Page 219 - DMTH502_LINEAR_ALGEBRA

Unit 20: Cyclic Decomposition and the Rational Form

Step 1: There exist non-zero vectors , ..., in V such that Notes
1 r
(a) V = W + Z ( ; T) + ... + Z( ; T);
0 1 r
(b) if 1 k r and
W = W + Z ( ; T) + ... + Z( ; T)
k 0 1 k
then the conductor p = s( ; W ) has maximum degree among all T-conductors into the subspace
k k k – 1
W , i.e., for every k
k–1
s
deg p = max deg ( ;W k 1 )
k in V
This step depends only upon the fact that W is an invariant subspace. If W is a proper T-invariant
0
subspace, then
0 max deg ( ; W ) dim V
s
and we can choose a vector so that deg s ( ; W) attains that maximum. The subspace W + Z
( ; T) is then T-invariant and has dimension larger than dim W. Apply this process to W = W to
0
obtain . If W = W + Z( ; T) is still proper, then apply the process to W to obtain . Continue
1 1 0 1 1 2
is that manner. Since dim W > dim W , we must reach W = V is not more than dim V steps.
k k – 1 r
Step 2: Let , .... be non-zero vectors which satisfy conditions (a) and (b) of Step 1. Fix k, 1 k
1 r
r. Let be any vector in V and let f = s ( ; W ). If
k–1
f 0 g i i , i in W i
1 i k
then f divides each polynomial g and = f , where is in W .
i 0 0 0 0
If k = 1, this is just the statement that W is T-admissible. In order to prove the assertion for k > 1,
0
apply the division algorithm:
g i fh i i , r r i 0 or deg r i deg . ...(2)
f
We wish to show that r = 0 for each i, Let
i
k 1
h i i . ...(3)
1
Since – is in W ,
k – 1
s ( ; W k 1 ) s ( ; W k 1 ) . f
Furthermore

k 1
f 0 r i i . ...(4)
1
Suppose that some r is different from 0. We shall deduce a contradiction. Let j be the largest
i
index i for which r 0. Then
i
j
f r , r 0 and deg r deg .
f
0 i i j j ...(5)
1
Let p = s ( ; W ). Since W contains W , the conductor f = s ( ; W ) must divide p:
j – 1 k 1 j 1 k 1
p = fg.

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