Explicit inverse of a generalized Vandermonde matrix

Explicit inverse of a generalized Vandermonde matrix

Applied Mathematics and Computation 146 (2003) 643–651 www.elsevier.com/locate/amc Explicit inverse of a generalized Vandermonde matrix Moawwad E.A. ...

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Applied Mathematics and Computation 146 (2003) 643–651 www.elsevier.com/locate/amc

Explicit inverse of a generalized Vandermonde matrix Moawwad E.A. El-Mikkawy Department of Mathematics, Faculty of Science, Mansoura University, Mansoura 35516, Egypt

Abstract In this paper the author gives an explicit closed form expression for the n  n inðkÞ ðkÞ verse matrix ðVG ðnÞÞ1 of the generalized n  n Vandermonde matrix VG ðnÞ by using the elementary symmetric functions. Symbolic and numerical results are presented. Ó 2002 Elsevier Inc. All rights reserved. Keywords: Algorithms; Vandermonde matrix; Stirling matrix; Symmetric functions; MAPLE

1. Introduction There are many special types of matrices which are of great importance in many scientific and engineering work. For instance matrices of type tridiagonal [1,2], pentadiagonal [3], Pascal [4], Vandermonde [5,6] and others. This paper will be concerned with matrices of Vandermonde type. These types of matrices frequently appear in many applications. For example in curve fitting, interpolation, scattering and in the derivation of explicit Runge–Kutta and Runge–Kutta–Nystrom [7–9] numerical methods. In this paper we are going to study the possibility of obtaining an explicit closed form of the Vandermonde inverse matrix. The main results are presented in Sections 2 and 3.

E-mail address: [email protected] (M.E.A. El-Mikkawy). 0096-3003/$ - see front matter Ó 2002 Elsevier Inc. All rights reserved. doi:10.1016/S0096-3003(02)00609-4

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2. The main results Let us begin this section by giving the following definitions Definition 2.1 [5]. If the n parameters c1 ; c2 ; . . . ; cn are distinct, then the eleðnÞ mentary symmetric functions ri;j in c1 ; c2 ; . . . ; cj1 ; cjþ1 ; . . . ; cn are defined for j ¼ 1ð1Þn by ðnÞ

r1;j ¼ 1 n n X X ðnÞ ri;j ¼ 

n X

i1 Y

crm

for i ¼ 2ð1Þn

ð2:1Þ

ri1 ¼ri2 þ1 m¼1 ri1 6¼j

r1 ¼1 r2 ¼r1 þ1 r1 6¼j r2 6¼j

Definition 2.2 [10]. The n  n Stirling matrix of the first kind is defined by 3 2 ð1Þ 0 0  0 r1;1 7 6 ð2Þ 6 rð2Þ r1;2 0  0 7 2;2 7 6 ð3Þ ð3Þ ð3Þ 6 r3;3 r2;3 r1;3 0 0 7 ð2:2Þ S1 ¼ 6 7 7 6 .. .. .. .. 7 6 4 . 0 5 . . . ðnÞ ðnÞ nþ1 ðnÞ nþ2 ðnÞ ð1Þ rn;n ð1Þ rn1;n    r2;n r1;n ðnÞ

in which the ri;j are calculated at cr ¼ r, r ¼ 1ð1Þn. Denote by rðGÞ ¼ ðbij Þ; i; j ¼ 1ð1Þn, where bij are given by n ðnÞ

bij ¼ ri;j

ð2:3Þ

The elements of the n  n matrix rnðGÞ in (2.3) may be calculated by using the following algorithm Algorithm 2.1. For n P 1 we may calculate the elements of the first column of in (2.3) as follows the n  n matrix rðGÞ n ð1Þ

Set r1;1 ¼ 1 For i ¼ 2; 3; . . . ; n ðiÞ ði1Þ ri;1 ¼ ri1;1 ci For j ¼ i  1; i  2; . . . ; 2 ðiÞ

ði1Þ

ði1Þ

rj;1 ¼ rj1;1 ci þ rj;1

ð2:4Þ

Next j Next i The elements in the remaining n  1 columns of rðGÞ may be obtained by n symmetry by using

M.E.A. El-Mikkawy / Appl. Math. Comput. 146 (2003) 643–651 ðnÞ

ri;k

ðnÞ ¼ ri;1

ck !c1

i ¼ 1ð1Þn; k ¼ 2ð1Þn

;

