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?



$C\lVert\sum_i a_{ii}\rVert \ge \lVert(a_{ij})\rVert$ for matrices with entries in a $C^*$-algebra
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
$$\|A\| \le n \lVert\operatorname{Diag}(A)\rVert=n \max\_{i=1}^n \|a\_{i,i}\| \le n \left\|\sum\_{i=1}^n a\_{i,i}\right\|.$$


