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Compact Operators on a Hilbert Space

  1. Jul 12, 2010 #1
    Hello, I hope I am asking this in the right area of the forums. My teacher wrote the following formula down at our last meeting, and I was wondering if it was true ( [tex] \mathcal{H} [/tex] is the infinite dimension separable Hilbert space):

    [tex] \mathcal{K} (\mathcal{H}) \approx \mathcal{K} (\mathcal{H} \oplus \mathcal{H} \oplus \mathcal{H} \oplus \mathcal{H})\approx M_{4} (\mathcal{K} (\mathcal{H})) [/tex]

    I'm pretty sure this is true, but I am worried I am crazy, because I don't understand how every compact operator could secretly be 16 compact operators.

    I think one formula that could aid in the proof of the above isomorphism is:

    [tex] \mathcal{H} \approx \mathcal{H} \oplus \mathcal{H} \oplus \mathcal{H} \oplus \mathcal{H} [/tex]

    which I believe to be true considering that [tex] \mathcal{H} \cong \mathpzc{l}^2 ( \mathbb{N} ) [/tex]. So let

    [tex] \mathcal{W} = \mathpzc{l}^2 (\mathbb{N}) \oplus \mathpzc{l}^2 (\mathbb{N}) \oplus \mathpzc{l}^2 (\mathbb{N}) \oplus \mathpzc{l}^2 (\mathbb{N}) \cong \mathcal{H} \oplus \mathcal{H} \oplus \mathcal{H} \oplus \mathcal{H} [/tex] .

    From this, we take the maximal orthonormal bases [tex] (e_n)_i \text{ of } \mathcal{H}_i = \mathcal{H} \text{ for } i = 1, 2, 3, 4 \text{ and set } (b_n) \subseteq \mathcal{W} [/tex] by

    [tex] (b_1) = (e_1)_1 , (b_2) = (e_1)_2, \ldots , b_4 = (e_1)_4 , b_5 = (e_2)_1 , \ldots [/tex]

    which is a countable maximal orthonormal basis for [tex] \mathcal{W} [/tex]. This shows

    [tex] \mathpzc{l}^2 (\mathbb{N}) \cong \mathcal{W} \cong \mathcal{H} \oplus \mathcal{H} \oplus \mathcal{H} \oplus \mathcal{H} \qedsymbol [/tex]

    but I can't find any proofs like this in any of my textbooks, which makes me feel like I am making a mistake. Can anyone please help me?
     
    Last edited: Jul 12, 2010
  2. jcsd
  3. Jul 12, 2010 #2

    Hurkyl

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    They sound believable. It looks like just another manifestation of the fact a countably infinite set can be split into two disjoint countably infinite sets.
     
  4. Jul 12, 2010 #3
    What you have stated is true, but the point is rather mysterious.

    You can, of course, break any infinite-dimensional separable Hilbert space into any finite (or countable) number of direct summands -- simply look at the closed span of some partition of a Hilbert space basis. This is essentially what you have above.

    The question is, "why ?".
     
  5. Jul 12, 2010 #4
    Yeah, the idea of partitioning the basis makes total sense. For some reason I just hadn't been able to find anyone saying that this statement was true, so I started to get worried. But the reason I was interested in a Hilbert space being equal to the direct sum of 4 had to do with the compact operator question I posted at the top of the page. Thanks for your help.
     
  6. Jul 20, 2010 #5
    that's not right...
     
  7. Jul 20, 2010 #6
    Which part isn't right?
     
  8. Oct 13, 2010 #7
    how to prove that the set of all compact operators on a hilbert space is a vector space.. help..
     
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