Determinant Of The Vandermonde Matrix [expository notes]

3 Pages·2012·0.14 MB·English

by  Paul Skoufranis

MATH 115A - Determinant Of The Vandermonde Matrix
Paul Skoufranis
August 10, 2012
InthisdocumentwewillusethePrincipleofMathematicalInductiontocomputethedeterminantofthe
Vandermonde Matrix.
Example) Prove that if x ,x ,...,x ∈F then
1 2 n
 
1 1 1 ··· 1
 x1 x2 x3 ··· xn 
det x...21 x...22 x...23 ·.·.·. x...2n =1≤i(cid:89)<j≤n(xj −xi)
xn−1 xn−1 xn−1 ··· xn−1
1 2 3 n
(cid:81)
foralln≥2. (Note: (x −x )meanstotaketheproductofalltermsx −x where1≤i<j ≤n.
1≤i<j≤n j i j i
(cid:81)
For example (x −x )=(x −x )(x −x )(x −x ).)
1≤i<j≤3 j i 3 1 3 2 2 1
Proof: To prove the result, we will proceed by induction on n using the known results of the determi-
nant from MATH 33A.
Base Case: n=2 For n=2
(cid:18)(cid:20) (cid:21)(cid:19)
1 1 (cid:89)
det =x −x = (x −x )
x x 2 1 j i
1 2
1≤i<j≤2
by the formula for the determinant of a 2×2 matrix.
Inductive Step Suppose that the result is true for some fixed n≥2: that is,
 
1 1 1 ··· 1
 x1 x2 x3 ··· xn 
det x...21 x...22 x...23 ·.·.·. x...2n =1≤i(cid:89)<j≤n(xj −xi)
xn−1 xn−1 xn−1 ··· xn−1
1 2 3 n
foranyx ,...,x ∈F. Wedesiretoshowthattheformulaholdsforallx ,...,x ∈F. Fixx ,x ,...,x ∈
1 n 1 n+1 1 2 n+1
F and consider the (n+1)×(n+1) matrix
 
1 1 1 ··· 1
 x1 x2 x3 ··· xn+1 
A= x21 x22 x23 ··· x2n+1 .
 ... ... ... ... ... 
xn xn xn ··· xn
1 2 3 n+1
By subtracting x times the nth row of this matrix to the (n+1)st row, we obtain the matrix
1
 
1 1 1 ··· 1
 x1 x2 x3 ··· xn+1 
 x21 x22 x23 ··· x2n+1 
 ... ... ... ... ... .
 xn−1 xn−1 xn−1 ··· xn−1 
1 2 3 n+1
0 xn−x xn−1 xn−x xn−1 ··· xn −x xn−1
2 1 2 3 1 3 n+1 1 n+1
By subtracting x times the (n−1)st row of this matrix to the nth row, we obtain the matrix
1
 
1 1 1 ··· 1
 x1 x2 x3 ··· xn+1 
 x21 x22 x23 ··· x2n+1 
 ... ... ... ... ... .
 xn−2 xn−2 xn−2 ··· xn−2 
 10 xn2−1−2x1xn2−2 xn3−1−3x1xn3−2 ··· xnn−+11−n+x11xnn−+12 
0 xn−x xn−1 xn−x xn−1 ··· xn −x xn−1
2 1 2 3 1 3 n+1 1 n+1
By continuing this procedure, we obtain the matrix
 
1 1 1 ··· 1
 0 x2−x1 x3−x1 ··· xn+1−x1 
 0 x22−x1x2 x23−x1x3 ··· x2n+1−x1xn+1 
B = ... ... ... ... ... .
 0 xn2−1−x1x2n−2 xn3−1−x1x3n−2 ··· xnn−+11−x1xnn−+12 
0 xn−x xn−1 xn−x xn−1 ··· xn −x xn−1
2 1 2 3 1 3 n+1 1 n+1
From MATH 33A, the determinant of A and B are equal. Moreover, by MATH 33A, we can expand the
determinant along the first column of B to obtain that the determinant of A is the same as the determinant
of the n×n matrix
 
x −x x −x ··· x −x
2 1 3 1 n+1 1
 x22−x1x2 x23−x1x3 ··· x2n+1−x1xn+1 
C = ... ... ... ... .
 xn2−1−x1xn2−2 xn3−1−x1xn3−2 ··· xnn−+11−x1xnn−+12 
xn−x xn−1 xn−x xn−1 ··· xn −x xn−1
2 1 2 3 1 3 n+1 1 n+1
However, we notice that
 
(x −x )1 (x −x )1 ··· (x −x )1
2 1 3 1 n+1 1
 (x2−x1)x2 (x3−x1)x3 ··· (xn+1−x1)xn+1 
C = ... ... ... ... .
 (x2−x1)xn2−2 (x3−x1)xn3−2 ··· (xn+1−x1)xnn−+12 
(x −x )xn−1 (x −x )xn−1 ··· (x −x )xn−1
2 1 2 3 1 3 n+1 1 n+1
Therefore, since the determinant is linear in each column by MATH 33A, the determinant of C is the same
(cid:81)
as (x −x ) times the determinant of the n×n matrix
1<j≤n+1 j 1
 
1 1 ··· 1
 x2 x3 ··· xn+1 
D = ... ... ... ... .
 xn−2 xn−2 ··· xn−2 
2 3 n+1
xn−1 xn−1 ··· xn−1
2 3 n+1
2
However, by the inductive hypothesis
(cid:89)
det(D)= (x −x )
j i
2≤i<j≤n+1
so
  
(cid:89) (cid:89) (cid:89)
det(A)= (xj −x1) (xj −xi)= (xj −xi)
1<j≤n+1 2≤i<j≤n+1 1≤i<j≤n+1
Whence we have completed the inductive step.
Therefore, by the Principle of Mathematical Induction, the result is true for all n≥2. (cid:3)
3