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Inner product-preserving map that isn't unitary?

  1. Feb 11, 2012 #1
    Suppose you've got a linear map U between two Hilbert spaces H1 and H2. If U preserves the inner product - that is, [itex](Ux,Uy)_2 = (x,y)_1[/itex] 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?
     
  2. jcsd
  3. Feb 11, 2012 #2

    morphism

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    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 [itex]\dim H_1 \neq \dim H_2[/itex] (e.g. take [itex]v \mapsto (v,0)[/itex]). If [itex]\dim H_1 = \dim H_2 < \infty[/itex] then U will necessary be onto, by the rank-nullity theorem. But there are counterexamples if [itex]\dim H_1 = \dim H_2 = \infty[/itex], e.g. the forward shift [itex]S \colon \ell^2 \to \ell^2[/itex] defined by [itex]S(x_1,x_2,\ldots) = (0,x_1,x_2,\ldots)[/itex].
     
  4. Feb 11, 2012 #3
    Very good! Letting U be the right shift operator seems to do what I want, since then

    [tex]
    (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),
    [/tex]

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