Gram行列式的妙用
Theorem 设 α 1 , α 2 , ⋯ , α m \alpha_1,\alpha_2,\cdots,\alpha_m α 1 , α 2 , ⋯ , α m n n n V V V G ( α 1 , α 2 , ⋯ , α m ) = ∣ ( α 1 , α 1 ) ( α 1 , α 2 ) ⋯ ( α 1 , α m ) ( α 2 , α 1 ) ( α 2 , α 2 ) ⋯ ( α 2 , α m ) ⋮ ⋮ ⋮ ( α m , α 1 ) ( α m , α 2 ) ⋯ ( α m , α m ) ∣ 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} G ( α 1 , α 2 , ⋯ , α m ) = ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ( α 1 , α 1 ) ( α 2 , α 1 ) ⋮ ( α m , α 1 ) ( α 1 , α 2 ) ( α 2 , α 2 ) ⋮ ( α m , α 2 ) ⋯ ⋯ ⋯ ( α 1 , α m ) ( α 2 , α m ) ⋮ ( α m , α m ) ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ G G G α 1 , α 2 , ⋯ , α m 线性无关 ⇔ G ( α 1 , α 2 , ⋯ , α m ) ≠ 0 \alpha_1,\alpha_2,\cdots,\alpha_m 线性无关\Leftrightarrow G(\alpha_1,\alpha_2,\cdots,\alpha_m) \neq 0 α 1 , α 2 , ⋯ , α m 线 性 无 关 ⇔ G ( α 1 , α 2 , ⋯ , α m ) = 0
（1）证明必要性：
设有实数 k 1 , k 2 , ⋯ , k m k_1,k_2,\cdots,k_m k 1 , k 2 , ⋯ , k m k 1 α 1 + k 2 α 2 + ⋯ + k m α m = 0 k_1 \alpha_1+k_2 \alpha_2+\cdots+k_m \alpha_m=0 k 1 α 1 + k 2 α 2 + ⋯ + k m α m = 0
∑ i = 1 m k i α i = 0 \sum_{i=1}^{m}k_i \alpha_i=0 i = 1 ∑ m k i α i = 0
两边对 α j \alpha_j α j
∑ i = 1 m ( α j , α i ) k i = 0 \sum_{i=1}^{m}(\alpha_j,\alpha_i)k_i=0 i = 1 ∑ m ( α j , α i ) k i = 0
写成矩阵乘法即：
[ ( α 1 , α 1 ) ( α 1 , α 2 ) ⋯ ( α 1 , α m ) ( α 2 , α 1 ) ( α 2 , α 2 ) ⋯ ( α 2 , α m ) ⋮ ⋮ ⋮ ( α m , α 1 ) ( α m , α 2 ) ⋯ ( α m , α m ) ] [ k 1 k 2 ⋮ k m ] = 0 \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 ⎣ ⎢ ⎢ ⎢ ⎢ ⎡ ( α 1 , α 1 ) ( α 2 , α 1 ) ⋮ ( α m , α 1 ) ( α 1 , α 2 ) ( α 2 , α 2 ) ⋮ ( α m , α 2 ) ⋯ ⋯ ⋯ ( α 1 , α m ) ( α 2 , α m ) ⋮ ( α m , α m ) ⎦ ⎥ ⎥ ⎥ ⎥ ⎤ ⎣ ⎢ ⎢ ⎢ ⎢ ⎡ k 1 k 2 ⋮ k m ⎦ ⎥ ⎥ ⎥ ⎥ ⎤ = 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 k_i=0 k i = 0
故 α 1 , α 2 , ⋯ , α m \alpha_1,\alpha_2,\cdots,\alpha_m α 1 , α 2 , ⋯ , α m
（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 , ⋯ , k m k_1,k_2,\cdots,k_m k 1 , k 2 , ⋯ , k m
