On the coincidence (or non-coincidence) of two norms defined on the quotient of a given Hilbert $ C^{\ast} $-module by a certain linear subspace Let $ A $ be a $ C^{\ast} $-algebra, $ I $ a closed two-sided ideal of $ A $, and $ \mathcal{E} $ a Hilbert $ A $-module. Let
$$
\mathcal{E}_{I}
\stackrel{\text{df}}{=}
\{ x \in \mathcal{E} \mid \langle x,x \rangle_{\mathcal{E}} \in I \}.
$$
Using the Cauchy-Schwarz Inequality for Hilbert $ C^{\ast} $-modules, it is not difficult to show that
$$
\mathcal{E}_{I} =
\{
x \in \mathcal{E} \mid
(\forall y \in \mathcal{E})(\langle x,y \rangle_{\mathcal{E}} \in I)
\}.
$$
This implies that $ \mathcal{E}_{I} $ is a linear subspace of $ \mathcal{E} $.
Furthermore, as the $ A $-valued inner product on $ \mathcal{E} $ is continuous, $ \mathcal{E}_{I} $ is a $ \| \cdot \|_{\mathcal{E}} $-closed subset of $ \mathcal{E} $.
All of this implies that $ \mathcal{E}_{I} $ is a Hilbert $ I $-module.
Now, the quotient space $ \mathcal{E} / \mathcal{E}_{I} $ is a Banach space w.r.t. the quotient norm $ \| \cdot \|_{\text{q}} $ defined by
$$
\forall x \in \mathcal{E}: \qquad
\| x + \mathcal{E}_{I} \|_{\text{q}}
\stackrel{\text{df}}{=}
\inf_{y \in \mathcal{E}_{I}} \| x + y \|_{\mathcal{E}}.
$$
It is also a right $ (A / I) $-module equipped with an $ (A / I) $-valued pre-inner product $ [\cdot,\cdot] $ defined by
$$
\forall x_{1},x_{2} \in \mathcal{E}: \qquad
[x_{1} + \mathcal{E}_{I},x_{2} + \mathcal{E}_{I}]
\stackrel{\text{df}}{=}
\langle x_{1},x_{2} \rangle_{\mathcal{E}} + I.
$$


Question. Is the norm on $ \mathcal{E} / \mathcal{E}_{I} $ that is induced by $ [\cdot,\cdot] $ the same as $ \| \cdot \|_{\text{q}} $, i.e.,
  $$
\forall x \in \mathcal{E}: \qquad
  \| x + \mathcal{E}_{I} \|_{\text{q}}
= \sqrt{\| \langle x,x \rangle_{\mathcal{E}} + I \|_{A / I}}?
$$

Thank you for your help!
 A: Here are a couple of supplements to Omar's answer.
The first thing to note is that the module $\mathcal{E}_I$ is equal to the module $\mathcal{E}I=\{ei\ |\ e\in \mathcal{E},\ i\in I\}$. The inclusion $\mathcal{E}I\subset \mathcal{E}_I$ follows easily from the linearity of the inner product and from the fact that $I$ is an ideal. For the reverse inclusion, let $i_\lambda$ be an approximate unit for $I$, and then show (by expressing the norm in terms of the inner product) that for each $e\in \mathcal{E}_I$ one has $\| e-ei_\lambda\|\to 0$ as $\lambda\to\infty$. This gives $\mathcal{E}_I\subset \overline{\mathcal{E}I}$, and the latter is equal to $\mathcal{E}I$ by the Cohen factorisation theorem. For full details see Lemma 3.23 in: 
Raeburn, Iain; Williams, Dana P., Morita equivalence and continuous-trace $C^*$-algebras, Mathematical Surveys and Monographs. 60. Providence, RI: American Mathematical Society (AMS). xiv, 327 p. (1998). ZBL0922.46050.
Now, turning to the inner product on $\mathcal{E}/\mathcal{E}_I$: here are three arguments showing that the norm defined by the inner product is equal to the quotient norm. 
(1) A short direct computation is given in Lemma 3.1 in:
Zettl, Heinrich H., Ideals in Hilbert modules and invariants under strong Morita equivalence of C*-algebras, Arch. Math. 39, 69-77 (1982). ZBL0498.46034.
(2) An argument based on the uniqueness of the $C^*$-norm on the linking algebra is given in Proposition 3.25 in the book of Raeburn-Williams cited above. (The authors attribute the argument to Siegfried Echterhoff.) 
