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=° »,
A2
’1 ’ x 1

˜ (0, 1, 0, 0)T .
C =

36
˜ ˜
A cyclic-vector computation shows that the system V = H V is equivalent to Z = KZ,
where
® 
0 1 0 0
 
 
 
0 0 1 0
 ,
K =  
 
0 0 0 1
° »
’g1 ’g2 ’g3 ’g4
x6 + 3x5 + 3x4 + 5x3 + 6x2 + 3x ’ 3
g1 = ,
x(x3 + x2 + 1)
x6 + 5x5 + 3x3 ’ 7x2 ’ 1

g2 = ,
x3 (x3 + x2 + 1)
2x3 ’ 2x2 + 1

g3 = ,
x2
x3 ’ 2

g4 = .
x(x3 + x2 + 1)
˜
The equivalence is given by the equation Z = B V , where
® 
1 0 0 0
 
 
 0
’x
0 1
B= .
 
 ’1 + x ’2x 
’x ’1
1
° »
3x3 ’x2 +2
3x ’ x2
1
2+ x 0
x2

˜ ˜ ˜
Therefore, the equation V = H V + C is equivalent to

˜
Z = KZ + B C. (3.32)

(The reader can verify that K = B B ’1 + BHB ’1 .) Since K is in companion-matrix form,
it is easy to convert (3.32) into the inhomogeneous scalar equation L(y) = ˆ where
ˆ b,

ˆ D4 + g4 D3 + g3 D2 + g2 D + g1
L= (gi as above),
x4 + 2x3 + x2 + 4x + 3
ˆ= .
b
x3 + x2 + 1
ˆ
Computations using the DFactor and ratsols commands in DEtools show that L is ir-
reducible over C(x) and that this equation admits no rational solutions. Thus, the vector
space W referred to in the third statement of Lemma 3.3.3 is all of C 4 . We conclude that
the Galois group GL is (SL2 (C)) — SL2 (C)) C4.


Example 3.3.5 Consider the equation L(y) = 0, where L = L1 —¦ L2 , L1 = D2 + x D +
1


1, L2 = D2 ’ D.

37
As in the previous example, the Galois group GL of L is an extension of GL , the group
˜

˜
of L = LCLM(L1 , L2 ). To calculate GL note that the Galois group GL1 of L1 is SL2 and
˜

the Galois group GL2 of L2 is the multiplicative group C — . The group GL is a subgroup of
˜

GL1 — GL2 that projects surjectively onto each factor. The Theorem of [Kol68] implies that,
in this case, GL = GL1 — GL2 .
˜

We now consider the equation

˜ ˜ ˜
V = H V + C,

where

’AT — I2 + I2 — A2 ,
H =
®1 
0 1
=° »,
A1
’1 ’ x 1
® 
01
=° »,
A2
01
˜ (0, 1, 0, 0)T .
C =

˜ ˜
A cyclic-vector computation shows that the system V = H V is equivalent to Z = KZ,
where
® 
0 1 0 0
 
 
 
0 0 1 0
= ,
K  
 
0 0 0 1
° »
’g2 ’g3 ’g4
’g1
10x4 + 5x3 ’ 6x2 + 6x + 3
= ,
g1
x2 (5x2 ’ 3)
10x3 + 15x2 + 9x + 12

g2 = ,
x(5x2 ’ 3)
5x2 + 5x + 2
g3 = 3 ,
5x2 ’ 3
5x2 + 5x ’ 3
’2
g4 = .
5x2 ’ 3
˜
The equivalence is given by the equation Z = B V , where
® 
1 0 0 0
 
 
0 0
1 1
B= .
 
 ’1 1 2
1
° »
x

’ x ’2 ’1 3 + x
1 3



38
˜ ˜ ˜
Therefore, the system V = H V + C is equivalent to

˜
Z = KZ + B C.

(The reader can again verify that K = B B ’1 + BHB ’1 .) Conversion to an inhomogeneous
scalar equation yields L(y) = ˆ where
ˆ b,

ˆ D4 + g4 D3 + g3 D2 + g2 D + g1
L= (gi as above),
2
ˆ = ’2 + 5x + 5x + 12 .
b
5x2 ’ 3

Using the eigenring command of DEtools, one sees that the dimension of the endomor-
ˆ ˆ ˆ
phism ring of D/DL is two. Since L is completely reducible, this implies that D/DL is the
ˆ
direct sum of two nonisomorphic irreducible D-modules. This furthermore implies that L
has exactly two nontrivial irreducible right (resp., left) factors. A computation using the
command endomorphism charpoly yields two di¬erent right factors. From these one cal-
culates the unique left factors and then one can show that for neither of these left factors
L does the equation L(y) = ˆ have a rational solution. Since L(y) = ˆ also has no rational
¯ ¯ ˆ
b b
solutions, we conclude that GL is (SL2 (C) — C — ) C4.


Example 3.3.6 Consider the equation L(y) = 0, where L = L1 —¦ L2 , L1 = LCLM(D ’
2x, D), L2 = D2 .
Here it is clear that GL , the group of L = LCLM(L1 , L2 ), is C — .
˜
˜

We now consider the equation

˜ ˜ ˜
V = H V + C,

where

’AT — I2 + I2 — A2 ,
H =
®1 
0 1
=° »,
A1
1
0 2x + x
® 
0 1
=° »,
A2
0 0
˜ (0, 1, 0, 0)T .
C =

˜ ˜
A cyclic-vector computation shows that the system V = H V is equivalent to Z = KZ,

39
where
® 
0 1 0 0
 
 
0 
0 1 0
 ,
K =  
0 
0 0 1
° »
’h1 ’h2
0 0
8x6 ’ 12x4 + 18x2 + 9
h1 = 2 ,
4x4 + 3
x(4x4 ’ 4x2 + 3)
h2 = 4 .
4x4 + 3
˜
The equivalence is given by the equation Z = B V , where
® 
’x ’x
0 0
 
 
 
’1 ’x

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