Exercise 6.2.4

• (1)

  $\mathit{det}(\mathit{LD}{L}^T)=\mathit{det}(L)\mathit{det}(D)\mathit{det}(L^T)=\mathit{det}(L)\mathit{det}(D)\mathit{det}(L)=\mathit{det}{(L)}^2\mathit{det}(D)$

$\mathit{det}(\mathit{Q\Lambda }{Q}^T)=\mathit{det}(Q)\mathit{det}(\Lambda )\mathit{det}(Q^T)=\mathit{det}(Q)\mathit{det}(\Lambda )\mathit{det}(Q)=\mathit{det}{(Q)}^2\mathit{det}(\Lambda )$

So we have $\mathit{det}{(L)}^2\mathit{det}(D)=\mathit{det}{(Q)}^2\mathit{det}(\Lambda )$, which means $\mathit{det}(D)$ and $\mathit{det}(\Lambda )$ have the same sign. If the determinant is negative then $S$ has 1 positive eigenvalue in $\Lambda$ and 1 positive pivot in $D$.

• (1)

  If the determinant is positive then $S$ could be positive definite or negative definite. Let $S=\left[\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill b\hfill & \hfill c\hfill \end{array}\right]$. If both eigenvalues are positive, their product equals to the determinant of $S$, i.e. $\mathit{det}(S)=\mathit{ac}-b^2>0$, which also indicate that $\mathit{ac}>b^2>0$. Also their sum equals to the trace of $S$, i.e. $\mathit{trace}(S)=a+c>0$. We conclude that $a>0$ and $c>0$.

If we work out the $S=\mathit{LD}{L}^T$ decomposition, we see that $S=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill \frac{b}{a}\hfill & \hfill 1\hfill \end{array}\right]\left[\begin{array}{cc}\hfill a\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill c-\frac{b^2}{a}\hfill \end{array}\right]\left[\begin{array}{cc}\hfill 1\hfill & \hfill \frac{b}{a}\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$

From the conclusion above, we see the pivots $a$ and $c-\frac{b^2}{a}$ are both larger than 0.