[ ( α 1 , α 1 ) ( α 1 , α 2 ) ⋯ ( α 1 , α m ) ( α 2 , α 1 ) ( α 2 , α 2 ) ⋯ ( α 2 , α m ) ⋮ ⋮ ⋮ ( α m , α 1 ) ( α m , α 2 ) ⋯ ( α m , α m ) ] [ k 1 k 2 ⋮ k m ] = 0 \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 ⎣ ⎢ ⎢ ⎢ ⎢ ⎡ ( α 1 , α 1 ) ( α 2 , α 1 ) ⋮ ( α m , α 1 ) ( α 1 , α 2 ) ( α 2 , α 2 ) ⋮ ( α m , α 2 ) ⋯ ⋯ ⋯ ( α 1 , α m ) ( α 2 , α m ) ⋮ ( α m , α m ) ⎦ ⎥ ⎥ ⎥ ⎥ ⎤ ⎣ ⎢ ⎢ ⎢ ⎢ ⎡ k 1 k 2 ⋮ k m ⎦ ⎥ ⎥ ⎥ ⎥ ⎤ = 0
故有
[ k 1 k 2 ⋮ k m ] T [ ( α 1 , α 1 ) ( α 1 , α 2 ) ⋯ ( α 1 , α m ) ( α 2 , α 1 ) ( α 2 , α 2 ) ⋯ ( α 2 , α m ) ⋮ ⋮ ⋮ ( α m , α 1 ) ( α m , α 2 ) ⋯ ( α m , α m ) ] [ k 1 k 2 ⋮ k m ] = 0 \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 ⎣ ⎢ ⎢ ⎢ ⎢ ⎡ k 1 k 2 ⋮ k m ⎦ ⎥ ⎥ ⎥ ⎥ ⎤ T ⎣ ⎢ ⎢ ⎢ ⎢ ⎡ ( α 1 , α 1 ) ( α 2 , α 1 ) ⋮ ( α m , α 1 ) ( α 1 , α 2 ) ( α 2 , α 2 ) ⋮ ( α m , α 2 ) ⋯ ⋯ ⋯ ( α 1 , α m ) ( α 2 , α m ) ⋮ ( α m , α m ) ⎦ ⎥ ⎥ ⎥ ⎥ ⎤ ⎣ ⎢ ⎢ ⎢ ⎢ ⎡ k 1 k 2 ⋮ k m ⎦ ⎥ ⎥ ⎥ ⎥ ⎤ = 0
这个和度量矩阵的形式是一样的，写成和式即为
∑ i = 1 m ∑ j = 1 m ( α i , α j ) x i x j = ∑ i = 1 m ∑ j = 1 m ( x i α i , x j α j ) = ( ∑ i = 1 m k i α i , ∑ i = 1 m k i α i ) = 0 \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 i = 1 ∑ m j = 1 ∑ m ( α i , α j ) x i x j = i = 1 ∑ m j = 1 ∑ m ( x i α i , x j α j ) = ( i = 1 ∑ m k i α i , i = 1 ∑ m k i α i ) = 0
所以 ∑ i = 1 m k i α i = 0 \sum_{i=1}^{m}k_i\alpha_i=0 ∑ i = 1 m k i α i = 0 α 1 , α 2 , ⋯ , α m \alpha_1,\alpha_2,\cdots,\alpha_m α 1 , α 2 , ⋯ , α m
Problem 若 n n n A A A a i j = 1 i + j − 1 a_{ij}=\frac{1}{i+j-1} a i j = i + j − 1 1 A A A
方法一（暴力）：
∣ A ∣ n = ∣ 1 1 2 1 3 ⋯ 1 n 1 2 1 3 1 4 ⋯ 1 n + 1 1 3 1 4 1 5 ⋯ 1 n + 2 ⋮ ⋮ ⋮ ⋮ 1 n 1 n + 1 1 n + 2 ⋯ 1 2 n − 1 ∣ = R i − R n ∣ n − 1 n n − 1 2 ( n + 1 ) n − 1 3 ( n + 2 ) ⋯ n − 1 n ( 2 n − 1 ) n − 2 2 n n − 2 3 ( n + 1 ) n − 2 4 ( n + 2 ) ⋯ n − 2 ( n + 1 ) ( 2 n − 1 ) n − 3 3 n n − 3 4 ( n + 1 ) n − 3 5 ( n + 2 ) ⋯ n − 3 ( n + 2 ) ( 2 n − 1 ) ⋮ ⋮ ⋮ ⋮ 1 n 1 n + 1 1 n + 2 ⋯ 1 2 n − 1 ∣ = ( n − 1 ) ! n ( n + 1 ) ⋯ ( 2 n − 1 ) ∣ 1 1 2 1 3 ⋯ 1 n 