645

ð2:5Þ ðnÞ

The notation in (2.5) means that for specific i and k, ri;k may be obtained from ðnÞ the algebraic expression of ri;1 by replacing each ck by c1 in the expression of ðnÞ ri;1 . For example for n ¼ 4 we have 2

3 1 1 1 1 6 c1 þ c3 þ c4 c2 þ c1 þ c4 c2 þ c3 þ c1 7 c2 þ c3 þ c4 ðGÞ 7 r4 ¼ 6 4 c2 c3 þ c2 c4 þ c3 c4 c1 c3 þ c1 c4 þ c3 c4 c2 c1 þ c2 c4 þ c1 c4 c2 c3 þ c2 c1 þ c3 c1 5 c2 c3 c4 c1 c3 c4 c2 c1 c4 c2 c3 c1

ð2:6Þ

Unless otherwise stated, we shall be concerned with the case when cr ¼ r, r ¼ 1ð1Þn. For this case the matrix rðGÞ will be denoted by rðSÞ n n . The capital ðSÞ letters G and S stand for general and special respectively. For example r4 is given by 2

ðSÞ

r4

1 6 9 ¼6 4 26 24

1 8 19 12

1 7 14 8

3 1 6 7 7 11 5 6

ð2:7Þ

Definition 2.3. For the n þ 1 real values k; c1 ; c2 ; . . . ; cn we define a generalized ðkÞ ðkÞ Vandermonde matrix VG ðnÞ by VG ðnÞ ¼ ðVij Þ with Vij ¼ ckþj1 ; i; j ¼ 1ð1Þn. i For the special case when cr ¼ r, r ¼ 1ð1Þn this matrix will be denoted by ðkÞ VS ðnÞ. Qn Qn Qi1 ðkÞ It is known that detðVG ðnÞÞ ¼ ð m¼1 ckm Þ i¼2 j¼1 ðci  cj Þ. Thus both ðkÞ ðkÞ VG ðnÞ and VS ðnÞ are non singular if and only if the n parameters c1 ; c2 ; . . . ; cn are distinct. ðkÞ ð0Þ For any k, the two matrices VG ðnÞ and VG ðnÞ satisfy ðkÞ

ð0Þ

VG ðnÞ ¼ DVG ðnÞ

ð2:8Þ

where D ¼ diagðck1 ; ck2 ; . . . ; ckn Þ

ð2:9Þ

For the matrix D in (2.9) we have k k D1 ¼ diagðck 1 ; c2 ; . . . ; cn Þ ðkÞ

ð2:10Þ ðkÞ

Therefore the inverse matrix ðVG ðnÞÞ1 of the matrix VG ðnÞ may be obtained ð0Þ once the inverse matrix ðVG ðnÞÞ1 is available.

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M.E.A. El-Mikkawy / Appl. Math. Comput. 146 (2003) 643–651

In what follows we are going to show how can we obtain the matrix ð0Þ 1 ðVG ðnÞÞ . Suppose we have n distinct values c1 ; c2 ; . . . ; cn and consider the Lagrange polynomials L1 ðxÞ; L2 ðxÞ; . . . ; Ln ðxÞ of degree (n  1) defined by n Q

ðx  cj Þ

j¼1 j6¼i

Li ðxÞ ¼ Q n

ð2:11Þ ðci  cj Þ

j¼1 j6¼i

Consider also the polynomial wðxÞ of degree n defined by wðxÞ ¼

n Y ðx  ci Þ

ð2:12Þ

i¼1

It is easy to show that Li ðxÞ ¼

ðx  ci Þ1 wðxÞ d wðxÞjx¼ci dx

for i ¼ 1ð1Þn

ð2:13Þ

Moreover by using partial fractions together with (2.13) we obtain n X

Li ðxÞcmi ¼ xm

for m ¼ 0ð1Þn  1

ð2:14Þ

i¼1

Thus 2

L1 ðxÞ L2 ðxÞ .. .

3

2

1 c1 c21 .. .

7 6 6 7 6 6 7 6 6 7¼6 6 7 6 6 4 Ln1 ðxÞ 5 4 c1n1 Ln ðxÞ

1 c2 c22 .. .

  

1 cn1 c2n1 .. .

1 cn c2n .. .

