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Let's say, that we have an $n \times n$ complex matrix called $X$, which is an isometry. That means that it fulfils the $X^\dagger X = \mathbf{1}$ condition. I was wondering, if it is possible for $X$ to not be unitary at the same time? That is $X X^\dagger \neq \mathbf{1}$?

I know, that it's possible to find such $n \times m$ matrices, where $n \neq m$, but all of the ones with equal dimensions that I've found so far were also unitary.

brzepkowski
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    For a square matrix $X,$ $X^\dagger X = \mathbf{1}\Rightarrow XX^\dagger= \mathbf{1}$ because [$YX=\mathbf{1}\Rightarrow Y=X^{-1}.$](https://math.stackexchange.com/questions/3852/if-ab-i-then-ba-i) – Anne Bauval Nov 12 '22 at 10:19

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