Is it possible to define the eigenvalues of a linear operator without first defining the matrix of the linear operator?

Question

Suppose you had a linear map $\phi\colon V \rightarrow V$, where $V$ is some finite dimensional vector space. Does saying a statement along the lines of "the eigenvalues of $\phi$ are..." making any 'sense' without first fixing a basis of $V$ and deducing the corresponding matrix of the map?

Answer 1

The definition of an eigenvalue and an eigenvector is completely independent from the basis we choose for our vector space, and thus from the matrix representation of the linear transform $\phi$ we are describing.

Indeed, consider any vector space $V$, which need not have finite dimension, over a field $\mathbb{K}$. Then $\lambda\in\mathbb{K}$ and $v\in V, v\not=0$ are called eigenvalues and eigenvalues of a linear map $\phi: V\rightarrow V$, if we have:

$$\phi v = \lambda v$$

Now, if $V$ is finite-dimensional with $\text{dim } V = n$, then, after a choice of basis for $V$, $\phi$ can be represented by a matrix $\Phi\in\mathbb{K}^{n\times n}$ and all the usual rules apply. If you change to a different basis, the matrix representation of $\phi$ will change, but its eigenvalues and -vectors won't (though the representation of $v$ as a linear combination of basis vectors das will also change), which also suggests that it is meaningful to talk about eigenvalues and eigenvectors without a matrix representation.

The same holds true in the infinite-dimensional case, except that here, a matrix representation of a linear map is generally impossible.

Still, this does not stop us from talking about eigenvalues and -vectors. Consider, for instance, the vector space of all real-valued smooth functions $C^\infty(\mathbb{R})$. The differentiation operator $\frac{d}{dx}$ is a linear map on this space with eigenvectors $v(x) = c\cdot e^{\lambda x}$ (with $c\in\mathbb{R}$) with corresponding eigenvalues $\lambda$ - completely analogous to the finite-dimensional case.