31 2 7 7 7 7 7 5

3 1 6 x 7 6 . 7 6 .. 7 6 7 4 xn2 5

 xn1    cn1 cn1 n1 n 3 3 2 2 1 1 6 x 7 6 x 7 7 7 6 6 ð0Þ ð0Þ T 1 6 .. 7 1 T 6 .. 7 ¼ ððVG ðnÞÞ Þ 6 . 7 ¼ ððVG ðnÞÞ Þ 6 . 7 4 xn2 5 4 xn2 5 c2n1

xn1 ð0Þ

ð2:15Þ

xn1

Now if ððVG ðnÞÞ1 ÞT ¼ ðwij Þ1 6 i; j 6 n then (2.15) and (2.11) together with (2.1) give

M.E.A. El-Mikkawy / Appl. Math. Comput. 146 (2003) 643–651 n Q

Li ðxÞ ¼

n X j¼1

ðx  cj Þ

j¼1 j6¼i

wij xj1 ¼ Q n

n P

¼ ðci  cj Þ

j¼1 j6¼i n P

¼

647

ðnÞ

ð1Þjþ1 rj;i xnj

j¼1 n Q

ðci  cj Þ

j¼1 j6¼i

nj ðnÞ rnjþ1;i xj1

ð1Þ

j¼1 n Q

ð2:16Þ ðci  cj Þ

j¼1 j6¼i

Thus (2.16) yields nj ðnÞ

ð1Þ rnjþ1;i ; wij ¼ Q n ðci  cj Þ

i; j ¼ 1ð1Þn

ð2:17Þ

j¼1 j6¼i

At this point it is now possible to formulate the following general result: ðkÞ

Algorithm 2.2. For any integer n P 1 the n  n matrix ðVG ðnÞÞ1 in Definition 2.3 is given by ðkÞ

ðVG ðnÞÞ

1

¼ MðnÞ

ð2:18Þ

where the n  n matrix MðnÞ is given by MðnÞ ¼ ðmij Þ;

i; j ¼ 1ð1Þn

mij ¼ ð1Þnþi

rniþ1;j ; n Q ckj ðcj  ci Þ

with ðnÞ

i; j ¼ 1ð1Þn

ð2:19Þ

i¼1 i6¼j

The following are special cases of Algorithm 2.2 ð0Þ

1

Algorithm 2.3. For any integer n P 1, the n  n matrix ðVS ðnÞÞ 2.3 is given by ð0Þ

ðVS ðnÞÞ1 ¼

1 V ðnÞ ðn  1Þ!

in Definition

ð2:20Þ

where the n  n matrix V ðnÞ is a matrix whose entries are all integers and given by

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M.E.A. El-Mikkawy / Appl. Math. Comput. 146 (2003) 643–651

V ðnÞ ¼ ðvij Þ;

i; j ¼ 1ð1Þn

with vij ¼ ð1Þ

iþj



n  1 ðnÞ rniþ1;j ; j1

i; j ¼ 1ð1Þn

ð2:21Þ

Note 2.1. Another approach to find the n  n matrix V ðnÞ in (2.20) is given in [11]. This approach depends on writing V ðnÞ as the product of two matrices of the form T ðnÞU ðnÞ where T ðnÞ is an upper triangular matrix related to the matrix S1 in (2.2). Also the matrices T ðnÞ and U ðnÞ satisfy some recurrence relations. For more details see [11]. ðkÞ

Algorithm 2.4. For any integer n P 1, the n  n matrix ðVS ðnÞÞ1 in Definition 2.3 is given by ðkÞ

ðVS ðnÞÞ

1

¼ W ðnÞ

ð2:22Þ

where the n  n matrix W ðnÞ is given by W ðnÞ ¼ ðwij Þ;

i; j ¼ 1ð1Þn

with iþj

wij ¼

ð1Þ ðn  1Þ!



n1 j1

ðnÞ

rniþ1;j ; jk

i; j ¼ 1ð1Þn

ð2:23Þ

Note 2.2. From (2.23) it is clear that for any integer n P 1 and any k, the elðkÞ 1 ements of the n  n matrix ðVS ðnÞÞ have a checkerboard sign pattern.

