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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

$$x^TA x=1$$

with $xin mathbbR^3$ 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 $3times3$ 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!

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 $Apreceq 0$).

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 $x^T cdot kIcdot x=1$ 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 $$x^TT^TATx=1iff x^Tcdot kIcdot x=1iff T^TAT=kI$$since $T$ is invertible (which arises as a necessary condition for change of coordinates) we can write$$A=k(T^T)^-1T^-1=k(Tcdot T^T)^-1$$The above equality means that $A$ must be symmetric and positive definite. Such a $T$ always exists by Cholesky Decomposition.

Start with the equation of the unit sphere $mathbf x^TImathbf x=1$. You have the (I hope invertible) point transformation $mathbf x'=Tmathbf x$. What do you get if you substitute $T^-1mathbf x'$ 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 $sqrtlambda_i$. In general, a quadric with matrix $A$ transforms as $T^-TAT^-1$ under the invertible point transformation $mathbf x'=Tmathbf x$ (verify this!), so in this case we would have $T^-TAT^-1=I$.