Exercise 3.2.10

We decompose $H$ into a format of $\mathit{LD}{L}^T$, that is $H=\left[\begin{array}{cc}\hfill S\hfill & \hfill C\hfill \\ \hfill C^T\hfill & \hfill 0\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill C^T{S}^{-1}\hfill & \hfill 1\hfill \end{array}\right]\left[\begin{array}{cc}\hfill S\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill -C^T{S}^{-1}C\hfill \end{array}\right]\left[\begin{array}{cc}\hfill 1\hfill & \hfill S^{-1}C\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$ So the matrix $D=\left[\begin{array}{cc}\hfill S\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill -C^T{S}^{-1}C\hfill \end{array}\right]$, whichhas $n$ positive pivots and $n$ negative pivots coming from $-C^T{S}^{-1}C$. According to prlbem 9, we see that the eigenvalues of $H$ match the signs of pivots in $H$, i.e. $D$.