How is the Hessian of the following function negative definite?

Question

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?

Answer 1

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 $16e^{2x}>0$.