Let $E$ be a complex Hilbert space and $\mathcal{L}(E)$ be the algebra of all 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 if the operators $A_k$ are commuting?

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 $A_1A_1^*+A_2A_2^*+A_3A_3^*=I$.