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Let $A$ be a $C^*$-algebra and $(a_{ij}) \in M_n(A)$ be a positive matrix. Does there exist a constant $C \ge 0$ (not depending on the $a_{ij}$) such that $$\lVert(a_{ij})\rVert \le C \Bigl\lVert\sum_i a_{ii}\Bigr\rVert?$$

Attempt: Choose a faithful representation $A \subseteq B(H)$. Then $M_n(A) \subseteq M_n(B(H)) = B(H^{n})$ so we have $$\lVert(a_{ij})\rVert = \sup_{(x_1, \dotsc, x_n) \in (H^n)_1} \sum_j \Bigl\lVert\sum_i a_{ji} x_i\Bigr\rVert.$$

How to proceed?

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  • $\begingroup$ If you don't care if $C$ depends on $n$, then yes. If you do care, then not. $\endgroup$
    – Diego Martinez
    Commented Oct 31, 2021 at 10:28
  • $\begingroup$ @DiegoMartínez $C$ can depend on $n$. What did you have in mind? $\endgroup$
    – Andromeda
    Commented Oct 31, 2021 at 10:30
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    $\begingroup$ The map $M_n(A) \to A$ given by $(a_{i,j}) \mapsto \sum a_{i,i}$ is completely positive with norm $n$. $\endgroup$
    – Jamie Gabe
    Commented Oct 31, 2021 at 11:45
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    $\begingroup$ Ah, I misread the the problem! Whoops! For the implication you want, you can use that $(a_{i,j})_{i,j=1}^n \leq 2^{n-1} \mathrm{diag}(a_1,\dots, a_n)$. The $2^{n-1}$ is definitely not optimal, but it works. To show the inequality, use the $n=2$ case over and over again (which is quite easy to prove) by considering matrices which "look like" $2\times 2$-matrices embedded in $n\times n$-matrices with a $n^2-4$ entries being zero. $\endgroup$
    – Jamie Gabe
    Commented Oct 31, 2021 at 17:37
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    $\begingroup$ Ah, yes, this exercise in Paulsen's book is the correct generalisation of the n=2 case I mentioned above (the estimate I got applied the n=2 case over and over again to get a (quite bad) upper bound). This is much better! $\endgroup$
    – Jamie Gabe
    Commented Oct 31, 2021 at 20:18

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This answer arose by a discussion with @JamieGabe in the comments.

One can prove that the map $$\Phi: M_n(A) \to M_n(A): A \mapsto n \operatorname{Diag}(A)-A$$ is completely positive [Paulsen, "Completely bounded maps and operator spaces", exercise 3.6].

In particular, it is positive. Hence, writing $A= (a_{i,j})$ as in the OP, we obtain $$A \le n \operatorname{Diag}(A)$$ and taking norms leads to $$\|A\| \le n \lVert\operatorname{Diag}(A)\rVert=n \max_{i=1}^n \|a_{i,i}\| \le n \left\|\sum_{i=1}^n a_{i,i}\right\|.$$

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    $\begingroup$ As you indicated, you are using @JamieGabe's hint, and it would be appropriate explicitly to acknowledge that. \\ Also, TeX note: $\lVert\operatorname{Tr}\rVert$ \lVert\operatorname{Tr}\rVert spaces better than $\|\operatorname{Tr}\|$ \|\operatorname{Tr}\|. I have edited accordingly. $\endgroup$
    – LSpice
    Commented Oct 31, 2021 at 14:47
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    $\begingroup$ @LSpice Thanks for the suggestions. I edited the post to give credit and I will keep your Latex suggestion in mind. $\endgroup$
    – Andromeda
    Commented Oct 31, 2021 at 15:14

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