Let $A$ be a bounded linear operator on a complex infinite dimensional Hilbert space $H$. If $A$ possesses this property $AA^{Tr}=A^{Tr}A=I$, where $A^{Tr}$ is the transpose of $A$ with respect to a fixed but arbitrary orthonormal basis of $H$, and $I$ is the identity operator. What do you call an operator $A$ such that $AA^{Tr}=A^{Tr}A=I$?

  • $\begingroup$ is this unitary? what i know about unitary is this. AA*=AA*=I. But here, we are dealing with the transpose of an operator. Are the transpose and adjoint the same thing for operators? for matrices, the two are different. $\endgroup$ – Dong North Jun 15 '17 at 7:44
  • $\begingroup$ Note that the transpose of an operator (as distinct from the adjoint) over a Hilbert space is not usually defined, so the burden is on you to provide a definition. Also: do you have a preferred term for matrices that satisfy $AA^{Tr} = A^{Tr}A = I$? Note that the term "orthogonal" is usually used on real vector spaces, where the definitions of transpose and adjoint coincide. $\endgroup$ – Omnomnomnom Jun 15 '17 at 8:01
  • $\begingroup$ as far as i know, an operator $A$ for which $AA^{\star}=I$ is orthogonal (over a field R) and an operator $A$ for which $AA^{\star}=I$ is unitary (over C). I really have no idea if an operator has a transpose. $\endgroup$ – Dong North Jun 15 '17 at 8:17
  • $\begingroup$ actually what i am looking for right now is the formal definition of a transpose of a linear operator (if there is). I've read an article about linear preservers and the author used the notation $A^{Tr}$ to denote the transpose of an operator with respect to a fixed but arbitrary orthonormal basis of $H$. I dont know if the transpose $A^{Tr}$ that he's talkin' about is actually an adjoint of an operator $A$, but he just simply used different terminology. I'm a bit confused. Can you please help me on this? $\endgroup$ – Dong North Jun 15 '17 at 8:45

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