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



> 
> Let $A\_1,\cdots,A\_d$ be pairwise **commuting** operators on $E$. Is the equality
>  $$\left\|\displaystyle\sum\_{k=1}^dA\_k^\*A\_k \right\|=\left\|\displaystyle\sum\_{k=1}^dA\_kA\_k^\* \right\|,$$
>  need not hold? 
> 
> 
> 


Recall that an operator $T\in \mathcal{L}(E)$ is said to be isometry if $T^\*T=I$.


I try to find two **commuting** isometries $A\_1$ and $A\_2$ such that
$A\_1A\_1^\*+A\_2A\_2^\*=I$. However, **this is not possible.**


Indeed, if $A\_1A\_1^\*+A\_2A\_2^\*=I$, multiplying on the LHS by $A\_1^\*$ you get
$$A\_1^\*A\_1A\_1^\*+A\_1^\*A\_2A\_2^\*=A\_1^\* \\
A\_1^\*+A\_1^\*A\_2A\_2^\*=A\_1^\* \\
A\_1^\*A\_2A\_2^\*=0 \\
$$
Multiply on the RHA by $A\_2$ to get 
$$A\_1^\*A\_2=0 \\
A\_1^\*A\_2A\_1 =0$$


Now, since $A\_1,A\_2$ commute you get 
$$A\_1^\*A\_1A\_2=0 \\
A\_2=0$$


Also the same idea shows that it is not possible for three commuting isometies $A\_1,A\_2,A\_3$ to get $A\_1A\_1^\*+A\_2A\_2^\*+A\_3A\_3^\*=I$.



$\sum_{k=1}^dA_k^*A_k$ and $\sum_{k=1}^dA_kA_k^*$ have the same norms if $A_k$ are commuting
No. 3 copies of Hilbert spaces $H\_1,H\_2,H\_3$. $A\_1$ a partial isomtry copying $H\_1$ to $H\_2$, and $A\_2$ a partial isometry copying $H\_1$ to $H\_3$. Then $A\_1 A\_2 =A\_2A\_1 =0$. But $\|A\_1^\* A\_1 + A\_2^\* A\_2\| = 2 \neq \|A\_1 A\_1^\* + A\_2 A\_2^\*\| =1$.


