Understanding the Spectrum of a Linear Operator

[member 428835]

Hi PF!

What is meant by the spectrum of a linear operator $A$? I read somewhere that if $0$ belongs in the spectrum, then $A$ is not invertible. Can anyone finesse this for me?

I read the wikipedia page, but this was tough for me to understand. Perhaps illustrating with a simple example, say $(Af)(x) = \sin x f(x)$, or $(Af)(x) = e^x f(x)$ or really anything that's easy to understand.

 

[member 587159]

The spectrum of a linear operator $A:V \to W$ between $\mathbb{F}$-vector spaces (in most cases $\mathbb{F} \in \{\mathbb{R}, \mathbb{C}\})$ is the set of all possible eigenvalues of $A$. Put concretely

$$\operatorname{spec}(A) := \{\lambda \in \mathbb{F}\mid \exists x \in V \setminus \{0\}: Ax = \lambda x\}$$

Now,

$A$ has eigenvalue $0$
$\iff \exists x \neq 0: Ax = 0x = 0$
$\iff \ker A \neq 0$
$\iff A$ is non-injective

thus an operator is not injective if and only if $0$ belongs to the spectrum.

 

[fresh_42]

Maybe this is of more help: https://www.physicsforums.com/insights/hilbert-spaces-relatives-part-ii/
However, a precise definition is a bit more than just saying: The spectrum are the eigenvalues. One has to distinguish the spectrum of an operator and its point spectrum which are the eigenvalues. E.g. (the closure of) the Dirac operator of the free electron has a spectrum $(-\infty,-1] \cup [1,\infty)$ but no eigenvalues.

A number is in the spectrum of an operator $\lambda\in \sigma(T)$ if the operator $\lambda−T$ is either not invertible or unbounded.

Your first example $A(f(x))=\sin x f(x)$ is bounded, but has no eigenvalues, so $\lambda - A$ must be invertible: $(\lambda -A)^{-1} (f(x)) := \dfrac{f(x)}{\lambda - \sin x}$, and the spectrum is empty.

Your second example is unbounded, and so are $\lambda -A$ for all $\lambda$. In this case all numbers are in the spectrum.

A non trivial example is the one I mentioned above, a simple one is $A(f) = cf$ with a spectrum $\{\,c\,\}$.

 

[member 428835]

Now,

$A$ has eigenvalue $0$
$\iff \exists x \neq 0: Ax = 0x = 0$
$\iff \ker A \neq 0$
$\iff A$ is non-injective

thus an operator is not injective if and only if $0$ belongs to the spectrum.

So you're talking about an operator being injective, which is only a necessary condition for invertibility, right? I don't see how a matrix $A$ being injective implies anything about invertibility.

Also, does $A$ have $0$ as an eigenvalue? In the first example we have $(\lambda - \sin x)f(x) = 0$ and $\lambda = 0 \implies -\sin x f(x) = 0 \implies f(x) = 0$, which seems wrong.

 

[fresh_42]

So you're talking about an operator being injective, which is only a necessary condition for invertibility, right? I don't see how a matrix $A$ being injective implies anything about invertibility.

I guess @Math_QED talked only about the point spectrum. This is a subset of the spectrum. All eigenvalues are in the spectrum, but not the other way around.

Also, does $A$ have $0$ as an eigenvalue? In the first example we have $(\lambda - \sin x)f(x) = 0$ and $\lambda = 0 \implies -\sin x f(x) = 0 \implies f(x) = 0$, which seems wrong.

It is wrong. An eigenvalue zero requires an eigenvector unequal zero. The spectrum is empty in this case.

 

[member 428835]

Thanks!