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This is more a conceptual question than any other kind. As far as I know, one can define matrices over arbitrary fields, and so do linear algebra in different settings than in the typical freshman-year course.

Now, how does the concept of eigenvalues translate when doing so? Of course, a matrix need not have any eigenvalues in a given field, that I know. But do the eigenvalues need to be numbers?

There are examples of fields such as that of the rational functions. If we have a matrix over that field, can we have rational functions as eigenvalues?

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    $\begingroup$ The possible eigenvalues are precisely elements of the base field. If your base field consists of rational functions over $\Bbb C$ (say) then the eigenvalues of the matrices with such entries will be rational functions. $\endgroup$ – Wojowu Dec 9 '15 at 17:47
  • $\begingroup$ @Wojowu Ok, thanks :) $\endgroup$ – A.Sh Dec 9 '15 at 17:53
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    $\begingroup$ As motivation into a well developed example: as I recall Spinors are eigenvalues of some equations in Quantum Mechanics and they are represented by matrices themselves. Please correct if my memory is serving me wrong. $\endgroup$ – rrogers Dec 16 '15 at 14:47
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Of course. The definition of an eigenvalue does not require that the field in question is that of the real or complex numbers. In fact, it doesn't even need to be a matrix. All you need is a vector space $V$ over a field $F$, and a linear mapping $$L: V\to V.$$

Then, $\lambda\in F$ is an eigenvalue of $L$ if and only if there exists a nonzero element $v\in V$ such that $L(v)=\lambda v$.

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    $\begingroup$ I compensated by upvoting this one :-) $\endgroup$ – Justpassingby Dec 9 '15 at 18:01
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    $\begingroup$ I upvoted both myself :) $\endgroup$ – A.Sh Dec 9 '15 at 18:02
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    $\begingroup$ @5xum Oh, wait, you were first…whoops…I'll change my verdict then. All for fairness. (I interpreted your timestamps backwards…) $\endgroup$ – A.Sh Dec 9 '15 at 18:10
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    $\begingroup$ And as a trivial explicit construction, for any $\lambda \in F$, the map that takes every $v$ to $\lambda v$ has eigenvalue $\lambda$, so every $\lambda \in F$ is an eigenvalue of some linear map. $\endgroup$ – user2357112 Dec 9 '15 at 18:15
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    $\begingroup$ Then I believe we're all done here : ) $\endgroup$ – A.Sh Dec 9 '15 at 18:47
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Eigenvalues need to be elements of the field. The most common examples of fields contain objects that we usually call numbers, but this is not part of the definition of an eigenvalue. As a counterexample, consider the field $\mathbb R(x)$ of rational expressions in a variable $x$ with real coefficients. The $2\times 2$ matrix over that field

$$\left(\begin{matrix} x & 0 \\ 0 & \frac1x \end{matrix}\right)$$

has eigenvalues $x$ and $1/x$: not unknown numbers, but known elements of the field $\mathbb R(x).$

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  • $\begingroup$ Thank you for clearing this up, much appreciated. $\endgroup$ – A.Sh Dec 9 '15 at 18:00
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    $\begingroup$ Umm…it seems I misinterpreted the timestamps, and 5xum was indeed first, so I've changed the verdict. All for fairness. Thanks anyway :) $\endgroup$ – A.Sh Dec 9 '15 at 18:13
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Very interesting question. Consider real valued column "vectors" which are matrices $2N \times 2$, each 2x2 block on the form $\left[\begin{array}{rr}a&-b\\b&a\end{array}\right]$. This is a famous representation for complex numbers $a+bi$. So if we have a block-matrix on the same form we can find "eigenvalues" which will be 2x2 blocks complex numbers (in the same block-representation). This idea can then be extended to more advanced types of "numbers". This leads to the idea that more advanced "numbers" themselves are matrices - if the elements are "simpler" numbers in some sense. Instead of an eigenvalue decomposition with respect to the simplest numbers, we get a block-eigenvalue decomposition. If you are interested in this I would encourage you to read more about Representation Theory where more advanced types of numbers are written as matrices with elements of a less complicated type of number. For instance the field of quaternions can in turn be written as a matrix of $2 \times 2$ complex numbers - and since those in turn are $2 \times 2$ real numbers we could see them as a matrix with $4 \times 4$ real numbers. So we can get a "hierarchy" of more or less "advanced" numbers, depending on which size of matrix blocks we look.

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    $\begingroup$ This looks awesome! I've been meaning to learn the subject for some time now, and I just got admitted to a course on representation theory for next semester. I can't wait! $\endgroup$ – A.Sh Dec 9 '15 at 22:31
  • $\begingroup$ Yes it's really exciting, although my knowledge on the subject is quite informal and fussy. It is kind of limited to "fooling around on the keyboard in octave and sometimes getting lucky". $\endgroup$ – mathreadler Dec 9 '15 at 23:45
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I think it's worth providing a motived example where we aren't dealing with real or complex numbers as our eigenvalues

The Schrödinger Equation is a PDE over the complex numbers that is used to talk about wave-forms in quantum mechanics. The Schrödinger Equation gives a description of a wave that permits many different steady-state solutions, each one corresponding to an eigenvalue/vector pair. The eigenvectors are the wave equations and the eigenvalues are the corresponding energy functions.

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  • $\begingroup$ That's a great example, thank you :) $\endgroup$ – A.Sh Feb 1 '16 at 15:42
  • $\begingroup$ I disagree that this example, while interesting, is relevant to the question. Eigenvalues in quantum mechanics are definitely numbers. The question was if eigenvalues are always numbers. $\endgroup$ – Justpassingby Feb 1 '16 at 15:44
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    $\begingroup$ The energy is a function of time. Although is basic applications, it's the constant function, we are often interested in a driven system, the QM equivalent of having a spring on the end of a motor. Introducing a wabble into this motor is one way to get a nonconstant energy. $\endgroup$ – Stella Biderman Feb 1 '16 at 15:48

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