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?

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- Jan 10th 2009, 04:56 PMebotProperties 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 PMRapha
I'm not sure about this:

I is Identity matrix

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

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

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

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

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

$\displaystyle = det(CA - I*t)$ - Jan 10th 2009, 09:38 PMNonCommAlg
suppose $\displaystyle v$ is an eigenvector of $\displaystyle A.$ so $\displaystyle Av=\lambda v,$ for some scalar $\displaystyle \lambda.$ let $\displaystyle V_{\lambda}$ be the eigenspace corresponding to $\displaystyle \lambda.$ since all eigenvalues of $\displaystyle A$ are distinct, $\displaystyle V_{\lambda}$ is one dimensional, i.e. $\displaystyle V_{\lambda}=<v>.$

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

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