We have just observed that we can write ** f where D is diagonalizable and N is nilpotent, and where D and N not only commute but are polynomials in T. Now suppose that we also have ** f where D' is diagonalizable, N' is nilpotent, and ** f. We shall prove that ** f.

Since D' and N' commute with one another and ** f, we see that D' and N' commute with T. Thus D' and N' commute with any polynomial in T; hence they commute with D and with N. Now we have ** f or ** f and all four of these operators commute with one another. Since D and D' are both diagonalizable and they commute, they are simultaneously diagonalizable, and ** f is diagonalizable. Since N and N' are both nilpotent and they commute, the operator ** f is nilpotent; for, using the fact that N and N' commute ** f and so when r is sufficiently large every term in this expression for ** f will be 0. (Actually, a nilpotent operator on an n-dimensional space must have its nth power 0; if we take ** f above, that will be large enough. It then follows that ** f is large enough, but this is not obvious from the above expression.) Now ** f is a diagonalizable operator which is also nilpotent. Such an operator is obviously the zero operator; for since it is nilpotent, the minimal polynomial for this operator is of the form ** f for some ** f; but then since the operator is diagonalizable, the minimal polynomial cannot have a repeated root; hence ** f and the minimal polynomial is simply x, which says the operator is 0. Thus we see that ** f and ** f.