I was wondering whether it is possible to extend the following standard theorem to infinite dimensional Hilbert spaces?

Let $M\in \mathbb{C}^{n\times n}$ arbitrary. The matrix $M$ is unitary if and only if it is an isometry: $$\forall v\in \mathbb{C}^n: \|v \| = 1 \Rightarrow \| M v\|=1$$

My question is if this holds true for $M$ being a linear operator in some infinite dimensional Hilbert space $\mathcal{H}$. Thank you for your help


EDIT: In my original answer, I forgot all about bijectivity! As such, I have updated my answer to fix this. (Thank you to Dimitar for pointing this out.)

The short answer is almost. The slightly longer answer is that an operator $U$ on a Hilbert space $H$ is unitary if it is surjective and it preserves the inner product---i.e. $\langle Ux,Uy\rangle = \langle x, y\rangle$ for all $x, y \in H$.

Since $\|Ux\|^2 = \langle Ux, Ux \rangle = \langle x, x \rangle = \|x\|^2$, it is clear that if an operator is unitary, then it is an isometry.

The converse is almost true, because of the polarization identity: $\langle x,y \rangle = \frac{1}{4}\left(\|x + y\|^2 - \|x - y\|^2 + i\|x + iy\|^2 - i\|x - iy\|^2\right)$. So, if an operator preserves the norm (which is the meaning of isometry), then it automatically preserves the inner product. The only problem is that surjectivity is not guaranteed---there are isometries that are not surjective!

Here is an example. Let your Hilbert space be the set of sequences of complex numbers $a_0, a_1, a_2, \ldots$ with the condition that $\sum_{n = 0}^\infty |a_n|^2 < \infty$. This is a Hilbert space with the inner product $\langle a_0, a_1, \ldots, b_0, b_1, \ldots\rangle = \sum_{n = 0}^\infty a_n \overline{b_n}$.

Now, consider the map $a_0, a_1, \ldots \mapsto 0, a_0, a_1, \ldots$---it is easy to check that this is an isometry, but it is also clear that it is not surjective!

  • $\begingroup$ Don't we also need to show that $U$ is surjective in the second part? $\endgroup$ – Dimitar Ho Jan 13 '15 at 1:05

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