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Easy question about linear operators - in physics (often) the terms spectrum and (set of) eigenvalues of an operator are used interchangeably.

I'd like a simple compare and contrast to know the difference according to mathematicians. Many thanks!

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The other answers adequately address the finite-dimensional case. If we consider a Hilbert space $\newcommand{\Hil}{\mathcal{H}}\Hil$ of countable dimension with orthonormal basis $(e_i)_{i \in \mathbb{N}}$, we can define an operator $T : \Hil \rightarrow \Hil$ by $$ T(e_n) = \frac{1}{n}e_n $$ It is clear from this definition that each $e_n$ is an eigenvector with eigenvalue $\frac{1}{n}$, and as $(e_i)_{i \in \mathbb{N}}$ is an orthonormal basis, this is a complete set of eigenvectors. So the set of eigenvalues is $\{ \frac{1}{n} \mid n \in \mathbb{N} \}$.

However, $T$ is not invertible (one way to see this is that if we try to define an inverse we get an unbounded operator, another is that the vector $\sum_{n=1}^\infty \frac{1}{n}e_n$ is in $\Hil$ but not in the image of $T$), so $0$ is an element of its spectrum that is not an eigenvalue. In fact $\{ 0 \} \cup \{ \frac{1}{n} \mid n \in \mathbb{N} \}$ is exactly the spectrum of $T$.

The spectrum is always a non-empty closed subset of $\mathbb{C}$, while the set of eigenvalues does not need to be. Also, the set of eigenvalues can be empty (for example, this is the case for the operator $S(e_n) = e_{n+1}$).

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  • $\begingroup$ +1 for great, simple example and explanation. Question: is the inverse of T unbounded because it would seem to require $T^{-1}(e_n) = n e_n$? Also I see why S has no eigenvalues but what is its spectrum then? (can you give me a hint to figure it out for myself?) $\endgroup$
    – lux
    Jul 13 '19 at 8:53
  • $\begingroup$ @lux Yes, the operator $T^{-1}(e_n) = ne_n$ is unbounded (and therefore partially defined - if you evaluate it on the vector $\sum_{n=1}^\infty \frac{1}{n}e_n$ you get a sum that doesn't converge in $\mathcal{H}$). $\endgroup$ Jul 13 '19 at 8:55
  • $\begingroup$ OK, thanks! But is it obvious that the vector you propose to act upon with the inverse operator converges in H? And can you help me with the spectrum of your operator S? $\endgroup$
    – lux
    Jul 13 '19 at 9:06
  • $\begingroup$ @lux To show $\sum_{n=1}^\infty \frac{1}{n}e_n \in \mathcal{H}$, use the fact that $\sum_{n=1}^\infty \frac{1}{n^2}$ converges (to $\frac{\pi^2}{6}$). This is enough to show that the original sum converges in norm in the Hilbert space $\mathcal{H}$. If it makes things easier, set $\mathcal{H} = \ell^2$ and let $(e_i)_{i \in \mathbb{N}}$ be the standard basis. $\endgroup$ Jul 13 '19 at 9:12
  • $\begingroup$ Regarding $S$, it is more usually known as the "unilateral right shift". Its spectrum is the unit disc (i.e. the set of all complex numbers $\lambda$ such that $|\lambda| \leq 1$). $\endgroup$ Jul 13 '19 at 9:14
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The spectrum of a linear operator $T$ on a finite dimensional space is exactly the set of its eigenvalues.

It's defined as $\{\lambda:T-\lambda I$ is not invertible$\}$, and $T-\lambda I$ is not invertible in a finite dimensional space iff its kernel is nonzero, meaning exactly that $\lambda$ is an eigenvalue of $T$.

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    $\begingroup$ OK great - what about on an infinite dimensional space? $\endgroup$
    – lux
    Jul 13 '19 at 8:35
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I'd like to add something further in the infinite-dimensional case than the other responses, which I think is important if we're going to talk about eigenvalues vs. spectrum. I'm going to specify how they're related.

As mentioned by @RobertFurber, we can have elements of the spectrum which are not eigenvalues. Let $T:X\rightarrow X$ be a bounded operator on a Banach space. The spectrum $\sigma(T)$ can be decomposed into three pieces:

  1. The point spectrum: this consists of $\lambda\in\sigma(T)$ so that $T-\lambda I$ is not injective. In this case, $\lambda$ is called an eigenvalue of $T$.

  2. The continuous spectrum: this consists of $\lambda\in \sigma(T)$ such that $T-\lambda I$ is one-to-one, and the range of $T-\lambda I$ is dense in $X$, but it is not equal to $X$.

  3. The residual spectrum: this consists of $\lambda\in \sigma(T)$ such that $T-\lambda I$ is one-to-one, and such that $T-\lambda I$ is not dense in $X$.

While the spectrum is always non-empty, we can have elements with empty point spectrum. As an example, if $X=L^2([0,1])$ and $M:X\rightarrow X$ is the multiplication operator, then $\sigma(M)=[0,1],$ and each $\lambda\in [0,1]$ is part of the continuous spectrum (this is an easy exercise, and you can also find it on this site).

An example where every element of the spectrum (except zero) is an eigenvalue is the case of a compact, self-adjoint operator on a Hilbert space.

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OK, so in mathematics the spectrum of a matrix is the set of its eigenvalues. I know it's confusing, so let me explain.

The spectrum of a matrix is basically a linear map of the matrix. You can imagine it as some kind of finite number line, but with eigenvalues as the numbers.

Actually, there is a much more better answer on Wikipedia: "Let V be a finite-dimensional vector space over some field K and suppose T: V → V is a linear map. The spectrum of T, denoted σT, is the multiset of roots of the characteristic polynomial of T. Thus the elements of the spectrum are precisely the eigenvalues of T, and the multiplicity of an eigenvalue λ in the spectrum equals the dimension of the generalized eigenspace of T for λ (also called the algebraic multiplicity of λ).

Now, fix a basis B of V over K and suppose M∈MatK(V) is a matrix. Define the linear map T: V→V point-wise by Tx=Mx, where on the right-hand side x is interpreted as a column vector and M acts on x by matrix multiplication. We now say that x∈V is an eigenvector of M if x is an eigenvector of T. Similarly, λ∈K is an eigenvalue of M if it is an eigenvalue of T, and with the same multiplicity, and the spectrum of M, written σM, is the multiset of all such eigenvalues." ( Taken from wikipedia, link here: https://en.wikipedia.org/wiki/Spectrum_of_a_matrix)

But it's full of technical terms, so if you're new to this field, I would suggest you think of it as a simple number line. That's it!

Hope this helped you! :) (Please upvote.)

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  • $\begingroup$ @lux: An operator on an infinite-dimensional space may have additional elements in its spectrum, and may have no eigenvalues. $\endgroup$
    – user688633
    Jul 13 '19 at 8:39
  • $\begingroup$ Hope this helped! $\endgroup$
    – user688633
    Jul 13 '19 at 8:40

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