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Given a vector space $V$ over scalar field $F$, and given a linear transformation $T : V → V$, the definition of eigenvalues and eigenvectors of $T$ is:

A nonzero vector $\mathbf{v} \in V$ is an eigenvector of $T$ iff there exists a scalar $\lambda \in F$ such that $T(\mathbf{v}) = \lambda \mathbf{v}$. $\lambda$ is then said to be an eigenvalue of $T$ corresponding to $\mathbf{v}$.

But it seems some eigenvalues escape this definition. For example, let $F = \mathbb{R}$, $V = \mathbb{R}^2$, and $T = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$. Then some computation reveals that the eigenvalues are $i$ and $-i$, which do not live in $F$.

If $V$ is finite-dimensional, I can at least confidently say that all eigenvalues live in $\overline{F}$, the algebraic closure of $F$. But what for infinite-dimensional cases?

Dannyu NDos
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  • In the infinite dimensional case your point (eigenvalue-) spectrum may be empty. – undefined Feb 11 '22 at 23:20
  • @undefined No worries even in that case; All eigenvalues vacuously live within $\overline{F}$. – Dannyu NDos Feb 11 '22 at 23:23
  • Some infinite dimensional operators do not have any eigenvalues. E.g $(a_1, a_2, \dots a_n , \dots ) \mapsto (0,a_1, a_3, \dots a_n , \dots ) $ is a linear operator on sequences over $\mathbb{F} $ with no eigenvalues. – Dionel Jaime Feb 11 '22 at 23:59
  • As for when eigenvalues do exist, I am not sure if they have to lie in $\bar{F} $. I would bet money that they don't. – Dionel Jaime Feb 12 '22 at 00:09

1 Answers1

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There are no problems with the definition. If your map $T$ is defined on $\mathbb{R^2}\to\mathbb{R^2}$ then it doesn't have any eigenvalues. It is clear that there is no vector $0\ne v\in\mathbb{R^2}$ such that $T(v)=iv$, so how is $i$ an eigenvalue? (not to mention that we don't even know what $i$ is, as we work over $\mathbb{R}$)

Now, if you work over $\mathbb{C}$ and define $T$ on $\mathbb{C^2}$ then the new transformation will have two eigenvalues, and they are indeed in $\mathbb{C}$.

Mark
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  • I guess that the question is if the eigenvalues belong to the algebraic closure of the field generated by the coefficients of the operator. – coudy Feb 11 '22 at 23:05
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    @coudy Again, existence of eigenvalues depend on the field. Over $\mathbb{R}$ this operator doesn't have any eigenvalues, so talking about "the eigenvalues" makes no sense. Over an algebraically closed field, any finite dimensional linear map has an eigenvalue. (which is clear, because the characteristic polynomial has a root) – Mark Feb 11 '22 at 23:09
  • The question deals with the infinite dimensional case. – coudy Feb 11 '22 at 23:10
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    @coudy When the dimension is infinite, there might be maps with no eigenvalues even over an algebraically closed field. See here: https://math.stackexchange.com/questions/1173511/counterexamples-for-every-linear-map-on-an-infinite-dimensional-complex-vector – Mark Feb 11 '22 at 23:12
  • This does not answer the question wether the eigenvalues are in the algebraic closure when they exist. Anyway, I am not the one that posted the question. – coudy Feb 11 '22 at 23:16
  • I also don't see how this answers the question. I didn't post it, but a good answer would be getting at whether the eigenvalues of an infinite dimensional operator necessarily lie in the algebraic closure of the field the space is over. – Dionel Jaime Feb 11 '22 at 23:55
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    @DionelJaime Again, what you mean by "the eigenvalues"? A linear map is defined over a specific field, and the eigenvalues are also defined over a specific field. Most of the time you can't "extend" the map to a different field. In finite dimensions we can at some sense do that, because a matrix over $\mathbb{R}$ can be viewed as a matrix over $\mathbb{C}$ as well. But in infinite dimensions we don't work with matrices anymore. So not sure how do you want to change the field. – Mark Feb 12 '22 at 10:58
  • The answer addresses the question. The eigenvalues of a linear transformation lie in whatever field the linear transformation is defined over. – user2357112 Feb 12 '22 at 10:58
  • Yes, you are right. – Dionel Jaime Feb 12 '22 at 18:55