范德蒙行列式的相关应用

范德蒙行列式的相关应用
（一）范德蒙行列式在行列式计算中的应用 范德蒙行列式的标准规范形式是：
1
22
22121
11112
111()n n n i j n i j n n n n
x x x D x x x x x x x x ≥>≥---==-∏L L L L L L L L
根据范德蒙行列式的特点，将所给行列式包括一些非范德蒙行列式利用各种方法将其化为范德蒙行列式，然后利用范德蒙行列式的结果，把它计算出来。

常见的化法有以下几种：
1.所给行列式各列（或各行）都是某元素的不同次幂，但其幂次数排列与范德蒙行列式不完全相同，需利用行列式的性质（如提取公因式，调换各行（或各列）的次序，拆项等）将行列式化为范德蒙行列式。

例1 计算
2221
1
1222333n
n n n
D n n n =
L
L L
L
解 n D 中各行元素都分别是一个数自左至右按递升顺序排列，但不是从0变到n r -。

而是由1递升至n 。

如提取各行的公因数，则方幂次数便从0变到1n -.
[]21
212
1
1
111
1222!!(21)(31)(1)(32)(2)(1)1
3331
n n n n D n n n n n n n
n n ---==-------L L L L L L L L L L L L
!(1)!(2)!2!1!n n n =--L
例2 计算
11
1
1(1)()(1)()1111
n n n n n n a a a n a a a n D a a a n ---+----=--L
L L L
L L
L L
解 本项中行列式的排列规律与范德蒙行列式的排列规律正好相反，为使
1n D +中各列元素的方幂次数自上而下递升排列，将第1n +列依次与上行交换直至
第1行，第n 行依次与上行交换直至第2行L 第2行依次与上行交换直至第n 行，于是共经过
(1)
(1)(2)212
n n n n n ++-+-+++=
L 次行的交换得到1n +阶范德蒙行列式：
[][](1)
2
1111(1)2
1
1
111(1)
(1)()(1)
()(1)
(1)(2)()2(1)((1))!
n n n n n n n n
n n n
k a a a n D a a a n a a a n a a a a a n a a a a n a n k ++---+=--=-----=--------------=∏L L L L L L L L
L L 若n D 的第i 行（列）由两个分行（列）所组成，其中任意相邻两行（列）均含相同分行（列）；且n D 中含有由n 个分行（列）组成的范德蒙行列式，那么将
n D 的第i 行（列）乘以-1加到第1i +行（列），消除一些分行（列）即可化成范
德蒙行列式： 例3 计算
1
2
3
4
2
2
2
2
11223344232323231122
3344
1
1111sin 1sin 1sin 1sin sin sin sin sin sin sin sin sin sin sin sin sin sin sin sin sin D +Φ+Φ+Φ+Φ=
Φ+ΦΦ+ΦΦ+ΦΦ+ΦΦ+ΦΦ+ΦΦ+ΦΦ+Φ
解 将D 的第一行乘以-1加到第二行得：
1
2
3
4
2
2
2
2
112233442323232311223344
1111sin sin sin sin sin sin sin sin sin sin sin sin sin sin sin sin sin sin sin sin ΦΦΦΦΦ+ΦΦ+ΦΦ+ΦΦ+ΦΦ+ΦΦ+ΦΦ+ΦΦ+Φ
再将上述行列式的第2行乘以-1加到第3行，再在新行列式中的第3行乘以-1加到第4行得：
1
2
3
4
2
2
2
2
14
123433341234
1111sin sin sin sin (sin sin )sin sin sin sin sin sin sin sin i j j i D ≤<≤ΦΦΦΦ=
=
Φ-ΦΦΦΦΦΦΦΦΦ∏
例4 计算
21112
222
2111111111n
n
n
n n n
x x x x x x D x x x ++++++=
+++L
L L L L
L
L
（1） 解 先加边，那么
2
21111112
22222222
21000111
111111
1
11111
1111
n n n
n n
n n n
n n
n
n
x x x x x x D x x x x x x x x x x x x ---+++=+++=+++L
L L L L L L L L
L L L L L L L L
L
再把第1行拆成两项之和，
221111112211112000111
1
n n n n n
n n
n
n n
x x x x x x D x x x x x x =-
L L L L L L L L L L L L
L
L
L
L
111
111
1
2()(1)
()
()[2(1)]
n
n k j i k j j k n
i j k n
n
n
k j i i j k n
i i x x x x x x x x x x x ≤<≤=≤<≤≤<≤===----=
---∏∏∏
∏∏∏L
2.加行加列法
各行（或列）元素均为某一元素的不同方幂，但都缺少同一方幂的行列式，可用此方法： 例5 计算
22212
33
312
12
111
3n n n n
n n
x x x D x x x x x x =L L L L L
L L L
解 作1n +阶行列式：
122222121333312121111n n
n n
n
n
n n
n
z x x x z x x x D z x x x z x x x +=
L L L
L
L L L L L
L
=1
()
()n
i j k i l k j n
x z x x =≤<≤--∏∏
由所作行列式可知z 的系数为D -，而由上式可知z 的系数为：
21
1211
(1)
()()n
n n j k i n j k l
i x x x x x x -=≥>≥--∑∏L
通过比较系数得：
1211
()()n
n j k i n j k l
i D x x x x x x =≥>≥=-∑∏L
3.拉普拉斯展开法
运用公式D =1122n n M A M A M A ++L 来计算行列式的值： 例6 计算
111111122122111000010010000
10010000
1
n n n n n n n n n
n
x x y y x x D y y x x y y ------=L L L L L
L L L L L L L L
解 取第1，3，L 21n -行，第1，3，21n -L 列展开得：
1111111122221111111
1
n n n n n n n n n n
x x y y x x y y D x x y y ------=
L L
L L L L
L L L
L
L L
L
L
=
()()j i j i n j i l
x x y y ≥>≥--∏
4.乘积变换法
例7 设1
2
1(0,1,22)n
k k k k k n
i i s x x x x k n ==+++==-∑L L ，计算行列式
01112122
n n
n n n s s s s s s D s s s ---=
L L L L L L L
解
11121
1
1
1222
1
1
1
n
n
n i
i
i i n
n
n n i
i
i
i i i n
n
n
n n n i
i
i i i n
x
x
x
x
x
D x
x
x
-=====--====
∑∑∑∑∑∑∑∑L L L L
L
L
L
2
1111122
12
22222
12
2
111122
1111
11
()n n n n n n n n n n
n
n
j i l i j n
x x x x x x x x x x x x x x x x x x x x -----≤<≤==
-∏
L L L L L
L L L
L L L L L L L
L 例8 计算行列式
000101011101()()()()()()()()()n n n n n
n n
n n
n n
n n n n a b a b a b a b a b a b D a b a b a b ++++++=
+++L L L L
L
L L
解 在此行列式中，每一个元素都可以利用二项式定理展开，从而变成乘积的和。

