I'm studying convex optimization (Boyd's course).
Apprently, an ellipsoid can be described using the set (see page 22 here): $$\mathcal{E}_1 = \{x: (x-c)^T P^{-1} (x-c)\le 1\}$$ where $P$ is a symmetric positive definite matrix.
Another representation is: $$\mathcal{E}_2 = \{c + A u : ||u|| \le 1\}$$
However, it seems that $\mathcal{E}_1$ is unique but not $\mathcal{E}_2$. I fail to see why! Boyd, says though that $\mathcal{E}_2$ can be made unique if $A$ was symmetric positive definite, but I have no clue to as why is that true. Any help is appreciated.
Edit: Trying to find the relationship between $\mathcal{E}_1$ and $\mathcal{E}_2$
For simplicity, assume $c=0$. If $\mathcal{E}_1 = \mathcal{E}_2$, then $x = A u, ||u|| \le 1$ verifies: $$u^T A^T P^{-1} A u \le 1 \Longleftrightarrow u^T A^T P^{-1/2}P^{-1/2} A u \le 1 \Longleftrightarrow ||P^{-1/2} A u||^2 \le 1.$$
However, for this latter inequality to be verified, we need $Q = P^{-1/2} A$ to be orthogonal (to preserve lengths) because we also have $||u||\le 1$.
We conclude that $$A = P^{1/2}Q.$$
Because $P$ is symmetric positive definite, we know $P^{1/2}$ is unique. Hence, $A$ is not unique (some would say $A$ is unique up to an orthogonal matrix!).
Now, assuming $Q = H^{1/2}$, where $H$ is symmetric, we have $A = P^{1/2}H^{1/2}$ and is then unique because $H^{1/2}$ is also unique. In particular, we can choose $H = I$, which gives $A = P^{1/2}$.