Can eigenvalues be outside of the scalar field of a vector space?

Question

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?

Answer 1

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.

Answer 2

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.