If $A\_i$ are commuting normal matrices, then


$$\displaystyle\sup\_{\|y\|=1}\bigg(\displaystyle\sum\_{i=1}^n|\langle A\_iy,y\rangle|^2\bigg)=\displaystyle\sup\_{\|y\|=1}\bigg(\displaystyle\sum\_{i=1}^n\|A\_iy\|^2\bigg)\;.$$



> 
> Is the converse of this result true?
> 
> 
> 



Converse of a result related to normal matrices
no, just take $A\_1$ and $A\_2$ a pair of unitary operators with one common eigenvector $x\_0$; the right-hand side of the equation equals $2$ for all $x$ and the left-hand side reaches this maximum of $2$ for $x=x\_0$, but there is no reason why $A\_1$ and $A\_2$ should commute.




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example:
$$A\_1=\left(
\begin{array}{ccc}
 1 & 0 & 0 \\
 0 & \cos \phi\_1 & \sin \phi\_1 \\
 0 & -\sin \phi\_1 & \cos \phi\_1 \\
\end{array}\right),\;\;A\_2=\left(
\begin{array}{ccc}
 1 & 0 & 0 \\
 0 & \cos \phi\_2 & \sin \phi\_2 \\
 0 & \sin \phi\_2 & -\cos \phi\_2 \\
\end{array}\right)$$
single common eigenvector is $(1,0,0)$ and they do not commute:
$$A\_1 A\_2 - A\_2 A\_1=\left(
\begin{array}{ccc}
 0 & 0 & 0 \\
 0 & 2 \sin \phi\_1 \sin \phi\_2 & -2 \cos \phi\_2 \sin \phi\_1 \\
 0 & -2 \cos \phi\_2 \sin \phi\_1 & -2 \sin \phi\_1\sin \phi\_2 \\
\end{array}
\right)$$