根据行列式的乘法规则，12D D D =g ，其中
1000111101n n
n n n n n n
n n n n n
n n n n C C a C a C C a C a D C C a C a =
L L L
L L L L
0111
101211
1
n n n n n n n n
b b b b b b D ---=L L
L L L L L
对2D 进行例2中的行的变换，就得到范德蒙行列式，于是
00121112111
n n
n n n n
n n n
a a a a D D D C C C a a ==L L g L L L
L
L
L
(1)2
(1)
n n +-g 01011
11n
n n n n
b b b b b b L L g
L L L L L
=(1)122
00()(1)
()n n n n n n
i j i j j i n
j i n
C C C
a a
b b +≤<≤≤<≤---∏
∏
L g
=120()()n
n n n
i j i j j i n
C C C a a b b ≤<≤--∏
L
5.升阶法
例9 计算行列式
221111222222
221111221111
n n n n n n n n n n n n n
n
n
n
x x x x x x x x D x x x x x x x x --------=L L L
L L L L L L L
解 将D 升阶为下面的1n +阶行列式
2
2
1
1111122122
2
2
2
1221111112212
2
1
111
11
n n n
n n n n n n n n n n n n n n n n n n n n n n n
x x x x x x x x x x x x x x x x x
x x x
x
x x
x
x
----+-----------=
L L L L L L L L L V L L L
即插入一行与一列，使1n +V 是关于12,,,n x x x x L 的1n +阶范德蒙行列式，此处x 是变数，于是1121()()()()n n i j j i n
x x x x x x x x +≤<≤=----∏
V L 故1n +V 是一个关于x 的n 次
多项式，它可以写成
{}11121()(1)()n n n i j n j i n
x x x x x x x -+≤<≤=
-+-++++∏
V L L
另一方面，将1n +V 按其第1n +行展开，即得
21111()(1)n n n n i j j i n
x x x Dx +-+≤<≤=
-+-+∏
V L
比较1n +V 中关于1n x -的系数，即得
121()
()n i j j i n
D x x x x x ≤<≤=+++-∏
L
(二) 范德蒙行列式在多项式理论中的应用
例1 设01(),n n f x c c x c x =++L 若()f x 至少有1n +个不同的根，则()0f x =。

