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Theorem 1. Let there be given indeterminates ui , vi , ki , xi , yi , li , for 1 ⩽ i ⩽ n. We define the followingn × nmatrices     u1 u2 . . . un v1 v2 ... vn  k 1 v1 k 2 v2 . . . k n v n  k1 u1 k2 u2 . . . kn un      Un = k2 u k2 u . . . k2 u  Vn =  k2 v 2 2  k v . . . k v n n 2 2 n n   1 1 2 2  1 1 .. .. .. .. . ... ... . . ... ... . (1) where the rows contain alternatively u’s andv’s. Similarly:     y1 y2 . . . yn x1 x2 . . . xn  l1 y1 l2 y2 . . . ln yn   l1 x1 l2 x2 . . . ln xn      X n = l 2 x l 2 x . . . l 2 x  Yn = l2 y l2 y . . . l2 y  n n n n  1 1 2 2  1 1 2 2 .. .. .. .. . ... ... . . ... ... . (2) There holds det (

1⩽i,j⩽n

u i y j – vi xj lj – k i

1 )= ∏ i,j (lj – ki )

U n X n Vn Yn 2n×2n

(3)

Proof. Let A, B, C, D be n × nmatrices, with A and C invertible. ( ) ( ) ( I A–1 B ) A 0 B Using A = we obtain 0 C C D I C–1 D A B –1 –1 (4) C D = |A||C||C D – A B| where vertical bars denote determinants. Let d(u) = diag(u1 , . . . , un ) ∏ and pu = 1⩽i⩽n ui . We define similarly d(v), d(x), d(y) and pv , px , py . From the previous identity we get Ad(u) Bd(x) = |A||C| pu pv d(v)–1 C–1 Dd(y) – d(u)–1 A–1 Bd(x) Cd(v) Dd(y) –1 –1 = |A||C| d(u)C Dd(y) – d(v)A Bd(x) (5) The special caseA = C, B = D, gives Ad(u) Bd(x) = det(A)2 det ((ui yj – vi xj )(A–1 B)ij ) 1⩽i,j⩽n Ad(v) Bd(y) 2n×2n

(6)

Let W(k) be the Vandermonde matrix with rows (1 . . . 1), (k1 . . . kn ), (k21 . . . k2n ), …, and ∆(k) = det W(k) its determinant. Let ∏ K(t) = (t – km ) (7) 1 ⩽m ⩽n

and letC b e t h en × nmatrix (cim )1⩽i,m⩽n , where thecim ’s are defined by the partial fraction expansions:

∑ cim ti–1 = K(t) 1⩽m⩽n t – km

1⩽i⩽n

(8)

We have the two matrix equations:

C·

(

1 lj – km

)

C = W(k) diag(K′ (k1 )–1 , . . . , K′ (kn )–1 )

(9a)

= W(l) diag(K(l1 )–1 , . . . , K(ln )–1 )

(9b)

1⩽m,j⩽n

This gives the (well-known) identity: ( ) 1 = diag(K′ (k1 ), . . . , K′ (kn ))W(k)–1 W(l) diag(K(l1 )–1 , . . . , K(ln )–1 ) lj – k m 1⩽m,j⩽n

(10) We can thus rewrite the determinant we want to compute as: u i yj – v i xj ∏ ∏ ′ –1 –1 = K (km ) K(lj ) (ui yj –vi xj )(W(k) W(l))ij lj – ki n ×n m j 1⩽i,j⩽n

(11) We shall now make use of(6) with A = W(k) and B = W(l). u i yj – v i xj ∏ ∏ –2 ′ –1 W(k)d(u) W(l)d(x) = ∆(k) K (km ) K(lj ) W(k)d(v) W(l)d(y) lj – ki m j 1⩽i,j⩽n

n(n–1)

(–1) 2 =∏ i,j (lj – ki )

W(k)d(u) W(l)d(x) W(k)d(v) W(l)d(y) 2n×2n

(12) [n ] 2

The sign (–1)n(n–1)/2 = (–1) is the signature of the permutation which exchanges rows i and n + ifor i = 2, 4, . . . , 2[ n2 ] and U n X n . transforms the determinant on the right-hand side into Vn Yn This concludes the proof.