Simultaneous diagonalization while having repeated eigenvalues

[McLaren Rulez]

Hi,

Can anyone help me prove that two commuting matrices can be simultaneously diagonalized? I can prove the case where all the eigenvalues are distinct but I'm stumped when it comes to repeated eigenvalues.

I came across this proof online but I am not sure how $$B'_{ab}=0$$ implies that B is block diagonal. Thank you.

http://www.mathematics.thetangentbundle.net/wiki/Linear_algebra/simultaneous_diagonalization_of_commuting_normal_matrices is the link for the proof.

 

[McLaren Rulez]

Okay I see the block diagonal bit but I still cannot see how to diagonalize those little blocks where all the diagonal elements are the same.

Basically, if v1 is an eigenvector of matrix P with eigenvalue a1 and we have PQ=QP, then PQ(v1) = QP(v1) = a1*Q(v1) which shows that Q(v1) is an eigenvector of P with eigenvalue a1. If the eigenvalues are distinct, then I can say that Q(v1) is proportional to v1 which makes v1 an eigenvector of Q as well. But this last step fails for repeated eigenvalues. So can anyone help me with this?

Thank you.

 

[McLaren Rulez]

Erm anybody?

 

[HallsofIvy]

First, two matrices cannot be "simultaneously diagonalized" (i.e. A and B are simultaneously diagonalizable if and only if there exist a specific matrix P such that both P^{-1}AP and P^{-1}BP are both diagonal) unless they are each diagonalizable separately. So you must assume, even though there are repeated eigenvalues, that there exist a basis consisting entirely of eigenvectors of, say, A.

 

[McLaren Rulez]

Ok let me try a concrete example. Say matrix A and B commute. Let's say A has only one eigenvalue, m. Let the eigenvectors be (1,0,0) (0,1,0) and (0,0,1). Now how do I know that each of these is also an eigenvector of B?

I assume that is the argument being used here i.e. the two matrices have a common set of eigenvectors. So when we change the basis to the one formed by the eigenvectors, both must be diagonalized.

Thank you

 

[Hurkyl]

Ok let me try a concrete example. Say matrix A and B commute. Let's say A has only one eigenvalue, m. Let the eigenvectors be (1,0,0) (0,1,0) and (0,0,1). Now how do I know that each of these is also an eigenvector of B?

Note that A=mI... (where I is the identity matrix)

 

[McLaren Rulez]

Oh I see it now! Just diagonalize B and A is unaffected since the identity matrix doesn't change when the basis is changed. Thank you!