506 B The inverse of a multivariate homogeneous polynomial
Continuation of Box.
subject to
a 22 = a
2
11 ,
a 23 = 2a 11 a 12 ,
a 24 = 2a 11 a 13 + a
2
12 ,
a 25 = 2a 11 a 14 + 2a 12 a 13 ,
etc.
a 33 = a
3
11 ,
a 34 = 3a
2
11 a 12 ,
a 35 = 3a
2
11 a 13 + 3a 11 a
2
12 ,
etc.
a 44 = a
4
11 ,
a 45 = 4a
3
11 a 12 ,
etc.
(B.5)
A =
2
6
6
6
6
6
6
6
4
a 11 a 12
a 13
· · ·
a 1n
0 a
2
11 2a 11 a 12
. . .
a 2n
.
.
...
...
.
0
0
. . .
a n−1n−1 a n−1n
0
0
. . .
0
a nn
3
7
7
7
7
7
7
7
5
,
(B.6)
B =
2
6
6
6
6
6
6
6
4
b 11 b 12 . . . b 1n−1
b 1n
0 b 22 . . . b 2n−1
b 2n
.
. ...
.
.
0 0 . . . b n−1n−1 b n−1n
0 0 . . .
0
b nn
3
7
7
7
7
7
7
7
5
.
(B.7)
Consult Box B.4 for the general representation of a mn .
Backward substitution:
A
B = I
⇔
(i) b 11 a 11 = 1
⇒
b 11 = a
−1
11 ,
(ii) b 11 a 12 + b 12 a 22 = 0
⇒
b 12 = −b 11 a 12 a
−1
22 = −a
−3
11 a 12 ,
(iii) b 11 a 13 + b 12 a 23 + b 13 a 33 = 0
⇒
b 13 = − (b 11 a 13 + b 12 a 23 ) a
−1
33 = a
−1
11
`
a 12 a
−1
22 a 23 − a 13
´
a
−1
33 ,
(iv) b 11 a 14 + b 12 a 24 + b 13 a 34 + b 14 a 44 = 0
⇒
b 14 = − (b 11 a 14 + b 12 a 24 + b 13 a 34 ) a
−1
44 = a
−1
11
ˆ
a 12 a
−1
22
`
a 24 − a 23 a
−1
33 a 34
´
+ a 13 a
−1
33 a 34 − a 14
˜
a
−1
44 .
(B.8)
Consult Box B.5 for the general representation of b 1n .
Continuation of Box.
subject to
a 22 = a
2
11 ,
a 23 = 2a 11 a 12 ,
a 24 = 2a 11 a 13 + a
2
12 ,
a 25 = 2a 11 a 14 + 2a 12 a 13 ,
etc.
a 33 = a
3
11 ,
a 34 = 3a
2
11 a 12 ,
a 35 = 3a
2
11 a 13 + 3a 11 a
2
12 ,
etc.
a 44 = a
4
11 ,
a 45 = 4a
3
11 a 12 ,
etc.
(B.5)
A =
2
6
6
6
6
6
6
6
4
a 11 a 12
a 13
· · ·
a 1n
0 a
2
11 2a 11 a 12
. . .
a 2n
.
.
...
...
.
0
0
. . .
a n−1n−1 a n−1n
0
0
. . .
0
a nn
3
7
7
7
7
7
7
7
5
,
(B.6)
B =
2
6
6
6
6
6
6
6
4
b 11 b 12 . . . b 1n−1
b 1n
0 b 22 . . . b 2n−1
b 2n
.
. ...
.
.
0 0 . . . b n−1n−1 b n−1n
0 0 . . .
0
b nn
3
7
7
7
7
7
7
7
5
.
(B.7)
Consult Box B.4 for the general representation of a mn .
Backward substitution:
A
B = I
⇔
(i) b 11 a 11 = 1
⇒
b 11 = a
−1
11 ,
(ii) b 11 a 12 + b 12 a 22 = 0
⇒
b 12 = −b 11 a 12 a
−1
22 = −a
−3
11 a 12 ,
(iii) b 11 a 13 + b 12 a 23 + b 13 a 33 = 0
⇒
b 13 = − (b 11 a 13 + b 12 a 23 ) a
−1
33 = a
−1
11
`
a 12 a
−1
22 a 23 − a 13
´
a
−1
33 ,
(iv) b 11 a 14 + b 12 a 24 + b 13 a 34 + b 14 a 44 = 0
⇒
b 14 = − (b 11 a 14 + b 12 a 24 + b 13 a 34 ) a
−1
44 = a
−1
11
ˆ
a 12 a
−1
22
`
a 24 − a 23 a
−1
33 a 34
´
+ a 13 a
−1
33 a 34 − a 14
˜
a
−1
44 .
(B.8)
Consult Box B.5 for the general representation of b 1n .
