Page 207 - DMTH502_LINEAR_ALGEBRA

Unit 18: The Primary Decomposition Theorem

Since p ,..., p are distinct prime polynomials, the polynomials f ,..., f are relatively prime. Thus Notes
1 k 1 k
there are polynomials g ,..., g such that
1 k
n
 f g  1
i
i
i 1
m r
Note also that if i  j, then f f is divisible by the polynomial p, because f f contains each p as a
i j i j m
factor. We shall show that the polynomials h = f g behave in the manner described in the first
i i i
paragraph of the proof.
Let E = h (T) = f (T)g (T). Since h + ... + h = 1 and p divides f f for i  j, we have
i i i i 1 k i j
E + ... + E = I
1 k
E E = 0, if i  j
i j
Thus the E are projections which correspond to some direct sum decomposition of the space V.
i
We wish to show that the range of E is exactly the subspace W . It is clear that each vector in the
i i
range of E is in W , for if  is in the range of E , then  = E  and so
i i i i
r
p (T) i  = p (T) i E 
r
i i i
r
= p (T) i f (T)g (T)
i i i
= 0
r
because p i f g is divisible by the minimal polynomial p. Conversely, suppose that  is in the null
i i
i r
r
space of p (T) i . If j  i, then f g is divisible by p and so f (T)g (T) = 0, i.e., E  = 0 for j  i. But then
i j j i j j j
it is immediate that E = , i.e., that  is in the range of E. This completes the proof of statement (i).
i i
It is certainly clear that the subspaces W are invariant under T. If T is the operator induced on W
i i i
ri
by T, then evidently p (T ) i = 0, because by definition p (T) is 0 on the subspace W . This shows
r
i i i i
i r
that the minimal polynomial for T divides p . Conversely, let g be any polynomial such that
i i
i r
g(T ) = 0. Then g(T)f (T) = 0. Thus gf is divisible by the minimal polynomial p of T, i.e., p f
i i i i i
i r
divides gf . It is easily seen that p divides g. Hence the minimal polynomial for T is p i r .
i i i i
Corollary: If E ,..., E are the projections associated with the primary decomposition of T, then
1 k
each E is a polynomial in T, and accordingly if a linear operator U commutes with T then U
i
commutes with each of the E , i.e., each subspace W is invariant under U.
i i
In the notation of the proof of Theorem 1, 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 p
i
r
is of the form p = x – c . Now the range of E is the null space W of (T – c I) i . Let us put D = c E +
i i i i i 1 1
... + c E . By Theorem 2 of unit 17, D is a diagonalizable operator which we shall call the
k k
diagonalizable part of T. Let us look at the operator N = T – D. Now
T = TE + ... + TE
1 k
D = c E + ... + c E
1 1 k k
so
N = (T – c I)E + ... + (T – c I)E
1 1 k k
The reader should be familiar enough with projections by now so that he sees that
N = (T – c I) E + ... + (T – c I) E
2
2
2
1 1 k k
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