In an answer to this Math.SE question, the constraint $A+B+C=1$ is used in the equation for an origin-centered quadric surface, Eq(1), in the development of the least-squares minimization that will result in an ellipsoid.
$$\tag{1} A x^2 + B y^2 + C z^2 + D xy + E xz + F yz + G = 0 $$
The purpose of this post is to ask why and how the constraint $A+B+C=1$ is justified for use to constrain the least-squares minimization problem described in the other post? From where does this constraint $A+B+C=1$ originate?
Some thoughts I have that may link the constraint $A+B+C=1$ to an ellipsoid are:
- the equation for an axis-aligned ellipsoid in 'standard form' is $$\tag{2} \frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$$ where $x,y,z$ are the cartesian coordinates of some dataset of points $\{ P_i = (x_i, y_i, z_i) , i = 1, N\}$. In 'quadric form', Eq(2) is $$\tag{3} Ax^2+By^2+Cz^2=1$$ where $A=1/a^2, B=1/b^2, C=1/c^2$. Thus, the values for $a,b,c$ in Eq(2) are equivalent to $a=\sqrt{1/A}; b=\sqrt{1/B}; c=\sqrt{1/C}$.
- The set of points $P_i$ will have components $x_i,y_i,z_i$ that are all real numbers (I believe this is notated as $(x_i, y_i, z_i) \in \Bbb R^3$). The magnitude of the position vector $\vec r$ for these points is $$\tag{4} |\vec r| = \sqrt{x^2+y^2+z^2}$$ and thus always a positive value ($\ge 0$).
- A non-axis-aligned ellipsoid is given by the equation $$\tag{5} \mathbf{r}^T Q \mathbf{r} = 1$$ where
$Q = \begin{bmatrix} A && D / 2 && E / 2 \\ D / 2 && B && F/2 \\ E/2 && F/2 && C \end{bmatrix}$
and
$r = [x, y, z]^T$
For an ellipsoid, the matrix $Q$ is always positive definite. I think positive definite means the diagonal coefficients $A,B,C$ of matrix $Q$ must be $>0$.