本系列主要记录一些矩阵计算理论题目比较好的解题思路，由于查不到答案，所以解题来源于自己的思考和与同学的交流。

题目：证明

$\left \| V\left ( x_{0},x_{1},\cdots ,x_{n} \right )^{-1} \right \|_{\infty }\leqslant \max_{0\leqslant k\leqslant n}\prod_{i=0,i\neq k}^{n}\frac{1+\left | x_{i} \right |}{\left | x_{k}-x_{i} \right |}$

其中，$V\left ( x_{0},x_{1},\cdots ,x_{n} \right )$表示$Vandermonde$矩阵。

证明：设$p\left ( x \right )=\left ( x+x_{1} \right )\left ( x+x_{2} \right )\cdots \left ( x+x_{n} \right )=a_{0}+a_{1}x+a_{2}x^{2}+\cdots +x^{n}$。

先证：

$\left | a_{0} \right |+\left | a_{1} \right |+\cdots +\left | a_{n-1} \right |+1\leqslant \prod_{i=1}^{n}\left ( 1+\left | x_{i} \right | \right )$

由韦达定理（根于系数的关系）有：

$\left\{\begin{matrix}a_{n-1}=x_{1}+x_{2}+\cdots +x_{n},\\ a_{n-2}=\sum_{1\leqslant i< j\leqslant n}^{}x_{i}x_{j},\; \; \; \; \; \; \,\\ \cdots,\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \;  \\ a_{0}=\prod_{i=1}^{n}x_{i}\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \end{matrix}\right.$

因此，$\left | a_{0} \right |+\left | a_{1} \right |+\cdots +\left | a_{n-1} \right |+1=\left | \prod_{i=1}^{n}x_{i} \right |+\cdots +\left | \sum_{i\leqslant i< j\leqslant n}^{}x_{i}x_{j} \right |+\left | x_{1}+x_{2}+\cdots +x_{n} \right |+1\leqslant \prod_{i=1}^{n}\left | x_{i} \right |+\cdots +\sum_{1\leqslant i< j\leqslant n}^{}\left | x_{i}\right |\left | x_{j} \right |+\left ( \left | x_{1} \right |+\left | x_{2} \right |+\cdots +\left | x_{n} \right | \right )+1= \prod_{i=1}^{n}\left ( 1+\left | x_{i} \right | \right )%

设$A=V\left ( x_{0},x_{1},\cdots ,x_{n} \right )^{-1}=\begin{bmatrix}a_{00} &\cdots   &a_{0n} \\ \vdots  &\ddots   &\vdots  \\  a_{n0}&\cdots   &a_{nn} \end{bmatrix}$

令$p_{i}\left ( x \right )=a_{i0}+a_{i1}x+\cdots +a_{in}x^{n}$，由$AV\left ( x_{0},x_{1},\cdots ,x_{n} \right )=I$，有

$p_{i}\left ( x_{j} \right )=\delta _{ij}=\left\{\begin{matrix}1,\; \; i=j\\ 0,\; \; i\neq j\end{matrix}\right.$

由$Lagrangen$插值多项式有

$p_{i}\left ( x \right )=\frac{\prod_{j=0,j\neq i}^{n}\left ( x-x_{i} \right )}{\prod_{j=0,j\neq i}^{n}\left ( x_{j}-x_{i} \right )}=a_{i0}+a_{i1}x+\cdots +a_{in}x^{n}$

那么就有

$\left | a_{i0} \right |+\left | a_{i1} \right |+\cdots +\left | a_{in} \right|\leqslant \left | \frac{1}{\prod_{j=0,j\neq i}^{n}\left ( x_{j}-x_{i} \right )} \right |\prod_{j=0,j\neq i}^{n}\left ( 1+\left | x_{i} \right | \right )\leqslant \frac{\prod_{j=0,j\neq i}^{n}\left ( 1+\left | x_{i} \right | \right )}{\prod_{j=0,j\neq i}^{n}\left | x_{j}-x_{i} \right |}$

\left\{\begin{matrix}a_{n-1}=x_{1}+x_{2}+\cdots +x_{n},\\ a_{n-2}=\sum_{1\leqslant i< j\leqslant n}^{}x_{i}x_{j},\; \; \; \; \; \; \, \\ \cdots,\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \;  \\ 
a_{0}=\prod_{i=1}^{n}x_{i}\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \end{matrix}\right.