Can we interpret the determinant of a $2×2$ Hessian matrix in terms of the area of a parallelogram?

Question

As we known, the determinant of a $2×2$ matrix is the area of a parallelogram? In this case, can we interpret the determinant of a $2×2$ Hessian matrix in terms of the area of a parallelogram? If so, what is that parallelogram? $$\begin{vmatrix} \frac{\partial^2f}{\partial x^2}&\frac{\partial^2f}{\partial x\partial y}\\ \frac{\partial^2f}{\partial x\partial y}&\frac{\partial^2f}{\partial y^2} \end{vmatrix}$$

Answer 1

I'm just going for intuition here. Also, I'm going to use $f_x$ for $\dfrac{\partial f}{\partial x}$, etc.

A tangent plane to the graph of $f(x,y)$ at a point $(a,b)$ passes through $(a,b,f(a,b))$ and has normal vector $\left\langle f_x(a,b),f_y(a,b),-1\right\rangle$. Now, if we move the point a little bit ($\Delta x$) in the $x$ direction to $(a+\Delta x,b)$, then the new height-$1$ downward normal is $\left\langle f_x(a+\Delta x,b),f_y(a+\Delta x,b),-1\right\rangle$. If we subtract the two normals, we get the horizontal vector $\left\langle f_x(a+\Delta x,b)-f_x(a,b),f_y(a+\Delta x,b)-f_y(a,b),0\right\rangle$. But this is very tiny when $\Delta x$ is, so we should standardize it by dividing by $\Delta x$, to get a vector representing how the normal vector changes when you move $(a,b)$ the tiniest bit in the $x$ direction: $$\dfrac{1}{\Delta x}\left\langle f_x(a+\Delta x,b)-f_x(a,b),f_y(a+\Delta x,b)-f_y(a,b),0\right\rangle\text{.}$$

Similarly, if we move the point in the $y$ direction by some $\Delta y$ and take a difference, we get a horizontal vector $\left\langle f_x(a,b+\Delta y)-f_x(a,b),f_y(a,b+\Delta y)-f_y(a,b),0\right\rangle$. But that gets tiny, so let's standardize it to $$\dfrac{1}{\Delta y}\left\langle f_x(a,b+\Delta y)-f_x(a,b),f_y(a,b+\Delta y)-f_y(a,b),0\right\rangle$$

The signed area of the horizontal parallelogram formed by these two standardized differences is the following determinant: $$\begin{vmatrix}\dfrac{f_x(a+\Delta x,b)-f_x(a,b)}{\Delta x}&\dfrac{f_x(a,b+\Delta y)-f_x(a,b)}{\Delta y}\\\dfrac{f_y(a+\Delta x,b)-f_y(a,b)}{\Delta x}&\dfrac{f_y(a,b+\Delta y)-f_y(a,b)}{\Delta y}\end{vmatrix}$$

If the second-order partials exist, then as $\Delta x$ and $\Delta y$ get very small, this is well-approximated by $$\begin{vmatrix}f_{xx}(a,b)&f_{xy}(a,b)\\f_{yx}(a,b)&f_{yy}(a,b)\end{vmatrix}$$

Assuming the second order partials are continuous, Clairaut's Theorem applies, and this equals the Hessian in the question: $$\begin{vmatrix}f_{xx}(a,b)&f_{yx}(a,b)\\f_{yx}(a,b)&f_{yy}(a,b)\end{vmatrix}$$

Obviously we could say "it's the area of the parallelogram formed by vectors of the second-order partials". But by the above argument, it's the limiting area of standardized parallelograms formed by slight changes in the height-$1$ downward normal vectors as the slight changes go to $0$.

Answer 2

If you think of the gradient of $f$ as a mapping unto itself, namely $F = \nabla f \colon \Bbb R^2\to\Bbb R^2$, then the determinant you are considering is the determinant of the derivative of the mapping $F$. As such it tells you the infinitesimal rate at which $F$ distorts area (that's where the parallelogram determined by $\dfrac{\partial F}{\partial x}$ and $\dfrac{\partial F}{\partial y}$ comes in).