Gram行列式的妙用

设 $\alpha_1,\alpha_2,\cdots,\alpha_m$ 是 $n$ 维欧式空间 $V$ 中的一组向量,记$$G(\alpha_1,\alpha_2,\cdots,\alpha_m)=\begin{vmatrix}(\alpha_1,\alpha_1)& (\alpha_1,\alpha_2) & \cdots &(\alpha_1,\alpha_m) \\ (\alpha_2,\alpha_1)& (\alpha_2,\alpha_2) & \cdots &(\alpha_2,\alpha_m) \\ \vdots & \vdots & & \vdots \\ (\alpha_m,\alpha_1)& (\alpha_m,\alpha_2) & \cdots &(\alpha_m,\alpha_m)\end{vmatrix}$$求证:$$\alpha_1,\alpha_2,\cdots,\alpha_m 线性无关\Leftrightarrow  G(\alpha_1,\alpha_2,\cdots,\alpha_m) \neq 0$$


(1)证明必要性:

设有实数$k_1,k_2,\cdots,k_m$使得$k_1 \alpha_1+k_2 \alpha_2+\cdots+k_m \alpha_m=0$,即$$\sum_{i=1}^{m}k_i \alpha_i=0$$

两边对$\alpha_j$做内积得到$$\sum_{i=1}^{m}(\alpha_j,\alpha_i)k_i=0$$

写成矩阵乘法即:$$\begin{bmatrix}(\alpha_1,\alpha_1)& (\alpha_1,\alpha_2) & \cdots &(\alpha_1,\alpha_m) \\ (\alpha_2,\alpha_1)& (\alpha_2,\alpha_2) & \cdots &(\alpha_2,\alpha_m) \\ \vdots & \vdots & & \vdots \\ (\alpha_m,\alpha_1)& (\alpha_m,\alpha_2) & \cdots &(\alpha_m,\alpha_m)\end{bmatrix} \begin{bmatrix}k_1\\ k_2\\ \vdots\\ k_m\end{bmatrix}=0$$

由于左边Gram矩阵行列式不为零,故该齐次线性方程组只有零解,所有的$k_i=0$

故$\alpha_1,\alpha_2,\cdots,\alpha_m$ 线性无关

(2)证明充分性

假设左边的Gram行列式不可逆,那么有不全为零的$k_1,k_2,\cdots,k_m$,使得$$\begin{bmatrix}(\alpha_1,\alpha_1)& (\alpha_1,\alpha_2) & \cdots &(\alpha_1,\alpha_m) \\ (\alpha_2,\alpha_1)& (\alpha_2,\alpha_2) & \cdots &(\alpha_2,\alpha_m) \\ \vdots & \vdots & & \vdots \\ (\alpha_m,\alpha_1)& (\alpha_m,\alpha_2) & \cdots &(\alpha_m,\alpha_m)\end{bmatrix} \begin{bmatrix}k_1\\ k_2\\ \vdots\\ k_m\end{bmatrix}=0$$

故有$$\begin{bmatrix}k_1\\ k_2\\ \vdots\\ k_m\end{bmatrix}^T\begin{bmatrix}(\alpha_1,\alpha_1)& (\alpha_1,\alpha_2) & \cdots &(\alpha_1,\alpha_m) \\ (\alpha_2,\alpha_1)& (\alpha_2,\alpha_2) & \cdots &(\alpha_2,\alpha_m) \\ \vdots & \vdots & & \vdots \\ (\alpha_m,\alpha_1)& (\alpha_m,\alpha_2) & \cdots &(\alpha_m,\alpha_m)\end{bmatrix} \begin{bmatrix}k_1\\ k_2\\ \vdots\\ k_m\end{bmatrix}=0$$

这个和度量矩阵的形式是一样的,写成和式即为$$\sum_{i=1}^{m}\sum_{j=1}^{m}(\alpha_i,\alpha_j)x_i x_j=\sum_{i=1}^{m}\sum_{j=1}^{m}(x_i\alpha_i,x_j\alpha_j)=(\sum_{i=1}^{m}k_i\alpha_i,\sum_{i=1}^{m}k_i\alpha_i)=0$$