证明 取121,,n x x x +L 为()f x 的1n +个不同的根，则有齐次线形方程组
2011211201222220112110,
0,
0,
n n n
n n n n n n c c x c x c x c c x c x c x c c x c x c x +++⎧++++=⎪++++=⎪⎨⎪⎪++++=⎩L L L L L L L L L L L L L L
L (2)
其中01,,n c c c L 看作未知量
因为方程组（2）的系数行列式D 是Vander monde 行列式，且
1()0,j i i j n
D x x ≤<≤=
-≠∏
所以方程组（2）只有零解，从而有010,n c c c ====L 即
()f x 是零多项式。

例2 设12,n a a a L 是数域F 中互不相同的数，12,,n b b b L 是数域F 中任一 组给定的不全为零的数，则存在唯一的数域F 上次数小于n 的多项式
()f x ，使()i f a =i b ，1,2,i n =L
证明 设1011(),n n f x c c x c x --=+++L 由条件()i i f a b =，1,2,i n =L 知
10111111
0121221011,,
,
n n n n n n n n n c c a c a b c c a c a b c c a c a b ------⎧+++=⎪+++=⎪⎨⎪⎪+++=⎩L L L L L L L L L L L L L （3）
因为12,,n a a a L 互不相同，所以方程组（3）的系数行列式
21
1121222
1211
1()01
n n n j i i j n
n n
n n
a a a a a a D a a a a a --≤<≤-=
=
-≠∏
L L L L L
L
L
L
则方程组（3）有唯一解，即唯一的次数小于n 的多项式
1011(),n n f x c c x c x --=++L
使得()i i f a b =，1,2,i n =L
例3设多项式1212(),n p p p n f x a x a x a x =+++L 0,1,2,i a i n ≠=L ，,i j p p ≠
,,1,2,i j i j n ≠=L ，则()f x 不可能有非零且重数大于1n -的根。

证明 反设0α≠是()f x 的重数大于1n -的根，则()f α=0，()0,f α'=L (1)()0,n f α-=进而1(1)()0,()0,()0.n n f f f ααααα--'===L 即
12121212112211
1122220,0,(1)(2)(1)(2)(1)(2)0
n n n p p p n p p p
n n p p p n n n n a a a p a p a p a p p p n a p p p n a p p p n a ααααααααα⎧+++=⎪++=⎪⎨
⎪⎪--++--+++--+=⎩L L L L L L L L L L L L L L L
L L L L
（4）
把（4）看成关于1212,,,n p p p n a a a αααL 为未知量的齐次线形方程组则（4）的系数行列式
1
2
1122111222111(1)
(1)(1)
(1)(2)
(1)(2)(1)(2)
n
n n n n n p p p D p p p p p p p p p n p p p n p p p n =
-----+--+--+L L
L
L L L
L
=
12222
12111112
111()0
n
n j i i j n
n n n n
p p p p p p p p p p p ≤<≤---=-≠∏L L L L L L L L
所以方程组（4）只有零解，从而0,1,2,i p i a i n α==L ，所以必有0,α= 这与0α≠矛盾，故()f x 没有非零且重数大于1n +的根。

