# Eigenvectors of commuting hermitian operators?

Didn't know if this belonged here or on the maths StackExchange, let me know if I should switch over.

Currently going through a quantum mechanics class and I'm reading the following theorem (translating from french, apologies for any mistakes):

if $$A$$ and $$B$$ are two commuting hermitian operators, there exists an orthonormal basis consisting of eigenvectors shared between the two operators.

This part I get. The problem is with a particular case of this theorem:

Let $$a_i$$ and $$b_i$$ be eigenvalues of $$A$$ and $$B$$ respectively. For 3D matrices, if $$a_1=a_2\equiv a_{12}$$ and $$b_2=b_3\equiv b_{23}$$. $$b_1$$ is not degenerate, thus $$|b_1\rangle$$ is determined. $$|b_1\rangle$$ is also an eigenvector of $$A$$ with eigenvalue $$a_{12}$$ such that $$|a_{12}^{(1)}\rangle\leftrightarrow|b_1\rangle$$

I get the example in 3D, but I'm having a hard time generalizing the theorem for more dimensions. Right now, for instance, I have two 4D operators with $$a_1=a_2\neq a_3\neq a_4$$ and $$b_1\neq b_2 \neq b_3 = b_4$$.

How does this theorem work exactly for 4+ dimensions? Is it that since neither $$b_1$$ or $$b_2$$ are degenerated, both correspond to a different eigenvector of $$A$$? Do they both have an eigenvalue of $$a_{12}$$ as well?

In the proof for :

If A and B are two commuting hermitian operators, there exists an orthonormal basis consisting of eigenvectors shared between the two operators.

the following lemma is often used :

Let $$A,B$$ be two commuting hermitian operators and $$E_\lambda$$ be an eigenspace of $$A$$. Then $$E_\lambda$$ is stable by $$B$$.

The eigenspace $$E_\lambda$$ is the set of kets $$|\psi\rangle$$ such that $$A|\psi\rangle = \lambda|\psi\rangle$$. That $$E_\lambda$$ is stable by $$B$$ means that if $$|\psi\rangle\in E_\lambda$$, then $$B|\psi\rangle\in E_\lambda$$.

If $$E_\lambda$$ has dimension $$\geq 2$$, we can use the spectral theorem on $$B$$ (restricted to $$E_\lambda$$) and find common eigenvectors of both $$A$$ and $$B$$.

If $$E_\lambda$$ has dimension $$1$$, then we take a non-zero ket $$|\psi\rangle \in E_\lambda$$ and $$B|\psi\rangle \in E_\lambda$$ means that $$B|\psi\rangle = \mu|\psi\rangle$$ for some $$\mu\in\mathbb C$$.

So the general result you are looking for is :

Let $$A,B$$ be two commuting hermitian operators and $$|\psi\rangle$$ be an eigenvector of $$A$$ for a non-degenerate eigenvalue. Then $$|\psi\rangle$$ is also an eigenvector of $$B$$.

However, we have no information on the other eigenvalue. It could be degenerate or not, anything really.