所以$\sum_{i=1}^{m}k_i\alpha_i=0$,这与$\alpha_1,\alpha_2,\cdots,\alpha_m$ 线性无关矛盾


例题:n阶方阵$A$满足$a_{ij}=\frac{1}{i+j-1}$,证明:$A$可逆

方法一(暴力):$$\begin{aligned}|A|_n&=\begin{vmatrix}1 & \frac{1}{2}&\frac{1}{3} & \cdots & \frac{1}{n}\\ \frac{1}{2} & \frac{1}{3} &\frac{1}{4}&\cdots &\frac{1}{n+1} \\ \frac{1}{3}&\frac{1}{4}&\frac{1}{5}&\cdots &\frac{1}{n+2}\\\vdots & \vdots &\vdots& & \vdots\\ \frac{1}{n} & \frac{1}{n+1}&\frac{1}{n+2} &\cdots &\frac{1}{2n-1} \end{vmatrix}\overset{R_i-R_n}{=}\begin{vmatrix}\frac{n-1}{n} & \frac{n-1}{2(n+1)}&\frac{n-1}{3(n+2)} & \cdots & \frac{n-1}{n(2n-1)}\\ \frac{n-2}{2n} & \frac{n-2}{3(n+1)} &\frac{n-2}{4(n+2)}&\cdots &\frac{n-2}{(n+1)(2n-1)} \\ \frac{n-3}{3n}&\frac{n-3}{4(n+1)}&\frac{n-3}{5(n+2)}&\cdots &\frac{n-3}{(n+2)(2n-1)}\\\vdots & \vdots &\vdots& & \vdots\\ \frac{1}{n} & \frac{1}{n+1}&\frac{1}{n+2} &\cdots &\frac{1}{2n-1} \end{vmatrix}\\&=\frac{(n-1)!}{n(n+1)\cdots(2n-1)}\begin{vmatrix}1 & \frac{1}{2}&\frac{1}{3} & \cdots & \frac{1}{n}\\ \frac{1}{2} & \frac{1}{3} &\frac{1}{4}&\cdots &\frac{1}{n+1} \\ \frac{1}{3}&\frac{1}{4}&\frac{1}{5}&\cdots &\frac{1}{n+2}\\\vdots & \vdots &\vdots& & \vdots\\ 1 & 1&1 &\cdots &1 \end{vmatrix}\overset{C_i-C_n}{=}\frac{(n-1)!}{n(n+1)\cdots(2n-1)}\begin{vmatrix}\frac{n-1}{n} & \frac{n-2}{2n}&\frac{n-3}{3n} & \cdots & \frac{1}{n}\\ \frac{n-1}{2(n+1)} & \frac{n-2}{3(n+1)} &\frac{n-3}{4(n+1)}&\cdots &\frac{1}{n+1} \\ \frac{n-1}{3(n+2)}&\frac{n-2}{4(n+2)}&\frac{n-3}{5(n+2)}&\cdots &\frac{1}{n+2}\\\vdots & \vdots &\vdots& & \vdots\\ 0 & 0&0 &\cdots &1 \end{vmatrix}\\&=\left [ \frac{(n-1)!}{n(n+1)\cdots(2n-1)}\right ]^2|A|_{n-1}\end{aligned}$$

由$|A|_1=1\neq 0$,递推得到$|A|_n\neq 0$

方法二(Gram行列式):

设$\alpha_i =x^{i-1}\in P[x]_n$,取向量的$C[a,b]$内积$$(\alpha_i,\alpha_j)=\int_{0}^{1}x^{i-1} x^{j-1}dx=\frac{1}{i+j-1}=a_{ij}$$

这样的话有$$|A|_n=\begin{vmatrix}(\alpha_1,\alpha_1)& (\alpha_1,\alpha_2) & \cdots &(\alpha_1,\alpha_n) \\ (\alpha_2,\alpha_1)& (\alpha_2,\alpha_2) & \cdots &(\alpha_2,\alpha_n) \\ \vdots & \vdots & & \vdots \\ (\alpha_n,\alpha_1)& (\alpha_n,\alpha_2) & \cdots &(\alpha_n,\alpha_n)\end{vmatrix}$$

而在$P[x]_n$上,显然有$\alpha_1,\alpha_2,\cdots,\alpha_n$线性无关

故左边的Gram行列式$|A|_n\neq 0$


注:在实线性空间$C[a,b]$中,对任意两个实连续函数$f(x),g(x)$规定内积$$(f(x),g(x))=\int_{a}^{b}f(x)g(x)dx$$由定积分的性质不难验证该内积符合内积的性质,故$C[a,b]$构成了欧式空间,我们把该内积称为$C[a,b]$内积

利用$C[a,b]$内积可以简化许多证明过程,例如:

(1)结合内积不等式$|(\alpha_i,\alpha_j)|\leq\left \| \alpha_i\right \|\left \| \alpha_j\right \|$可以证明柯西-施瓦茨不等式:$$\left | \int_{a}^{b}f(x)g(x)dx\right |\leq \sqrt{\int_{a}^{b}f(x)^2dx \int_{a}^{b}g(x)^2dx}$$

(2)结合三角不等式$\left \| \alpha_i +\alpha_j \right \| \leq \left \| \alpha_i  \right \|+\left \| \alpha_j \right \|$可以证明闵可夫斯基不等式:$$\sqrt{\int_{a}^{b}[f(x)+g(x)]^2 dx}\leq \sqrt{\int_{a}^{b}f(x)^2 dx}+\sqrt{\int_{a}^{b}g(x)^2 dx}$$

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