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.
