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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  k1 v1 k2 v 2 . . . k n v n  k1 u1 k2 u2 . . . kn un      Un = k2 u k2 u . . . k2 u  Vn =  k2 v k2 v . . . k2 v  1 2 n 1 2 n n n 2 2  1   1  .. .. .. .. . ... ... . . ... ... . (1) where the rows contain alternatively u’s and v’s. Similarly:     y1 y2 . . . yn x1 x 2 . . . xn  l1 y 1 l 2 y 2 . . . ln y n  l1 x1 l2 x2 . . . ln xn      X n = l2 x l2 x . . . l 2 x  Y n =  l 2 y l 2 y . . . l2 y  1 2 n 1 2 n n n 2 2  1  1 .. .. .. .. . ... ... . . ... ... . (2) There holds Un Xn ui yj – v i x j 1 det ( )= ∏ (3) 1⩽i,j⩽n (lj – ki ) Vn Yn lj – k i i,j

2n×2n

Proof. (Let )A, B, n × n matrices, with A and C invertible. ( AC, )D( Ibe –1 B ) 0 A B A Using C D = 0 C I C–1 D we obtain 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 ) (6) Ad(v) Bd(y) 1⩽i,j⩽n 2n×2n 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:

∑ cim ti–1 = K(t) 1⩽m⩽n t – km

1⩽i⩽n

(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: ∏ ∏ ui yj – vi xj –1 –1 ′ K(lj ) (ui yj –vi xj )(W(k) W(l))ij = K (km ) lj – k i n×n 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 –1 W(k)d(u) W(l)d(x) –2 ′ K(l ) = ∆ (k) K (k ) j m lj – k i W(k)d(v) W(l)d(y) 1⩽i,j⩽n m j n(n–1) W(k)d(u) W(l)d(x) (–1) 2 =∏ W(k)d(v) W(l)d(y) (l – k ) i i,j j 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[ 2n ] and trans Un Xn . This forms the determinant on the right-hand side into Vn Yn concludes the proof.