I need your expertise in understanding and hopefully solving the following problem:
Let $C \subset \mathbb{R}^d$, and lets assume that $C$ is not centrally symmetric, thus when computing the $MVEE(C)$ (i.e the minimum volume enclosing ellipsoid), they use the following "lifting step": $$ C^\prime = \left\lbrace \pm \begin{bmatrix} x\\ 1 \end{bmatrix} \middle| x \in C \right\rbrace,$$ where $C^\prime$ is centrally symmetric.
It is proven that for any $\varepsilon > 0$, $MVEE(C^\prime) \cap \Pi$ is a $(1 + \varepsilon)$-approximation to $MVEE(C) \times \left\lbrace 1 \right\rbrace$, where $\Pi = \left\lbrace x \in \mathbb{R^{d+1}} \middle| x_{d+1} = 1 \right\rbrace$.
Let $G \in \mathbb{d \times d}$ be an invertible matrix, let $ v \in \mathbb{R}^d$ such that the pair $(G,v)$ represents the matrix and the center of $MVEE(C)$ respectively. Let $\hat{G} \in \mathbb{R}^{(d + 1) \times (d + 1)}$ be an invertible matrix such that the pair $(\hat{G},\overset{\to}{0})$ represents the matrix and center of $MVEE(C^\prime)$ respectively.
Then, $\forall x \in MVEE(C)$ $$ \left| \left| \left[ x , 1 \right] \hat{G} \begin{bmatrix} x\\1 \end{bmatrix} \right| \right|_2 = (1 + \varepsilon) \cdot \left| \left| (x - v) G (x - v)^T \right| \right|_2 .$$
Regarding $x \not\in MVEE(C) $, then it is easy to see that the above term is not satisfied and moreover the right term is bigger or equal to the left term.
What I need to know is whether it possible the bound the right term by the left term (maybe because $MVEE(C) \times \left\lbrace 1 \right\rbrace \subseteq MVEE(C^\prime)$), i.e. $$ poly(d) \cdot \left| \left| \left[ x , 1 \right] \hat{G} \begin{bmatrix} x\\1 \end{bmatrix} \right| \right|_2 \geq (1 + \varepsilon) \cdot \left| \left| (x - v) G (x - v)^T \right| \right|_2 \ ?$$
Please Do correct if I am wrong and thanks in advance.