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[原创]问题解答51

1. 高等代数问题解答51

Example 1. %
(西南大学2012,浙江大学2003)设$V$是$n$维欧氏空间,$\sigma$是$V$的正交变换.$V_{1}=\{\alpha\in V|\sigma(\alpha)=\alpha\},V_{2}=\{\alpha-\sigma(\alpha)\},$证明:

(1)$V_{1},V_{2}$是$V$的子空间;

(2)$V=V_{1}\oplus V_{2}.$

\textbf{证明:}(1)略.

(2)(法1)只需证明$V_{1}=V_{2}^{\bot}.$

先证明$V_{1}\subseteq V_{2}^{\bot}.$$\forall \alpha\in V_{1},\beta\in V_{2},$则存在$\gamma\in V$使得$\beta=\gamma-\sigma(\gamma),$且$\sigma(\alpha)=\alpha,$注意到$\sigma$是正交变换,则
$$\begin{aligned}%
(\alpha,\beta)&=(\alpha,\gamma-\sigma(\gamma))\\
&=(\alpha,\gamma)-(\alpha,\sigma(\gamma))\\
&=(\alpha,\gamma)-(\sigma(\alpha),\sigma(\gamma))\\
&=(\alpha,\gamma)-(\alpha,\gamma)\\
&=0.
\end{aligned}$$
即$V_{1}\bot V_{2},$从而$V_{1}\subseteq V_{2}^{\bot}.$

再证明$V_{2}^{\bot}\subseteq V_{1}.$$\forall\alpha\in V_{2}^{\bot},$由于$\alpha-\sigma(\alpha)\in V_{2},$故$(\alpha,\alpha-\sigma(\alpha))=0.$于是
$$\begin{aligned}%
(\alpha-\sigma(\alpha),\alpha-\sigma(\alpha))&=(\alpha,\alpha-\sigma(\alpha))-(\sigma(\alpha),\alpha-\sigma(\alpha))\\
&=(\sigma(\alpha),\sigma(\alpha))-(\sigma(\alpha),\alpha)\\
&=(\alpha,\alpha)-(\alpha,\sigma(\alpha))\\
&=(\alpha,\alpha-\sigma(\alpha))\\
&=0.
\end{aligned}$$
从而$\sigma(\alpha)-\alpha=0,$即$\sigma(\alpha)=\alpha,$故$\alpha\in V_{1}.$即证明了$V_{2}^{\bot}\subseteq V_{1}.$

(法2)$\forall\alpha\in V_{1}\cap V_{2},$则$\sigma(\alpha)=\alpha,$且存在$\beta\in V$使得$\alpha=\beta-\sigma(\beta),$于是由$\sigma$是正交变换有
$$(\alpha,\alpha)=(\alpha,\beta-\sigma(\beta))=(\alpha,\beta)-(\alpha,\sigma(\beta))=(\alpha,\beta)-(\sigma(\alpha),\sigma(\beta))=0,$$
故$\alpha=0$,于是$V_{1}\cap V_{2}=\{0\}.$


$$V_{1}=ker(I-\sigma),V_{2}=Im(I-\sigma),$$
于是
$$dim(V_{1}+V_{2})=dimV_{1}+dimV_{2}-dim(V_{1}\cap V_{2})=dimV=n,$$
而$V_{1}+V_{2}$显然是$V$的子空间,故$V=V_{1}\oplus V_{2}.$

(法3)先证明$V_{1}\subseteq V_{2}^{\bot}.$$\forall \alpha\in V_{1},\beta\in V_{2},$则存在$\gamma\in V$使得$\beta=\gamma-\sigma(\gamma),$且$\sigma(\alpha)=\alpha,$注意到$\sigma$是正交变换,则
$$\begin{aligned}%
(\alpha,\beta)&=(\alpha,\gamma-\sigma(\gamma))\\
&=(\alpha,\gamma)-(\alpha,\sigma(\gamma))\\
&=(\alpha,\gamma)-(\sigma(\alpha),\sigma(\gamma))\\
&=(\alpha,\gamma)-(\alpha,\gamma)\\
&=0.
\end{aligned}$$
即$V_{1}\bot V_{2},$从而$V_{1}\subseteq V_{2}^{\bot}.$


$$dimV_{1}=dimker(I-\sigma)=n-dimIm(I-\sigma)=dimV_{2}^{\bot}.$$
从而$V_{1}=V_{2}^{\bot}.$

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