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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 . . . un v1 v2 . . . vn  k v k v . . . kn vn   k u k u . . . k n un  2 2  1 1 2 2   1 1  Un = k2 u k2 u . . . k2 u  Vn = k2 v k2 v . . . k2 v  (1) n n n n  1 1 2 2  1 1 2 2 .. .. .. .. . ... ... . . ... ... . where the rows contain alternatively u’s and v’s. Similarly:    y1 y2 x1 x2 . . . xn l x l x  l y l y . . . ln yn   1 1 2 2 1 1 2 2 Yn = l2 y l2 y Xn = l2 x l2 x . . . l2 x  n n 1 1 2 2 1 1 2 2  .. .. .. . ... . ... ... . There holds det (

1⩽i,j⩽n

ui y j – v i x j lj – k i

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

 yn ln xn   l2n yn   .. .

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

Proof. Let A, B, C, D be n × n matrices, with A and C invertible. Using ( A 0 ) ( I A–1 B ) we obtain 0 C I C–1 D A B –1 –1 C D = |A||C||C D – A B|

(2)

(3) (

A B C D

)

=

(4)

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) (5) –1 –1 = |A||C| d(u)C Dd(y) – d(v)A Bd(x) 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 = W(k) diag(K′ (k1 )–1 , . . . , K′ (kn )–1 ) C·

(

1 ) = W(l) diag(K(l1 )–1 , . . . , K(ln )–1 ) 1⩽m,j⩽n lj – km

(9a) (9b)

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

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

1⩽i,j⩽n

(11)

j

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

(10)

(12)

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[ 2n ] and transforms the determinant on the Un Xn . This concludes the proof. right-hand side into Vn Yn