Help with proving: If $X$ is a Hilbert $A$-$B$-module, then $ \| _A \langle x,x \rangle \| = \| \langle x,x \rangle _B \| $ for all $x\in X $. Sorry, I posted a related question last week on here, but I'm still having trouble and this is a little different, I hope it's OK.  Thank you!
( proof that this is an isometric map (on a $C^*$-module) )

Notation: ${}_A\langle \cdot ,\cdot \rangle $ and $ \langle \cdot ,\cdot \rangle _B$ denotes the left and right inner product.  
Notation: $\theta_{x,y}(z) = x\langle y,z \rangle _B $
Notation: $\mathcal{K}(X) = \{ \theta _{x,y} : x,y \in X\} $
Definition:
A (right/left) inner product A-module $V$ is called a (right/left Hilbert $A$-module if $E$ is complete with respect to the norm defined by $\| x \| = \| \langle x,x \rangle \| ^{1/2} $ $(x \in V) $. 
Definition:
Let $A$ and $B$ be $C^*$-algebras. A Hilbert $A$-$B$-module V is both a left Hilbert $A$-module and a right Hilbert $B$-module such that 
$$
(ax)b=a(xb) \ \text{ and } \ _A \langle x,y \rangle z = x\langle y, z \rangle _B
$$
for all $a\in A$, $b\in B$, and $x,y,z\in V$.
Proposition:
If $X$ is a Hilbert $A$-$B$-module, then 
$$
\| _A \langle x,x \rangle \| = \| \langle x,x \rangle _B \| \ \text{ for all } x\in X.
$$
Proof:
Let $x\in X$ and $a= {}_A \langle x,x \rangle $. In the following, note that $\| x \| = \| \langle x,x \rangle _B \| ^{1/2}_B $. So, when one considers $\| \lambda _a \| _{\mathcal{K}(X)}$, one has to consider $\| az \| = \| az \| _B $ with $ \| z \| = \| z \| _B \leq 1$ (i.e., the norm on $X$ is taken from the inner product $\langle \cdot , \cdot \rangle _B $). Now we have
$$
\theta _{x,x} (z) = x\langle x,z \rangle _B = {}_A\langle x,x \rangle z = az = \lambda _a (z) \ (z\in X),
$$
where $\lambda _a$ is the map $z\mapsto az$ on $X$.
The following part of the proof is causing me trouble, but here is my attempt
Moreover, $ \lambda $ is bounded since for all $z\in X$ such that $\| z \| _A \leq 1 $, we have 
$$
\| a z \| _A \leq \| a \| _A \| z \| _A \leq \| a \| _A .
$$
Therefore, to show that $\| \lambda _a \| _{\mathcal{K}(X)} = \| a \| _A$, it suffices to show that $\lambda $ has trivial kernel.
Assume $\lambda _a = 0 $. Let $e_\alpha = {}_A\langle u_\alpha , v_\alpha \rangle $ be an approximate identity of ${}_A\langle X,X \rangle $. Then,
\begin{align*}
ae_\alpha a^* &= a{}_A \langle u_\alpha , v_\alpha \rangle a^* \\
&= {}_A \langle au_\alpha , au_\alpha \rangle \\
&= {}_A \langle \lambda _a (u_\alpha ), \lambda _a (v_\alpha ) \rangle \\
&= 0.
\end{align*}
Therefore, $aa^*=0$ and hence $a=0$. So, $\| \lambda _a \| _{\mathcal{K}(X)} = \| a \| _A$.
It follows that 
$$
\| _A\langle x,x \rangle \| _A = \| \lambda _a \| _{\mathcal{K}(X)} = \| \theta _{x,x} \| _{\mathcal{K}(X)} = \| \langle x,x \rangle _B \| _B.
$$

