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I was reading the question A Finite Dimensional non-Unitary Isometry?, which gives an example of a non unitary isometry which is a map $T: R \rightarrow R^2 $. This question is based on a previous question Difference between an isometric operator and a unitary operator on a Hilbert space, in which there is an example of non-unitary isometry in an infinite-dimensional Hilbert space.

Are there any examples of operators on finite dimensional Hilbert spaces $V: H_A \rightarrow H_A$ which have $V^\dagger V = \mathbb{I}$ but $V V^\dagger \neq \mathbb{I}$, or does isometry imply unitarity in this special case?

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If $V : H \to H$ is an isometry, it is in particular injective. Since $H$ is finite-dimensional, it follows that $V$ is also surjective. Hence $V$ is invertible so multiplying the relation $V^*V= I$ with $V^{-1}$ yields $V^{-1} = V^*$.

Therefore $V^*V = VV^* = I$ so $V$ is unitary.

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  • $\begingroup$ Thank you, this is what I was looking for $\endgroup$ – Olivier Mar 12 at 22:40
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For all linear operators $A,B$ on a finite dimensional vectorspace $AB=I$ implies $BA=I$. See here.

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Very generally, if $X$ is a finite-dimensional vector space and $A,B:X\to X$ are linear maps such that $AB=1$, then $BA=1$. Indeed, if $AB=1$, then $A$ and $B$ must be invertible (consider their determinants), and so $$BA=BA(BB^{-1})=B(AB)B^{-1}=BB^{-1}=1.$$

From a different perspective, a linear isometry $T:X\to Y$ between two Hilbert spaces is just map that is unitary onto its image, i.e. $T$ is unitary when considered as a map $X\to T(X)$. This implies $T(X)$ has the same dimension as $X$. So if $X$ and $Y$ have the same finite dimension, $T(X)$ must be all of $Y$ and so $T$ must actually be a unitary. What's going on in infinite dimensions is that $Y$ can have a proper subspace of the same dimension, but that can't happen in finite diimensions.

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