From a previous course in linear algebra, I was taught that a function is positive definite if it satisfies $$\left< \vec x \mid v(\vec x) \mid \vec x\right> > 0 $$ In simpler notation, $$\vec x^Tv(\vec x)\vec x> 0$$
But today in lecture the prof defined that a function is positive definite if $$v(\vec x) > 0$$ except at equilibrium points $\bar x$ where $$v(\bar x) = 0$$
This violates what I knew about positive definiteness in two folds, first we are forgetting about the quadratic form, second we allow this function to vanish at some equilibrium points.
Can someone check whether this new function still indeed satisfies the definition of positive definiteness?