Is commutativity transitive for non degenerate eigenvalues?

[bob900]

Suppose we have three hermitian operators A,B,C each with only non degenerate eigenvalues.

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?

 

[Bill_K]

Suppose [X,Y] = 0 and [Y,Z] = 0, and suppose transitivity fails. That means that within some eigenspace of Y, [X,Z] ≠ 0, implying that the eigenspace is at least two-dimensional.