Matrices with entries in a $C^*$-algebra Let $\mathcal{A}$ be a $C^\ast$-algebra. Consider vector space of matrices of size $n\times n$ whose entries in $\mathcal{A}$. Denote this vector space $M_{n,n}(\mathcal{A})$. We can define involution on $M_{n,n}(\mathcal{A})$ by equality
$$
[a_{ij}]^*=[a_{ji}^*],\qquad\text{where}\quad [a_{ij}]\in M_{n,n}(\mathcal{A}).
$$
Thus we have an involutive algebra $M_{n,n}(\mathcal{A})$. It is well known  that there exist at most one norm on $M_{n,n}(\mathcal{A})$ making it a $C^\ast$-algebra. This norm does exist. Indeed take universal representation $\pi:\mathcal{A}\to\mathcal{B}(H)$ and define linear injective $*$-homomorphism
$$
\Pi:M_{n,n}(\mathcal{A})\to\mathcal{B}\left(\bigoplus\limits_{k=1}^n H\right):[a_{ij}]\mapsto\left((x_1,\ldots,x_n)\mapsto\left(\sum\limits_{j=1}^n\pi(a_{1j})x_j,\ldots,\sum\limits_{j=1}^n\pi(a_{nj})x_j\right)\right)
$$
Hence we can define norm on $M_{n,n}(\mathcal{A})$ as $\left\Vert[a_{ij}]\right\Vert_{M_{n,n}(\mathcal{A})}=\Vert\Pi([a_{ij}])\Vert$. At first sight this definition depends on the choice of representation, but in fact it does not.
My question. This norm on $M_{n,n}(\mathcal{A})$ can be defined internally. Namely
$$
\Vert[a_{ij}]\Vert_{M_{n,n}(\mathcal{A})}=\sup\left\Vert\sum\limits_{i=1}^n\sum\limits_{j=1}^n x_i a_{ij}y_j^*\right\Vert
$$
where supremum is taken over all tuples $\{x_i\}_{i=1}^n\subset\mathcal{A}$, $\{y_i\}_{i=1}^n\subset\mathcal{A}$ such that  $\left\Vert\sum\limits_{i=1}^n x_i x_i^*\right\Vert\leq 1$, $\left\Vert\sum\limits_{i=1}^n y_i y_i^*\right\Vert\leq 1$. 
Is there a proof of this fact without usage of structural theorem for $C^*$-algebras, 
a straightforward proof which can be made by simple checking axioms of $C^*$-algebras?
The same question on math.stackexchange.com
 A: For $x=(x_i)_{i=1}^n, y=(y_i)_{i=1}^n \subseteq A$ define $(x,y) = \sum_i x_i y_i^* \in A$, and set $\|x\| = \|(x,x)\|^{1/2}$.

Lemma: We have that $(x,y)^* (x,y) \leq \|x\|^2 (y,y)$ the order in the C$^*$-algebra sense.
Proof: (Copied from Lance's Hilbert C$^*$-module book).  Wlog $\|x\|=1$.  For $a\in A$ let $a\cdot x = (ax_i)$.  Then
  \begin{align*} 0 &\leq (a\cdot x-y, a\cdot x-y) \\
&= a (x,x) a^* - (y,x)a^* - a(x,y) + (y,y) \\
&\leq aa^*- (y,x)a^* - a(x,y) + (y,y) \end{align*}
  The claim that $a(x,x)a^* \leq aa^*$ follows as if $c\in A^+$ then always $aca^* \leq \|c\|aa^*$.  Now set $a=(x,y)^*=(y,x)$ and the claim follows.

In particular, $\|(x,y)\|^2 \leq \|x\|^2 \|y\|^2$ and so $\sup\{ \|(x,y)\| : \|x\| \leq 1 \} = \|y\|$.
So you define
$$ \| a \| = \sup \{ (x, ay) : \|x\|\leq 1, \|y\|\leq 1 \} $$
where $(ay)_i = \sum_j y_j a_{ij}^*$.  Then from the observation above, 
\begin{align*} \|a\|^2 &= \sup \{ \|ay\|^2 : \|y\|\leq 1 \}
= \sup\{ (ay,ay) : \|y\|\leq 1 \}  \\
&= \sup\Big\{ \Big\| \sum y_j a_{ij}^* a_{ik} y_k^* \Big\| : \|y\|\leq 1 \Big\} \\
&= \sup\{ \|(y, (a^*a)y)\| : \|y\|\leq 1 \} \\
&\leq \sup\{ \|(a^*a)y\| : \|y\|\leq 1 \} \\
&= \sup\{ \|(z,(a^*a)y)\| : \|y\|\leq 1, \|z\|\leq 1 \} = \|a^*a\|.
\end{align*}
However, for any $z$,
$$ \|az\| = \sup\{ \|(x,az)\| : \|x\|\leq 1 \}
\leq \sup\{ \|(x,ay)\| : \|x\|\leq 1, \|y\|\leq \|z\| \}
= \|a\|\|z\|. $$
It follows from this that the norm on $M_n(A)$ is an algebra norm (i.e. submultiplicative).  From the very definition, the involution is an isometry on $M_n(A)$.  So we have the usual trick: $$\|a\|^2 \leq \|a^*a\| \leq \|a^*\| \|a\| = \|a\|^2$$
and so we have equality throughout, establishing the C$^*$-identity for $M_n(A)$.
The idea is to define a generalised Hilbert space of rows of $A$ with an $A$-valued inner-product, and then copy the usual proof that operators on a Hilbert space are a C$^*$-algebra.
