Norms on $M_n(A)$ where A is a unital C* algebra.

Question

Let $H$ be a Hilbert space,and let $(a_{i,j})$ be in $M_n(B(H))$. Prove that $\|(a_{i,j})\|\leq(\sum_{i,j}\|a_{i,j}\|^2)^{1/2} \leq n\|(a_{i,j})\|$.

By definition $\|(a_{i,j})\|^2=sup \{\|(a_{i,j})(h_j)\|^2: \|(h_j)\|^2\leq1, (h_j)\in H^{\oplus n}\} =sup \{\|(\sum_ja_{i,j}h_j)\|^2: \|(h_j)\|^2\leq1, (h_j)\in H^{\oplus n}\}=sup \{\sum_i\|\sum_ja_{i,j}h_j\|^2: \|(h_j)\|^2\leq1, (h_j)\in H^{\oplus n}\}\leq sup \{\sum_i(\sum_j\|a_{i,j}h_j\|)^2: \|(h_j)\|^2\leq1, (h_j)\in H^{\oplus n}\}$. (by triangle inequality)

$=sup \{\sum_i(\sum_j\|a_{i,j}\|^2)(\sum_j\|h_j\|^2): \|(h_j)\|^2\leq1, (h_j)\in H^{\oplus n}\}$ (by Cauchy-Schwarz inequality) $=\sum_{i,j}\|a_{i,j}\|^2.$

Hence $\|(a_{i,j})\|\leq(\sum_{i,j}\|a_{i,j}\|^2)^{1/2}$

How to show the right-hand side inequality? This result will imply that if $\phi:B(H)\mapsto B(H)$ is bounded, then $\phi_n:M_n(B(H))\mapsto M_n(B(H))$ defined by $\phi_n((a_{i,j}))=(\phi(a_{i,j}))$ is also bounded and $\|\phi_n\|\leq n\|\phi\|.$

Answer 1

You have a mistake in your argument, you wrote as an equality the step where you are using Cauchy-Schwarz (which you don't mention either).

As for the other inequality, the proof is exactly the same as the better known case (where $\dim H=1$). I'll write $A$ for the matrix. You always have $$ \|a_{ij}\|\leq\|A\|. $$ Then \begin{align} \sum_i\sum_j\|a_{ij}\|^2\leq\sum_{i,j}\|A\|^2=n^2\|A\|^2. \end{align}