Suppose that has this property that if is an eigenvalue of then is not an eigenvalue of . Show that if and only if for any . In other words the centralizer of equals the centralizer of .
It is clear that implies for any . Now suppose that for some and set . We want to prove that . We have
and so . It now follows that for any integer and thus for any and any integer we have
where is the identity matrix. Now let be a generalized eigenvector corresponding to an eigenvalue of . Then for some integer and thus, by we have . Therefore, since we are assuming that is not an eigenvalue of , we must have . So, since every element of is a linear combination of some generalized eigenvectors of , we get for all , i.e. and hence .
The exercise can also be found here.