For each i, let ** f. Since ** f are distinct prime polynomials, the polynomials ** f are relatively prime (Theorem 8, Chapter 4). Thus there are polynomials ** f such that ** f. Note also that if ** f, then ** f is divisible by the polynomial p, because ** f contains each ** f as a factor. We shall show that the polynomials ** f behave in the manner described in the first paragraph of the proof.

Let ** f. Since ** f and p divides ** f for ** f, we have ** f. Thus the ** f are projections which correspond to some direct sum decomposition of the space V. We wish to show that the range of ** f is exactly the subspace ** f. It is clear that each vector in the range of ** f is in ** f for if |a is in the range of ** f, then ** f and so ** f because ** f is divisible by the minimal polynomial p. Conversely, suppose that |a is in the null space of ** f. If ** f, then ** f is divisible by ** f and so ** f, i.e., ** f. But then it is immediate that ** f, i.e., that |a is in the range of ** f. This completes the proof of statement (a).

It is certainly clear that the subspaces ** f are invariant under T. If ** f is the operator induced on ** f by T, then evidently ** f, because by definition ** f is 0 on the subspace ** f. This shows that the minimal polynomial for ** f divides ** f. Conversely, let g be any polynomial such that ** f. Then ** f. Thus ** f is divisible by the minimal polynomial p of T, i.e., ** f divides ** f. It is easily seen that ** f divides g. Hence the minimal polynomial for ** f is ** f.