题目

5. 已知alpha_{1),alpha_(2)}是R^2的一组基，求从基alpha_(1)+2alpha_(2),3alpha_(1)+5alpha_(2)到基-alpha_(1)+alpha_(2),alpha_(2)的过渡矩阵。

5. 已知$\{\alpha_{1},\alpha_{2}\}$是$R^{2}$的一组基，求从基$\alpha_{1}+2\alpha_{2},3\alpha_{1}+5\alpha_{2}$到基$-\alpha_{1}+\alpha_{2},\alpha_{2}$的过渡矩阵。

题目解答

答案

为了找到从基$\alpha_1 + 2\alpha_2, 3\alpha_1 + 5\alpha_2$到基$-\alpha_1 + \alpha_2, \alpha_2$的过渡矩阵，我们需要将新基中的每个向量表示为旧基中向量的线性组合。设旧基为$\mathcal{B} = \{\alpha_1 + 2\alpha_2, 3\alpha_1 + 5\alpha_2\}$，新基为$\mathcal{C} = \{-\alpha_1 + \alpha_2, \alpha_2\}$。我们需要找到系数$a_{ij}$使得： \[ -\alpha_1 + \alpha_2 = a_{11}(\alpha_1 + 2\alpha_2) + a_{12}(3\alpha_1 + 5\alpha_2) \] \[ \alpha_2 = a_{21}(\alpha_1 + 2\alpha_2) + a_{22}(3\alpha_1 + 5\alpha_2) \] 首先，让我们表示$-\alpha_1 + \alpha_2$： \[ -\alpha_1 + \alpha_2 = a_{11}(\alpha_1 + 2\alpha_2) + a_{12}(3\alpha_1 + 5\alpha_2) = (a_{11} + 3a_{12})\alpha_1 + (2a_{11} + 5a_{12})\alpha_2 \] 通过比较$\alpha_1$和$\alpha_2$的系数，我们得到方程组： \[ a_{11} + 3a_{12} = -1 \] \[ 2a_{11} + 5a_{12} = 1 \] 我们可以通过消元法解这个方程组。将第一个方程乘以2： \[ 2a_{11} + 6a_{12} = -2 \] 从第二个方程中减去这个： \[ (2a_{11} + 5a_{12}) - (2a_{11} + 6a_{12}) = 1 - (-2) \] \[ -a_{12} = 3 \] \[ a_{12} = -3 \] 将$a_{12} = -3$代回第一个方程： \[ a_{11} + 3(-3) = -1 \] \[ a_{11} - 9 = -1 \] \[ a_{11} = 8 \] 因此，我们有$a_{11} = 8$和$a_{12} = -3$。现在，让我们表示$\alpha_2$： \[ \alpha_2 = a_{21}(\alpha_1 + 2\alpha_2) + a_{22}(3\alpha_1 + 5\alpha_2) = (a_{21} + 3a_{22})\alpha_1 + (2a_{21} + 5a_{22})\alpha_2 \] 通过比较$\alpha_1$和$\alpha_2$的系数，我们得到方程组： \[ a_{21} + 3a_{22} = 0 \] \[ 2a_{21} + 5a_{22} = 1 \] 我们可以通过消元法解这个方程组。将第一个方程乘以2： \[ 2a_{21} + 6a_{22} = 0 \] 从第二个方程中减去这个： \[ (2a_{21} + 5a_{22}) - (2a_{21} + 6a_{22}) = 1 - 0 \] \[ -a_{22} = 1 \] \[ a_{22} = -1 \] 将$a_{22} = -1$代回第一个方程： \[ a_{21} + 3(-1) = 0 \] \[ a_{21} - 3 = 0 \] \[ a_{21} = 3 \] 因此，我们有$a_{21} = 3$和$a_{22} = -1$。从基$\alpha_1 + 2\alpha_2, 3\alpha_1 + 5\alpha_2$到基$-\alpha_1 + \alpha_2, \alpha_2$的过渡矩阵是： \[ \begin{pmatrix} 8 & 3 \\ -3 & -1 \end{pmatrix} \] 因此，答案是： \[ \boxed{\begin{pmatrix} 8 & 3 \\ -3 & -1 \end{pmatrix}} \]

