Exercise 3.2.11

Since $S$ is positive definite (symmetric), we have $S=Q\mathrm{\Sigma }{Q}^T$, and $Q^{-1}=Q^T$, all diagonal entries in $\mathrm{\Sigma }$ are positive. So we have $S^{-1}=Q{\mathrm{\Sigma }}^{-1}{Q}^T$, $S^{-1}$ is still a positive definite matrix. So for any $x\ne 0$, we have $-x^{T}A{S}^{-1}{A}^{T}x=-{(A^{T}x)}^T{S}^{-1}(A^{T}x)=-b^T{S}^{-1}b0$, so $-A{S}^{-1}{A}^T$ is negative definite. Note we used the constraint that $\mathit{Ax}=b$

Then the last $m$ pivots of $H$ are negative.