基变换与坐标变换

基变换

若 $\boldsymbol \varepsilon_1, \boldsymbol \varepsilon_2, \dots \boldsymbol \varepsilon_n$ 和 $\boldsymbol \eta_1, \boldsymbol \eta_2, \dots \boldsymbol \eta_n$ 是 $n$ 维空间 $V$ 的两组基满足

$$\begin{cases} \boldsymbol \eta_1 = c_{11} \boldsymbol \varepsilon_1 + c_{21} \boldsymbol \varepsilon_2 + \cdots + c_{n1} \boldsymbol \varepsilon_n \\ \boldsymbol \eta_2 = c_{12} \boldsymbol \varepsilon_1 + c_{22} \boldsymbol \varepsilon_2 + \cdots + c_{n2} \boldsymbol \varepsilon_n \\ \vdots \\ \boldsymbol \eta_n = c_{1n} \boldsymbol \varepsilon_1 + c_{2n} \boldsymbol \varepsilon_2 + \cdots + c_{nn} \boldsymbol \varepsilon_n \end{cases}$$

令

$$\mathbf C = \begin{pmatrix} c_{11} & c_{12} & \cdots & c_{1n} \\ c_{21} & c_{22} & \cdots & c_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ c_{n1} & c_{n2} & \cdots & c_{nn} \end{pmatrix}$$

则基变换方程可表示为

$$(\boldsymbol \eta_1, \boldsymbol \eta_2, \dots \boldsymbol \eta_n) = (\boldsymbol \varepsilon_1, \boldsymbol \varepsilon_2, \dots \boldsymbol \varepsilon_n) \mathbf C$$

注意

注意和方程的下标是转置的

称 $\mathbf C=(c_{ij})_{n \times n}$ 为过渡矩阵

坐标变换

若 $\boldsymbol \xi$ 在 $\boldsymbol \eta_1, \boldsymbol \eta_2, \dots \boldsymbol \eta_n$ 下坐标为

$$\mathbf x = (x_1 \quad x_2 \quad \cdots \quad x_n)^T$$

在 $\boldsymbol \varepsilon_1, \boldsymbol \varepsilon_2, \dots \boldsymbol \varepsilon_n$ 下坐标为

$$\mathbf x' = (x_1' \quad x_2' \quad \cdots \quad x_n')^T$$

又

$$(\boldsymbol \eta_1, \boldsymbol \eta_2, \dots \boldsymbol \eta_n) = (\boldsymbol \varepsilon_1, \boldsymbol \varepsilon_2, \dots \boldsymbol \varepsilon_n) \mathbf C$$

则

$$\mathbf x = \mathbf C \mathbf x'$$

$$\mathbf x' = \mathbf C^{-1} \mathbf x$$

性质

若已知 $\boldsymbol \varepsilon_1, \boldsymbol \varepsilon_2, \dots \boldsymbol \varepsilon_n$ 是一组基, 则 $\mathbf C$ 可逆 $\iff$ $\boldsymbol \eta_1, \boldsymbol \eta_2, \dots \boldsymbol \eta_n$ 是一组基

求解过渡矩阵

$(\boldsymbol \alpha_1 \quad \boldsymbol \alpha_2 \quad \cdots \quad \boldsymbol \alpha_n)$ 到 $(\boldsymbol \beta_1 \quad \boldsymbol \beta_2 \quad \cdots \quad \boldsymbol \beta_n)$ 的过渡矩阵为

$$(\boldsymbol \beta_1 \quad \boldsymbol \beta_2 \quad \cdots \quad \boldsymbol \beta_n) = (\boldsymbol \alpha_1 \quad \boldsymbol \alpha_2 \quad \cdots \quad \boldsymbol \alpha_n) \mathbf C$$

$$\mathbf C = (\boldsymbol \alpha_1 \quad \boldsymbol \alpha_2 \quad \cdots \quad \boldsymbol \alpha_n)^{-1} (\boldsymbol \beta_1 \quad \boldsymbol \beta_2 \quad \cdots \quad \boldsymbol \beta_n)$$

坐标变换的不动点

若 $\boldsymbol \rho$ 在两组基下坐标相同

$$\boldsymbol \rho = \mathbf A \mathbf x = \mathbf B \mathbf x$$

$$\implies (\mathbf A - \mathbf B) \mathbf x = 0$$

若 $|\mathbf A - \mathbf B| = 0$, 则有非零解 若 $|\mathbf A - \mathbf B| \neq 0$, 则只有零解, 即只有原点不动