3. Numerical results By writing a MAPLE [12] program based on Algorithms 2.1 and 2.4, the following results are obtained as sample output for the following three cases Case 1: n ¼ 5 and k ¼ 0 In this case the MAPLE program gives 2 3 1 1 1 1 1 61 2 4 8 16 7 6 7 ð0Þ 6 VS ð5Þ ¼ 6 1 3 9 27 81 7 7 4 1 4 16 64 256 5 1 5 25 125 625

ð3:1Þ

M.E.A. El-Mikkawy / Appl. Math. Comput. 146 (2003) 643–651

and

2 ð0Þ

ðVS ð5ÞÞ1

5

10

10

5

107 6 59 6 13 6 1 6

39 2 49 4

61 6 41 6 11 6 1 6

6 77 6 12 6 71 ¼6 6 24 6 7 4 12 1 24

3 1 4

1

649

3

25 7 12 7 7 35 7 24 7 5 7 12 5 1 24

ð3:2Þ

Case 2: n ¼ 8 and k ¼ 1=2 For this case the program yields ð1=2Þ

VS

2 6 6 6 6 6 6 6 6 6 6 ¼6 6 6 6 6 6 6 6 6 4

ð8Þ 1 pffiffiffi 2 pffiffiffi 3

1 pffiffiffi 2 2 pffiffiffi 3 3

1 pffiffiffi 4 2 pffiffiffi 9 3

1 pffiffiffi 8 2 pffiffiffi 27 3

1 pffiffiffi 16 2 pffiffiffi 81 3

1 pffiffiffi 32 2 pffiffiffi 243 3

1 pffiffiffi 64 2 pffiffiffi 729 3

1 pffiffiffi 128 2 pffiffiffi 2187 3

2 8 32 128 512 2048 8192 32768 pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi 5 5 5 25 5 125 5 625 5 3125 5 15625 5 78125 5 pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi 6 6 6 36 6 216 6 1296 6 7776 6 46656 6 279936 6 pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi 7 7 7 49 7 343 7 2401 7 16087 7 117649 7 823543 7 pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi 2 2 16 2 128 2 1024 2 8192 2 65536 2 524288 2 4194304 2

3 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 5

ð3:3Þ

and ð1=2Þ

ðVS

2

ð8ÞÞ1 8

6 6 481 6 35 6 6 349 6 6 36 6 6 329 6 90 ¼6 6 115 6 144 6 6 6 73 6 720 6 6 1 6 144 4 1 5040

pffiffiffi pffiffiffi 3 pffiffiffi pffiffiffi pffiffiffi pffiffiffi 56 56 8 14 2 5  143 6 7  14 2 3 35 3 5 7 pffiffiffi pffiffiffi 7 pffiffiffi 691 pffiffiffi 2143 pffiffiffi pffiffiffi 7 621 363  141 5 7 3 6  103 2  2003 2 7 20 45 8 5 180 35 560 7 pffiffiffi pffiffiffi 7 pffiffiffi pffiffiffi pffiffiffi pffiffiffi 797 1457 4891 187 527 469 7  18353 3 5  6 7  2 2 720 20 18 180 16 180 720 7 7 p ffiffi ffi p ffiffi ffi p ffiffi ffi p ffiffi ffi p ffiffi ffi p ffiffi ffi 7 15289 268 10993 1193 2803 67 967  5 7 3 6  2  2 7 1440 15 288 90 480 45 2880 pffiffiffi pffiffiffi pffiffiffi pffiffiffi 7 pffiffiffi pffiffiffi 7 179 71 179 2581 13 61 7  72 2 5 8 6 7  72 2 7 3 16 18 720 144 7 pffiffiffi pffiffiffi 7 pffiffiffi 209 pffiffiffi pffiffiffi pffiffiffi 239 149 391 61 49 23 7  5 7 3 6  2  2 7 720 240 144 720 240 720 1440 7 p ffiffi ffi p ffiffi ffi p ffiffi ffi p ffiffi ffi p ffiffi ffi p ffiffi ffi 7 17 11 1 31 1 29 1  720 2 5  7  3 6 2 7 240 9 720 48 5040 720 5 p ffiffi ffi p ffiffi ffi p ffiffi ffi p ffiffi ffi p ffiffi ffi p ffiffi ffi 1 1 1 1 1 1 1  720 5 1440 6  5040 7 20160 2 2  720 3 288 1440

ð3:4Þ

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M.E.A. El-Mikkawy / Appl. Math. Comput. 146 (2003) 643–651

