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标题: 蓝以中上册线性空间与线性变换 270页习题二27 解答 [打印本页]

作者: castelu 时间: 2016-5-29 22:44
标题: 蓝以中上册线性空间与线性变换 270页习题二27 解答
习题二27：
　　设$M$为线性空间$V$的一个子空间。在$M$内取定一组基$\epsilon_1,\epsilon_2,\cdots,\epsilon_r$，用两种方式扩充为$V$的基
$$\epsilon_1,\epsilon_2,\cdots,\epsilon_r,\epsilon_{r+1},\cdots,\epsilon_n$$
$$\epsilon_1,\epsilon_2,\cdots,\epsilon_r,\eta_{r+1},\cdots,\eta_n$$
　　这两组基之间的过渡矩阵为$T$，即
$$(\epsilon_1,\epsilon_2,\cdots,\epsilon_r,\eta_{r+1},\cdots,\eta_n)=(\epsilon_1,\epsilon_2,\cdots,\epsilon_r,\epsilon_{r+1},\cdots,\epsilon_n)T$$
　　其中
$$T=\left( {\begin{array}{*{20}{c}}
{E_r}&{*}\\
{0}&{T_0}
\end{array}} \right)$$
　　证明：$V/M$内两组基
$$\overline {\epsilon}_{r+1}=\epsilon_{r+1}+M,\overline {\epsilon}_{r+2}=\epsilon_{r+2}+M,\cdots,\overline {\epsilon}_n=\epsilon_n+M$$
$$\overline {\eta}_{r+1}=\eta_{r+1}+M,\overline {\eta}_{r+2}=\eta_{r+2}+M,\cdots,\overline {\eta}_n=\eta_n+M$$
　　之间的过渡矩阵为
$$(\overline {\eta}_{r+1},\cdots,\overline {\eta}_n)=(\overline {\epsilon}_{r+1},\cdots,\overline {\epsilon}_n)T_0$$

解：
　　注意到
$$\epsilon_{r+1},\cdots,\epsilon_n$$
$$\eta_{r+1},\cdots,\eta_n$$
　　这两组基之间的过渡矩阵为$T_0$，即
$$(\eta_{r+1},\cdots,\eta_n)=(\epsilon_{r+1},\cdots,\epsilon_n)T_0$$
　　它们是商空间$V/M$内两组特殊基
　　设
$$T_0=\left(t_{ij}\right)_{[n-(r+1) \times n-(r+1)]}$$
　　成立线性方程组
$$\left\{ \begin{array}{l}
\eta_{r+1}=t_{11}\epsilon_{r+1}+t_{21}\epsilon_{r+2}+\cdots+t_{n1}\epsilon_n\\
\eta_{r+2}=t_{12}\epsilon_{r+1}+t_{22}\epsilon_{r+2}+\cdots+t_{n2}\epsilon_n\\
\cdots\\
\eta_n=t_{1n}\epsilon_{r+1}+t_{2n}\epsilon_{r+2}+\cdots+t_{nn}\epsilon_n
\end{array} \right.$$
　　对$\forall \epsilon \in M$，考虑商空间$V/M$内两组一般基
$$\overline {\epsilon}_{r+1}=\epsilon_{r+1}+\epsilon,\overline {\epsilon}_{r+2}=\epsilon_{r+2}+\epsilon,\cdots,\overline {\epsilon}_n=\epsilon_n+\epsilon$$
$$\overline {\eta}_{r+1}=\eta_{r+1}+\epsilon,\overline {\eta}_{r+2}=\eta_{r+2}+\epsilon,\cdots,\overline {\eta}_n=\eta_n+\epsilon$$
　　这两组基之间的过渡矩阵为
$$S_0=\left(s_{ij}\right)_{[n-(r+1) \times n-(r+1)]}$$
　　成立线性方程组
$$\left\{ \begin{array}{l}
\eta_{r+1}+\epsilon=s_{11}(\epsilon_{r+1}+\epsilon)+s_{21}(\epsilon_{r+2}+\epsilon)+\cdots+s_{n1}(\epsilon_n+\epsilon)\\
\eta_{r+2}+\epsilon=s_{12}(\epsilon_{r+1}+\epsilon)+s_{22}(\epsilon_{r+2}+\epsilon)+\cdots+s_{n2}(\epsilon_n+\epsilon)\\
\cdots\\
\eta_n+\epsilon=s_{1n}(\epsilon_{r+1}+\epsilon)+s_{2n}(\epsilon_{r+2}+\epsilon)+\cdots+s_{nn}(\epsilon_n+\epsilon)
\end{array} \right.$$
　　整理后可得
$$\left\{ \begin{array}{l}
(t_{11}-s_{11})\epsilon_{r+1}+(t_{21}-s_{21})\epsilon_{r+2}+\cdots+(t_{n1}-s_{n1})\epsilon_n=(-1+\sum\limits_{i=1}^ns_{i1})\epsilon\\
(t_{12}-s_{12})\epsilon_{r+1}+(t_{22}-s_{22})\epsilon_{r+2}+\cdots+(t_{n2}-s_{n2})\epsilon_n=(-1+\sum\limits_{i=1}^ns_{i2})\epsilon\\
\cdots\\
(t_{1n}-s_{1n})\epsilon_{r+1}+(t_{2n}-s_{2n})\epsilon_{r+2}+\cdots+(t_{nn}-s_{nn})\epsilon_n=(-1+\sum\limits_{i=1}^ns_{in})\epsilon
\end{array} \right.$$
　　由于$\eta_{r+1},\cdots,\eta_n$与$\epsilon$线性无关，且$\epsilon_{r+1},\cdots,\epsilon_n$与$\epsilon$线性无关
　　所以
$$t_{ij}=s_{ij}$$
　　故$V/M$内两组基之间的过渡矩阵为
$$(\overline {\eta}_{r+1},\cdots,\overline {\eta}_n)=(\overline {\epsilon}_{r+1},\cdots,\overline {\epsilon}_n)T_0$$

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