(+)분류 : 가져온 문서/오메가
방데르몽드 행렬식(Vandermonde determinant) 또는 방데르몽드 다항식(Vandermonde polynomial)은 정사각행렬인 방데르몽드 행렬의 행렬식이다.
1. 계산 ✎ ⊖
n\\times n 방데르몽드 행렬
V_n=\\begin {bmatrix}1 & \\alpha_1 & \\alpha_1^2 & \\cdots & \\alpha_1^{n-1} \\\\1 & \\alpha_2 & \\alpha_2^2 & \\cdots & \\alpha_2^{n-1} \\\\\\vdots & \\vdots & \\vdots & \\ddots & \\vdots\\\\1 & \\alpha_n & \\alpha_n^2 & \\cdots & \\alpha_n^{n-1}\\end {bmatrix}
에 대해
\\det(V_n)=\\prod_{1\\le j < i \\le n}(\\alpha_i-\\alpha_j)
이다.
V_n=\\begin {bmatrix}1 & \\alpha_1 & \\alpha_1^2 & \\cdots & \\alpha_1^{n-1} \\\\1 & \\alpha_2 & \\alpha_2^2 & \\cdots & \\alpha_2^{n-1} \\\\\\vdots & \\vdots & \\vdots & \\ddots & \\vdots\\\\1 & \\alpha_n & \\alpha_n^2 & \\cdots & \\alpha_n^{n-1}\\end {bmatrix}
에 대해
\\det(V_n)=\\prod_{1\\le j < i \\le n}(\\alpha_i-\\alpha_j)
이다.
1.1. 증명 ✎ ⊖
V_n의 행렬식
\\det(V_n)=\\begin{vmatrix}1 & \\alpha_1 & \\alpha_1^2 & \\cdots & \\alpha_1^{n-1} \\\\1 & \\alpha_2 & \\alpha_2^2 & \\cdots & \\alpha_2^{n-1} \\\\\\vdots & \\vdots & \\vdots & \\ddots & \\vdots\\\\1 & \\alpha_n & \\alpha_n^2 & \\cdots & \\alpha_n^{n-1}\\end{vmatrix}
의 2,3,\\dots,n행에서 1행을 빼는 기본행연산을 수행하면,
\\det(V_n)=\\begin{vmatrix}1 & \\alpha_1 & \\alpha_1^2 & \\cdots & \\alpha_1^{n-1} \\\\0 & \\alpha_2 - \\alpha_1 & \\alpha_2^2 -\\alpha_1^2 & \\cdots & \\alpha_2^{n-1}-\\alpha_1^{n-1} \\\\\\vdots & \\vdots & \\vdots & \\ddots & \\vdots\\\\0 & \\alpha_n-\\alpha_1 & \\alpha_n^2-\\alpha_1^2 & \\cdots & \\alpha_n^{n-1}-\\alpha_1^{n-1}\\end{vmatrix}
을 얻는다. j\\in\\{1,2,\\dots,n-1\\}에 대해, (j+1)열에서 j항의 \\alpha_1배를 빼는 기본열연산을 내림차순으로 수행하면,
\\begin{aligned}\\det(V_n)&=\\begin{vmatrix}1 & 0 & 0 & \\cdots & 0 \\\\0 & \\alpha_2 - \\alpha_1 & \\alpha_2(\\alpha_2-\\alpha_1) & \\cdots & \\alpha_2^{n-2}(\\alpha_2-\\alpha_1) \\\\\\vdots & \\vdots & \\vdots & \\ddots & \\vdots\\\\0 & \\alpha_n-\\alpha_1 & \\alpha_n(\\alpha_n-\\alpha_1) & \\cdots & \\alpha_n^{n-2}(\\alpha_n-\\alpha_1)\\end{vmatrix}\\\\ &=\\prod_{i=2}^n (\\alpha_i -\\alpha_1) \\begin{vmatrix} 1 & \\alpha_2 & \\cdots & \\alpha_2^{n-2} \\\\ 1 & \\alpha_3 & \\cdots &\\alpha_3^{n-2} \\\\ \\vdots & \\vdots & \\ddots & \\vdots \\\\ 1 & \\alpha_n & \\cdots & \\alpha_n^{n-2} \\end{vmatrix}\\end{aligned}
위의 계산과정을 반복하면,
\\det(V_n)= \\prod_{k=1}^{n-1}\\left(\\prod_{i=k+1}^n (\\alpha_i-\\alpha_k)\\right) =\\prod_{1\\le j < i \\le n}(\\alpha_i-\\alpha_j)
을 얻는다.
\\det(V_n)=\\begin{vmatrix}1 & \\alpha_1 & \\alpha_1^2 & \\cdots & \\alpha_1^{n-1} \\\\1 & \\alpha_2 & \\alpha_2^2 & \\cdots & \\alpha_2^{n-1} \\\\\\vdots & \\vdots & \\vdots & \\ddots & \\vdots\\\\1 & \\alpha_n & \\alpha_n^2 & \\cdots & \\alpha_n^{n-1}\\end{vmatrix}
의 2,3,\\dots,n행에서 1행을 빼는 기본행연산을 수행하면,
\\det(V_n)=\\begin{vmatrix}1 & \\alpha_1 & \\alpha_1^2 & \\cdots & \\alpha_1^{n-1} \\\\0 & \\alpha_2 - \\alpha_1 & \\alpha_2^2 -\\alpha_1^2 & \\cdots & \\alpha_2^{n-1}-\\alpha_1^{n-1} \\\\\\vdots & \\vdots & \\vdots & \\ddots & \\vdots\\\\0 & \\alpha_n-\\alpha_1 & \\alpha_n^2-\\alpha_1^2 & \\cdots & \\alpha_n^{n-1}-\\alpha_1^{n-1}\\end{vmatrix}
을 얻는다. j\\in\\{1,2,\\dots,n-1\\}에 대해, (j+1)열에서 j항의 \\alpha_1배를 빼는 기본열연산을 내림차순으로 수행하면,
\\begin{aligned}\\det(V_n)&=\\begin{vmatrix}1 & 0 & 0 & \\cdots & 0 \\\\0 & \\alpha_2 - \\alpha_1 & \\alpha_2(\\alpha_2-\\alpha_1) & \\cdots & \\alpha_2^{n-2}(\\alpha_2-\\alpha_1) \\\\\\vdots & \\vdots & \\vdots & \\ddots & \\vdots\\\\0 & \\alpha_n-\\alpha_1 & \\alpha_n(\\alpha_n-\\alpha_1) & \\cdots & \\alpha_n^{n-2}(\\alpha_n-\\alpha_1)\\end{vmatrix}\\\\ &=\\prod_{i=2}^n (\\alpha_i -\\alpha_1) \\begin{vmatrix} 1 & \\alpha_2 & \\cdots & \\alpha_2^{n-2} \\\\ 1 & \\alpha_3 & \\cdots &\\alpha_3^{n-2} \\\\ \\vdots & \\vdots & \\ddots & \\vdots \\\\ 1 & \\alpha_n & \\cdots & \\alpha_n^{n-2} \\end{vmatrix}\\end{aligned}
위의 계산과정을 반복하면,
\\det(V_n)= \\prod_{k=1}^{n-1}\\left(\\prod_{i=k+1}^n (\\alpha_i-\\alpha_k)\\right) =\\prod_{1\\le j < i \\le n}(\\alpha_i-\\alpha_j)
을 얻는다.
