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Questions in category: 行列式 (Determinant).

柯西行列式(Cauchy determinant)

Posted by haifeng on 2022-04-05 23:25:25 last update 2022-04-05 23:43:27 | Answers (1) | 收藏


设 $A_n=(\frac{1}{x_i+y_j})_{1\leqslant i,j\leqslant n}$, 则其行列式为
\[
\det(A_n)=
\begin{vmatrix}
\frac{1}{x_1+y_1} & \frac{1}{x_1+y_2} & \cdots & \frac{1}{x_1+y_n}\\
\frac{1}{x_2+y_1} & \frac{1}{x_2+y_2} & \cdots & \frac{1}{x_2+y_n}\\
\vdots & \vdots & &\vdots\\
\frac{1}{x_n+y_1} & \frac{1}{x_n+y_2} & \cdots & \frac{1}{x_n+y_n}\\
\end{vmatrix}
\]
证明:
\[
\det(A)=\dfrac{\prod\limits_{1\leqslant i < j\leqslant n}(x_i-x_j)(y_i-y_j)}{\prod\limits_{1\leqslant i,j\leqslant n}(x_i+y_j)}
\]