# How is the Hessian of the following function negative definite?

I'm trying to find if the Hessian matrix of the function:

$$f(x,y) = e^{2x} - 2x + 2y^2 + 3$$

I found the second order partial differentials as:

1. $f_{xx} = 4 e^{2x}$
2. $f_{yy} = 4$
3. $f_{xy}= f_{yx} = 0$

Which gives the Hessian as:

$$\begin{matrix} 4 e^{2x} & 0 \\ 0 & 4 \\ \end{matrix}$$

Isn't this matrix positive definite? My reference book says that this matrix is negative definite. Is that a mistake or am I missing something here?

• Yes you are correct. – gimusi Mar 27 '18 at 20:56

I think it's a mistake of your book and it's positive definite, because the principal minors of the matrix are $4>0$ and $16e^{2x}>0$.