An ellipsoid is the hypersurface of the ellipse in $\mathbb{R}^3$ which belongs to the family of quadrics. It is defined by a quadratic form equating to 1 in its canonical (unrotated) form.
$\vec{x}^T D \vec{x} = 1$
where $D$ is a diagonal, positive definite matrix. In non-vector form it is represented by,
$\frac{(x-x_0)^2}{a^2} + \frac{(y-y_0)^2}{b^2} + \frac{(z-z_0)^2}{c^2} = 1$
where $x$, $y$, $z$ are the Cartesian coordinates of a point on its surface, $x_0$, $y_0$ and $z_0$ are the coordinates of its centroid and $a$, $b$ and $c$ are its semi-axis lengths.
The Hessian matrix is the matrix which holds all possible second order derivative information for a multi-variable scalar function $f$.
\begin{bmatrix} \frac{\partial^2 f}{\partial x^2} & \frac{\partial^2 f}{\partial x \partial y} & \frac{\partial^2 f}{\partial x \partial z} \\ \frac{\partial^2 f}{\partial y \partial x} & \frac{\partial^2 f}{\partial y^2} & \frac{\partial^2 f}{\partial y \partial z} \\ \frac{\partial^2 f}{\partial z \partial x} & \frac{\partial^2 f}{\partial z \partial y} & \frac{\partial^2 f}{\partial z^2} \end{bmatrix}
Here's my question: I expect the Hessian to vary along the surface of the ellipsoid but when you actually calculate the Hessian matrix of a canonical ellipsoid, it is invariate i.e doesn't depend on the coordinate variables. Why?
Thanks in advance, MotiveHunter