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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 k1 v1 k2 v2 Un = k 2 u k 2 u 1 1 2 2 .. . ...
un kn vn kn2 un .. ... .
... ... ...
v1 v2 k1 u1 k2 u2 Vn = k 2 v k 2 v 1 1 2 2 .. . ...
... vn . . . kn un . . . kn2 vn ...
.. . (1)
where the rows contain alternatively u’s and v’s. Similarly:
x1 x2 l 1 y1 l 2 y2 Xn = l 2 x l 2 x 1 1 2 2 .. . ...
xn l n yn ln2 xn .. ... . ... ... ...
y1 y2 l1 x1 l2 x2 Yn = l 2 y l 2 y 1 1 2 2 .. . ...
There holds ui yj − vi xj 1 det ( )=∏ 1⩽i ,j ⩽n lj − ki i,j ( lj − ki )
Un Xn Vn Yn
yn l n xn ln2 yn .. ... . (2) ... ... ...
(3) 2n ×2n
Proof. Let A, B, C , D( be n ×) n matrices, with A and C invertible. ( ) Using AC DB = ( A0 C0 ) II CA−−11DB we obtain 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 ) = |A||C | pu pv d ( v) −1 C −1 Dd ( y) − d ( u) −1 A −1 Bd ( x ) Cd ( v) Dd ( y) −1 −1 = |A||C | d ( u)C Dd ( y) − d ( v)A 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 ), ( k12 . . . kn2 ), . . . , 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: ∑ t i −1 cim 1⩽i ⩽n (8) = K ( t ) 1⩽m ⩽n t − km We have the two matrix equations: C = W ( k ) diag ( K ′ ( k1 ) −1 , . . . , K ′ ( kn ) −1 ) (
1
)
= W ( l ) diag ( K ( l1 ) −1 , . . . , K ( ln ) −1 ) lj − km 1⩽m,j ⩽n This gives the (well-known) identity: C·
(
1 lj − km
(9a) (9b)
)
= 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 y − v x i j i j lj − ki
= 1⩽i ,j ⩽n
∏
K ′ ( km )
∏
m
K ( lj ) −1 ( ui yj −vi xj )( W ( k ) −1 W ( l ))ij
n ×n
j
(11) We shall now make use of (6) with A = W ( k ) and B = W ( l ). u y − v x i j i j lj − ki
= ∆( k ) 1⩽i ,j ⩽n
−2
∏ m
′
K ( km )
∏ j
K ( lj )
−1
W ( k )d ( u) W ( l )d ( x ) W ( k )d ( v) W ( l )d ( y)
W ( k )d ( u) W ( l )d ( x ) ( −1) =∏ W ( k )d ( v) W ( l )d ( y) ( l − k ) j i i ,j 2n ×2n n( n −1) 2
(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 U X the determinant on the right-hand side into n n . This concludes Vn Yn the proof.
