# Thread: what happens if a second order det of hessian is negative?

1. ## what happens if a second order det of hessian is negative?

Hessian matrix
$H=\left(\begin{array}{cc}\alpha(\alpha-1)x^{\alpha-2}y^{\beta}&\alpha\beta x^{\alpha-1}y^{\beta-1}\\\alpha\beta x^{\alpha-1}\beta^{\beta-1}&\beta(\beta-1)x^{\alpha}y^{\beta-2}\end{array}\right)$

if $det(H)<0$

what can we say about it?

2. I guess you are using it to determine the nature of an extrema?
Anyway, matrix is negative definite when you take its 1x1, 2x2, 3x3 ... nxn determinants and the sign goes - + - + - ...
You must take the determinant from the upper left.
If the "second order" (2x2) determinant is negative then the matrix is neither positive or negative definite. You have a saddle point.