6.4.

The primary decomposition theorem

We are trying to study a linear operator T on the finite-dimensional space V, by decomposing T into a direct sum of operators which are in some sense elementary.
We can do this through the characteristic values and vectors of T in certain special cases, i.e., when the minimal polynomial for T factors over the scalar field F into a product of distinct monic polynomials of degree 1.
What can we do with the general T?
If we try to study T using characteristic values, we are confronted with two problems.
First, T may not have a single characteristic value;
this is really a deficiency in the scalar field, namely, that it is not algebraically closed.
Second, even if the characteristic polynomial factors completely over F into a product of polynomials of degree 1, there may not be enough characteristic vectors for T to span the space V.
This is clearly a deficiency in T.
The second situation is illustrated by the operator T on Af (F any field) represented in the standard basis by Af.
The characteristic polynomial for A is Af and this is plainly also the minimal polynomial for A (or for T).
Thus T is not diagonalizable.
One sees that this happens because the null space of Af has dimension 1 only.
On the other hand, the null space of Af and the null space of Af together span V, the former being the subspace spanned by Af and the latter the subspace spanned by Af and Af.

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 Af where Af are distinct elements of F, then we shall show that the space V is the direct sum of the null spaces of Af.
The diagonalizable operator is the special case of this in which Af 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.

Theorem 12.

(primary decomposition 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, Af, where the Af, are distinct irreducible monic polynomials over F and the Af are positive integers.
Let Af be the null space of Af.
Then (A) Af;
(B) each Af is invariant under T;
(C) if Af is the operator induced on Af by T, then the minimal polynomial for Af is Af.
Proof.

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

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

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

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

Corollary.

If Af are the projections associated with the primary decomposition of T, then each Af is a polynomial in T, and accordingly if a linear operator U commutes with T then U commutes with each of the Af, i.e., each subspace Af is invariant under U.

In the notation of the proof of Theorem 12, let us take a look at the special case in which the minimal polynomial for T is a product of first-degree polynomials, i.e., the case in which each Af is of the form Af.
Now the range of Af is the null space Af of Af.
Let us put Af.
By Theorem 10, D is a diagonalizable operator which we shall call the diagonalizable part of T.
Let us look at the operator Af.
Now Af Af so Af.
The reader should be familiar enough with projections by now so that he sees that Af and in general that Af.
When Af for each i, we shall have Af, because the operator Af will then be 0 on the range of Af.

Definition.

Let N be a linear operator on the vector space V.
We say that N is nilpotent if there is some positive integer R such that Af.

Theorem 13.

Let T be a linear operator on the finite-dimensional vector space V over the field F.
Suppose that the minimal polynomial for T decomposes over F into a product of linear polynomials.
Then there is a diagonalizable operator D on V and a nilpotent operator N in V such that (A) Af, (b) Af.
The diagonalizable operator D and the nilpotent operator N are uniquely determined by (A) and (B) and each of them is a polynomial in T.
Proof.

We have just observed that we can write Af 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 Af where D' is diagonalizable, N' is nilpotent, and Af.
We shall prove that Af.

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

Corollary.

Let V be a finite-dimensional vector space over an algebraically closed field F, e.g., the field of complex numbers.
Then every linear operator T in V can be written as the sum of a diagonalizable operator D and a nilpotent operator N which commute.
These operators D and N are unique and each is a polynomial in T.

From these results, one sees that the study of linear operators on vector spaces over an algebraically closed field is essentially reduced to the study of nilpotent operators.
For vector spaces over non-algebraically closed fields, we still need to find some substitute for characteristic values and vectors.
It is a very interesting fact that these two problems can be handled simultaneously and this is what we shall do in the next chapter.

In concluding this section, we should like to give an example which illustrates some of the ideas of the primary decomposition theorem.
We have chosen to give it at the end of the section since it deals with differential equations and thus is not purely linear algebra.

Example 11.

In the primary decomposition theorem, it is not necessary that the vector space V be finite dimensional, nor is it necessary for parts (A) and (B) that P be the minimal polynomial for T.
If T is a linear operator on an arbitrary vector space and if there is a monic polynomial P such that Af, then parts (A) and (B) of Theorem 12 are valid for T with the proof which we gave.

Let N be a positive integer and let V be the space of all N times continuously differentiable functions F on the real line which satisfy the differential equation Af where Af are some fixed constants.
If Af denotes the space of N times continuously differentiable functions, then the space V of solutions of this differential equation is a subspace of Af.
If D denotes the differentiation operator and P is the polynomial Af then V is the null space of the operator p (,), because Af simply says Af.
Let us now regard D as a linear operator on the subspace V.
Then Af.

If we are discussing differentiable complex-valued functions, then Af and V are complex vector spaces, and Af may be any complex numbers.
We now write Af where Af are distinct complex numbers.
If Af is the null space of Af, then Theorem 12 says that Af.
In other words, if F satisfies the differential equation Af, then F is uniquely expressible in the form Af where Af satisfies the differential equation Af.
Thus, the study of the solutions to the equation Af is reduced to the study of the space of solutions of a differential equation of the form Af.
This reduction has been accomplished by the general methods of linear algebra, i.e., by the primary decomposition theorem.

To describe the space of solutions to Af, one must know something about differential equations;
that is, one must know something about D other than the fact that it is a linear operator.
However, one does not need to know very much.
It is very easy to establish by induction on R that if F is in Af then Af;
that is, Af, etc.
Thus Af if and only if Af.
A function G such that Af, i.e., Af, must be a polynomial function of degree Af or less: Af.
Thus F satisfies Af if and only if F has the form Af.
Accordingly, the 'functions' Af span the space of solutions of Af.
Since Af are linearly independent functions and the exponential function has no zeros, these R functions Af, form a basis for the space of solutions.