附件：(外文资料原文)
New proof of the Vander monde determinant and
some applications
(A): a new method of proof: mathematical induction
We on the n for that the inductive method. （1）When 2n =错误!未找到引用源。

，错误!未找到引用源。

When the result is right.
（2）The Vander monde determinant conclusion assumptions for the class, now look at the level of .in
错误!未找到引用源。

错误!未找到引用源。

，Subtracting the 1n - rows 1x times, the first 1n - rows by subtracting 2n - 1x times, that is, a bottom-up sequentially subtracted from each row on row 1x timeshare 有
2131
12
2212313
112
12
122123131111
1000n n n
n n n n n n n n
x x x x x x d x x x x x x x x x x x x x x x x x x ---------=------L
L L M M M
M L
21
31122212313
1121212
2123131n n n n n n n n n n n x x x x x x x x x x x x x x x x x x x x x x x x ------------=
---L L
M M
M L
22321311222
3111()()()
n n n n n n x x x x x x x x x x x x ---=---L L
L M M M L
The latter determinant is a 1n - Van Dear Mind determinant, according to the induction assumption, it is equal to all possible difference i j x x -()2j i n ≤<≤；Contains
i x difference all appear in front of the consequent conclusion van dear Mend determinant
of the level n the establishment of mathematical induction, the proof is completed •
This result can be abbreviated as even the multiplication sign
2322223111112
111()
n n i j j i n
n n n n
x x x x x x x x x x x ≤<≤---=-∏L L L M M M L
Immediately by the results obtained necessary and sufficient condition for van der Mond determinant is zero
123,,,,n x x x x L
At least two equal number n
The application of the Vander monde determinant
（一）Vander monde determinant in the determinant calculation
The Vander monde determinant standards form ：
1
22
2212
1
1
1112111()n n n i j n i j n n n n
x x x D x x x x x x x x ≥>≥---==-∏
L L L
L L L L L
According to the characteristics of the Vander monde determinant given determinant
using various methods, including some non-Vander monde determinant into the Vander monde determinant, and then use the results of the Vander monde determinant, it calculated. The common method of following ：
1. Given determinant of the columns (or rows) are different powers of an element, but the number of power arrangement with the Vander monde determinant is not exactly the same, the need to use the nature of the determinant (such as extraction common divisor, change each line (or column) order, the dissolution of items, etc.) as the determinant of the Vander monde determinant.
Example 1.
2221
1
1222333n
n n n
D n n n =
L
L L
L
Solutions of n D elements of each row are a number from left to right in ascending order, but not from 0 to n . But by a delivery rosé. The common factor, such as extraction of each line number of a power from zero change to 1n -..
[]21
212
1
1
111
1
222!!(21)(31)(1)(32)(2)(1)1
3331
n n n n D n n n n n n n
n n ---==-------L L L L L L L L L L L L
!(1)!(2)!2!1!n n n =--L
Example 2
11
1
1(1)()(1)()111
1
n n n n n n a a a n a a a n D a a a n ---+----=--L
L L L
L L
L L
Solution The law of the law of the determinant arranged with the arrangement of the Vander monde determinant on the contrary, to make the columns in the 1n D + elements of a power of frequency from top to bottom in ascending order, the 1n + columns uplink switch in turn until the first line, n rows sequentially exchanged with the uplink until the second row 2nd row sequentially with uplink switch until the first n rows, so after a total of
(1)
(1)(2)212
n n n n n ++-+-+++=
L Sub-line exchange Vander monde determinant of order 1n +：
[][](1)2
1111(1)2
1
1
111(1)
(1)()(1)
()(1)
(1)(2)()2(1)((1))!
n n n n n n n n
n n n
k a a a n D a a a n a a a n a a a a a n a a a a n a n k ++---+=--=-----=--------------=∏L L L L L L L L
