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[原创]高等代数问题解答83

1. 高等代数问题解答83

Example 1. (陕西师范大学,2002)证明:(1)若$\bm{A}$为实反对称矩阵,则$\bm{Q}=(\bm{E}-\bm{A})(\bm{E}+\bm{A})^{-1}$是正交矩阵;

(2)若$\bm{Q}$为正交矩阵且$\bm{E}+\bm{Q}$可逆,则存在实反对称矩阵$\bm{A}$使得
$$\bm{Q}=(\bm{E}-\bm{A})(\bm{E}+\bm{A})^{-1}.$$

\textbf{分析:}(1)只需证明$\bm{Q}^\mathrm{T}\bm{Q}=\bm{E}$即可,计算过程中可知需要考虑$(\bm{E}+\bm{A})$与$(\bm{E}-\bm{A})$的可换性.

(2)首先试试能否找到反对称矩阵$\bm{A},$由
$$\bm{Q}=(\bm{E}-\bm{A})(\bm{E}+\bm{A})^{-1}$$
可得
$$\bm{Q}(\bm{E}+\bm{A})=\bm{E}-\bm{A},$$

$$(\bm{Q}+\bm{E})\bm{A}=\bm{E}-\bm{Q},$$
于是
$$\bm{A}=(\bm{E}+\bm{Q})^{-1}(\bm{E}-\bm{Q}).$$
下面判断$\bm{A}$的反对称性.由于
$$\begin{aligned}
\bm{A}^\mathrm{T}&=(\bm{E}-\bm{Q}^\mathrm{T})[(\bm{E}+\bm{Q})^\mathrm{T}]^{-1}=(\bm{E}-\bm{Q}^\mathrm{T})(\bm{E}+\bm{Q}^\mathrm{T})^{-1}\\
&=(\bm{E}-\bm{Q}^\mathrm{T})(\bm{Q}\bm{Q}^\mathrm{T}+\bm{Q}^\mathrm{T})^{-1}=(\bm{E}-\bm{Q}^\mathrm{T})[(\bm{Q}+\bm{E})\bm{Q}^\mathrm{T})]^{-1}\\
&=(\bm{E}-\bm{Q}^\mathrm{T})\bm{Q}(\bm{Q}+\bm{E})^{-1}=(\bm{Q}-\bm{E})(\bm{Q}+\bm{E})^{-1}.
\end{aligned}$$
$$\begin{aligned}
-\bm{A}=(\bm{E}+\bm{Q})^{-1}(\bm{Q}-\bm{E}),
\end{aligned}$$
所以,只需判断
$$(\bm{Q}-\bm{E})(\bm{Q}+\bm{E})^{-1}=(\bm{E}+\bm{Q})^{-1}(\bm{Q}-\bm{E}),$$
也就是
$$(\bm{Q}+\bm{E})(\bm{Q}-\bm{E})=(\bm{Q}-\bm{E})(\bm{Q}+\bm{E}),$$
这是显然的.

\textbf{证:}(1)由于$(\bm{E}+\bm{A})(\bm{E}-\bm{A})=(\bm{E}-\bm{A})(\bm{E}+\bm{A})=\bm{E}-\bm{A}^{2},$于是
$$\begin{aligned}%
\bm{Q}^\mathrm{T}\bm{Q}=&[(\bm{E}-\bm{A})(\bm{E}+\bm{A})^{-1}]^\mathrm{T}(\bm{E}-\bm{A})(\bm{E}+\bm{A})^{-1}\\
=&[(\bm{E}+\bm{A})^\mathrm{T}]^{-1}(\bm{E}-\bm{A})^\mathrm{T}(\bm{E}-\bm{A})(\bm{E}+\bm{A})^{-1}\\
=&(\bm{E}-\bm{A})^{-1}(\bm{E}+\bm{A})(\bm{E}-\bm{A})(\bm{E}+\bm{A})^{-1}\\
&=(\bm{E}-\bm{A})^{-1}(\bm{E}-\bm{A})(\bm{E}+\bm{A})(\bm{E}+\bm{A})^{-1}\\
=&\bm{E}.
\end{aligned}$$
从而$\bm{Q}$是正交矩阵.

