How to formulate the definition of an ellipsoid using $\| Ax-b\|_2$?

An ellipsoid in $$\mathbf{R}^n$$ has the following form

$$$$\label{ellipsoid} \mathcal E = \left\{ x \mid (x - x_c)^T P^{-1} (x - x_c) \leq 1 \right\} = \left\{ x_c + Ax \mid \left \|x \right \|_2 \leq 1 \right\}$$$$

where $$P = P^T \succ 0$$ and matrix $$A$$ is square and nonsingular ($$A=P^{1/2}$$). The ellipsoid can also be defined as follows:

$$$$\label{ellipsoid1} \mathcal E = \left\{ x \mid (x - x_c)^T P^{-1} (x - x_c) \right\} = \left\{ x \mid (x-x_c)^T E (x-x_c) \leq 1 \right\}$$$$

Here it suggests to formulate the ellipsoid equation as the following form:

$$$$\mathcal E = \left\{ x \in \mathbb{R}^n \mid \left \| Ax-b \right \|_2 \leq 1 \right\}$$$$ where $$A = E^{1/2}$$ and $$b = E^{1/2}x_c$$.

Problem: After expanding I was able to prove $$A = E^{1/2}$$. However, I feel like $$b = E^{1/2}x_c$$ seems to be wrong, could someone help me to find out what went wrong?

$$(x-x_c)^tE(x-x_c) = x^tEx - x^tEx_c - x_c^tEx + c^tEc$$

$$(Ax-b)^t(Ax-b) = x^tA^tAx -x^tA^tb-b^tAx+b^tb$$

• $\| Ax-b\|_2$ is l2 norm, could put it as answer, I could not prove it myself. Apr 2, 2022 at 8:30