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?

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