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=ï£° ï£»,
A2
âˆ’1 âˆ’ x 1

Ëœ (0, 1, 0, 0)T .
C =

36
Ëœ Ëœ
A cyclic-vector computation shows that the system V = H V is equivalent to Z = KZ,
where
ï£® ï£¹
0 1 0 0
ï£¯ ï£º
ï£¯ ï£º
ï£¯ ï£º
0 0 1 0
ï£¯ ï£º,
K = ï£¯ ï£º
ï£¯ ï£º
0 0 0 1
ï£° ï£»
âˆ’g1 âˆ’g2 âˆ’g3 âˆ’g4
x6 + 3x5 + 3x4 + 5x3 + 6x2 + 3x âˆ’ 3
g1 = ,
x(x3 + x2 + 1)
x6 + 5x5 + 3x3 âˆ’ 7x2 âˆ’ 1
âˆ’
g2 = ,
x3 (x3 + x2 + 1)
2x3 âˆ’ 2x2 + 1
âˆ’
g3 = ,
x2
x3 âˆ’ 2
âˆ’
g4 = .
x(x3 + x2 + 1)
Ëœ
The equivalence is given by the equation Z = B V , where
ï£® ï£¹
1 0 0 0
ï£¯ ï£º
ï£¯ ï£º
ï£¯ 0ï£º
âˆ’x
0 1
B=ï£¯ ï£º.
ï£¯ ï£º
ï£¯ âˆ’1 + x âˆ’2x ï£º
âˆ’x âˆ’1
1
ï£° ï£»
3x3 âˆ’x2 +2
3x âˆ’ x2
1
2+ x 0
x2

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

Ëœ
Z = KZ + B C. (3.32)

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

Ë† D4 + g4 D3 + g3 D2 + g2 D + g1
L= (gi as above),
x4 + 2x3 + x2 + 4x + 3
Ë†= .
b
x3 + x2 + 1
Ë†
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)) Ã— SL2 (C)) C4.

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

1, L2 = D2 âˆ’ D.

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

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

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

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

We now consider the equation

Ëœ Ëœ Ëœ
V = H V + C,

where

âˆ’AT âŠ— I2 + I2 âŠ— A2 ,
H =
ï£®1 ï£¹
0 1
=ï£° ï£»,
A1
âˆ’1 âˆ’ x 1
ï£® ï£¹
01
=ï£° ï£»,
A2
01
Ëœ (0, 1, 0, 0)T .
C =

Ëœ Ëœ
A cyclic-vector computation shows that the system V = H V is equivalent to Z = KZ,
where
ï£® ï£¹
0 1 0 0
ï£¯ ï£º
ï£¯ ï£º
ï£¯ ï£º
0 0 1 0
=ï£¯ ï£º,
K ï£¯ ï£º
ï£¯ ï£º
0 0 0 1
ï£° ï£»
âˆ’g2 âˆ’g3 âˆ’g4
âˆ’g1
10x4 + 5x3 âˆ’ 6x2 + 6x + 3
= ,
g1
x2 (5x2 âˆ’ 3)
10x3 + 15x2 + 9x + 12
âˆ’
g2 = ,
x(5x2 âˆ’ 3)
5x2 + 5x + 2
g3 = 3 ,
5x2 âˆ’ 3
5x2 + 5x âˆ’ 3
âˆ’2
g4 = .
5x2 âˆ’ 3
Ëœ
The equivalence is given by the equation Z = B V , where
ï£® ï£¹
1 0 0 0
ï£¯ ï£º
ï£¯ ï£º
ï£¯0 0ï£º
1 1
B=ï£¯ ï£º.
ï£¯ ï£º
ï£¯ âˆ’1 1 2ï£º
1
ï£° ï£»
x

âˆ’ x âˆ’2 âˆ’1 3 + x
1 3

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

Ëœ
Z = KZ + B C.

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

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

Using the eigenring command of DEtools, one sees that the dimension of the endomor-
Ë† Ë† Ë†
phism ring of D/DL is two. Since L is completely reducible, this implies that D/DL is the
Ë†
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 diï¬€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) = Ë† have a rational solution. Since L(y) = Ë† also has no rational
Â¯ Â¯ Ë†
b b
solutions, we conclude that GL is (SL2 (C) Ã— C âˆ— ) C4.

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

We now consider the equation

Ëœ Ëœ Ëœ
V = H V + C,

where

âˆ’AT âŠ— I2 + I2 âŠ— A2 ,
H =
ï£®1 ï£¹
0 1
=ï£° ï£»,
A1
1
0 2x + x
ï£® ï£¹
0 1
=ï£° ï£»,
A2
0 0
Ëœ (0, 1, 0, 0)T .
C =

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

39
where
ï£® ï£¹
0 1 0 0
ï£¯ ï£º
ï£¯ ï£º
ï£¯0 ï£º
0 1 0
ï£¯ ï£º,
K = ï£¯ ï£º
ï£¯0 ï£º
0 0 1
ï£° ï£»
âˆ’h1 âˆ’h2
0 0
8x6 âˆ’ 12x4 + 18x2 + 9
h1 = 2 ,
4x4 + 3
x(4x4 âˆ’ 4x2 + 3)
h2 = 4 .
4x4 + 3
Ëœ
The equivalence is given by the equation Z = B V , where
ï£® ï£¹
âˆ’x âˆ’x
0 0
ï£¯ ï£º
ï£¯ ï£º
ï£¯ ï£º
âˆ’1 âˆ’x
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