# Examples of non-unitary isometries on finite dimensional Hilbert spaces?

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?

## 3 Answers

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.

• Thank you, this is what I was looking for – Olivier Mar 12 at 22:40

For all linear operators $$A,B$$ on a finite dimensional vectorspace $$AB=I$$ implies $$BA=I$$. See here.

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.