# Infinite Coupled Masses, symmetry, and the simultaneous diagonal theorem for infinite dimensional vector spaces

In The Physics of Waves by Georgi, in Chapter 4, we show that, in a coupled system of masses connected by springs, a transformation that preserves some symmetry $$S$$ commutes with $$K^{-1}M$$. From my linear algebra knowledge, this means they share an eigenvector basis. Moreover, the eigenvalues of both operators are distinct, so we can map eigenvalues from $$S$$ to eigenvalues of $$K^{-1}M$$. This means we can solve the eigenvalue problem for $$S$$ and then use the map to find the normal frequencies (i.e. eigenvalues of $$K^{-1}M$$) of the system. I'm pretty sure I understand this finite-case very well (unless something in the above reasoning is incorrect, in which case please let me know!)

My problem is how the textbook handles the case where we have infinitely-many coupled masses, like so, We take $$S$$ to be a relabeling of the $$j$$th mass to the $$(j-1)$$th mass, which clearly preserves the dynamics of the system. My first question: $$S$$ and $$K^{-1}M$$ commute, so do they share an eigenvector basis? If this were a finite-dimensional vector space, surely it would, but now we're dealing with infinite-dimensional vectors and matrices, so I don't know if the theorem still applies. The book assumes (without justification) that it does.

Suppose $$A_j^{(\beta)}$$ is the $$j$$th component of the eigenvector of $$S$$ with eigenvalue $$\beta$$. You can show with some simple algebra that $$A_j^{(\beta)}=\beta^j$$. This implies that each eigenspace is one-dimensional, because any two eigenvectors with the same eigenvalue $$\beta$$ are scalar multiples of each other.

Under the assumption that $$S$$ and $$M^{-1}K$$ share the same eigenvector basis and a few steps that I don't understand later (equations 5.14-5.16 on page 112 of the textbook), we arrive at $$\omega^2=2B-C\beta-C\beta^{-1}$$. By this equation, $$\beta$$ and $$\beta^{-1}$$, both eigenvalues of $$S$$, correspond the same eigenvalue $$\omega^2$$ of $$M^{-1}K$$. But this is a contradiction if the two operators share the same eigenvector basis. My second question: Is this actually a contradiction, or is there some mathematical thing happening that I'm not seeing?