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[原创]高等代数问题解答63

1. 高等代数问题解答63

Example 1. %
(大连海事大学2010,重庆大学2007,湖南大学2010)设$\alpha_{1},\alpha_{2},\cdots,\alpha_{n}$与$\beta_{1},\beta_{2},\cdots,\beta_{n}$是欧氏空间$V$的两组标准正交基,若$V$的一个正交变换$\sigma$使得$\sigma(\alpha_{1})=\beta_{1}.$证明:
$$L(\sigma(\alpha_{2}),\cdots,\sigma(\alpha_{n}))=L(\beta_{2},\cdots,\beta_{n}).$$

\textbf{证明:} (法1)首先由于$\sigma$是正交变换,从而$\sigma(\alpha_{1}),\sigma(\alpha_{2}),\cdots,\sigma(\alpha_{n})$也是$V$的标准正交基,这样
$$\begin{aligned}%
V=&L(\sigma(\alpha_{1}),\sigma(\alpha_{2}),\cdots,\sigma(\alpha_{n}))\\
=&L(\sigma(\alpha_{1}))\oplus L(\sigma(\alpha_{2}),\cdots,\sigma(\alpha_{n}))\\
=&L(\beta_{1})\oplus L(\sigma(\alpha_{2}),\cdots,\sigma(\alpha_{n}))\\
\end{aligned}
$$
其中易知$L(\beta_{1})^{\bot}=L(\sigma(\alpha_{2}),\cdots,\sigma(\alpha_{n})),$又
$$
\begin{aligned}
V=&L(\beta_{1},\beta_{2},\cdots,\beta_{n})\\
=&L(\beta_{1})\oplus L(\beta_{2},\cdots,\beta_{n})
\end{aligned}
$$
显然$L(\beta_{1})^{\bot}=L(\beta_{2},\cdots,\beta_{n}),$由非平凡子空间的正交补是唯一的,有
$$L(\sigma(\alpha_{2}),\cdots,\sigma(\alpha_{n}))=L(\beta_{2},\cdots,\beta_{n}).$$

(法2)参见 王品超.高等代数新方法(下册).546-547.

由条件可设
$$\sigma(\alpha_{1},\alpha_{2},\cdots,\alpha_{n})=(\beta_{1},\beta_{2},\cdots,\beta_{n})
\begin{pmatrix}%
1&a_{12}&\cdots&a_{1n}\\
0&a_{22}&\cdots&a_{2n}\\
\vdots&\vdots&\cdots&\vdots\\
0&a_{n2}&\cdots&a_{nn}
\end{pmatrix},$$
由$\sigma$是正交变换,$\alpha_{1},\alpha_{2},\cdots,\alpha_{n}$与$\beta_{1},\beta_{2},\cdots,\beta_{n}$是欧氏空间$V$的两组标准正交基,于是
$$1+a_{12}^{2}+\cdots+a_{1n}^{2}=1,$$

$$a_{12}=\cdots=a_{1n}=0.$$
从而
$$\sigma(\alpha_{1},\alpha_{2},\cdots,\alpha_{n})=(\beta_{1},\beta_{2},\cdots,\beta_{n})
\begin{pmatrix}%
1&0\\
0&A_{1}
\end{pmatrix},$$
其中
$$A_{1}=\begin{pmatrix}%
a_{22}&\cdots&a_{2n}\\
\vdots&\cdots&\vdots\\
a_{n2}&\cdots&a_{nn}
\end{pmatrix}$$
是可逆矩阵,于是
$$\sigma(\alpha_{2},\cdots,\alpha_{n})=(\beta_{2},\cdots,\beta_{n})A_{1}.$$

$$L(\sigma(\alpha_{2}),\cdots,\sigma(\alpha_{n}))=L(\beta_{2},\cdots,\beta_{n}).$$

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