방데르몽드 행렬 Vandermonde matrix
- Alexandre-Théophile Vandermonde : 1735 – 1796
- 각 행이 등비수열
a geometric progression in each row
\[V =
\left(\begin{array}{cc}
1 & \alpha_1 & \alpha_1^2 & \cdots & \alpha_1^{n-1} \\
1 & \alpha_2 & \alpha_2^2 & \cdots & \alpha_2^{n-1} \\
1 & \alpha_3 & \alpha_3^2 & \cdots & \alpha_3^{n-1} \\
\vdots & \vdots & \vdots & & \vdots \\
1 & \alpha_m & \alpha_m^2 & \cdots & \alpha_m^{n-1} \\
\end{array}\right)\]
방데르몽드 행렬식 Vandermonde determinant
\[\displaystyle \det(V) = \prod_{1\le i\le j\le n}(\alpha_j - \alpha_i)\]
- 따라서 같은 공비를 가진 행이 있는 경우 행렬식 = 0
-
예시
\[\begin{align*}
\begin{vmatrix}
\;1 & \alpha_1 \; \\
\;1 & \alpha_2 \; \\
\end{vmatrix} \;&=\;\;
\alpha_2 - \alpha_1 \\ \\
\begin{vmatrix}
\;1 & \alpha_1 & \alpha_1^2 \; \\
\;1 & \alpha_2 & \alpha_2^2 \; \\
\;1 & \alpha_3 & \alpha_3^2 \; \\
\end{vmatrix} \;&=\;
(\alpha_3 - \alpha_2)(\alpha_3 - \alpha_1)(\alpha_2 - \alpha_1)
\end{align*}\]
3차 방정식 복소수 근의 응용
- $x^3=1$ 의 복소수 근을 $\omega$ 라고 하면
$1+\omega+\omega^2=0$ 이므로
\[\begin{align*}
W = &\left(\begin{array}{cc}
1 & 1 & 1 \\
1 & \omega & \omega^2 \\
1 & \omega^2 & \omega^4 \\
\end{array} \right) \\ \\
= &\left(\begin{array}{cc}
1 & 1 & 1 \\
1 & \omega & \omega^2 \\
1 & \omega^2 & \omega \\
\end{array} \right) \\ \\
W^{-1} =\; \frac{\;1\;}{3}
&\left(\begin{array}{cc}
1 & 1 & 1 \\
1 & \omega^2 & \omega \\
1 & \omega & \omega^2 \\
\end{array} \right)
\end{align*}\]
4차 방정식 복소수 근의 응용
- $x^4=1$ 의 복소수 근을 $\omega$ 라고 하면
$1+\omega+\omega^2+\omega^3=0$ 이므로
또는 $\omega = \pm \;i$ 이므로,
\[\begin{align*}
W = &\left(\begin{array}{cc}
1 & 1 & 1 & 1 \\
1 & \omega & \omega^2 & \omega^3 \\
1 & \omega^2 & \omega^4 & \omega^6 \\
1 & \omega^3 & \omega^6 & \omega^9 \\
\end{array} \right) \\ \\
= &\left(\begin{array}{cc}
1 & 1 & 1 & 1 \\
1 & \omega & \omega^2 & \omega^3 \\
1 & \omega^2 & 1 & \omega^2 \\
1 & \omega^3 & \omega^2 & \omega \\
\end{array} \right) \\ \\
W^{-1} =\; \frac{\;1\;}{4}
&\left(\begin{array}{cc}
1 & 1 & 1 & 1 \\
1 & \omega^3 & \omega^2 & \omega \\
1 & \omega^2 & 1 & \omega^2 \\
1 & \omega & \omega^2 & \omega^3 \\
\end{array} \right)
\end{align*}\]
- Note : $x^4=1$ 의 실수근($\pm 1$)의 경우는
방데르몽드 행렬식 = 0 으로 역행렬 없음
5차 방정식 복소수 근의 응용
- $x^5=1$ 의 복소수 근을 $\omega$ 라고 하면
$1+\omega+\omega^2+\omega^3+\omega^4=0$ 이므로
\[\begin{align*}
W = &\left(\begin{array}{cc}
1 & 1 & 1 & 1 & 1 \\
1 & \omega & \omega^2 & \omega^3 & \omega^4 \\
1 & \omega^2 & \omega^4 & \omega^6 & \omega^8 \\
1 & \omega^3 & \omega^6 & \omega^9 & \omega^{12} \\
1 & \omega^4 & \omega^8 & \omega^{12} & \omega^{16} \\
\end{array} \right) \\ \\
=\; &\left(\begin{array}{cc}
1 & 1 & 1 & 1 & 1 \\
1 & \omega & \omega^2 & \omega^3 & \omega^4 \\
1 & \omega^2 & \omega^4 & \omega & \omega^3 \\
1 & \omega^3 & \omega & \omega^4 & \omega^2 \\
1 & \omega^4 & \omega^3 & \omega^2 & \omega \\
\end{array} \right) \\ \\
W^{-1} =\; \frac{\;1\;}{5}
&\left(\begin{array}{cc}
1 & 1 & 1 & 1 & 1 \\
1 & \omega^4 & \omega^3 & \omega^2 & \omega \\
1 & \omega^3 & \omega & \omega^4 & \omega^2 \\
1 & \omega^2 & \omega^4 & \omega & \omega^3 \\
1 & \omega & \omega^2 & \omega^3 & \omega^4 \\
\end{array} \right)
\end{align*}\]