How can we prove that the Cholesky decomposition of an ellipsoid transforms that ellipsoid onto the unit sphere

Say I have a collection of points $$x_i$$ that define the surface of a fully general ellipsoid in three dimensions, except let's assume that ellipsoid is centered at the origin. I know that the ellipsoid can be represented by

$$v^T A v = 1$$

where $$v=[x\ y\ z]^T$$ and $$A$$ is a matrix formed from the coefficients of the quadric representation of the ellipsoid. The Cholesky decomposition gives a matrix $$L$$ such that

$$A = LL^T$$

I read and believe that I can transform my points onto the unit sphere by

$$y_i = Lx_i$$

I am struggling to understand why this should be the case, and haven't managed to find a good explanation as to why this works.

If an intuitive geometric interpretation exists, that would be particularly valuable. The problem at hand is sensor calibration. I have written a working C++ function to calibrate noisy sensor data, but I made the classic blunder of not properly recording my reference material when I first wrote it, and now I can't find it again. I have found many papers following a very similar approach but everything I've found treats this step as obvious. I'm now trying to document this in a way that can be understood by other software engineers.

My own background is in mathematical physics, so if an intuitive geometric interpretation does not exist, a more formal treatment is fine.

After this answer has been accepted, I found a mistake. The correct formula for $$y_i$$ should be $$y_i:=L^\top x_i$$. Consider the counter-example in 2D space \begin{align*} A&:=\begin{bmatrix}1 & 1 \\ 1 & 2 \end{bmatrix} \\ L&:=\begin{bmatrix}1 & 0 \\ 1 & 1 \end{bmatrix} \\ x&:=\begin{bmatrix}1 \\ 0 \end{bmatrix} \end{align*} Then \begin{align*} LL^\top &= A \\ x^\top A x &= 1 \\ Lx&=\begin{bmatrix}1 \\ 1 \end{bmatrix}\not\in S^1 \\ L^\top x &= \begin{bmatrix}1 \\ 0 \end{bmatrix}\in S^1 \end{align*}
To prove that a vector $$y_i$$ is on the unit sphere, simply prove that $$\lVert y_i \rVert=1$$, or equivalently $$\lVert y_i \rVert^2=1$$.
For all $$x_i$$ on the ellipsoid, we indeed have \begin{align*} 1 &= x_i^\top A x_i \\ &= x_i^\top L L^\top x_i \\ &= (L^\top x_i)^\top(L^\top x_i) \\ &= \lVert L^\top x_i \rVert^2 \\ &= \lVert y_i \rVert^2, \\ \end{align*} so all the corresponding $$y_i$$ are on the unit sphere.
• @ColinBroderick Actually I just found a mistake. The correct definition of $y_i$ should be $L^\top x_i$. I edited my answer. Commented Dec 20, 2022 at 18:13