# Does in invertibility of Hessian matrix $H_{F(X)}$ implies $v^tH_{f(X)}v\neq 0$?

Let $$U$$ be an open convex subset of $$\mathbb{R}^n$$, and $$f: U\rightarrow\mathbb{R}$$ be a function having continuous first and second partial derivatives on $$U$$. Let $$H_{f(X)}$$ denote the Hessian of $$f$$ at $$X\in U$$. If $$\det(H_{f(X)})\neq 0$$ for all $$X\in U$$, can we conclude that $$v^tH_{f(X)}v\neq0$$ for all $$X\in U$$ and $$v\neq 0$$, $$v\in\mathbb{R}^n$$.

I found this in the discussion of absolute maxima or absolute minima of functions on convex open sets (page number 43), in the book $$`$$Multivariate Calculus and Geometry(Third Edition)' by Sean Dineen.

It seems to me the answer is "no".

Let

$$f = x^2 + y^2 - z^2; \tag 1$$

then

$$H_f = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & -2 \end{bmatrix} = \text{diag}(2, 2, -2); \tag 2$$

we have

$$\det H_f = -8 \tag 3$$

everywhere in $$U$$, but with

$$v = \begin{pmatrix} 1 \\ 1 \\ \sqrt 2 \end{pmatrix} \tag 4$$

we find

$$v^T H_f v = \begin{pmatrix} 1 & 1 & \sqrt 2 \end{pmatrix} \begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & -2 \end{bmatrix}\begin{pmatrix} 1 \\ 1 \\ \sqrt 2 \end{pmatrix} = 0, \tag 5$$

as easy computations affirm.