**Problem**. 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 cenralizer of equals the centralizer of

**Solution**. It is clear that implies for any Now suppose that for some and put 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 eigenectors of we get for all i.e. and hence

**Example**. The converse of the result given in the above problem is trivially false. For example, consider

Then and so if and only if but is an eigenvalue of