2024高等代数【南昌大学】
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已知 f ( x ) = 1 + x + x 2 + ⋯ + x n − 1 f(x) = 1 + x + x^2 + \cdots + x^{n-1} f(x)=1+x+x2+⋯+xn−1，证明： f ( x ) ∣ [ f ( x ) + x n ] 2 − x n f(x) \mid \left[f(x) + x^n \right]^2 - x^n f(x)∣[f(x)+xn]2−xn。
x f ( x ) = x + x 2 + x 3 + ⋯ + x n xf(x) = x + x^2 + x^3 + \cdots + x^n xf(x)=x+x2+x3+⋯+xn
x f ( x ) − f ( x ) = x n − 1 xf(x) - f(x) = x^n - 1 xf(x)−f(x)=xn−1
现在计算 [ f ( x ) + x n ] 2 − x n \left[f(x) + x^n\right]^2 - x^n [f(x)+xn]2−xn：
[ f ( x ) + x n ] 2 − x n = f 2 ( x ) + 2 x n f ( x ) + x n ( x n − 1 ) = f 2 ( x ) + 2 x n f ( x ) + x n ( x − 1 ) f ( x ) = f ( x ) [ f ( x ) + x n + x n + 1 ] \begin{align*} \left [f(x) + x^n\right]^2 - x^n &= f^2(x) + 2x^n f(x) + x^n(x^n - 1) \\ &= f^2(x) + 2x^n f(x) + x^n(x - 1)f(x) \\ &= f(x) \left [f(x) + x^n + x^{n+1}\right] \end{align*} [f(x)+xn]2−xn=f2(x)+2xnf(x)+xn(xn−1)=f2(x)+2xnf(x)+xn(x−1)f(x)=f(x)[f(x)+xn+xn+1]
因此可以得出结论， [ f ( x ) + x n ] 2 − x n \left[f(x) + x^n\right]^2 - x^n [f(x)+xn]2−xn 是 f ( x ) f(x) f(x) 的倍数。
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计算 n n n 阶行列式：
D n = ∣ x 1 2 − 2 x 1 x 2 ⋯ x 1 x n x 2 x 1 x 2 2 − 2 ⋯ x 2 x n ⋮ ⋮ ⋱ ⋮ x n x 1 x n x 2 ⋯ x n 2 − 2 ∣ D_n = \left| \begin{array}{cccc} x_1^2 - 2 & x_1 x_2 & \cdots & x_1 x_n \\ x_2 x_1 & x_2^2 - 2 & \cdots & x_2 x_n \\ \vdots & \vdots & \ddots & \vdots \\ x_n x_1 & x_n x_2 & \cdots & x_n^2 - 2 \end{array} \right| Dn= x12−2x2x1⋮xnx1x1x2x22−2⋮xnx2⋯⋯⋱⋯x1xnx2xn⋮xn2−2
记向量：
α = [ x 1 x 2 ⋮ x n ] \alpha = \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix} α= x1x2⋮xn
计算 D n D_n Dn 的值：
D n = ∣ α α T − 2 E n ∣ = ( − 1 ) n ∣ 2 E n − α α T ∣ = ( − 1 ) n 2 n − 1 ∣ 2 E 1 − α T α ∣ = ( − 2 ) n ( 1 − 1 2 ∑ k = 1 n x k 2 ) \begin{align*} D_n &= \left| \alpha \alpha^T - 2E_n \right| \\ &= (-1)^n \left| 2E_n - \alpha \alpha^T \right| \\ &= (-1)^n 2^{n-1} \left| 2E_1 - \alpha^T \alpha \right| \\ &= (-2)^n \left( 1 - \frac{1}{2} \sum_{k = 1}^n x_k^2 \right) \end{align*} Dn= ααT−2En =(−1)n 2En−ααT =(−1)n2n−1 2E1−αTα =(−2)n(1−21k=1∑nxk2)
进一步展开计算：
D n = ∣ 1 x 1 x 2 ⋯ x n 0 x 1 2 − 2 x 1 x 2 ⋯ x 1 x n 0 x 2 x 1 x 2 2 − 2 ⋯ x 2 x n 0 ⋮ ⋮ ⋱ ⋮ 0 x n x 1 x n x 2 ⋯ x n 2 − 2 ∣ = ∣ 1 x 1 x 2 ⋯ x n − x 1 − 2 0 ⋯ 0 − x 2 0 − 2 ⋯ 0 ⋮ ⋮ ⋮ ⋱ ⋮ − x n 0 0 ⋯ − 2 ∣ = 1 2 ∣ 2 x 1 x 2 ⋯ x n − 2 x 1 − 2 0 ⋯ 0 − 2 x 2 0 − 2 ⋯ 0 ⋮ ⋮ ⋮ ⋱ ⋮ − 2 x n 0 0 ⋯ − 2 ∣ = ( − 2 ) n − 1 ( 1 − 1 2 ∑ k = 1 n x k 2 ) \begin{align*} D_n &= \left| \begin{array}{ccccc} 1 & x_1 & x_2 & \cdots & x_n \\ 0 & x_1^2 - 2 & x_1 x_2 & \cdots & x_1 x_n \\ 0 & x_2 x_1 & x_2^2 - 2 & \cdots & x_2 x_n \\ 0 & \vdots & \vdots & \ddots & \vdots \\ 0 & x_n x_1 & x_n x_2 & \cdots & x_n^2 - 2 \end{array} \right| \\ &= \left| \begin{array}{ccccc} 1 & x_1 & x_2 & \cdots & x_n \\ -x_1 & -2 & 0 & \cdots & 0 \\ -x_2 & 0 & -2 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ -x_n & 0 & 0 & \cdots & -2 \end{array} \right| \\ &= \frac{1}{2} \left| \begin{array}{ccccc} 2 & x_1 & x_2 & \cdots & x_n \\ -2x_1 & -2 & 0 & \cdots & 0 \\ -2x_2 & 0 & -2 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ -2x_n & 0 & 0 & \cdots & -2 \end{array} \right| \\ &= (-2)^{n-1} \left( 1 - \frac{1}{2} \sum_{k = 1}^n x_k^2 \right) \end{align*} Dn= 10000x1x12−2x2x1⋮xnx1x2x1x2x22−2⋮xnx2