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?
Best Answer
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 $\mathbb{C}$.