Contents - Student subdomain for University of Bath

218 APPENDIX A. ALGEBRAIC BACKGROUND
Notation 30 In the context of V (k 1 , . . . , k n ), let P (z) =
n∏
(z − k i ).
i=1
i≠j
n∏
(z−k i ) and P j (z) =
Proposition 76 The inverse of a Vandermonde matrix has a particularly simple
form: V (k 1 , . . . , k n ) −1 = (m i,j ) with
m i,j = coeff(P j(z), z i )
∏
k≠j (k j − k k ) = coeff(P j(z), z i )
. (A.9)
P j (k j )
i=1
For example
⎛
V (k 1 , . . . , k 3 ) −1 = ⎜
⎝
k 2k 3
(−k 2+k 1)(−k 3+k 1)
−
k 2+k 3
(−k 2+k 1)(−k 3+k 1)
−
k 1k 3
(−k 2+k 1)(k 2−k 3)
k 1+k 3
(−k 2+k 1)(k 2−k 3)
1
(−k 2+k 1)(−k 3+k 1)
−
1
(−k 2+k 1)(k 2−k 3)
k 1k 2
(k 2−k 3)(−k 3+k 1)
−
k 1+k 2
(k 2−k 3)(−k 3+k 1)
1
(k 2−k 3)(−k 3+k 1)
⎞
⎟
⎠
Corollary 25 If all the k i are distinct, then V (k 1 , . . . , k n ) is invertible.
Equation (A.9) and the fact that the P j can be rapidly computed form p suggest
a rapid way of computing the elements of the inverse of a n × n Vandermonde
matrix in O(n 2 ) time. In fact, we can solve a system of Vandermonde linear
equations in O(n 2 ) time and O(n) space.
Algorithm 43 (Vandermonde solver)
Input: Vandermonde matrix V (k 1 , . . . , k n ), right-hand side w.
Output: Solution x to V (k 1 , . . . , k n )x = w
x := 0
P := ∏ n
i=1 (z − k i) #O(n 2 )
for i := 1 to n
Q := P/(z − k i ) #nO(n)
D := P (k i ) #nO(n) by Horner’s rule
for j := 1 to n
coeff(Q,z
x j := x j + w j−1 )
i D
In section 4.3.2 we will want to solve a slightly different system. Algorithm 43
solves a system of the form
x 1 + k 1 x 2 + k1x 2 3 + · · · + k1 n−1 x n = w 1
x 1 + k 2 x 2 + k2x 2 3 + · · · + k2 n−1 x n = w 1
(A.10)
.
x 1 + k n x 2 + knx 2 3 + · · · + kn n−1 x n = w 1