# Equality between two norms on $\mathcal{L}(E)^n$

Let $$E$$ be a complex Hilbert space and $$\mathcal{L}(E)$$ the algebra of all bounded linear operators on $$E$$.

On $$\mathcal{L}(E)^n$$, we have two equivalent norms: $$\begin{eqnarray*} N_1({\bf A}) &=&\sup\left\{\bigg(\displaystyle\sum_{k=1}^n\|A_kx\|^2\bigg)^{\frac{1}{2}},\;x\in E,\;\|x\|=1\;\right\}, \end{eqnarray*}$$ and $$N_2({\bf A})=\bigg(\displaystyle\sum_{k=1}^n\|A_k\|^2\bigg)^{1/2},$$ for every $${\bf A} = (A_1,...,A_n) \in \mathcal{L}(E)^n$$

In general $$N_1\neq N_2$$. If $$A_iA_j=A_jA_i$$ for all $$i,j$$, is $$N_1=N_2?$$ If the claim is false, under which conditions we have $$N_1=N_2?$$

As quite an obvious counterexample, take $$A_k$$ to be the orthoprojector $$E\ni x:=(x_1,\dots,x_n)\mapsto x_ke_k\in E$$ on $$E:=\mathbb{C}^n$$. Then $$A_iA_j=A_jA_i=\delta_{ij}A_i$$ but $$N_1(A)=1$$ and $$N_2(A)=\sqrt{n}$$.
On the positive side, I think $$A_k:=f_k(A)$$ with $$A$$ self-adjoint, and increasing functions $$f_k:\mathbb{R}\mapsto \mathbb{R}$$ should work, even if they don't necessarily commute between them.
• You may take $n=2$ in the above example. Commented Nov 28, 2019 at 11:33