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Suppose V is a vector space over the scalar field F, and T is a linear operator on V. Can T have an eigenvalue l such that l is not an element of F?

I am working through Linear Algebra Done Right, 3 ed, by Sheldon Axler. Definition 5.5 on page 134 defines eigenvalues of T as elements of the scalar field F that V exists over. The definition seems to suggest that eigenvalues must always be elements of this field.

When checking my work on a problem in this section, I read the following question:

Find eigenvalues of $T(x,y)=(-3y,x)$

In the above question, the accepted answer claims that a linear operator on a vector space over the real numbers can have a complex eigenvalue.

Is this correct? Is the idea of a complex eigenvalue well defined for such a linear operator?

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  • $\begingroup$ Take for example a 90 degree rotation on $\mathbb{R}^2$. It is correct to say that this operator has no real eigenvalues. However, an $n \times n$ matrix will have all of its eigenvalues if you go up to an algebraically closed field (like $\mathbb{C}$) $\endgroup$
    – Joppy
    Sep 26 '18 at 0:21
  • $\begingroup$ Just as complex numbers appear as intermediate results when finding real roots of cubics, it’s convenient to allow complex eigenvalues and eigenvectors of real matrices and then use them to construct real-valued solutions to various problems. $\endgroup$
    – amd
    Sep 26 '18 at 0:49
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Eigenvalues are always in the field over which we are working. But since real numbers are also complex numbers, you can see your linear transformation as a linear transformation of vector spaces over $\mathbb{C}$, hence it makes sense to talk about complex Eigenvalues.

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  • $\begingroup$ Alright, thank you. If my understanding is correct, this means that when viewing R as the scalar field, it is a valid answer to say that T has no eigenvalues. Is that right? $\endgroup$ Sep 26 '18 at 0:27
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    $\begingroup$ Yes, that is right $\endgroup$
    – Pedro
    Sep 26 '18 at 0:29
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The eigenvalues live in the space where the solution of characteristic polynomials of a linear operator with coefficients from the field over the vector space you are considering. So technically yes. That means you need a vector space over a field that is algebraically closed to have eigenvalues in that field. Otherwise you may have eigenvalues that are outside the field over which the vector space is defined.

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  • $\begingroup$ I'm not familiar with all of the terminology you used, but does this mean that it would never make sense to define the eigenvalue of a linear operator with complex coefficients in terms of a further extended number system? $\endgroup$ Sep 26 '18 at 0:32
  • $\begingroup$ $F$ is algebraically closed if every non constant polynomial with coefficients from $F$ has a solution in $F$. As you know the eigenvalues of $T$ are solution to the characteristic polynomial of $T$. So if you want all eigenvalues of all linear operator defined over $\mathbb{R}$ , then you have to consider $\mathbb{C}$ which is the algebraic closure of $R\mathbb{R}$ . So if you have a linear operator with complex coefficients then the eigenvalues will always be in $\mathbb{C}$ , since $\mathbb{C}$ is algebraically closed. Hope that helps. $\endgroup$ Sep 26 '18 at 0:45

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