L L If n D i th row (column) consists of two branches (column), any two adjacent lines (columns) contain the same branch (column); and contains the Vander monde ranks n branches (column)type, then the n D i-th row (column) multiplied by -1 added to the 1i + -row (column) to eliminate some of the branches (column) into the Vander monde determinant ：
Example3
1
2
3
4
2222112233442323232311223344
11111sin 1sin 1sin 1sin sin sin sin sin sin sin sin sin sin sin sin sin sin sin sin sin D +Φ+Φ+Φ+Φ=
Φ+ΦΦ+ΦΦ+ΦΦ+ΦΦ+ΦΦ+ΦΦ+ΦΦ+Φ
Solution will be the first line of the d multiplied by -1 to the second line we have:
1234
2222112233442323232311223344
1111sin sin sin sin sin sin sin sin sin sin sin sin sin sin sin sin sin sin sin sin ΦΦΦΦΦ+ΦΦ+ΦΦ+ΦΦ+ΦΦ+ΦΦ+ΦΦ+ΦΦ+Φ
Then the second row of the above determinant is multiplied by -1 added to line 3, the new determinant line 3 line 4 was multiplied by -1 added:
1234
222214
123433341234
1111sin sin sin sin (sin sin )sin sin sin sin sin sin sin sin i j j i D ≤<≤ΦΦΦΦ=
=
Φ-ΦΦΦΦΦΦΦΦΦ∏
2 plus line to add law
Each row (or column) the different square a power of the whose elements are all the of an element, but are the lack of the the determinant of the same party a power of, available this method:
Example 4
22
212
33
312
12
11
13n n n n
n n
x x x D x x x x x x =L L L L L
L L L
Solution for order determinant ：
122222121333312121111n n
n n
n
n
n n
n
z x x x z x x x D z x x x z x x x +=
L L L
L
L L L L L
L
=
1
()()n
i
j k i l k j n
x z x x =≤<≤--∏∏
z seen by made determinant coefficient D -, by the coefficients of the above equation
z :
21
1211
(1)
()()n
n n j k i n j k l
i x x x x x x -=≥>≥--∑∏L
By comparing the coefficients obtained:
1211
()()n
n j k i n j k l
i D x x x x x x =≥>≥=-∑∏L
3 Laplace expansion methods
Apply the formula D = 1122n n M A M A M A ++L to calculate the value of the determinant ：
Example 5
111111
122122111000010010000
10010000
1
n n n n n n n n n
n
x x y y x x D y y x x y y ------=L L L L L L L L L L L L L
Dereference lines 1，3，L 21n -, 1，3，21n -L series expansion ：
1111111122221111111
1
n n n n n n n n n n
x x y y x x y y D x x y y ------=
L L
L L L L
L L L
L
L L
L
L
=
()()j i j i n j i l
x x y y ≥>≥--∏
4 product transformation method Example 6
Set up 1
2
1
(0,1,22)n
k k k k k n
i
i s x x x x
k n ==+++=
=-∑L L ，Compute the determinant
1112122
n n
n n n s s s s s s D s s s ---=
L L L L L L L
Solution
11121
1
1
1222
1
1
1
n
n
n i
i
i i n
n
n n i
i
i
i i i n
n
n
n n n i
i
i i i n
x
x
x
x
x
D x
x
x
-=====--====
∑∑∑∑∑∑∑∑L L L L
L
L
L
21
1
11122
12
22222
12
2
111122
111111
()n n n n n n n n n n n
n
j i l i j n
x x x x x x x x x x x x x x x x x x x x -----≤<≤==
-∏
L L L L L
L L L L
L L L L L L
L
. Ascending Order
Example 7 compute the determinant
221111222222221111221111
n n n n n n n n n n n n n
n
n
n
x x x x x x x x D x x x x x x x x --------=L L L
L L L L L L L
The solution D l order for the following 1n +-order determinant
2
2
1
1111122122
2
2
2
1221111112212
2
1
111
11
n n n
n n n n n n n n n n n n n n n n n n n n n n n
x x x x x x x x x x x x x x x x x
x x x
x
x x
x
x
----+-----------=
L L L L L L L L L V L L L
Insert a row with a 1n +V ,12,,,n x x x x L , 1n + order Vander monde determinant, where
x is the variable, so therefore 1121()()()
()n n i j j i n
x x x x x x x x +≤<≤=----∏
V L 1n +V G H
polynomial, it can be written as
{}11121()(1)()n n n i j n j i n
x x x x x x x -+≤<≤=
-+-++++∏
V L L
On the other hand, the 1n +V started their first 1n +-line, that was
21111()(1)n n n n i j j i n
x x x Dx +-+≤<≤=
-+-+∏
V L
Compare 1n +V 1
n x
- factor that was
121()
()n i j j i n
D x x x x x ≤<≤=+++-∏
L
(B) The Vander monde determinant polynomial theory
Example 1 01(),n n f x c c x c x =++L set ()f x is at least 1n + type root,
()0f x =。