(2)令
$$\bm{A}=(\bm{E}+\bm{Q})^{-1}(\bm{E}-\bm{Q}),$$
注意到
$$(\bm{Q}+\bm{E})(\bm{Q}-\bm{E})=(\bm{Q}-\bm{E})(\bm{Q}+\bm{E}),$$

$$(\bm{Q}-\bm{E})(\bm{Q}+\bm{E})^{-1}=(\bm{E}+\bm{Q})^{-1}(\bm{Q}-\bm{E}),$$
于是
$$\begin{aligned}
\bm{A}^\mathrm{T}&=(\bm{E}-\bm{Q}^\mathrm{T})[(\bm{E}+\bm{Q})^\mathrm{T}]^{-1}=(\bm{E}-\bm{Q}^\mathrm{T})(\bm{E}+\bm{Q}^\mathrm{T})^{-1}\\
&=(\bm{E}-\bm{Q}^\mathrm{T})(\bm{Q}\bm{Q}^\mathrm{T}+\bm{Q}^\mathrm{T})^{-1}=(\bm{E}-\bm{Q}^\mathrm{T})[(\bm{Q}+\bm{E})\bm{Q}^\mathrm{T})]^{-1}\\
&=(\bm{E}-\bm{Q}^\mathrm{T})\bm{Q}(\bm{Q}+\bm{E})^{-1}=(\bm{Q}-\bm{E})(\bm{Q}+\bm{E})^{-1}\\
&=-\bm{A}.
\end{aligned}$$

$$(\bm{E}+\bm{Q})\bm{A}=\bm{E}-\bm{Q},$$

$$\bm{Q}\bm{A}+\bm{Q}=\bm{E}-\bm{A},$$
于是
$$\bm{Q}=(\bm{E}-\bm{A})(\bm{E}+\bm{A})^{-1}.$$
故结论成立.

Example 2. 设$\bm{A}$是一个实反对称矩阵,证明:

(1)$\bm{A}$的非零特征值是纯虚数;

(2)若$\bm{A}$可逆,则$\bm{A}^2+\bm{A}^{-1}$可逆,且$\bm{B}=(\bm{A}^2-\bm{A}^{-1})(\bm{A}^2+\bm{A}^{-1})^{-1}$是正交矩阵.

\textbf{分析\quad}(1)略.

(2)先证明$\bm{A}^2+\bm{A}^{-1}$可逆.结合(1),应该考虑证明$\bm{A}^2+\bm{A}^{-1}$的特征值都不是0,显然考虑反证法即可.若$\bm{\alpha}$是$\bm{A}^2+\bm{A}^{-1}$的属于特征值0的特征向量,则
$$(\bm{A}^2+\bm{A}^{-1})\bm{\alpha}=0,$$
于是等式两边左乘$\bm{A}$可得
$$\bm{A}^3\bm{\alpha}=-\bm{\alpha},$$
此式表明$-1$是$\bm{A}^3$的特征值,注意到$\bm{A}^3$仍然是反对称矩阵,所以矛盾.

再证明$\bm{B}$是正交矩阵.由于
$$\begin{aligned}
\bm{B}^\mathrm{T}=&[(\bm{A}^2+\bm{A}^{-1})^{-1}]^\mathrm{T}(\bm{A}^2-\bm{A}^{-1})^\mathrm{T}\\
=&[(\bm{A}^2+\bm{A}^{-1})^\mathrm{T}]^{-1}(\bm{A}^2-\bm{A}^{-1})^\mathrm{T}\\
=&[(\bm{A^\mathrm{T}})^2+(\bm{A}^\mathrm{T})^{-1}]^{-1}[(\bm{A^\mathrm{T}})^2-(\bm{A}^\mathrm{T})^{-1}]\\
=&(\bm{A}^2-\bm{A}^{-1})^{-1}(\bm{A}^2+\bm{A}^{-1})
\end{aligned}
$$
于是
$$\bm{B}^\mathrm{T}\bm{B}=(\bm{A}^2-\bm{A}^{-1})^{-1}(\bm{A}^2+\bm{A}^{-1})(\bm{A}^2-\bm{A}^{-1})(\bm{A}^2+\bm{A}^{-1})^{-1},$$
计算
$$(\bm{A}^2+\bm{A}^{-1})(\bm{A}^2-\bm{A}^{-1})=(\bm{A}^2-\bm{A}^{-1})(\bm{A}^2+\bm{A}^{-1})$$
是否成立即可.

