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[原创]高等代数问题解答85

1. 高等代数问题解答85

Example 1. 设$W_1,W_2$是$n$维欧氏空间$V$的两个$m$维子空间,证明:存在正交变换$\sigma$使得$\sigma(W_1)=W_2.$

\textbf{分析\quad}首先,应该想到下面的结论:

结论1:设$\alpha_1,\alpha_2,\cdots,\alpha_n$是线性空间$V$的基,$\sigma$是$V$的线性变换,则
$$\textrm{Im}\sigma=L(\sigma(\alpha_1),\sigma(\alpha_2),\cdots,\sigma(\alpha_n)).$$

结论2: 设$\sigma$是线性空间$V$的线性变换,$\alpha_1,\alpha_2,\cdots,\alpha_s\in V,$则
$$\sigma(L(\alpha_1,\alpha_2,\cdots,\alpha_s))=L(\sigma(\alpha_1),\sigma(\alpha_2),\cdots,\sigma(\alpha_s)).$$

这样,如果设
$$\alpha_1,\alpha_2,\cdots,\alpha_m$$

$$\beta_1,\beta_2,\cdots,\beta_m$$
分别为$W_1$与$W_2$的标准正交基,将其分别扩充为$V$的标准正交基
$$\alpha_1,\alpha_2,\cdots,\alpha_m,\alpha_{m+1},\cdots,\alpha_n$$

$$\beta_1,\beta_2,\cdots,\beta_m,\beta_{m+1},\cdots,\beta_n.$$
令线性变换$\sigma$满足
$$\sigma(\alpha_i)=\beta_i,i=1,2,\cdots,m,m+1,\cdots,n.$$
下面验证$\sigma$是否满足要求.

$\forall\alpha,\beta\in V,$设
$$\alpha=(\alpha_1,\alpha_2,\cdots,\alpha_n)\bm{X},\beta=(\alpha_1,\alpha_2,\cdots,\alpha_n)\bm{Y},$$

$$\sigma(\alpha)=(\beta_1,\beta_2,\cdots,\beta_n)\bm{X},\sigma(\beta)=(\beta_1,\beta_2,\cdots,\beta_n)\bm{Y},$$
于是
$$(\sigma(\alpha),\sigma(\beta))=\bm{X}^\mathrm{T}\bm{Y}=(\alpha,\beta),$$
即$\sigma$是正交变换.而
$$\sigma(W_1)=\sigma(L(\alpha_1,\alpha_2,\cdots,\alpha_m))=L(\sigma(\alpha_1),\sigma(\alpha_2),\cdots,\sigma(\alpha_m))=L(\beta_1,\beta_2,\cdots,\beta_m)=W_2.$$
所以结论成立.

\textbf{证\quad}(1)若$m=0\mbox{或}n,$取$\sigma=I$(恒等变换)即可.

(2)若$0<m<n,$设
$$\alpha_1,\alpha_2,\cdots,\alpha_m$$

$$\beta_1,\beta_2,\cdots,\beta_m$$
分别为$W_1$与$W_2$的标准正交基,将其分别扩充为$V$的标准正交基
$$\alpha_1,\alpha_2,\cdots,\alpha_m,\alpha_{m+1},\cdots,\alpha_n$$

$$\beta_1,\beta_2,\cdots,\beta_m,\beta_{m+1},\cdots,\beta_n.$$
令线性变换$\sigma$满足
$$\sigma(\alpha_i)=\beta_i,i=1,2,\cdots,m,m+1,\cdots,n.$$
下面验证$\sigma$是否满足要求.

$\forall\alpha,\beta\in V,$设
$$\alpha=(\alpha_1,\alpha_2,\cdots,\alpha_n)\bm{X},\beta=(\alpha_1,\alpha_2,\cdots,\alpha_n)\bm{Y},$$

$$\sigma(\alpha)=(\beta_1,\beta_2,\cdots,\beta_n)\bm{X},\sigma(\beta)=(\beta_1,\beta_2,\cdots,\beta_n)\bm{Y},$$
于是
$$(\sigma(\alpha),\sigma(\beta))=\bm{X}^\mathrm{T}\bm{Y}=(\alpha,\beta),$$
即$\sigma$是正交变换.而
$$\sigma(W_1)=\sigma(L(\alpha_1,\alpha_2,\cdots,\alpha_m))=L(\sigma(\alpha_1),\sigma(\alpha_2),\cdots,\sigma(\alpha_m))=L(\beta_1,\beta_2,\cdots,\beta_m)=W_2.$$
所以结论成立.

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