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.
Can a Linear operator on infinite dimensional vector space have infinite eigenvalues
eigenvalues-eigenvectorslinear algebravector-spaces
Related Solutions
You won't find anything like that. If $V$ is an infinite dimensional vector spaces, then it has an infinite basis. Any proper subset of that basis spans a proper subspace whose dimension is the cardinality of the subset. So, since an infinite set has both finite and infinite subsets, every infinite dimensional vector space has both finite and infinite proper subspaces.
There is essentially no condition on the space on which your functions live that will ensure existence of any eigenfunctions. For example, for $X=[a,b]$ and $V$ the space of real-valued (or complex-valued, hardly matters...) functions on $X$, and $L$ the multiplication-by-$x$ operator... there are no eigenfunctions at all (unless $a=b$) [EDIT: among continuous or $L^2$ or $L^p$ functions]. That is, unless the physical space is a finite set (or has a weird-enough topology so that there aren't very many continuous functions), some very simple, non-pathological operators fail to have any eigenvalues at all (EDIT: among continuous or $L^2$ or...].
Another non-pathological example that shows that it is not generally reasonable to expect eigenvalues is the Laplacian on the real line. Fourier inversion shows that everything in $L^2$ is a superposition of generalized eigenvalues [EDIT: oop, eigenfunctions] (the exponentials), but not a sum, and those eigenvalues are not in the space itself. EDIT: Fourier inversion expresses (e.g.) a Schwartz function $f$ as $$ f(x) \;=\; \int_{\mathbb R} e^{2\pi i\xi x}\;\widehat{f}(\xi)\; d\xi $$ where $\widehat{f}$ is the Fourier transform of $f$ Thus, $f$ is a superposition (=integral) of eigenfunctions (the functions $x\to e^{2\pi i\xi x}$ for $\Delta=\partial^2/\partial x^2$.
Yes, in the context of Sturm-Liouville problems (see also Fredholm alternative), the point is that the inverse of the differential operator (with boundary conditions) is a compact self-adjoint operator on a Hilbert space of functions, and the eigenvalues are in bijection by $\lambda \leftrightarrow \lambda^{-1}$, etc. The only general class of operators on infinite-dimensional spaces with a clear, simple, and happy spectral theory are compact self-adjoint, or unbounded operators which have compact self-adjoint operators as inverses/resolvents...
Thus, for example, the Laplace-Beltrami operator on a compact Riemannian manifold does provably have compact resolvent, so $L^2$ has an orthonormal basis of eigenfunctions.
EDIT: I'd also add that one almost surely wants a reasonable topology on the space of functions, related to the topology on the underlying physical space. And we'd want some sort of completeness on the space of functions, else we'd potentially lose eigenfunction/values for silly reasons.
Best Answer
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.