Creighton Hogg <wchogg@hep.wisc.edu> wrote in message

This sounds like a homework problem. Anyway, if I remember correctly two
diagonalizable matrices A and B can be simultaneously diagnonalized if and
only if they commute.

Thanks for your response, here is my concern however. Consider the
following problem of a double pendulum that has two identical masses
both on rods of length l. Suppose we use the usual general
coordinates: theta (for the upper rod) and phi (for the lower rod).
Then we have a kinetic energy matrix:

|2 1| |2 0|
M = m(l^2)| | and a potential energy matrix K = mgl| |
|1 1| |0 1|

The normal mode vectors are |1>=(1, root(2)) and |2>=(-1, root(2)).
I've used ket notation since vectors are hard to represent in text and
the vectors are intended to be columns as well.

Now we can form the modal matrix

A=(|1> |2>) we haven't yet normalize the normal modes, but this
doesn't change the diagonalization in this problem just what the
resulting diagonal entries are.

Now the kinetic energy matrix M can be diagonalized by

Transpose(A)*M*A and the potential energy matrix can be diagonalized
as well
by Transpose(A)*K*A. (if the modal matrix contained normalized
(relative to M) eigenvectors then M would be diagonalized to the
identity).

But, consider the following products

|2 1| * |2 0| |4 1| versus |2 0| * |2 1| |4 2|
|1 1| |0 1| = |2 1| |0 1| |1 1| = |1 1|

So the matrices clearly don't commute, but we have just simultaneously
diagonalized them. I buy the fact the if two hermitian operators
commute then they can be simultaneously diagonalized (seen the proof).
But I would like to know what simultaneous diagonalization implies
(since it doesn't seem to imply commutation).

Any thoughts?