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 Post subject: questionPosted: Sun, 28 Mar 2010 16:22:09 UTC
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Joined: Sun, 28 Mar 2010 16:11:46 UTC
Posts: 1
if matrix A and matrix B are commute
then prove that A & B have the same eigenvectors and vice reverse
plz replay before 12 am today

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 Post subject: Re: questionPosted: Sun, 28 Mar 2010 17:49:43 UTC
 S.O.S. Oldtimer

Joined: Sun, 4 Nov 2007 12:08:30 UTC
Posts: 245
Location: Bratislava, Slovakia
ibrahim_commu wrote:
if matrix A and matrix B are commute
then prove that A & B have the same eigenvectors and vice reverse
plz replay before 12 am today

I think some assumptions are missing.
Clearly
and do commute, they both have only one eingevalue 1, but the eigenspace of B is only one-dimensional, whereas every vector is eigenvector of A.

However, you can show:
If AB=BA and A has n distinct eigenvalues, then B has the same eigenvectors as A.

First, try to show the following easy lemma: If D is a diagonal matrix with distinct diagonal elements and DX=XD, then X is diagonal as well.

Now, you have for some diagonal matrix. The columns of P are precisely the eigenvectors of A. From you get

Now, using the above lemma, the matrix is diagonal. Let us denote it .
We have:

which implies that B has the same eigenvectors.

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 Post subject: Re: questionPosted: Sun, 28 Mar 2010 18:01:14 UTC
 S.O.S. Oldtimer

Joined: Sun, 4 Nov 2007 12:08:30 UTC
Posts: 245
Location: Bratislava, Slovakia
ibrahim_commu wrote:
if matrix A and matrix B are commute
then prove that A & B have the same eigenvectors and vice reverse
plz replay before 12 am today

The other direction is easy, if A and B are diagonalizable. In this case you have and for some diagonal matrices D and D'. (You can use the same B, because in both case you have the same basis consisting of eigenvectors.)
Now, since for diagonal matrices holds, you get easily:
I did not find a counterexample that without additional assumptions the claim fails.

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