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Suppose that $A_1, \dots , A_n, B_1, \dots , B_n \in \Bbb C^{d \times d}$ and that, for every $x \in \mathbb{C}^n$, the following holds $$\left( \sum_i x_i A_i \right) \left( \sum_i x_i A_i \right)^{\dagger} = \left( \sum_i x_i B_i \right) \left( \sum_i x_i B_i \right)^{\dagger} $$

where $\dagger$ denotes the adjoint (i.e., transpose conjugate) of a matrix. Is it true that there exists a unitary matrix $\Sigma \in {\rm U}(d, \mathbb{C})$ such that $A_i = \Sigma B_i$ for all $i$?

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  • $\begingroup$ Maybe $B=\Sigma A$? $\endgroup$
    – Fedor Petrov
    Commented May 2, 2023 at 10:13
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    $\begingroup$ When we set $x_i=1$ and $x_j=0$ for $j\neq i$, we should get $A_iA_i^*=B_iB_i^*$. This implies that $A_i=B_iU_i$ for some unitary $U_i$. $\endgroup$
    – Joseph Van Name
    Commented May 2, 2023 at 16:46
  • $\begingroup$ @JosephVanName I understand this, but I would like to get the same $\Sigma$ ($U$ in your notation) for all $i$. This is stronger, but I think it is likely to be true: for $d=1$ (i.e., when $A_i, B_i \in \mathbb{C}$) it can be seen easily. $\endgroup$
    – user493645
    Commented May 3, 2023 at 6:51

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