This will be more or less our general method for the second problem. If (remember this is an assumption) the minimal polynomial for T decomposes ** f where ** f are distinct elements of F, then we shall show that the space V is the direct sum of the null spaces of ** f. The diagonalizable operator is the special case of this in which ** f for each i. The theorem which we prove is more general than what we have described, since it works with the primary decomposition of the minimal polynomial, whether or not the primes which enter are all of first degree. The reader will find it helpful to think of the special case when the primes are of degree 1, and even more particularly, to think of the proof of Theorem 10, a special case of this theorem.

Let T be a linear operator on the finite dimensional vector space V over the field F. Let p be the minimal polynomial for T, ** f where the ** f are distinct irreducible monic polynomials over F and the ** f are positive integers. Let ** f be the null space of ** f. Then (a) ** f (b) each ** f is invariant under T (c) if ** f is the operator induced on ** f by T, then the minimal polynomial for ** f is ** f.

The idea of the proof is this. If the direct sum decomposition (a) is valid, how can we get hold of the projections ** f associated with the decomposition? The projection ** f will be the identity on ** f and zero on the other ** f. We shall find a polynomial ** f such that ** f is the identity on ** f and is zero on the other ** f, and so that ** f, etc..