Case 3: n ¼ 5 and any k In this case the program gives ðkÞ

ðVG ð5ÞÞ1  c2 c3 c4 c5 c1 c3 c4 c5 ¼ ; ; ðc2  c1 Þðc3  c1 Þðc4  c1 Þðc5  c1 Þck1 ðc2  c1 Þðc2  c3 Þðc2  c4 Þðc2  c5 Þck2 c2 c1 c4 c5 c2 c3 c1 c5 ; ; ðc2  c3 Þðc3  c1 Þðc3  c4 Þðc3  c5 Þck3 ðc2  c4 Þðc3  c4 Þðc4  c1 Þðc4  c5 Þck4  c2 c3 c4 c1  ðc2  c5 Þðc3  c5 Þðc4  c5 Þðc5  c1 Þck5  c5 c4 c3 þ c5 c4 c2 þ c5 c2 c3 þ c2 c3 c4 ;   ðc2  c1 Þðc3  c1 Þðc4  c1 Þðc5  c1 Þck1



c5 c4 c3 þ c5 c4 c1 þ c5 c1 c3 þ c1 c3 c4 c5 c4 c1 þ c5 c4 c2 þ c5 c2 c1 þ c2 c1 c4 ; ; ðc2  c1 Þðc2  c3 Þðc2  c4 Þðc2  c5 Þck2 ðc2  c3 Þðc3  c1 Þðc3  c4 Þðc3  c5 Þck3  c5 c1 c3 þ c5 c2 c1 þ c5 c2 c3 þ c2 c3 c1 c1 c3 c4 þ c2 c1 c4 þ c2 c3 c1 þ c2 c3 c4 ;  ðc2  c4 Þðc3  c4 Þðc4  c1 Þðc4  c5 Þck4 ðc2  c5 Þðc3  c5 Þðc4  c5 Þðc5  c1 Þck5  c5 c4 þ c5 c3 þ c5 c2 þ c4 c3 þ c4 c2 þ c2 c3 c5 c4 þ c5 c3 þ c5 c1 þ c4 c3 þ c4 c1 þ c1 c3 ; ;  ðc2  c1 Þðc3  c1 Þðc4  c1 Þðc5  c1 Þck1 ðc2  c1 Þðc2  c3 Þðc2  c4 Þðc2  c5 Þck2 

c5 c4 þ c5 c1 þ c5 c2 þ c4 c1 þ c4 c2 þ c2 c1 c5 c1 þ c5 c3 þ c5 c2 þ c1 c3 þ c2 c1 þ c2 c3 ; ; ðc2  c3 Þðc3  c1 Þðc3  c4 Þðc3  c5 Þck3 ðc2  c4 Þðc3  c4 Þðc4  c1 Þðc4  c5 Þck4  c4 c1 þ c1 c3 þ c2 c1 þ c4 c3 þ c4 c2 þ c2 c3  ðc2  c5 Þðc3  c5 Þðc4  c5 Þðc5  c1 Þck5  c5 þ c4 þ c3 þ c2 ;   ðc2  c1 Þðc3  c1 Þðc4  c1 Þðc5  c1 Þck1 

c5 þ c4 þ c3 þ c1 c5 þ c4 þ c1 þ c2 ; ; ðc2  c1 Þðc2  c3 Þðc2  c4 Þðc2  c5 Þck2 ðc2  c3 Þðc3  c1 Þðc3  c4 Þðc3  c5 Þck3  c5 þ c1 þ c3 þ c2 c1 þ c4 þ c3 þ c2 ;  k k ðc2  c4 Þðc3  c4 Þðc4  c1 Þðc4  c5 Þc4 ðc2  c5 Þðc3  c5 Þðc4  c5 Þðc5  c1 Þc5  1 1 ; ;  ðc2  c1 Þðc3  c1 Þðc4  c1 Þðc5  c1 Þck1 ðc2  c1 Þðc2  c3 Þðc2  c4 Þðc2  c5 Þck2 

1 1 ; ; ðc2  c3 Þðc3  c1 Þðc3  c4 Þðc3  c5 Þck3 ðc2  c4 Þðc3  c4 Þðc4  c1 Þðc4  c5 Þck4  1 ð3:5Þ  ðc2  c5 Þðc3  c5 Þðc4  c5 Þðc5  c1 Þck5



Note that by putting k ¼ 0 and cr ¼ r, r ¼ 1ð1Þ5 in (3.5) we obtain (3.2).

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651

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