1 2 1 3 1 4 ⋯ 1 n + 1 1 3 1 4 1 5 ⋯ 1 n + 2 ⋮ ⋮ ⋮ ⋮ 1 1 1 ⋯ 1 ∣ = C i − C n ( n − 1 ) ! n ( n + 1 ) ⋯ ( 2 n − 1 ) ∣ n − 1 n n − 2 2 n n − 3 3 n ⋯ 1 n n − 1 2 ( n + 1 ) n − 2 3 ( n + 1 ) n − 3 4 ( n + 1 ) ⋯ 1 n + 1 n − 1 3 ( n + 2 ) n − 2 4 ( n + 2 ) n − 3 5 ( n + 2 ) ⋯ 1 n + 2 ⋮ ⋮ ⋮ ⋮ 0 0 0 ⋯ 1 ∣ = [ ( n − 1 ) ! n ( n + 1 ) ⋯ ( 2 n − 1 ) ] 2 ∣ A ∣ n − 1 \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} ∣ A ∣ n = ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ 1 2 1 3 1 ⋮ n 1 2 1 3 1 4 1 ⋮ n + 1 1 3 1 4 1 5 1 ⋮ n + 2 1 ⋯ ⋯ ⋯ ⋯ n 1 n + 1 1 n + 2 1 ⋮ 2 n − 1 1 ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ = R i − R n ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ n n − 1 2 n n − 2 3 n n − 3 ⋮ n 1 2 ( n + 1 ) n − 1 3 ( n + 1 ) n − 2 4 ( n + 1 ) n − 3 ⋮ n + 1 1 3 ( n + 2 ) n − 1 4 ( n + 2 ) n − 2 5 ( n + 2 ) n − 3 ⋮ n + 2 1 ⋯ ⋯ ⋯ ⋯ n ( 2 n − 1 ) n − 1 ( n + 1 ) ( 2 n − 1 ) n − 2 ( n + 2 ) ( 2 n − 1 ) n − 3 ⋮ 2 n − 1 1 ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ = n ( n + 1 ) ⋯ ( 2 n − 1 ) ( n − 1 ) ! ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ 1 2 1 3 1 ⋮ 1 2 1 3 1 4 1 ⋮ 1 3 1 4 1 5 1 ⋮ 1 ⋯ ⋯ ⋯ ⋯ n 1 n + 1 1 n + 2 1 ⋮ 1 ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ = C i − C n n ( n + 1 ) ⋯ ( 2 n − 1 ) ( n − 1 ) ! ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ n n − 1 2 ( n + 1 ) n − 1 3 ( n + 2 ) n − 1 ⋮ 0 2 n n − 2 3 ( n + 1 ) n − 2 4 ( n + 2 ) n − 2 ⋮ 0 3 n n − 3 4 ( n + 1 ) n − 3 5 ( n + 2 ) n − 3 ⋮ 0 ⋯ ⋯ ⋯ ⋯ n 1 n + 1 1 n + 2 1 ⋮ 1 ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ = [ n ( n + 1 ) ⋯ ( 2 n − 1 ) ( n − 1 ) ! ] 2 ∣ A ∣ n − 1
由 ∣ A ∣ 1 = 1 ≠ 0 |A|_1=1\neq 0 ∣ A ∣ 1 = 1 = 0 ∣ A ∣ n ≠ 0 |A|_n\neq 0 ∣ A ∣ n = 0
方法二（Gram行列式）：
设 α i = x i − 1 ∈ P [ x ] n \alpha_i =x^{i-1}\in P[x]_n α i = x i − 1 ∈ P [ x ] n C [ a , b ] C[a,b] C [ a , b ]
