Claim:
$\mathcal{E}(x_c, 1) = \{x|(x - x_c)^T P^{-1} (x-x_c) \leq 1\}, P \in S^n_{++}$
has an alternative representation as:
$\mathcal{E}(x_c, 1) = \{x|\|Ax+b\|^2 \leq 1\}, A \in S^n_{++}$
What is a way to show this starting from
$\mathcal{E}(x_c, 1) = \{x|\|Ax+b\|^2 \leq 1\}, A \in S^n_{++}$?
Attempt:
the constraint is rewritten as:
$x^TA^TAx + 2b^TAx + b^Tb \leq 1$
Let $P^{-1} = A^TA, x_c = b$
Then we have $x^TP^{-1}x + 2b^TAx + x_c^Tx_c$
How should you proceed from here to get into $(x - x_c)^T P^{-1} (x-x_c)$ and how to deal with term in the middle?
Let $A x^* = -b$, then the alternative representation is $\|A(x-x^*)\|^2 \le 1$. That is, $(x-x^*)^T A^TA (x-x^*) \le 1$.
From this we see that we have equivalence when $P^{-1} = A^T A$ and $x_c = x^*$.