# Properties of Commutative Matrix

• Jan 10th 2009, 04:56 PM
ebot
Properties of Commutative Matrix
If A and C are both nxn matrices that have n distinct different eigenvalues.

AC=CA.

How do you show that the eigenvectors of A and C are the same? How do the eigenvalues relate?
• Jan 10th 2009, 07:22 PM
Rapha
Quote:

Originally Posted by ebot
If A and C are both nxn matrices that have n distinct different eigenvalues.

AC=CA.

How do you show that the eigenvectors of A and C are the same? How do the eigenvalues relate?

I is Identity matrix

$det(AC - t I) = det ( A(C-A^{-1}t))$

$= det(A) * det((C-A^{-1}t))$

$= det(C-A^{-1}t) * det(A)$

$= det ( (C-A^{-1}t)*A )$

$= det(CA - A^{-1}A t)$

$= det(CA - I*t)$
• Jan 10th 2009, 09:38 PM
NonCommAlg
Quote:

Originally Posted by ebot
If A and C are both nxn matrices that have n distinct different eigenvalues and AC=CA, how do you show that the eigenvectors of A and C are the same? How do the eigenvalues relate?

suppose $v$ is an eigenvector of $A.$ so $Av=\lambda v,$ for some scalar $\lambda.$ let $V_{\lambda}$ be the eigenspace corresponding to $\lambda.$ since all eigenvalues of $A$ are distinct, $V_{\lambda}$ is one dimensional, i.e. $V_{\lambda}=.$

now we have $\lambda Cv=CAv=ACv.$ thus $Cv \in V_{\lambda}=.$ hence $Cv=\mu v,$ for some scalar $\mu.$ that means $v$ is also an eigenvector of $C.$ so we've proved that every eigenvector of $A$ is an

eigenvector of $C.$ a similar argument shows that every eigenvector of $C$ is an eigenvector of $A. \ \Box$