解析

本题考查线性空间中基变换的过渡矩阵的求解。解题思路是先明确过渡矩阵的定义，即从旧基到新基的过渡矩阵的列向量是新基向量在旧基下的坐标。然后分别将新基中的向量用旧基向量线性表示，通过比较系数得到线性组合的系数，这些系数构成过渡矩阵的列向量。

设旧基$\mathcal{B} = \{\beta_1,\beta_2\}$，其中$\beta_1=\alpha_1 + 2\alpha_2$，$\beta_2=3\alpha_1 + 5\alpha_2$；新基$\mathcal{C} = \{\gamma_1,\gamma_2\}$，其中$\gamma_1=-\alpha_1 + \alpha_2$，$\gamma_2=\alpha_2$。过渡矩阵$P$满足$(\gamma_1,\gamma_2)=(\beta_1,\beta_2)P$，设$P=\begin{pmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{pmatrix}$，则有：
$\gamma_1 = a_{11}\beta_1 + a_{12}\beta_2$，$\gamma_2 = a_{21}\beta_1 + a_{22}\beta_2$。

求$\gamma_1=-\alpha_1 + \alpha_2$在旧基下的坐标：
将$\beta_1=\alpha_1 + 2\alpha_2$，$\beta_2=3\alpha_1 + 5\alpha_2$代入$\gamma_1 = a_{11}\beta_1 + a_{12}\beta_2$可得：
$-\alpha_1 + \alpha_2 = a_{11}(\alpha_1 + 2\alpha_2) + a_{12}(3\alpha_1 + 5\alpha_2)=(a_{11} + 3a_{12})\alpha_1 + (2a_{11} + 5a_{12})\alpha_2$
比较$\alpha_1$和$\alpha_2$的系数，得到方程组$\begin{cases}a_{11} + 3a_{12} = -1\\2a_{11} + 5a_{12} = 1\end{cases}$
- 由第一个方程$a_{11} + 3a_{12} = -1$两边同时乘以$2$得$2a_{11} + 6a_{12} = -2$。
- 用$2a_{11} + 5a_{12} = 1$减去$2a_{11} + 6a_{12} = -2$可得：
  $(2a_{11} + 5a_{12})-(2a_{11} + 6a_{12})=1-(-2)$
  $-a_{12}=3$，解得$a_{12} = -3$。
- 将$a_{12} = -3$代入$a_{11} + 3a_{12} = -1$得：
  $a_{11}+3\times(-3)=-1$
  $a_{11}-9=-1$，解得$a_{11} = 8$。
求$\gamma_2=\alpha_2$在旧基下的坐标：
将$\beta_1=\alpha_1 + 2\alpha_2$，$\beta_2=3\alpha_1 + 5\alpha_2$代入$\gamma_2 = a_{21}\beta_1 + a_{22}\beta_2$可得：
$\alpha_2 = a_{21}(\alpha_1 + 2\alpha_2) + a_{22}(3\alpha_1 + 5\alpha_2)=(a_{21} + 3a_{22})\alpha_1 + (2a_{21} + 5a_{22})\alpha_2$
比较$\alpha_1$和$\alpha_2$的系数，得到方程组$\begin{cases}a_{21} + 3a_{22} = 0\\2a_{21} + 5a_{22} = 1\end{cases}$
- 由第一个方程$a_{21} + 3a_{22} = 0$两边同时乘以$2$得$2a_{21} + 6a_{22} = 0$。
- 用$2a_{21} + 5a_{22} = 1$减去$2a_{21} + 6a_{22} = 0$可得：
  $(2a_{21} + 5a_{22})-(2a_{21} + 6a_{22})=1 - 0$
  $-a_{22}=1$，解得$a_{22} = -1$。
- 将$a_{22} = -1$代入$a_{21} + 3a_{22} = 0$得：
  $a_{21}+3\times(-1)=0$
  $a_{21}-3=0$，解得$a_{21} = 3$。

所以过渡矩阵$P=\begin{pmatrix}8&3\\-3&-1\end{pmatrix}$。