Ellipsoid and linear transformation

Question

I get confused in my algebra about this simple problem.

The equation of a 3D centred ellipsoid can be written on a compact form as

$$x^{T}A x=1$$

with $x\in \mathbb{R}^{3}$ et $A$ a matrix that contains the coefficients of the cartesian equation.
(Am I correct so far?)

Now I want to define my ellipsoid in a different way, saying that it is what becomes the unit sphere when I apply a linear transformation to it (of $3\times3$ matrix $T$). By that I mean that if a vector $x$ belongs to the unit sphere, then $xT$ belongs to my ellipsoid (sorry for bad vocabulary).

My question is, what is the explicit relation between $A$ and $T$??

If someone can help me with that, I would really appreciate it!

Answer 1

First of all, note that $A$ must be positive semi-definite (even positive definite is we are looking for a neat ellipsoid!) in $x^TAx=1$ representation, unless it might give you a hyperbola or nothing (say, in case that $A\preceq 0$).

Problem

If points $x$ belong to some sphere, does there exist some $T$ such that $Tx$ belongs to an ellipsoid?

Answer

Indeed there exists!

To see why, let a sphere be described as $x^T \cdot kI\cdot x=1$ where $I$ and $k$ are identity matrix of proper dimension and some arbitrary positive real respectively. Now if we set $y^TAy=1$ as an ellipsoid with $y=Tx$ we must have by substitution $$x^TT^TATx=1\iff x^T\cdot kI\cdot x=1\iff T^TAT=kI$$since $T$ is invertible (which arises as a necessary condition for change of coordinates) we can write$$A=k(T^T)^{-1}T^{-1}=k(T\cdot T^T)^{-1}$$The above equality means that $A$ must be symmetric and positive definite. Such a $T$ always exists by Cholesky Decomposition.

Answer 2

Start with the equation of the unit sphere $\mathbf x^TI\mathbf x=1$. You have the (I hope invertible) point transformation $\mathbf x'=T\mathbf x$. What do you get if you substitute $T^{-1}\mathbf x'$ for $\mathbf x$?

Going in the other direction, the principal axis theorem gives us an orthogonal basis in which the ellipsoid is axis-aligned—the principal axes—and tells us that the associated semi-axis lengths are the reciprocal square roots of the respective eigenvalues $\lambda_i$ (which are all positive for an ellipsoid). The ellipsoid can therefore be transformed into the unit sphere via a transformation $T$ that scales each principal axis by $\sqrt{\lambda_i}$. In general, a quadric with matrix $A$ transforms as $T^{-T}AT^{-1}$ under the invertible point transformation $\mathbf x'=T\mathbf x$ (verify this!), so in this case we would have $T^{-T}AT^{-1}=I$.