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?
Best Answer
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.