# When is the determinant of a Hessian matrix positive?

Let $f:\mathbb{R}^n\to \mathbb{R}^n$ be a $C^2$-function and let $H=\left(\frac{\partial^2f}{\partial x_i \partial x_j}\right)_{1\le i,j\le n}$ be its Hessian matrix. Suppose I know that $\det H(x_1,\ldots,x_n)\ge 0$ for all $x=(x_1,\ldots,x_n)$.

Does this have any geometric meaning for $f$?

e.g., when $n=1$, this means that $f$ is convex. This no longer holds for $n\ge 2$, but when $f$ is convex the Hessian determinant is certainly positive, so perhaps one could wonder if a weaker property holds.

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The title and the last paragraph have "positive", the first paragraph has $\ge0$. –  joriki Sep 7 '11 at 9:37
@trony: when judging the convexity of a function, we need to check if $H$ is positive definite, but not if $\det H$ is positive. –  Shiyu Sep 7 '11 at 11:58
@Shiyu: Please read my post again. I am not asking whether H is positive definite, I even said this is wrong, I wondered whether it had some other implications to the geometry of $f$. –  Bonanza Sep 7 '11 at 13:58
@trony: I've read your post again, but I still didn't see much meaning from$\det H$. There are many situations in which $\det H>0$. I don't think $\det H>0$ can imply anything. In stead, checking the eigenvalues of $H$ might be meaningful. –  Shiyu Sep 7 '11 at 14:56

The determinant is the product of the eigenvalues. In two dimensions, the product being positive means that the two eigenvalues have the same sign, so $f$ is either concave or convex, and if the first derivative vanishes the point is an extremum. In higher dimensions, there's no such conclusion, since you could have any number of both positive and negative eigenvalues and have the determinant come out positive or negative.