\textbf{证\quad}(1)略.

(2)先证明$\bm{A}^2+\bm{A}^{-1}$可逆.反证法,若$\bm{A}^2+\bm{A}^{-1}$不可逆,则其必有特征值0,设对应的特征向量为$\bm{\alpha},$则有
$$(\bm{A}^2+\bm{A}^{-1})\bm{\alpha}=0,$$
于是等式两边左乘$\bm{A}$可得
$$\bm{A}^3\bm{\alpha}=-\bm{\alpha},$$
此式表明$-1$是$\bm{A}^3$的特征值,注意到$\bm{A}^3$仍然是反对称矩阵,所以矛盾.同理可知$\bm{A}^2-\bm{A}^{-1}$可逆

再证明$\bm{B}$是正交矩阵.

(法1)由于
$$\begin{aligned}
\bm{B}^\mathrm{T}=&[(\bm{A}^2+\bm{A}^{-1})^{-1}]^\mathrm{T}(\bm{A}^2-\bm{A}^{-1})^\mathrm{T}\\
=&[(\bm{A}^2+\bm{A}^{-1})^\mathrm{T}]^{-1}(\bm{A}^2-\bm{A}^{-1})^\mathrm{T}\\
=&[(\bm{A^\mathrm{T}})^2+(\bm{A}^\mathrm{T})^{-1}]^{-1}[(\bm{A^\mathrm{T}})^2-(\bm{A}^\mathrm{T})^{-1}]\\
=&(\bm{A}^2-\bm{A}^{-1})^{-1}(\bm{A}^2+\bm{A}^{-1})
\end{aligned}
$$
于是注意到
$$(\bm{A}^2+\bm{A}^{-1})(\bm{A}^2-\bm{A}^{-1})=(\bm{A}^2-\bm{A}^{-1})(\bm{A}^2+\bm{A}^{-1}),$$

$$\bm{B}^\mathrm{T}\bm{B}=(\bm{A}^2-\bm{A}^{-1})^{-1}(\bm{A}^2+\bm{A}^{-1})(\bm{A}^2-\bm{A}^{-1})(\bm{A}^2+\bm{A}^{-1})^{-1}=\bm{E},$$
所以结论成立.

(法2)由于
$$\bm{A}^2\pm\bm{A}^{-1}=\bm{A}^{-1}(\bm{A}^3\pm\bm{E})=(\bm{A}^3\pm\bm{E})\bm{A}^{-1},$$
所以
$$\bm{B}=(\bm{A}^3-\bm{E})\bm{A}^{-1}[(\bm{A}^3+\bm{E})\bm{A}^{-1}]^{-1}=(\bm{A}^3-\bm{E})(\bm{A}^3+\bm{E})^{-1},$$
于是
$$\begin{aligned}
\bm{B}^\mathrm{T}&=[(\bm{A}^3+\bm{E})^{-1}]^\mathrm{T}(\bm{A}^3-\bm{E})^\mathrm{T}\\
&=(-\bm{A}^3+\bm{E})^{-1}(-\bm{A}^3-\bm{E})\\
&=[-(\bm{A}^3-\bm{E})]^{-1}[-(\bm{A}^3+\bm{E})]\\
&=(\bm{A}^3-\bm{E})^{-1}(\bm{A}^3+\bm{E}).
\end{aligned}$$
注意到
$$(\bm{A}^3+\bm{E})(\bm{A}^3-\bm{E})=(\bm{A}^3-\bm{E})(\bm{A}^3+\bm{E}),$$
所以$\bm{B}^\mathrm{T}\bm{B}=\bm{E}.$

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