哈尔滨工程大学 2018年高等代数第5题

在欧氏空间 $\mathbb{R}[x]_3$ 中，基为 $1, x, x^2$，记 $e_1=1, e_2=x, e_3=x^2$。度量矩阵 $G$ 的元素 $g_{ij} = (e_i, e_j)$，其中内积定义为 $(f(x), g(x)) = \int_{-1}^1 f(x)g(x) dx$。

$(1,1) = \int_{-1}^1 1 \cdot 1 \, dx = \int_{-1}^1 1 \, dx = [x]_{-1}^1 = 1 - (-1) = 2$

$(1,x) = \int_{-1}^1 1 \cdot x \, dx = \int_{-1}^1 x \, dx = \left[\frac{x^2}{2}\right]_{-1}^1 = \frac{1}{2} - \frac{1}{2} = 0$。由对称性，$(x,1) = 0$。

$(1,x^2) = \int_{-1}^1 1 \cdot x^2 \, dx = \int_{-1}^1 x^2 \, dx = \left[\frac{x^3}{3}\right]_{-1}^1 = \frac{1}{3} - \left(-\frac{1}{3}\right) = \frac{2}{3}$。由对称性，$(x^2,1) = \frac{2}{3}$。

$(x,x) = \int_{-1}^1 x \cdot x \, dx = \int_{-1}^1 x^2 \, dx = \frac{2}{3}$。

$(x,x^2) = \int_{-1}^1 x \cdot x^2 \, dx = \int_{-1}^1 x^3 \, dx = \left[\frac{x^4}{4}\right]_{-1}^1 = \frac{1}{4} - \frac{1}{4} = 0$。由对称性，$(x^2,x) = 0$。

$(x^2,x^2) = \int_{-1}^1 x^2 \cdot x^2 \, dx = \int_{-1}^1 x^4 \, dx = \left[\frac{x^5}{5}\right]_{-1}^1 = \frac{1}{5} - \left(-\frac{1}{5}\right) = \frac{2}{5}$。

将计算结果按顺序填入矩阵： $g_{11}=2, g_{12}=0, g_{13}=\frac{2}{3}$ $g_{21}=0, g_{22}=\frac{2}{3}, g_{23}=0$ $g_{31}=\frac{2}{3}, g_{32}=0, g_{33}=\frac{2}{5}$ 因此度量矩阵为 \[ \begin{pmatrix} 2 & 0 & \frac{2}{3} \\ 0 & \frac{2}{3} & 0 \\ \frac{2}{3} & 0 & \frac{2}{5} \end{pmatrix} \]