Prove that the number of similarity classes of matrices over which have a given characteristic polynomial in is the same as the number of similarity classes having characteristic polynomial in , where is any extension of . Demonstrate that the analogous result does not hold for matrices.

Let be a degree 3 polynomial, and suppose is the characteristic polynomial of a matrix over . There are five distinct possibilities for the irreducible factorization of .

- for some . Now is the irreducible factorization of over any extension of . Over both and , the possible lists of invariant factors whose product is are , , ; , , and . Thus over both and there are 3 similarity classes of matrices having characteristic polynomial .
- for some distinct . Similarly, this is the irreducible factorization of over any extension of . Over both fields, the possible lists of invariant factors whose product is are , and . Thus over both and there are 2 similarity classes of matrices having characteristic polynomial .
- for some distinct . Again this is the irreducible factorization of over any extension of . Over both fields, there is only one possible list of invariant factors whose product is , and that is . Thus over both and there is only one similarity class of matrices having characteristic polynomial .
- for some . Now over any extension of , the polynomial is either irreducible or factors as a product of linear polynomials. We claim that is not a perfect square. To see this, suppose we have for some . Now , but comparing coefficients, we see that is rational- a contradiction since is irreducible over . So in both cases, over both and any list of invariant factors whose product is must include, as divisors of its largest term, all factors of as well as . Thus over both and there is only one similarity class of matrices having characteristic polynomial .
- is irreducible in . There is then only one similarity class of matrices over having characteristic polynomial . Now let be an extension of . There are five possibilities for the irreducible factorization of over .
- is irreducible in . In this case, there is only one similarity class of matrices over having characteristic polynomial , as desired.
- factors into one linear and one irreducible quadratic. In any list of irreducible factors whose product is , both of these factors must appear in the largest term. Thus there is only one similarity class of matrices over having characteristic polynomial , as desired.
- factors into three distinct linear terms. Again, in any list of irreducible factors whose product is , all three factors must appear in the largest term. So there is only one similarity class of matrices over having characteristic polynomial , as desired.
- for some distinct . We claim that in fact . To see this, note that by comparing coefficients, we have (1) , (2) , and (3) . Substituting (1) into (2) for , we have (I) . Substituting (1) into (3) for , we have (II) . Multiplying (2) by and then substituting (3), we have (III) . Now , and thus . So , and thus . But then is reducible over , a contradiction.
- for some . Comparing coefficients, we see that is rational, a contradiction since is irreducible over .

Now consider . Over , factors into irreducibles as , but over , it factors into irreducibles as . Over there are two similarity classes of matrices having characteristic polynomial , but over there are 4 such classes.