Classification of critic points

Question

I'm trying to find and classify the critical points of the following function

$f(x,y,z) = x^2 y -y^2+z^2+yz $

I already found that the only critical point is $(0,0,0)$. The I found the Hessian matrix (matrix of second derivatives) of in this point and then I found the eigenvalues of the matrix which are $0$, $\sqrt{5}$ and $-\sqrt{5}$.

Now my question is how should I classify this critical point: for one side we have one zero eigenvalue (so we should use other technique) and for the other side we have two eigenvalues with opposite signs (so this point is NOT an extreme point of the function). What criteria is valid here?

Answer 1

To classify the critical points, once you got the Hessian, just remember that:

(I will call $\Delta$ the determinant of the Hessian matrix and $a$ the first element of the principal diagonal)

• If $\Delta >0$ and $a>0$ then that point is a local minimum;

• If $\Delta >0$ and $a <0$ then that point is a local maximum;

• If $\Delta <0$ that point is a saddle point;

• If $\Delta = 0$ the test is inconclusive and you need further methods.

Your Case

Can you confirm your Hessian is this ?

$$\left( \begin{array}{ccc} 2 y & 2 x & 0 \\ 2 x & -2 & 1 \\ 0 & 1 & 2 \\ \end{array} \right) $$

Once you got that the critical point is $0, 0, 0)$ you can easily find what the associated Hessian is.

Now, the problem is that the first element $a$ will be $0$, hence the rules I wrote above don't apply.

In this case, you can proceed with the eigenvalues method:

• If all the eigenvalues are positive, then the point is a local minimum;

• If all the eigenvalues are negative, then the point is a local maximum;

• If you have a medley of $+$ and $-$, then the point is a saddle point.

Therefore, $0, 0, 0$ is a saddle point.

To show you it's correct, here is the plot: