Determinants and Their Applications in Mathematical Physics

4.1 Alternants 53

When any two of thexrare equal,Xnhas two identical rows and therefore

vanishes. Hence, very possible difference of the form (xs−xr) is a factor

ofXn, that is,

Xn= K(x 2 −x 1 )(x 3 −x 1 )(x 4 −x 1 )···(xn−x 1 )

(x 3 −x 2 )(x 4 −x 2 )···(xn−x 2 )

(x 4 −x 3 )···(xn−x 3 )

······

(xn−xn− 1 )

=K

∏

1 ≤r<s≤n

(xs−xr),

which is the product ofKand

1

2

n(n−1) factors. One of the terms in the

expansion of this polynomial is the product ofKand the first term in each

factor, namely

Kx 2 x

2

3

x

3

4

···x

n− 1

n

.

Comparing this term with (4.1.4), it is seen thatK= 1 and the theorem

is proved.

Second Proof. Perform the column operations

C

′

j

=Cj−xnCj− 1

in the orderj=n, n− 1 ,n− 2 ,..., 3 ,2. The result is a determinant in which

the only nonzero element in the last row is a 1 in position (n,1). Hence,

Xn=(−1)

n− 1

Vn− 1 ,

whereVn− 1 is a determinant of order (n−1). The elements in rowsof

Vn− 1 have a common factor (xs−xn). When all such factors are removed

fromVn− 1 , the result is

Xn=Xn− 1

n− 1

∏

r=1

(xn−xr),

which is a reduction formula forXn. The proof is completed by reducing

the value ofnby 1 repeatedly and noting thatX 2 =x 2 −x 1.

Exercises

1.Let

An=

∣

∣

∣

∣

(

j− 1

i− 1

)

(−xi)

j−i

∣

∣

∣

∣

n

=1.