Eigenvalue of an abstract linear map?

Question

Suppose that there is a linear map $L$ such that:

$$ L:(U,F)\to (U,F) $$

Where $(U,F)$ is a linear space. Now, matrices are merely representations of a linear map with respect to a given basis - for example matrix $A\in F^{\dim U \times \dim U}$ could be the matrix representation of $L$ with respect to the canonical basis. We know how to find eigenvalues of the matrix $A$.

However, is there such a thing as eigenvalues of the actual linear map $L$ and not of its matrix representation? If so, how do you find them?

Specifically: we can find eigenvalues of a matrix by finding the zeros of the characteristic polynomial. Is it possible to find the characteristic polynomial of a linear map without needing to first define a basis with respect to which we can write a matrix representation? If not, is it possible to find the eigenvalues with such an algorithmic method, without having to define a basis first?

Thanks for your insight!

Answer 1

Not sure this is exactly what you want, but it's a linear transformation without explicitly specifying a matrix, or action on a basis, and also finding (real) eigenvalues.

Given a vector $v$ in ${\bf R}^3$, define the linear transformation $T_v:{\bf R}^3\to{\bf R}^3$ by $T_v(w)=v\times w$ (the cross product).

Now suppose $T_v(w)=\lambda w$. We know $T_v(w)$ is orthogonal to $w$, so we have to have $\lambda w$ orthogonal to $w$, which (for $w\ne0$) can only happen if $\lambda=0$. So, we have found the only real eigenvalue, zero (and a corresponding eigenvector is $v$).

Answer 2

Another example: consider the linear transformation on the vector space of infinitely differentiable functions given by $T(f)=f'$. You get the eigenvalues (and eigenvectors) by solving the differential equation $${dy\over dx}=\lambda y$$ Every real number $\lambda$ is an eigenvalue, with corresponding eigenvector $e^{\lambda x}$.