# Hessian Matrix and critical point

Consider $f(x,y,z)\in C^2$. Suppose that $(0,0,0)$ is a critical point of $f$ and the Hessian Matrix of $f$ in $(0,0,0)$ is given by $\left(\begin{array}{ccc} 1 & 0 & \pi\\ 0 & \omega & 43\\ \pi & 43 & -13 \end{array}\right)$, where $\omega \in \mathbb{R}$.

What kind of critical point we have? Maximum, minimum or both?

• look at the eigenvalues of the matrix, specifically are they positive, negative, a mixture of both? – Thoth Jun 6 '15 at 4:46

Fortunately, there is a quick way to decide: we use Sylvester's Criterion, since the matrix is symmetric. It will be positive definite if and only if the principal minors are all positive.

One minor is $1 > 0$, ok. The second one is $\omega$. The last one is $-13\omega - 43^2-\pi^2\omega$.

If $\omega > 0$, the third minor is negative. So we can't have a local minimum. If $\omega < 0$, then the first and the second minors will have opposite signs, so we can't have a local minimum either. If $\omega = 0$, the first and third minors will have opposite signs, problem again.

The matrix will be negative definite if its opposite is positive definite, so you repeat that analysis.

So we have neither maximum nor minimum.

• If we have opposite signs, clearly the principal minors aren't all positive, so the matrix isn't positive definite. But, why it implies that we can't have a local maximum? – Let DC Jun 6 '15 at 14:33
• Oops, I meant minimum. The point is a local maximum if the Hessian is negative definite and is a local minimum is the Hessian is positive definite. – Ivo Terek Jun 6 '15 at 15:24
• You are Brazilian, right? Awesome, me too. rsrsrs I'm confused, because the Theorem ensures that the Hessian is positive definite then the critical point is a local minimum. Thus, if the critical point isn't a local minimum, then the Hessian isn't a positive definite. In this case, we have that Hessian isn'n positive definite, but this does not imply, by Theorem, that point isn't a local minimum. Right? – Let DC Jun 6 '15 at 16:31
• I can take a look, which page is it? – Ivo Terek Jun 6 '15 at 16:32
• Right, but look at the following corollary (corollary 6, page 73). That's what I used. – Ivo Terek Jun 6 '15 at 16:51