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2nd step (forward substitution):
x
x 2 = b 11 b 12
a 11 a 12
0 a 22
x
x 2 + β 13
= b 11 a 11 b 11 a 12 + b 12 a 22
β 23
0 b 22
0
b 22 a 22
x
x 2 + β 13
,
(B.10)
β 23
subject to a 22 = a 11 .
Both the matrices A :=
and B :=
are upper triangular such that
A
B
1
B =I 2
B =
A .
(B.11)
A
3rd step (backward substitution):
=( a 11 a 22 ) 1 a 22 −a 12
= a 1
−a 3
11 a 12
1
⇒ b 11 = a 1
11
11 ,b 12
B =
a 2
0 a 11
0
A
11
a 3
=
11 a 12 ,
(B.12)
or
b 22 a 22 = b 22 a 11 =1
b 22 = a 1
22 = a 2
b 11 = a 1
11 , 11 a 11 =1
11 ,
(B.13)
a 1
a 3
b 11 a 12 + b 12 a 22 =0
b 12 =
22 a 12 b 11 =
11 a 12 ,
x ( y )= a 1
a 3
11 a 12 y 2 .
11 y
(B.14)
End of Example.
Example B.2 (Inversion of an univariate homogeneous polynomial of degree n =3).
Assume the univariate homogeneous polynomial y ( x )= a 11 x + a 12 x 2 + a 13 x 3 to be given and find
the inverse univariate homogeneous quadratic polynomial x ( y )= b 11 y + b 12 y 2 + b 13 y 3 by the GKS
algorithm.
1st step:
x ( y )= b 11 y + b 12 y 2 + b 13 y 3
= b 11 a 11 x +( b 11 a 12 + b 12 a 11 ) x 2 +( b 11 a 13 +2 b 12 a 11 a 12 + b 13 a 11 ) x 3
+ β 14 ,
x 2 ( y )= b 22 y 2 + b 23 y 3 + β 24
(B.15)
= b 22 a 11 x 2 +(2 b 12 a 11 a 12 + b 23 a 11 ) x 3 + β 24 ,
x 3 ( y )= b 33 y 3 + β 34
= b 33 a 11 x 3 + β 34 .
2nd step (forward substitution):
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