Exercise 6.4.2

If $F(X)=-\log <!--\; FUNCTION\; APPLICATION\; -->(\mathit{ad}-\mathit{bc})$, we have $\nabla <!--\; FUNCTION\; APPLICATION\; -->F=-\frac{1}{\mathit{det}(X)}\nabla <!--\; FUNCTION\; APPLICATION\; -->\mathit{det}(X)=-X^{-T}=\frac{1}{\mathit{ad}-\mathit{bc}}\left[\begin{array}{cc}\hfill -d\hfill & \hfill b\hfill \\ \hfill c\hfill & \hfill -a\hfill \end{array}\right]$, If we have $b=c$, then $\frac{\mathit{\partial F}}{\mathit{\partial a}}=-\frac{d}{\mathit{ad}-b^2},\frac{\mathit{\partial F}}{\mathit{\partial b}}=\frac{2b}{\mathit{ad}-b^2},\frac{\mathit{\partial F}}{\mathit{\partial d}}=-\frac{a}{\mathit{ad}-b^2}$

Take second derivatives, we have

So we have the 2nd derivative matrix ${\nabla <!--\; FUNCTION\; APPLICATION\; -->}^{2}F=\frac{1}{{(\mathit{ad}-b^2)}^2}\left[\begin{array}{ccc}\hfill d^2\hfill & \hfill -2\mathit{bd}\hfill & \hfill b^2\hfill \\ \hfill -2\mathit{bd}\hfill & \hfill 2\mathit{ad}+2b^2\hfill & \hfill -2\mathit{ab}\hfill \\ \hfill b^2\hfill & \hfill -2\mathit{ab}\hfill & \hfill a^2\hfill \end{array}\right]$

Since we assume the matrix $X$ is positive definite, so $\mathit{ad}-b^2>0$, $a>0$, so $d>0$ as well. It’s easy to check the leading determinants of the 2nd derivative matrix, we have $\mathit{det}(D_1)=d^2>0$, ${(\mathit{ad}-b^2)}^2\mathit{det}(D_2)=\frac{2d^2}{\mathit{ad}-b^2}>0$ and

So $\mathit{det}(D_2)=2(\mathit{ad}-b^2)>0$. So the second derivative matrix if positive definite as well.