A vector space with countable and uncountable basis at the same time

Question

Let $V$ be a vector space over $\mathbb{C}$. Two self-adjoint, commutable linear operators $\xi$ and $\eta$ act on it. Both of their eigenvectors form a complete set of $V$, but $\xi$'s eigenvalues are discrete, while $\eta$'s are continuous. How is this possible? I mean how is it possible that the same vector space has both an uncountably infinite basis and a countably infinite basis? This is taken from Dirac's Principles of Quantum Mechanics, he uses these kinds of operators without considering this problem, which buggles me.

While thinking about this, I've realized, that the problem of Fourier-series is an example of the same vector space having two bases with different cardinality. A function which equals its Fourier-series can be viewed as a vector expanded in a continous basis in time domain and in discrete basis in frequency domain.

So this problem is definitely not a contradiction, there exist vector spaces with countable and uncountable basis at the same time. Still, if a vector space is countably finite dimensional there shouldn't be uncountably many independent vectors in it, and conversely, if it is uncountably infinite dimensional, then a countably infinite set of vectors shouldn't be enough to span it.

Last but not least we are considering a closed space equipped with a scalar product, so infinite sums of vectors are allowed.

UPDATE: I've asked the same question on physics.stackexchange.com and posted an answer for myself there with the help of the answers I got here and from one of my teachers (link). It contains the same info about the "denseness" Asaf Karagila suggested. However, there is an interesting part (at least for me) namely the Fourier series expansion of a function derived as a kind of basis transformation.

Answer 1

You should differentiate between a Schauder basis and a Hamel basis. The former only spans a dense subspace, whereas the latter spans the entire basis.

Assuming the axiom of choice we can show that every two Hamel bases have the same cardinality. Of course this is not the case with Schauder bases (because a linear basis is always a topological basis too).

Perhaps the crux of the difference here, is that a Hamel basis, and the notion of linear span, only permits finite sums. In the topological case this is not the case anymore, and because this is a normed space infinite sums suffice for talking about convergence.

Answer 2

What you describe in the first paragraph is impossible, assuming you are talking about self-adjoint operators on Hilbert space. Eigenvectors for distinct eigenvalues of a self-adjoint operator are orthogonal. Hence the number is distinct eigenvalues is bounded above by the maximal size of an orthonormal set, a.k.a. the Hilbert space dimension. Any two complete orthonormal sets have the same cardinality, so you can't have one countable and the other uncountable.

However, if at least one of the operators need not be self-adjoint, the situation you describe can easily happen. E.g., consider the Hilbert space $\ell^2$ of absolute-square-summable sequences of complex numbers. Let $(a_n)$ be a bounded sequence of real numbers, and define the operator $A:\ell^2\to\ell^2$ by $A(x_0,x_1,x_2,\ldots)=(a_0x_0,a_1x_1,a_2x_2,\ldots)$. Then $A$ is a self adjoint operator with $\{a_n\}$ as a set of eigenvalues, and it has a (countable) orthonormal basis of eigenvectors.

On the same space, consider the operator $S$ defined by $S(x_0,x_1,x_2,\ldots)=(x_1,x_2,x_3,\ldots)$. This operator has all complex numbers with absolute value less than $1$ as eigenvalues, and correspondingly, an uncountable linearly independent set of eigenvectors. Namely, for $|\lambda|<1$, $\lambda$ is an eigenvalue for $A$ with eigenvector $(1,\lambda,\lambda^2,\lambda^3,\ldots)$. There is no contradiction here, because these vectors are not orthogonal, so the size is not constrained by the (countable) Hilbert space dimension. This shows that the vector space dimension (size of a maximal linearly independent set) can be greater than the Hilbert space dimension (size of a maximal orthonormal set).