Exercise 6.8.20

Observe that $D$ is the determinant of the Hessian matrix $A$. Here we denote ${\lambda }_1,{\lambda }_2$ to be the two eigenvalues of $A$, which exist since $A$ is real symmetric.

1.

If $D>0$, we know that ${\lambda }_1$ and ${\lambda }_2$ could not be zero. Since $\frac{{\partial }^{2}f(p)}{\partial t_1^2}>0$, we have $\frac{{\partial }^{2}f(p)}{\partial t_2^2}>0$ otherwise we’ll have $D\le 0$. Hence the trace of $A$ is positive. Thus we have

$${\lambda }_1+{\lambda }_2=-\mathrm{tr}<!--\; FUNCTION\; APPLICATION\; -->(A)<0$$

and

$${\lambda }_1{\lambda }_2=D>0.$$

This means that both of them are negative. Hence $p$ is a local minimum by the Second Derivative Test.

2.

If $D<0$, we know that ${\lambda }_1$ and ${\lambda }_2$ could not be zero. Since $\frac{{\partial }^{2}f(p)}{\partial t_1^2}<0$, we have $\frac{{\partial }^{2}f(p)}{\partial t_2^2}<0$ otherwise we’ll have $D\ge 0$. Hence the trace of $A$ is negative. Thus we have

$${\lambda }_1+{\lambda }_2=-\mathrm{tr}<!--\; FUNCTION\; APPLICATION\; -->(A)>0$$

and

$${\lambda }_1{\lambda }_2=D<0.$$

This means that both of them are positive. Hence $p$ is a local maximum by the Second Derivative Test.

3.

If $D<0$, we know that ${\lambda }_1$ and ${\lambda }_2$ could not be zero. Also, we have

$${\lambda }_1{\lambda }_2=D<0.$$

This means that they cannot be both positive or both negative. Again, by the Second Derivative Test, it’s a saddle point.

4.

If $D=0$, then one of ${\lambda }_1$ and ${\lambda }_2$ should be zero. Apply the Second Derivative Test.