the $C^\ast$-algebra $M_n(A)$, understanding the $C^\ast$-norm on $M_n(A)$

Question

Let $A$ be a $C^\ast$-algebra. I want to understand $M_n(A)$, the vector space of $n\times n$-matrices with entries in $A$, as a $C^\ast$-algebra. On $M_n(A)$ you can define an involution $$(a_{ij})^\ast=(a_{ji}^\ast)$$ for $(a_{ij})\in M_n(A)$, and as multiplication you take the usual multiplication of matrices. If you want to define a norm on $M_n(A)$, take universal representation $\pi:A\to L(H)$ of $A$ and define a homomorphism $$\phi :M_n(A)\to L(\bigoplus\limits_{i=1}^nH_i),\; (a_{ij})\mapsto \phi (a_{ij})$$ with $ \phi (a_{ij})(x_1,..,x_n)=(\pi(a_{11})x_1+..+\pi(a_{1n})x_n,...,\pi(a_{n1})x_1+..+\pi(a_{nn})x_n)$. I'm not sure, why you take $\|(a_{ij})\|=\|\phi(a_{ij})\|$ as $C^\ast$-norm on $M_n(A)$. More explicit, $\|(a_{ij})\|=\|\phi(a_{ij})\|$ is
$$\|(a_{ij})\|^2=\|\phi(a_{ij})\|^2=\sup\limits_{\|x_1\|^2+..+\|x_n\|^2=1}\sum\limits_{l=1}^n\|\sum\limits_{k=1}^n\pi(a_{lk})x_k\|^2,$$ is it correct? I think only necessary is to prove $\|(a_{sr})(b_{rp})\|\le \|(a_{sr})\|\|(b_{rp})\|$ and $\|(a_{rp})^\ast (a_{rp})\|=\|a_{rp}\|^2$. I would try this with the formula above if the formula is correct. Regards

Answer 1

What you need to show is that $\phi $ is one-to-one, and that it is a $*$-homomorphism. That lets you define the norm on $M_n (A) $ and the two properties you want are inherited automatically from the codomain.

Then you need to show that $M_n (A) $ is closed. This is done by showing that norm convergence is equivalent to entrywise norm convergence, and as $A $ is closed, that's it.