I understand that the intuition behind a unitary operator is that it preserves the length of the vector it acts upon. Also $U^\dagger U = I$. Doesn't that just mean that $U$ is just an invertible operator which preserves distance? And that $U^\dagger$ is the inverse?

If not, where am I making my mistake in reasoning? How should I rewire my thinking to avoid this flawed paradigm?

If so, does that then mean that all distance preserving and invertible operators are also normal operators?


This is one of the definitions of a unitary $-$ a (bounded) linear surjective map that preserves distances (the inner product). From this it can be deduced that $U$ is invertible and $U^{-1} = U^\dagger$.

Also, it's enough to require the preservation of the lengths of all vectors (instead of all inner products) due to Polarization identity

Unitaries are normal because $U^\dagger U = U U^\dagger$, clearly.

  • 1
    $\begingroup$ Nice getting the requirement of surjectivity - people often miss that. But doesn't boundedness follow automatically from the norm-preserving property? Is that why you put "(bounded)" in parentheses? $\endgroup$ – tparker Jun 9 '20 at 12:51

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