I also found a proof of this proposition, which I am having a hard time understanding...
http://books.google.ca/books?id=8LddfdeBbHgC&pg=PA1154&lpg=PA1154&dq=Quasi-multipliers+and+Embeddings+of+Hilbert+C*-bimodules&source=bl&ots=QGa0CfW5Rh&sig=QEmke3JZLCZe392l998-Pw5NiHo&hl=en&sa=X&ei=V0lGUdyxHaOg2gXd5IGADQ&ved=0CE0Q6AEwAw
(in particular, they don't show why $\lambda $ is bounded... and I don't understand why $\sum _i {}_A \langle \lambda _a x_i,x_i \rangle = 0 $ if $\lambda _a = 0 $)
(proposition 1.10 and cor 1.11)
so... my attempt is a little different, so I don't know if it is correct or not...
Thank you.
 A: First of all, $ {_{A}} \langle X,X \rangle $ is defined as follows:
$$
{_{A}} \langle X,X \rangle \stackrel{\text{def}}{=} ~
\overline{\text{Span}(\{ {_{A}} \langle x_{1},x_{2} \rangle ~|~ x_{1},x_{2} \in X \})}^
{\| \cdot \|_{A}}.
$$
Notice that this is a C$ ^{*} $-subalgebra of $ A $. Notice also that $ \mathcal{K}(X) $ is a closed two-sided ideal of $ \mathcal{L}(X) $, which is the C$ ^{*} $-algebra of adjointable operators on $ X $. Hence, $ \mathcal{K}(X) $ is a C$ ^{*} $-subalgebra of $ \mathcal{L}(X) $.
Next, define a continuous $ ^{*} $-homomorphism $ \Theta: {_{A}} \langle X,X \rangle \to \mathcal{K}(X) $ by first declaring
$$
\forall x_{1},x_{2} \in X: \quad
\Theta({_{A}} \langle x_{1},x_{2} \rangle) \stackrel{\text{def}}{=}
\theta_{x_{1},x_{2}} := (z \longmapsto {_{A}} \langle x_{1},x_{2} \rangle z)
$$
and then extending this initial definition by


*

*linearity and involution and

*finally by continuity, which is possible because
$$
  \Theta: \text{Span}(\{ {_{A}} \langle x_{1},x_{2} \rangle ~|~ x_{1},x_{2} \in X \})
  \to \text{Span}(\{ \theta_{x_{1},x_{2}} ~|~ x_{1},x_{2} \in X \})
  $$
is a bounded mapping. More precisely,
$$
  \forall {_{A}} \langle x_{1,1},x_{2,1} \rangle,\ldots,{_{A}} \langle x_{1,n},x_{2,n}
  \rangle: \quad
  \left\| \theta_{\sum_{i=1}^{n} {_{A}} \langle x_{1,i},x_{2,i} \rangle} \right\|_
  {\mathcal{K}(X)} \leq
  \left\| \sum_{i=1}^{n} {_{A}} \langle x_{1,i},x_{2,i} \rangle \right\|_{A}.
  $$
Now, one error that you have made about how approximate identities of $ {_{A}} \langle X,X \rangle $ look like is in assuming that $ e^{\alpha} = {_{A}} \langle u^{\alpha},v^{\alpha} \rangle $. In general, $ e^{\alpha} $ is of the form
$$
e^{\alpha} = \sum_{i=1}^{n} {_{A}} \langle u_{i}^{\alpha},v_{i}^{\alpha} \rangle,
$$
as explained in the linked article (using an argument of Dixmier). This is only a minor issue however, and once you have established that $ \text{Ker}(\Theta) $ is trivial, then you have obtained an injective $ ^{*} $-homomorphism from $ {_{A}} \langle X,X \rangle $ to $ \mathcal{K}(X) $. As an injective $ ^{*} $-homomorphism from one C$ ^{*} $-algebra to another is necessarily an isometry, and as $ \text{Range}(\Theta) $ is clearly dense in $ \mathcal{K}(X) $, we conclude that
$$
{_{A}} \langle X,X \rangle \cong \mathcal{K}(X).
$$
Therefore,
$$
\forall x \in X: \quad
\| {_{A}} \langle x,x \rangle \|_{A} = \| \theta_{x,x} \|_{\mathcal{K}(X)}
                                     = \| \langle x,x \rangle_{B} \|_{B}.
$$
