is peculiar to infinite-dimensional spaces. A simple example is the shift operator on $\ell^2(\mathbb{N})$: $$ U(a_0,a_1,a_2,\cdots) = (0,a_0,a_1,a_2,\cdots) $$ For $a=(a_0,a_1,a_2,\cdots)$, $\|Ua\|=\|a\|$. So $U$ is injective, and its range is a proper closed subspace of $\ell^2(\mathbb{N})$. You can't do this with a finite shift.

Case (3) is also peculiar to infinite-dimensional spaces. A nice example of this is the multiplication operator $(Mf)(x)=xf(x)$ on $L^{2}[0,1]$. Every $\lambda \in [0,1]$ is an approximate eigenvalue. The simplest case is $\lambda=0$. The functions $$ f_n = \sqrt{n}\chi_{[0,1/n]},\;\;\; n=1,2,3,\cdots, $$ are unit vectors because $$ \int_{0}^{1}|f_n|^2dx = \int_{0}^{1/n}ndx = 1. $$ And $$ \|Mf_n\|^2 = \int_{0}^{1}x^{2}|f_n|^2dx = \int_{0}^{1/n}n xdx= \frac{n}{2}x^{2}|_{0}^{1/n}= \frac{1}{n}\rightarrow 0. $$ So $0$ is an approximate eigenvalue of $M$. The range is dense because $M=M^{\star}$ means $M^{\star}$ does not have eigenvalue $0$ either. You can see how $\chi_{[\lambda-1/n,\lambda+1/n]}$ is very close to being an eigenfunction of $M$ with eigenvalue $\lambda$, if $\lambda \in [0,1]$, but it just can't quite get there.

Answer 2

For $T$ to have $\lambda$ as an eigenvalue, $T-\lambda I$ must be non-injective. For $\lambda$ to be in the spectrum of $T$, it must only be non-invertible. These are equivalent when $T$ is an operator on a finite-dimensional space, but not in general!

For example, let $T$ be the shift operator $(x_0,x_1,\dots) \mapsto (0,x_0,x_1,\dots)$ on your favorite subspace of $\Bbb{R}^\infty$.

Then $T-\lambda I$ maps $x=(x_0,x_1,\dots)$ to $(-\lambda x_0, x_0-\lambda x_1, x_1-\lambda x_2,\dots)$, which only vanishes when $x=0$. So the point spectrum of $T$ is empty. On the other hand, $T$ itself is not surjective and hence not invertible, so the spectrum of $T$ contains $0$.

Answer 3

Consider an example is the shift operator on ℓ2(N): U(a0,a1,a2,⋯)=(0,a0,a1,a2,⋯). The spectrum contains 0. Is there is spectrum contains any other elements?