According to Hessian matrix, It describes the local curvature of a function.
AFAIK, for one-variable function $f(x)$, its local curvature is
$$\kappa = \frac{|f''|}{(1 + f'^2)^{3/2}},$$
and its Hessian matrix is
$$\mathcal{Hess}(f) = [f''],$$
right? And here is my problem, I think the local curvature is not just described by its Hessian matrix, because $f'$ also has its role in it, doesn't it?
And furthermore, for 2-variable function $f(x,y)$, its Hessian matrix is
$$\mathcal{Hess}(f) = \left[ \begin{array}{cc}
f_{xx}'' & f_{xy}'' \\
f_{xy}'' & f_{yy}''
\end{array} \right].$$
How does it relate to the local curvature of $f(x,y)$?
Best Answer
The Hessian matrix describes the curvature just when the gradient of f vanishes ("Further informal discussion" section in wikipedia), at this time they share the same form as $f' = 0$
The relation between the Hessian matrix and curvature in 2-variable function shows as
$$K = {{{f_{xx}} \cdot {f_{yy}} - f_{xy}^2} \over {{{(1 + f_x^2 + f_y^2)}^2}}} = {{\det (Hess)} \over {{{(1 + f_x^2 + f_y^2)}^2}}}$$
I encountered the problem yesterday as well and hope it helpful for you.