( α i , α j ) = ∫ 0 1 x i − 1 x j − 1 d x = 1 i + j − 1 = a i j (\alpha_i,\alpha_j)=\int_{0}^{1}x^{i-1} x^{j-1}dx=\frac{1}{i+j-1}=a_{ij} ( α i , α j ) = ∫ 0 1 x i − 1 x j − 1 d x = i + j − 1 1 = a i j
这样的话有
∣ A ∣ n = ∣ ( α 1 , α 1 ) ( α 1 , α 2 ) ⋯ ( α 1 , α n ) ( α 2 , α 1 ) ( α 2 , α 2 ) ⋯ ( α 2 , α n ) ⋮ ⋮ ⋮ ( α n , α 1 ) ( α n , α 2 ) ⋯ ( α n , α n ) ∣ |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} ∣ A ∣ n = ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ( α 1 , α 1 ) ( α 2 , α 1 ) ⋮ ( α n , α 1 ) ( α 1 , α 2 ) ( α 2 , α 2 ) ⋮ ( α n , α 2 ) ⋯ ⋯ ⋯ ( α 1 , α n ) ( α 2 , α n ) ⋮ ( α n , α n ) ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣
而在 P [ x ] n P[x]_n P [ x ] n α 1 , α 2 , ⋯ , α n \alpha_1,\alpha_2,\cdots,\alpha_n α 1 , α 2 , ⋯ , α n
故左边的 Gram 行列式 ∣ A ∣ n ≠ 0 |A|_n\neq 0 ∣ A ∣ n = 0
注： C [ a , b ] C[a,b] C [ a , b ] f ( x ) , g ( x ) f(x),g(x) f ( x ) , g ( x )
( f ( x ) , g ( x ) ) = ∫ a b f ( x ) g ( x ) d x (f(x),g(x))=\int_{a}^{b}f(x)g(x)dx ( f ( x ) , g ( x ) ) = ∫ a b f ( x ) g ( x ) d x
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 ] C[a,b] C [ a , b ] C [ a , b ] C[a,b] C [ a , b ]
利用 C [ a , b ] C[a,b] C [ a , b ]
（1）结合内积不等式 ∣ ( α i , α j ) ∣ ≤ ∥ α i ∥ ∥ α j ∥ |(\alpha_i,\alpha_j)|\leq\left \| \alpha_i\right \|\left \| \alpha_j\right \| ∣ ( α i , α j ) ∣ ≤ ∥ α i ∥ ∥ α j ∥ 柯西-施瓦茨不等式
∣ ∫ a b f ( x ) g ( x ) d x ∣ ≤ ∫ a b f ( x ) 2 d x ∫ a b g ( x ) 2 d x \left | \int_{a}^{b}f(x)g(x)dx\right |\leq \sqrt{\int_{a}^{b}f(x)^2dx \int_{a}^{b}g(x)^2dx} ∣ ∣ ∣ ∣ ∣ ∣ ∫ a b f ( x ) g ( x ) d x ∣ ∣ ∣ ∣ ∣ ∣ ≤ ∫ a b f ( x ) 2 d x ∫ a b g ( x ) 2 d x
（2）结合三角不等式 ∥ α i + α j ∥ ≤ ∥ α i ∥ + ∥ α j ∥ \left \| \alpha_i +\alpha_j \right \| \leq \left \| \alpha_i \right \|+\left \| \alpha_j \right \| ∥ α i + α j ∥ ≤ ∥ α i ∥ + ∥ α j ∥ 闵可夫斯基不等式
∫ a b [ f ( x ) + g ( x ) ] 2 d x ≤ ∫ a b f ( x ) 2 d x + ∫ a b g ( x ) 2 d x \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} ∫ a b [ f ( x ) + g ( x ) ] 2 d x ≤ ∫ a b f ( x ) 2 d x + ∫ a b g ( x ) 2 d x