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?


1 Answer 1


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, 2020 at 12:51

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