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Prove two commutative Hermitian matrices have the same eigenvectors 
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#1
Feb1113, 01:56 AM

P: 7

Hi,
Does anyone know how to prove that two commutative Hermitian matrices can always have the same set of eigenvectors? i.e. AB  BA=0 A and B are both Hermitian matrices, how to prove A and B have the same set of eigenvectors? Thanks! 


#2
Feb1113, 01:30 PM

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PF Gold
P: 9,347




#3
Feb1113, 03:03 PM

P: 7

What I would say is that this is a common mathematical theorem which is one of the mathematical basis in quantum mechanics, but not a textbook style problem. I ask here because my text book (which is not written in English) only gives a simplified version of proof (assuming there is no duplicated eigenvalue). I am curious about a relatively more robust way to prove it. 


#4
Feb1113, 03:43 PM

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PF Gold
P: 9,347

Prove two commutative Hermitian matrices have the same eigenvectors
Textbook problems often ask the reader to prove a theorem that wasn't proved in the text. So this certainly could be a textbook problem or a small part of a homework assignment.



#5
Feb1113, 06:31 PM

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P: 18,231

I have moved this to homework. xuphys, please make an attempt when asking a question. What do you think of the problem? Is there something you can do?



#6
Feb1213, 12:27 AM

P: 7

Then AB is diagonalizable:
Thanks! 


#7
Feb1213, 11:13 AM

P: 24

Hermitian matrices are diagonalizable, so we have n eigenpairs (a_i, v_i) for A and eigenpairs (b_i, u_i) for B (where v_i and u_i may be chosen linearly independent).
It follows that A*v_i = a_i*v_i. This implies B*A*v_i = a_i*B*v_i. But this equals A*B*v_i, so B*v_i is an eigenvector to A with eigenvalue a_i. So B*v_i = k_i*v_i if eigenvalues are unique. The argument is symmetric, so it follows that A and B have the same eigenvectors. The argument is more difficult if eigenvalues aren't unique, but you get invariant subspaces and block matrices, for which you can choose diagonal bases, roughly. 


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