Nuclearity of finite dimensional $C^*$-algebras What is an elegant way of realizing that a finite dimensional $C^*$-algebra (in particular, $M_n(\mathbb{C})$) is nuclear?
For instance, can one easily deduce that the norm equivalence on a finite dimensional $A$ carries over to all of its tensor products, or does one need to involve nuclear operators?
Thank you very much!
 A: I'll do $M_n(\mathbb C)$ for simplicity, as any other finite-dimensional C$^*$-algebra is just a direct sum of those.
If one defines nuclearity in the sense that the identity is a nuclear map, the statement is completely trivial. So let us focus on the case where we define nuclearity as "unique norm on any tensor product".
The key here is to notice that $M_n(\mathbb C)\odot A\simeq M_n(A)$ is a C$^*$-algebra; so there is no completion to be taken. The isomorphism is trivial, as one maps $\sum_{kj}E_{kj}\otimes a_{kj}\longmapsto \big[a_{kj}\big]$. We can represent $M_n(A)\subset B(H^n)$, for a representation $A\subset B(H)$. So all one needs is to show that $M_n(A)$ is complete in $B(H^n)$.
Let $\{R_s \}_s \subset M_n(A)$ be a Cauchy sequence. Write $R_s =[a_{kj,s }]$. Given any $\xi,\eta\in H$,
$$
|\langle a _{kj,s }\xi,\eta\rangle|=\Bigg|\Bigg\langle R_s \begin{bmatrix} \xi\\ 0\\ \vdots\\ 0\end{bmatrix},\begin{bmatrix} \eta\\ 0\\ \vdots\\ 0\end{bmatrix}\Bigg\rangle\Bigg|\leq\|R_s \|\,\|\xi\|\,\|\eta\|.
$$
It follows that $\|a_{kj,s }\|\leq\|A_s \|$. As the sequence $\{A_s\}$ is Cauchy, we deduce that each entry $\{a_{kj,s}\}$ forms a Cauchy sequence. By the completeness of $A$, there exists $a_{kj}\in A$ with $a_{kj}=\lim_sa_{kj,s}$. Let $R=[a_{kj}]\in M_n(A)$. Fix $\varepsilon>0$; there exists $s_0$ such that for all $s\geq s_0$, $\|a_{kj,s}-a_{kj}\|<\varepsilon$ for all $k,j$. Then, for $\tilde\xi\in H^n$, with $\tilde\xi=(\xi_1,\ldots,\xi_m)^T$,
\begin{align}
\|(R_s-R)\xi\|^2&=\sum_{k,j,\ell}\langle (a_{\ell k,s}^*-a_{\ell k}^*)(a_{\ell j,s}-a_{\ell j})\xi_j,\xi_k\rangle
\leq \varepsilon^2\,n\,\sum_{k,j}\|\xi_j\|\,\|\xi_k\|\\[0.3cm]
&=\varepsilon^2\,n\,\Big(\sum_{k}\|\xi_j\|\Big)^2
\leq \varepsilon^2\,n^2\,\sum_{k}\|\xi_j\|^2=\varepsilon^2\,n^2\,\|\xi\|^2. 
\end{align}
Thus $R_s\to R$ and $M_n(A)$ is complete.
