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](https://math.stackexchange.com/questions/104303/matrices-with-entries-in-c-algebra)



Matrices with entries in a $C^*$-algebra
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.


