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

<br /> (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),<br />

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