Determinants and Their Applications in Mathematical Physics

52 4. Particular Determinants

If thex’s are not distinct, the determinant has two or more identical rows. If

they’s are not distinct, the determinant has two or more identical columns.

In both cases, the determinant vanishes.

Illustration.The Wronskian |D

j− 1

x (fi)|n is an alternant. The double

Wronskian|D

j− 1

x D

i− 1

y (f)|nis a double alternant,Dx=∂/∂x, etc.

Exercise.Define two third-order alternantsφandψ in column vector

notation as follows:

φ=|c(x 1 )c(x 2 )c(x 3 )|,

ψ=|C(x 1 )C(x 2 )C(x 3 )|.

Apply l’Hopital’s formula to prove that

lim

(

φ

ψ

)

=

|c(x)c

′

(x)c

′′

(x)|

|C(x)C

′

(x)C

′′

(x)|

,

where the limit is carried out asxi→x,1≤i≤3, provided the numerator

and denominator are not both zero.

4.1.2 Vandermondians....................

The determinant

Xn=|x

j− 1

i |n

=

∣ ∣ ∣ ∣ ∣ ∣ ∣

1 x 1 x

2

1 ··· x

n− 1

1

1 x 2 x

2

2 ··· x

n− 1

2

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

1 xn x

2

n ··· x

n− 1

n

∣ ∣ ∣ ∣ ∣ ∣ ∣ n

=V(x 1 ,x 2 ,...,xn) (4.1.3)

is known as the alternant of Vandermonde or simply a Vandermondian.

Theorem.

Xn=

∏

1 ≤r<s≤n

(xs−xr).

The expression on the right is known as a difference–product and contains

(n/2) =

1

2

n(n−1)factors.

First Proof. The expansion of the determinant consists of the sum ofn!

terms, each of which is the product ofnelements, one from each row and

one from each column. Hence,Xnis a polynomial in thexrof degree

0+1+2+3+···+(n−1) =

1

2

n(n−1).

One of the terms in this polynomial is the product of the elements in the

leading diagonal, namely

+x 2 x

2

3

x

3

4

···x

n− 1

n

. (4.1.4)