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Unit 18: The Primary Decomposition Theorem
spaces over non-algebraically closed fields, we still need to find some substitute for characteristic Notes
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 units.
In concluding this section, we should like to give examples, which illustrate 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 1: In the primary decomposition theorem, it is not necessary that the vector
space V be finite dimensional, nor is it necessary for parts (i) and (ii) 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 p(T) = 0, then parts (i) and (ii) of Theorem 1 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.
n
d f d n 1 f d j
a n 1 a 1 a f 0 ...(1)
0
dt n dt n 1 dt
where a ,..., a are some fixed constants. If C denotes the space of n times continuously
0 n–1 n
differentiable functions, then the space V of solutions of this differential equation is a subspace
of C . If D denotes the differentiation operator and p is the polynomial
n
n–1
p = x + a x + ... + a x + a
n
n–1 1 0
then V is the null space of the operator p(D), because (1) simply says p(D)f = 0. Therefore, V is
invariant under D. Let us now regard D as a linear operator on the subspace V. Then p(D) = 0.
If we are discussing differentiable complex-valued functions, then C and V are complex vector
n
spaces, and a ,..., a may be any complex numbers. We now write
0 n–1
r
p = (x – c ) 1 ... (x – c ) k r
1 k
r
where c ,..., c are distinct complex numbers. If W is the null space of (*D – c I) i , then Theorem 1
1 k j j
says that
V = W ... W
1 k
In other words, if f satisfies the differential equation (1), then f is uniquely expressible in the
form
f = f + ... + f
1 k
r
where f satisfies the differential equation (D – c I) j f = 0. Thus, the study of the solutions to the
j j j
equation (1) is reduced to the study of the space of solutions of a differential equation of the form
(D – cI) f = 0 ...(2)
r
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 (2), 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 C then
r
r
(D – cI) f = e D (e f )
–ct
ct
r
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