# Why does the concept of a unitary matrix seem so similar to that of an invertible matrix?

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

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.

• 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? – tparker Jun 9 '20 at 12:51