can a Linear operator on infinite dimensional vector space have infinite eigenvalues?

Question

I know that for a finite dimensional vector space, the number of eigenvalues is at most the dimension of the vector space. Is there an example of an infinite dimensional vector space and a linear operator such that the linear operator allows infinite eigenvalues? I cannot think of such an example, but I was wondering if it is possible to construct one since I was also unable to prove why the number of eigenvalues would necessarily be finite in such a case.

Answer 1

Of course there is, as a linear map is given by the images of a basis.
So, let $(B_i \mid i \in \mathbb{N})$ be a basis of an infinite dimensional $\mathbb{R}$-vector space $V$. Define a map $f : V \to V$ by putting $f(B_i) = i\cdot B_i$ and you have infinitely many eigenvalues.

Answer 2

In $\ell^2$, let$$A(x_1,x_2,x_3,x_4,\ldots)=\left(x_1,\frac{x_2}2,\frac{x_3}3,\ldots\right).$$Then, for each $n\in\mathbb N$, $\frac1n$ is an eigenvalue of $A$.

Answer 3

Another example from a different area just for the sake of building intuition. If you let $C^\infty(\mathbb{R})$ be the vector space of infinitely differentiable functions $f:\mathbb{R} \to \mathbb{R}$, and let $L$ be the derivative operator, mapping $L(f) = f'$, then the eigenvalues of $L$ satisfy $$ kf = L(f) = f', $$ in other words any real $k$ is an eigenvalue with the eigenvector $$ f(x) = e^{kx}. $$

So here is an example with not just infinite, but uncountable eigenvectors/eigenvalues.