Proof of 121,,n x x x +L ()f x 1n + different root, homogeneous linear equations
20112112012222201121
10,0,
0,
n n n
n n n n n n c c x c x c x c c x c x c x c c x c x c x +++⎧++++=⎪++++=⎪⎨
⎪⎪++++=⎩L L L L L L L L L L L L L L
L (2)
Where 01,,n c c c L as unknown amount
Vander monde Determinant D is because the coefficients of the equations (2), and
1()0,
j i i j n
D x x ≤<≤=
-≠∏
Equations (2) only the zero solution, thus
010,n c c c ====L is ()f x zero polynomial.
Example 2 Let 12,n a a a L number different from each other in the number field F,
12,,n b b b L is any number field F Group given not all zero number, then there exists a
unique number field F number of less than polynomial in n ()f x ，Make ()i f a =i b ，
1,2,i n =L
Proof Let 1011(),n n f x c c x c x --=+++L , by the condition ()i i f a b = , 1,2,i n =L
know
10111111
0121221011,,
,
n n n n n n
n n n c c a c a b c c a c a b c c a c a b ------⎧+++=⎪+++=⎪⎨⎪⎪+++=⎩L L L L L L L L L L L L L （3）
12,,n a a a L different from each other, so the coefficients of the equations (3) determinant
21
1121222
1211
1()01
n n n j i i j n
n n
n n
a a a a a a D a a a a a --≤<≤-=
=
-≠∏
L L L L L
L
L
L
The equations (3) has a unique solution, that is only the number of times less than a polynomial in n
1011(),n n f x c c x c x --=++L
Make ()i i f a b =，1,2,i n =L
Example 3 Let polynomial 1212(),n p p p n f x a x a x a x =+++L 0,1,2,i a i n ≠=L ，
,i j p p ≠
,,1,2,i j i j n ≠=L ，()f x impossible to have a non-zero weight is greater than the root
of 1n -.
Proof Anti weight greater than ()f x 0α≠ 1n - root, ()f α=0，()0,f α'=L
(1)
()0,n f
α-= And then 1(1)()0,()0,()0.n n f f f ααααα--'===L That
12121212112211
1122220,0,
(1)(2)(1)(2)(1)(2)0
n n n p p p n p p p
n n p p p n n n n a a a p a p a p a p p p n a p p p n a p p p n a ααααααααα⎧+++=⎪++=⎪⎨
⎪
⎪--++--+++--+=⎩L L L L L L L L L L L L L L L L L L L （4） (4) as the unknown quantity in 12
12,,,n p p
p n a a a αααL homogeneous system of linear
equations (4) coefficient determinant
1
2
11221112221
11(1)
(1)(1)
(1)(2)
(1)(2)(1)(2)
n
n n n n n p p p D p p p p p p p p p n p p p n p p p n =
-----+--+--+L L
L
L L L
L
= 1
22221
211111
2
111()0
n n j i i j n
n n n n
p p p p p p p p p p p ≤<≤---=-≠∏
L L L L L L L L
Equations (4) only the zero solution, thus 0,1,2,i
p i a i n α
==L must 0,α=
This is 0α≠ contradiction, it is non-zero root of multiplicity greater than 1n + ()f x no。