Consider the following two representations of ellipsoid:
$$E_1 = \{x \mid x^TSx\leq 1, \, S \succ 0\}$$ and $$E_2 = \{y \mid y = Ax, \, \|x\|\leq 1, \, \det(A) \neq 0\}$$
If I want $E_1=E_2$, I can do
$$x^TSx = x^TS^{1/2}S^{1/2}x = \|S^{1/2}x\|^2_2 \leq 1$$ and $$y=Ax \longrightarrow A^{-1}y = x \longrightarrow \|A^{-1}y\| = \|x\| \le1 \longrightarrow \|A^{-1}y\|^2_2\leq 1$$
So comparing both equations,
$$A^{-1} = S^{1/2}$$
In this case, $A$ has to be symmetric.
However, if I do the following
$$\|A^{-1}y\|^2_2\leq 1 \longrightarrow y^T(AA^T)^{-1}y \leq 1$$
and let $(AA^T)^{-1}=S$ with $S = Q\Lambda Q^T$, $A$ can be chosen as
$$A = Q\Lambda^{-1/2}$$
In this case, $A$ does not have to be symmetric.
What's wrong with the first approach? Or are both correct?
$S$ does not have to be symmetric, and neither does $A$.
Consider for instance: $$x^2+2y^2+2xy= \begin{pmatrix}x&y\end{pmatrix}\begin{pmatrix}1&2\\0&2\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix} =\begin{pmatrix}x&y\end{pmatrix}\begin{pmatrix}1&1\\1&2\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix} $$ It's just that we usually choose $S$ to be symmetric, since then we can do a useful eigenvalue decomposition.