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 $N\_1=N\_2?$
> 
> 
> 



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


