방데르몽드 행렬식(Vandermonde determinant) 또는 방데르몽드 다항식(Vandermonde polynomial)은 정사각행렬인 방데르몽드 행렬의 행렬식이다.
n × n {\displaystyle n\times n} 방데르몽드 행렬
V n = [ 1 α 1 α 1 2 ⋯ α 1 n − 1 1 α 2 α 2 2 ⋯ α 2 n − 1 ⋮ ⋮ ⋮ ⋱ ⋮ 1 α n α n 2 ⋯ α n n − 1 ] {\displaystyle 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 ) = ∏ 1 ≤ j < i ≤ n ( α i − α j ) {\displaystyle \det(V_{n})=\prod _{1\leq j<i\leq n}(\alpha _{i}-\alpha _{j})}
이다.
Vn의 행렬식
det ( V n ) = | 1 α 1 α 1 2 ⋯ α 1 n − 1 1 α 2 α 2 2 ⋯ α 2 n − 1 ⋮ ⋮ ⋮ ⋱ ⋮ 1 α n α n 2 ⋯ α n n − 1 | {\displaystyle \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 , … , n {\displaystyle 2,3,\dots ,n} 행에서 1행을 빼는 기본행연산을 수행하면,
det ( V n ) = | 1 α 1 α 1 2 ⋯ α 1 n − 1 0 α 2 − α 1 α 2 2 − α 1 2 ⋯ α 2 n − 1 − α 1 n − 1 ⋮ ⋮ ⋮ ⋱ ⋮ 0 α n − α 1 α n 2 − α 1 2 ⋯ α n n − 1 − α 1 n − 1 | {\displaystyle \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 ∈ { 1 , 2 , … , n − 1 } {\displaystyle j\in \{1,2,\dots ,n-1\}} 에 대해, (j+1)열에서 j항의 α 1 {\displaystyle \alpha _{1}} 배를 빼는 기본열연산을 내림차순으로 수행하면,
det ( V n ) = | 1 0 0 ⋯ 0 0 α 2 − α 1 α 2 ( α 2 − α 1 ) ⋯ α 2 n − 2 ( α 2 − α 1 ) ⋮ ⋮ ⋮ ⋱ ⋮ 0 α n − α 1 α n ( α n − α 1 ) ⋯ α n n − 2 ( α n − α 1 ) | = ∏ i = 2 n ( α i − α 1 ) | 1 α 2 ⋯ α 2 n − 2 1 α 3 ⋯ α 3 n − 2 ⋮ ⋮ ⋱ ⋮ 1 α n ⋯ α n n − 2 | {\displaystyle {\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 ) = ∏ k = 1 n − 1 ( ∏ i = k + 1 n ( α i − α k ) ) = ∏ 1 ≤ j < i ≤ n ( α i − α j ) {\displaystyle \det(V_{n})=\prod _{k=1}^{n-1}\left(\prod _{i=k+1}^{n}(\alpha _{i}-\alpha _{k})\right)=\prod _{1\leq j<i\leq n}(\alpha _{i}-\alpha _{j})}
을 얻는다.