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eorem 1. Let there be given indeterminates ui , vi , ki , xi , yi , li , for 1 ⩽ i ⩽ n. We de ne the following n × n matrices     v2 . . . vn  u2 . . . un   v1  u1     k1 u1 k2 u2 . . . kn un   k1 v1 k2 v2 . . . kn vn  Vn =  k2 v k2 v . . . k2 v  (1) Un = k2 u k2 u . . . k2 u  n n n n  1 1 2 2  1 1 2 2  ..  ..   ..  .. . ... ... . . ... ... . where the rows contain alternatively u’s and v’s. Similarly:    y2  x1 x2 . . . xn   y1    l1 y1 l2 y2 . . . ln yn  l1 x1 l2 x2 Xn = l2 x l2 x . . . l2 x  Yn = l2 y l2 y n n  1 1 2 2  1 1 2 2 ..   ..  .. . ... ... . . ... ere holds ui yj – vi xj 1 )= ∏ det ( 1⩽i,j⩽n lj – ki i,j (lj – ki )

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

 yn   ln xn   l2n yn  ..  .

Un Xn V Y n n 2n×2n

(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 de ne similarly d(v), d(x), d(y) and pv , px , py . From the previous identity we get Ad(u) Bd(x) –1 –1 –1 –1 = |A||C| p p d(v) C Dd(y) – d(u) A Bd(x) Cd(v) Dd(y) u v (5) –1 –1 = |A||C| d(u)C Dd(y) – d(v)A Bd(x) e 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 de ned 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 = W(k) diag(K′ (k1 )–1 , . . . , K′ (kn )–1 ) C·

(

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

(9a) (9b)

is 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: ∏ ∏ ui yj – vi xj ′ –1 –1 = K (k ) K(l ) (u y – v x )(W(k) W(l)) i j i j m j ij n×n lj – ki 1⩽i,j⩽n m

j

(11) We shall now make use of (6) with A = W(k) and B = W(l). ∏ ∏ ui yj – vi 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 1⩽i,j⩽n m j n(n–1) W(k)d(u) W(l)d(x) (–1) 2 =∏ i,j (lj – ki ) W(k)d(v) W(l)d(y) 2n×2n (12) n

e 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 determinant U X n . is concludes the proof. □ on the right-hand side into n Vn Yn