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A unitary operator is a surjective linear operator between complex inner product spaces, which preserves the inner product.

  1. What is the name of the analogue for the real case? Orthogonal operator does not necessarily have to be surjective as far as I know.

  2. Is there a name for an operator of a form $\lambda U$, where $\lambda$ is a scalar and $U$ is a unitary/real-case-unitary? I was thinking about a term "conformal"...

  3. Are there any standard (but not trivial) cryteria for an operator to be of the type described in the question 2 (preferably a reference with a list of equivalent conditions)? It is not very difficult to find some geometric conditions, which rely on the fact that any other operator maps some circle into an ellipse, and this ruins the relations between angles, lengths of vectors etc, but it was a bit cumbersome to do essentially the same thing several times.

Thank you.

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  1. A short computation shows that a linear map $A$ between real Hilbert spaces is surjective and preserves inner products if and only if its complexification defined by $A(v + iw) = Av + iAw$ has the same property. In both settings an equivalent condition is that $A^*A = AA^* = I$. So the correct term in the real case is of course "real unitary".

  2. I would just call it "scalar multiple of a unitary".

  3. $A$ is a scalar multiple of a unitary if and only if $A^*A = AA^* = \mu I$ for some $\mu \in [0,\infty)$. (Then $A = \lambda U$ for some unitary $U$ and $\lambda = u\sqrt{\mu}$ for some $|u| = 1$.)

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  • $\begingroup$ Thank you for 1. As for 2, "scalar multiple of a unitary" is a bit too mouthful; I am looking for something concise, since I am writing a paper filled with operators of this type. As for 3, I was hoping for some reference with a list of cryteria; off-course I was aware of the one that you have stated. Probably I should make it more clear in my question. Thank you again. $\endgroup$
    – erz
    Commented Jun 7, 2016 at 0:29

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