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In algebra, I learned that if $\lambda$ is an eigenvalue of a linear operator $T$, I can have \begin{equation} Tx = \lambda x \tag{1} \end{equation} for some $x\neq 0$, which is equivalent to $\lambda I-T$ not being invertible.

In functional analysis, it is said that if $\lambda$ is an element of a spectrum of the linear operator $T$, then $\lambda I - T$ is not invertible. However, my Professor never mentioned $(1)$.

Is the definition/concept in functional analysis the same as $(1)$ in linear algebra? Can I use $(1)$ in functional analysis too? Does it depend on which spaces we are in?

For example, suppose $\lambda$ is in the spectrum of $T$, where $T$ is a linear operator on $E$, a Banach space. I want to show $\lambda^n$ is in the spectrum of $T^n$. Would this problem is equivalent to showing if $\lambda$ is an eigenvalue of a linear operator $T$, then $\lambda^n$ is an eigenvalue of $T^n$?

Thank you.

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    $\begingroup$ I think your issue is adressed here pretty well: en.wikipedia.org/wiki/Spectrum_%28functional_analysis%29.. Look at the section regarding the spectrum. It provides an example showing that in infinite-dimensional linear spaces $T-\lambda I$ not invertible does not mean that $\lambda$ is an eigenvalue. However, for finite-dimensional linear spaces, it holds true that $T -\lambda I $invertible $\Leftrightarrow$ $\lambda $ is an eigenvalue. $\endgroup$
    – Dimitar Ho
    Apr 21, 2015 at 23:10
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    $\begingroup$ Addendum: all this boils down to the fact that in finite dimensions we have $$T \text{ bijective }\iff T \text{ injective } \iff T \text{surjective.}$$ So the spectrum consists only of eigenvalues (cf. injective). In infinite dimensions an non-bijective operator can be not injective (point spectrum) or not surjective (continuous and residual spectrum). $\endgroup$ Mar 10, 2020 at 22:12

2 Answers 2

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Spectral theory in infinite-dimensional spaces is quite a bit more complicated than in the finite-dimensional case. In particular, we have to distinguish between the spectrum $\sigma(A)$ of an operator and its eigenvalues. Let $A$ be a linear operator on a Banach space $X$ over the scalar field $C$. We have $$ \sigma(A) = \{ \lambda \in C: (\lambda I - A) \text{ does not have a bounded inverse} \}. $$ An eigenvalue $\lambda$ of $A$ is a value such that there exists a nonzero eigenvector $x \in X$ such that $$ A x = \lambda x, $$ or equivalently, $\operatorname{ker}(\lambda I - A) \neq \emptyset$. We then call $\dim \operatorname{ker}(\lambda I - A)$ the geometric multiplicity of the eigenvalue $\lambda$.

An eigenvalue is always in the spectrum, as you can see from the definition, but not every element of the spectrum is an eigenvalue in general.

In increasing order of "complicatedness", we could say:

  • Matrices (linear bounded operators on finite-dimensional vector spaces): the spectrum is finite, and each of its elements is an eigenvalue.
  • Compact self-adjoint operators on a Hilbert space: almost as nice as matrices. The spectrum is a compact set and countable, and it is contained in the reals. Every nonzero element of the spectrum is an eigenvalue with finite multiplicity. There is a spectral decomposition of the operator much as one would have for a matrix.
  • Bounded operators: the spectrum is still compact, but may be uncountable. In fact, for any compact set in $C$, you can find an operator which has this set as its spectrum. Yet it's still quite possible that no element of the spectrum is an eigenvalue; see the example of T.A.E. given in another answer.
  • Unbounded operators: the spectrum is in general unbounded.

This list could of course be refined with more specific conditions. I'm still studying the theory myself and will go back and add details as I learn about them. (Suggestions are welcome.) If you want to learn more, I found some relatively digestible lecture notes by E. Kowalski, ETH Zürich.

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  • $\begingroup$ I think you meant $\operatorname{ker}(\lambda I - A) \neq \{0 \}$, not $\operatorname{ker}(\lambda I - A) \neq \emptyset$. $\endgroup$
    – Hamilton
    Jan 18, 2023 at 14:28
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The multiplication operator $(Mf)(x)=xf(x)$ on $L^{2}[0,1]$ is a classical example of an operator with no eigenvalues, but its spectrum is $[0,1]$.

$M$ has no eigenvalue because $Mf=\lambda f$ gives $(x-\lambda)f=0$, which forces $f(x)=0$ a.e..

To see that $[0,1]\subseteq\sigma(M)$, note that the constant function $1$ is not in the range of $(M-\lambda I)$ for any $\lambda \in [0,1]$ because $(x-\lambda)g = 1$ would force $g = 1/(x-\lambda)$ a.e., which is not in $L^{2}$ for such $\lambda$. For any other $\lambda$, $(M-\lambda I)$ is invertible.

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  • $\begingroup$ I only noticed because it helped in answering this question :) math.stackexchange.com/questions/1769392/calculating-spectrum/… $\endgroup$
    – Math1000
    May 3, 2016 at 11:22
  • $\begingroup$ @Math1000 : Very good. By the way, you left off the closing bracket on one of the last intervals in your post. $\endgroup$ May 3, 2016 at 12:05
  • $\begingroup$ @DisintegratingByParts How do you show that $(M-\lambda I)$ is invertible for $\lambda \notin [0,1]$? $\endgroup$ May 16, 2019 at 7:36
  • $\begingroup$ @ASlowLearner : The inverse is multiplication by $\frac{1}{x-\lambda}$, which is a bounded multiplication operator on $L^2[0,1]$ for $\lambda\notin[0,1]$. $\endgroup$ May 16, 2019 at 7:55

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