(3) Here is a third proof, using ideas around operator modules and the Haagerup tensor product. All of the necessary background can be found in 
Blecher, David P.; Le Merdy, Christian, Operator algebras and their modules -- an operator space approach, London Mathematical Society Monographs. New Series 30;  Oxford Science Publications. Oxford: Oxford University Press (ISBN 0-19-852659-8/hbk). x, 387~p. (2004). ZBL1061.47002.
Consider $I\to A\to A/I$ as an exact sequence of operator $A$-bimodules. Taking the Haagerup tensor product with $\mathcal E$ (viewed as a right operator module over $A$) gives 
$$
\mathcal E\otimes_A I \to \mathcal E\otimes_A A \to \mathcal E\otimes_A (A/I).
$$
By the exactness property of the Haagerup tensor product over a $C^*$-algebra (a theorem of Anantharaman-Delaroche and Pop), the first map in the display is a (completely) isometric embedding, and the second map induces a (completely) isometric isomorphism
$$ 
(\mathcal E\otimes_A A)/ (\mathcal E\otimes_A I) \cong \mathcal E\otimes_A (A/I)\qquad (*)
$$
where the left-hand side carries its canonical quotient operator space structure (and in particular, the usual Banach-space quotient norm). 
The module action gives a (completely) isometric isomorphism $\mathcal E\otimes_A A \to \mathcal E$, which restricts to an isomorphism $\mathcal E\otimes_A I \to \mathcal E I=\mathcal E_I$. Making these identifications, the isomorphism $(*)$ is given by the formula
$$
\mathcal E/\mathcal E_I\to \mathcal E\otimes_A (A/I),\qquad (ea+\mathcal E_I)\mapsto e\otimes(a+I).\qquad (**)
$$
Now, $A/I$ is a (right) Hilbert $C^*$-module over itself, and the quotient map $A\to A/I$ gives a (left) action of $A$ on this $C^*$-module by adjointable operators. We can thus form the Hilbert $C^*$-module tensor product $\mathcal E\otimes^{C^*}_A (A/I)$, which will be a Hilbert $C^*$-module over $A/I$. A theorem of Blecher asserts that the identity map on the algebraic tensor products extends to a (completely) isometric isomorphism between $\mathcal E\otimes^{C^*}_A (A/I)$ and the Haagerup tensor product $\mathcal E\otimes_A (A/I)$. A straightforward computation shows that the map $(**)$ is isometric with respect to the inner product $[\cdot,\cdot]$ on $\mathcal E/\mathcal E_I$, and the canonical inner product on $\mathcal E\otimes^{C^*}_A (A/I)$. Since $(**)$ is also isometric for the quotient norm, the latter must coincide with the norm induced by $[\cdot,\cdot]$.
A: Yes, I think this is true. Here is an argument to show why. First for $E=A$, then this is just the standard theorem saying that the quotient norm on $A/I$ is the $C^*$-algebra norm. (any ref on $C^*$-algebras has a proof)
Then you argue directly that the case $E=A$ implies that it is also true for $E=l^2(\mathbb{N})\otimes A$.
Then you could use Kasparov stabilization theorem to prove it for coutably generated $C^*$-modules.
To prove it for all modules, you might argue that both norms calcuated at a fixed element, only needed a countably generated submodule to be computed on. (Both are infimum of something so you take the submodules generated by a sequence reaching the infimum)
Edit: Here some more details,
I will use $E'$ to denote the module $E\oplus l^2(\mathbb{N})\otimes A$. We directly see that $E'_I=E_I\oplus l^2(\mathbb{N})\otimes I$, hence $E'/E'_I=E/E_I\oplus l^2(\mathbb{N})\otimes A/I.$ Hence the inclusion map $i:E/E_I\to E'/E'_I$ preserves the two norms defined in the question. 
Theorem(Kasparov's Specialization theorem)(reference in a comment) Let $E$ be a countably generated $A$-module, then $E\oplus l^2(\mathbb{N})\otimes A\simeq l^2(\mathbb{N})\otimes A$. Here the isomorphism is a unitary isomorphism.
Since the module $E'$ is untarly equivalent to $l^2(\mathbb{N})\otimes A$ it follows that the two norms are equal on $E'/E'_I$ and hence on $E/E_I$.
