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几个特殊矩阵的行列式计算
在计算几个特殊几何体的体积中,需反复用到些特殊矩阵的行列式值的计算。经繁琐的推导后,得到了如下简洁的结果。
1.
$\det\left[\begin{array}{cccc}
\frac{1}{1-\lambda_{1}^{2}} & \frac{1}{1-\lambda_{1}\lambda_{2}} & \cdots & \frac{1}{1-\lambda_{1}\lambda_{n}}\\
\frac{1}{1-\lambda_{1}\lambda_{2}} & \frac{1}{1-\lambda_{2}^{2}} & \cdots & \frac{1}{1-\lambda_{2}\lambda_{n}}\\
\vdots & \vdots & \ddots & \vdots\\
\frac{1}{1-\lambda_{1}\lambda_{n}} & \frac{1}{1-\lambda_{2}\lambda_{n}} & \cdots & \frac{1}{1-\lambda_{n}^{2}}
\end{array}\right]=\left[\prod_{1\leq j_{1} 2. $\det\left[\begin{array}{cccc}
\frac{1}{2\lambda_{1}} & \frac{1}{\lambda_{1}+\lambda_{2}} & \cdots & \frac{1}{\lambda_{1}+\lambda_{n}}\\
\frac{1}{\lambda_{1}+\lambda_{2}} & \frac{1}{2\lambda_{2}} & \cdots & \frac{1}{\lambda_{2}+\lambda_{n}}\\
\vdots & \vdots & \ddots & \vdots\\
\frac{1}{\lambda_{1}+\lambda_{n}} & \frac{1}{\lambda_{2}+\lambda_{n}} & \cdots & \frac{1}{2\lambda_{n}}
\end{array}\right]=\left[\prod_{1\leq j_{1} 3. $\sum_{(k_{1},k_{2},\cdots,k_{n})\in\Omega_{0,\infty}^{n}}\det\left(\left[\begin{array}{cccc}
\lambda_{1}^{k_{1}} & \lambda_{1}^{k_{2}} & \cdots & \lambda_{1}^{k_{n}}\\
\lambda_{2}^{k_{1}} & \lambda_{2}^{k_{2}} & \cdots & \lambda_{2}^{k_{n}}\\
\vdots & \vdots & \ddots & \vdots\\
\lambda_{n}^{k_{1}} & \lambda_{n}^{k_{2}} & \cdots & \lambda_{n}^{k_{n}}
\end{array}\right]\right)=\left(\prod_{1\leq j_{1} 其中 $\Omega_{0,\infty}^{n}$ 指所有由自然数序列 $\{0,1,2,\cdots\}$ 中任取 $n$ 个数并按从小到大排序而得到的 $(k_{1},k_{2},\cdots,k_{n})$ 组成的集合。 相类似的算式没有查到,只好自己推。
https://blog.sciencenet.cn/blog-3343777-1067962.html
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