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Theorem 1. Let there be given indeterminates ui , vi , ki , xi , yi , li , for 1 ⩽ i ⩽ n. We define the following n × n matrices     u1 u2 . . . u n v1 v2 . . . vn  k 1 v1 k2 v 2 . . . k n v n   k1 u 1 k2 u2 . . . kn u n      Un = k2 u k2 u . . . k2 u  Vn = k2 v k2 v . . . k2 v   1 1 2 2  1 1 2 2 n n n n .. .. . .. . . ... ... . . ... ... . (1) where the rows contain alternatively u’s and v’s. Similarly:



x1 x2  l1 y 1 l 2 y 2  Xn =  2 2 l1 x1 l2 x2 .. . ...

... ... ... ...

 xn ln y n   l2n xn   .. .

There holds det (

1⩽i,j⩽n

u i yj – v i xj lj – ki



y1 y 2  l1 x1 l2 x 2  Yn = l2 y l2 y 1 1 2 2 .. . ...

... ... ... ...

 yn ln xn   l2n yn   .. .

U n X n )= ∏ Vn Yn (l – k ) j i 2n×2n i,j 1

(2)

(3)

Proof. Let A, B, C, D be n × n matrices, with A and C invertible. Using ( A B ) ( A 0 ) ( I A–1 B ) we obtain C D = 0 C 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) = |A||C| d(u)C–1 Dd(y) – d(v)A–1 Bd(x) (5) The special case A = C, B = D, gives

Ad(u) Bd(x) = det(A)2 det ((ui yj – vi xj )(A–1 B)ij ) Ad(v) Bd(y) 1⩽i,j⩽n 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 let C be the n × n matrix (cim )1⩽i,m⩽n , where the cim ’s are defined by the partial fraction expansions: 1⩽i⩽n

∑ cim ti–1 = K(t) 1⩽m⩽n t – km

(8)

We have the two matrix equations:

C·

(

1 lj – k m

)

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 – km 1⩽m,j⩽n

(10)

We can thus rewrite the determinant we want to compute as: u i y j – v i xj ∏ ∏ = K′ (km ) K(lj )–1 (ui yj – vi xj )(W(k)–1 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 ∏ ∏ W(k)d(u) W(l)d(x) = ∆(k)–2 K′ (km ) K(lj )–1 W(k)d(v) W(l)d(y) lj – ki m 1⩽i,j⩽n

j

n(n–1) 2 (–1) W(k)d(u) W(l)d(x) =∏ W(k)d(v) W(l)d(y) i,j (lj – ki ) 2n×2n

(12)

n

The sign (–1)n(n–1)/2 = (–1)[ 2 ] is the signature of the permutation which exchanges rows i and n + i for i = 2, 4, . . . , 2[ n2 ] and transforms the

Un Xn . This concludes the determinant on the right-hand side into Vn Y n proof.