# if $AA^*=BB^*$ what are the relations between A and B [closed]

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.

## closed as off-topic by Eevee Trainer, Leucippus, Shailesh, Cesareo, NikunjJan 13 at 11:03

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• 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 $B=AV$. – Sangchul Lee Jan 13 at 6:00
• @SangchulLee great! I think I've got my answer. thanks. – 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.