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'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 ) $
$\displaystyle = det(CA - A^{-1}A t)$
$\displaystyle = det(CA - I*t)$
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 $\displaystyle A. \ \Box$