Suppose we have three hermitian operators A,B,C each with only non degenerate eigenvalues.(adsbygoogle = window.adsbygoogle || []).push({});

If A and B commute, then for each eigenvector of A we can find an eigenvector of B, and because the eigenvalues are non degenerate the mapping is one to one. If B and C commute we can do the same. This implies that A and C also have a one to one mapping of their eigenbases, and so A and C commute.

But in general we know that the commutation relation is not transitive for some arbitrary set of operators {X,Y,Z}. Is this because some of them might have degenerate eigenvalues?

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# Is commutativity transitive for non degenerate eigenvalues?

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