ñòð. 9 |

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

ñòð. 9 |