证明引理 2. 1. 

 

证明:

 

(1)  先证明存在正交阵 ${\bf P},{\bf Q}$ 及对角阵 ${\bf D}$ 使得 $$\bex {\bf F}={\bf P}{\bf D}{\bf Q}. \eex$$ 事实上, 由 ${\bf F}$ 可逆知 ${\bf F}^T{\bf F}$ 正定, 而存在正交阵 ${\bf Q}$, 使得 $$\bex {\bf F}^T{\bf F}={\bf Q}^T\diag(\lm_1,\cdots,\lm_n){\bf Q},\quad(\lm_i>0). \eex$$ 取 $$\bex {\bf D}=\diag(\sqrt{\lm_1},\cdots,\sqrt{\lm_n}),\quad {\bf P}={\bf F}{\bf Q}^T{\bf D}^{-1}, \eex$$ 则可直接验证 ${\bf P},{\bf Q},{\bf D}$ 适合要求.

 

(2)  取 $$\bex {\bf R}={\bf P}{\bf Q},\quad {\bf U}={\bf Q}^T{\bf D}{\bf Q},\quad {\bf V}={\bf P}{\bf D}{\bf P}^T \eex$$ 即满足条件.