Inner product-preserving map that isn't unitary?

[AxiomOfChoice]

Suppose you've got a linear map U between two Hilbert spaces H1 and H2. If U preserves the inner product - that is, $(Ux,Uy)_2 = (x,y)_1$ for all x and y in H1 - is it necessarily unitary? Or are there inner product-preserving linear mappings that aren't one-to-one or onto?

 

[morphism]

If U preserves inner products, it will preserve norms, so it will always be one-to-one.

Thus the question is: must U be onto? The answer is no, and it's easy to give examples if $\dim H_1 \neq \dim H_2$ (e.g. take $v \mapsto (v,0)$). If $\dim H_1 = \dim H_2 < \infty$ then U will necessary be onto, by the rank-nullity theorem. But there are counterexamples if $\dim H_1 = \dim H_2 = \infty$, e.g. the forward shift $S \colon \ell^2 \to \ell^2$ defined by $S(x_1,x_2,\ldots) = (0,x_1,x_2,\ldots)$.

 

[AxiomOfChoice]

If U preserves inner products, it will preserve norms, so it will always be one-to-one.

Thus the question is: must U be onto? The answer is no, and it's easy to give examples if $\dim H_1 \neq \dim H_2$ (e.g. take $v \mapsto (v,0)$). If $\dim H_1 = \dim H_2 < \infty$ then U will necessary be onto, by the rank-nullity theorem. But there are counterexamples if $\dim H_1 = \dim H_2 = \infty$, e.g. the forward shift $S \colon \ell^2 \to \ell^2$ defined by $S(x_1,x_2,\ldots) = (0,x_1,x_2,\ldots)$.

Very good! Letting U be the right shift operator seems to do what I want, since then

$$(Ux,Uy) = (\{0,x_1,x_2,\ldots\},\{0,y_1,y_2,\ldots\}) = 0 + \sum_{n = 1}^\infty x_n y_n^* = (x,y),$$

but the mapping certainly isn't onto, since (e.g.) $\{1,0,0,\ldots\} \notin {\rm Ran} U$. Thanks!