Page 220 - DMTH502_LINEAR_ALGEBRA

Linear Algebra

Notes Apply g(T) to both sides of (5):

p gf gr g gr ...(6)
j j 0 i i
1 i j
By definition p is i W , and the last two terms on the right side of (6) are in W . Therefore, gr
j 1 j 1 j j
is in W . Now we use condition (b) on Step 1:
j 1
deg (gr ) deg ( ; W )
s
j j j 1
= deg p j

= deg ( ;s W j 1 )

= deg p
= deg (fg).
Thus deg r j deg , and that contradicts the choice of j. We now know that f divides each g and
f
i
hence that 0 f . Since W is T-admissible, 0 f 0 where is in W . We remark in passing
0
0
0
that Step 2 is a strengthened form of the assertion that each of the subspaces W W 2 , ...W is
,
r
1
T-admissible.
Step 3: There exist non-zero vectors 1 ,.... r in V which satisfy conditions (i) and (ii) of
Theorem 1.
Start with vectors 1 ,....., r as in Step 1. Fix , 1k k . r We apply Step 2 to the vector k and
the T-conductor f = p . We obtain
k
p k k p k 0 p h ...(7)
k i i
1 i k
where is in W and h ,....,h 1 are polynomials. Let
0 0 i k

h . ...(8)
k k 0 i i
1 i k
Since is in W ,
k k k – 1
( s k ; W k 1 ) s ( k ; W k 1 ) p k ...(9)

and since p k k 0, we have

W k 1 ( Z k ; ) {0}. ...(10)
T
Because each  satisfies (9) and (10), it follows that
k
W W ( Z ; ) ... Z ( ; )
T
T
k 0 1 k
and that p is the T-annihilator of . In other words, the vectors , ...., define the same
k k 1 r
sequence of subspaces W , W , ... as do the vectors , ..., and the T-conductors p = s ( , W )
1 2 1 r k k k–1
have the same maximality properties (condition (b) of Step 1). The vectors , ...., have the
1 r
additional property that the subspaces W , Z ( ; T), Z( ; T), ... are independent. It is therefore
0 1 2
easy to verify condition (ii) in Theorem 1. Since p = 0 for each i, we have the trivial relation
i i
p 0 p ... p .
k k 1 1 k 1 k 1
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