question

ibrahim_commu · **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

kompik · **Posted:** Sun, 28 Mar 2010 17:49:43 UTC

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
$A=\begin{pmatrix} 1 & 0 \\ 0 & 1\end{pmatrix}$ and $B=\begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix}$ 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 $A=PDP^{-1}$ for some diagonal matrix. The columns of P are precisely the eigenvectors of A. From AB=BA

you get
$PDP^{-1}B=BPDP^{-1}$
$DP^{-1}BP=P^{-1}BPD$
Now, using the above lemma, the matrix $P^{-1}BP$ is diagonal. Let us denote it

.
We have:
$D'=P^{-1}BP$
$PD'P^{-1}=B$
which implies that B has the same eigenvectors.

kompik · **Posted:** Sun, 28 Mar 2010 18:01:14 UTC

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 $A=PDP^{-1}$ and $B=PD'P^{-1}$ 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 DD'=D'D

holds, you get easily: $AB=PDP^{-1}PD'P^{-1}=PDD'P^{-1}=PD'DP^{-1}=PD'P^{-1}PDP^{-1}=BA$
I did not find a counterexample that without additional assumptions the claim fails.

S.O.S. Mathematics CyberBoard

question

Who is online