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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 1 v 1 k 2 v2 . . . k n vn k1 u1 k2 u2 . . . kn un Un = k2 u1 k2 u2 . . . k2 un Vn = k2 v1 k2 v2 . . . k2 vn n n 1 1 2 2 .. .. .. .. . ... ... . . ... ... . (1) where the rows contain alternatively u’s and v’s. Similarly:
x1 x2 l 1 y 1 l2 y 2 X n = l2 x l2 x 1 1 2 2 .. . ...
... ... ... ...
xn ln y n l2n xn .. .
y1 y2 l1 x1 l2 x2 Y n = l2 y 1 l2 y 2 2 1 .. . ...
There holds det (
1⩽i,j⩽n
u i yj – vi x j lj – k i
1 )= ∏ i,j (lj – ki )
... ... ... ...
Un Xn Vn Yn 2n×2n
yn ln x n l2n yn .. . (2)
(3)
Proof. Let A, B, C, D be n × n (matrices, with A and C ( A B ) ( A 0 ) I A–1 B ) invertible. Using C D = 0 C we obtain I C–1 D
A B –1 –1 C D = |A||C||C D – A B|
(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) –1 –1 –1 –1 = |A||C| pu pv d(v) C Dd(y) – d(u) A 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) Ad(v) Bd(y)
= det(A)2 det ((ui yj – vi xj )(A–1 B)ij ) 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 ) (9a) C·
(
) 1 = W(l) diag(K(l1 )–1 , . . . , K(ln )–1 ) 1 ⩽ m,j ⩽ n lj – k m
(9b)
This gives the (well-known) identity:
(
1 lj – k m
)
= diag(K′ (k1 ), . . . , K′ (kn ))W(k)–1 W(l) diag(K(l1 )–1 , . . . , K(ln )–1 ) 1⩽m,j⩽n
(10) We can thus rewrite the determinant we want to compute as:
u i yj – vi x j lj – k i
= 1⩽i,j⩽n
∏ m
K′ (km )
∏
K(lj )–1 (ui yj –vi xj )(W(k)–1 W(l))ij
j
(11)
n×n
We shall now make use of (6) with A = W(k) and B = W(l). u i yj – vi x j W(k)d(u) W(l)d(x) ∏ ∏ = ∆(k)–2 K′ (km ) K(lj )–1 lj – k i W(k)d(v) W(l)d(y) m 1⩽i,j⩽n j n(n–1) W(k)d(u) W(l)d(x) (–1) 2 =∏ 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 determinant on Un Xn . This concludes the the right-hand side into Vn Yn proof.
