I'm wondering if we have two linear operators $A, B \in \ell(V)$. and we know that $AA^*=BB^*$. then what informations can this give to us about relationships between $A$ and $B$?

I think they have some strong relations.


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    $\begingroup$ In view of polar decomposition, it tells that $A=PU$ and $B=PU'$ for $P = \sqrt{AA^*} = \sqrt{BB^*}$ and $U, U'$ some unitary operators. So, there exists a unitary operators $V$ such that $B=AV$. $\endgroup$ – Sangchul Lee Jan 13 at 6:00
  • $\begingroup$ @SangchulLee great! I think I've got my answer. thanks. $\endgroup$ – Peyman mohseni kiasari Jan 13 at 6:06

in finite dimension by polar decomposition, $A = \sqrt{AA^*}U_a, B = \sqrt{BB^*}U_b$, where the $U_a$ and $U_b$ are unitary operators. so we heve:

$\sqrt{BB^*} = \sqrt{AA^*} = AU^*_a = BU^*_b \Rightarrow A = BU^*_bU_a$

$U^*_bU_a(U^*_bU_a)^* = U^*_bU_aU^*_aU_b = I \Rightarrow U = U^*_bU_a$ is unitary

$\Rightarrow A = BU$ where $U$ is an unitary operator.

notice that if $A = BU$ then $A^* = U^*B^*$ so if we had $A^*A = B^*B$ then $A = U'B$ where $U'$ is an unitary operator.

  • $\begingroup$ This is not true in general, only in finite dimension. In general you cannot put a unitary in the polar decomposition, just a partial isometry. $\endgroup$ – Martin Argerami Jan 13 at 23:53
  • $\begingroup$ @MartinArgerami thank you, it is edited. actually, my studies are still in finite dimension and I didn't know that. $\endgroup$ – Peyman mohseni kiasari Jan 13 at 23:56

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