A norm making $M_n(A)$ a $C^*$-algebra when $A$ is a $C^*$-algebra

Question

Show that there is an unique norm on $M_n(A)$ making it a $C^*$-algebra, where $A$ is a $C^*$-algebra itself.

It is enough to show this when $A$ is a concrete C^*-algebra. Suppose we are able to find a norm $\|.\|_1$ for such a case. Because for any $C^*$-algebra $B$, there is an isometric isomorphism $\pi$ from $B$ to a concrete $C^*$-algebra. Then for $X \in M_n(B)$, we define $\|X\|=\|\pi(X)\|_1$, where $$\pi(X)=\begin{bmatrix} \pi(b_{11})&\pi(b_{12})&\ldots&\pi(b_{1n})\\ \vdots&\vdots& \vdots &\vdots\\ \pi(b_{n1})&\pi(b_{n2})&\ldots&\pi(b_{nn})\\ \end{bmatrix}$$

when $$X=\begin{bmatrix} (b_{11})&(b_{12})&\ldots&(b_{1n})\\ \vdots &\vdots& \vdots &\vdots\\ (b_{n1})&(b_{n2})&\ldots&(b_{nn})\\ \end{bmatrix}$$

Then $$\|X^*X\|=\|\pi(X^*X)\|_1=\|(\pi(X))^*\pi(X)\|_1=\|\pi(X)\|_1^2=\|X\|^2$$

So the problem now boils down to finding a norm on $M_n(A$) when $A$ is a concrete C^*-algebra. Let $X \in M_n(A)$. Suppose that $X=\begin{bmatrix} (b_{11})&(b_{12})&\ldots&(b_{1n})\\ \vdots &\vdots& \vdots &\vdots\\ (b_{n1})&(b_{n2})&\ldots&(b_{nn})\\ \end{bmatrix}$

Then for $$a=(a_1,a_2,\ldots,a_n) \in H^n, Xa=(\sum_{j=1}^n b_{1j}a_j,\ldots,\sum_{j=1}^nb_{nj}a_j)$$

Naturally I think $$\|Xa\|^2=\langle \sum_{j=1}^nb_{1j}a_j, \sum_{j=1}^nb_{1j}a_j \rangle + \ldots + \langle \sum_{j=1}^nb_{nj}a_j, \sum_{j=1}^nb_{nj}a_j \rangle=\sum_{i=1}^n \left\|\sum_{j=1}^nb_{ij}a_j\right\|^2$$ So I should define $$\|X\|=\sup_{\|a\| \le 1}\|Xa\|$$

where $$\|a\|=\left(\sum_{i=1}^n\|a_i\|^2\right)^{1/2}$$

I am not sure. Is there any other norm? Thanks for the help!!

Answer 1

What you did is correct (and canonical).

As for uniqueness: yes, that's the only possible norm. Once you have a C$^*$-algebra (not just a $*$-algebra) the C$^*$-norm is unique. This is due to $$ \|x\|^2=\|x^*x\|=\text{spr}\,(x ^*x), $$ which shows that the algebraic structure determines the norm. So this only depends on showing that $M_n(A)$ is closed when $A$ is a C$^*$-algebra.