题目：
\[\text { 设 } \boldsymbol{A} \text { 是 } n \text { 阶正定矩阵, } \boldsymbol{B} \text { 是 } n \text { 阶实矩阵. } \\(1)证明 \boldsymbol{E}-\boldsymbol{B}^{\mathrm{T}} \boldsymbol{B} 与 \boldsymbol{E}-\boldsymbol{B} \boldsymbol{B}^{\mathrm{T}} 合同. \\(2) 若 \boldsymbol{E}-\boldsymbol{A} 正定, 证明 \boldsymbol{E}-\boldsymbol{A}^{-1}负定\]解答：
(1)由于 \(\left(\begin{array}{cc}E & 0 \\ B^{\top} & E\end{array}\right)\left(\begin{array}{cc}E & 0 \\ 0 & E-B T B\end{array}\right)\left(\begin{array}{ll}E & B \\ 0 & E\end{array}\right)=\left(\begin{array}{ll}E & B \\ B^{T} & E\end{array}\right) \\\left(\begin{array}{ll}E & B \\ 0 & E\end{array}\right)\left(\begin{array}{cc}E-B B^{T} & 0 \\ 0 & E\end{array}\right)\left(\begin{array}{ll}E & 0 \\ B^{T} & E\end{array}\right)=\left(\begin{array}{ll}E & B \\ B^{T} & E\end{array}\right)\) 故 \(\left(\begin{array}{cc} E & 0 \\ 0 & E-B^{\top} B \end{array}\right) 与 \left(\begin{array}{cc} E-B B^T & 0 \\ 0 & E \end{array}\right)\) 合同
从而可知$ E-B^TB与E-BB^T$有相同的正负惯性指数，故它们合同
(2)
A正定，则存在可逆矩阵P使得$A=P^TP$
由E-A正定（即$E-P^TP$正定）可知$E-PP^T$正定，从而$PP^T-E$负定
又$PA^{-1}P^T=E$,从而有$P(E-A^{-1})P^T=PP^T-E$，所以$E-A^{-1}$负定