例 9.129

依赖于

• 例 9.126

被以下题目直接调用

• 例 9.130

例 9.129

设 $A_1$ 为 $n$ 阶正定实对称矩阵，$A_2,\cdots ,A_m$ 是 $n$ 阶实反对称矩阵，且对任意的 $2\le i<j\le m$，$A_i{A}_1^{-1}{A}_j$ 都是对称矩阵。求证：存在可逆矩阵 $C$，使得

$$\begin{array}{rl}{C}^{\prime }{A}_{1}C& =I_n,\\ C^{\prime }{A}_{i}C& =\mathrm{diag}\left\{\left(\begin{array}{cc}0& b_{i1}\\ -b_{i1}& 0\end{array}\right),\cdots ,\left(\begin{array}{cc}0& b_{ir}\\ -b_{ir}& 0\end{array}\right),0,\cdots ,0\right\},\\ & 2\le i\le m.\end{array}$$

解答

证明 由 $A_1$ 正定可知 $A_1^{-1/2}{A}_1{A}_1^{-1/2}=I_n$，由 $A_i{A}_1^{-1}{A}_j$ 对称可知 $A_i{A}_1^{-1}{A}_j=A_j{A}_1^{-1}{A}_i$，从而

$$(A_1^{-1/2}{A}_i{A}_1^{-1/2})(A_1^{-1/2}{A}_j{A}_1^{-1/2})=(A_1^{-1/2}{A}_j{A}_1^{-1/2})(A_1^{-1/2}{A}_i{A}_1^{-1/2}),$$

即实反对称矩阵 $A_1^{-1/2}{A}_i{A}_1^{-1/2}(2\le i\le m)$ 两两乘法可交换。 由例 9.126 可知，存在正交矩阵 $P$，使得

$$\begin{array}{rl}{P}^{\prime }{A}_1^{-1/2}{A}_i{A}_1^{-1/2}P& =\mathrm{diag}\left\{\left(\begin{array}{cc}0& b_{i1}\\ -b_{i1}& 0\end{array}\right),\cdots ,\left(\begin{array}{cc}0& b_{ir}\\ -b_{ir}& 0\end{array}\right),0,\cdots ,0\right\},\\ & 2\le i\le m.\end{array}$$

此时 $P^{\prime }{A}_1^{-1/2}{A}_1{A}_1^{-1/2}P=I_n$，故只要令 $C=A_1^{-1/2}P$ 即得结论。$\square$