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

  • $\begingroup$ I fixed the rendering issues. Nothing to say about the question though. $\endgroup$ Feb 1, 2012 at 13:14
  • $\begingroup$ @Vladimir Dotsenko, Great, thanks. But workaround looks like dirty hack! $\endgroup$
    – Norbert
    Feb 1, 2012 at 13:21
  • $\begingroup$ Well, if I remember correctly, the norm of an element in a C*-algebra is determined by its spectral radius (by some simple formula which I can't seem to recall right now), so perhaps one can independently prove that this formula indeed gives a C*-algebra norm in this case. $\endgroup$
    – Mark
    Feb 1, 2012 at 16:15
  • 3
    $\begingroup$ I think it would be good practice to link to the copy of this question that you posted on math.stackexchange.com $\endgroup$
    – Yemon Choi
    Feb 1, 2012 at 16:18
  • 3
    $\begingroup$ The post at MSE is: math.stackexchange.com/questions/104303/… There Jonas's gives a very similar answer to mine below, but with less detail, but better references to the surrounding ideas. $\endgroup$ Feb 1, 2012 at 16:53

1 Answer 1


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.

  • $\begingroup$ Cf. Jonas's answer on MSE $\endgroup$
    – Yemon Choi
    Feb 1, 2012 at 16:44
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    $\begingroup$ Indeed! As Jonas's says over at MSE, we appear to have independently come up with almost exactly the same idea (complete with reference to Lance's book). As you say, it would be really helpful for people to not cross-post from MSE without saying so!! $\endgroup$ Feb 1, 2012 at 16:49
  • $\begingroup$ Great, thanks! This is exactly what I meant saying straightforward proof. $\endgroup$
    – Norbert
    Feb 1, 2012 at 17:07
  • $\begingroup$ Matt, I left you a comment over at MSE $\endgroup$
    – Yemon Choi
    Feb 1, 2012 at 18:46

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