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Math Help - Hilbert spaces

  1. #1
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    Hilbert spaces

    Let H1 and H2 be Hilbert spaces..If T:H1-->H2 is a bounded linear transformation with M1 closed subspace of H1 and M2 closed subspace of H2 then show that T(M1) is contained in M2 if and only if T*(M2) is contained in M1..

    Any help appreciated.
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  2. #2
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    Quote Originally Posted by rishirich View Post
    Let H_1 and H_2 be Hilbert spaces. If T:H_1\to H_2 is a bounded linear transformation with M_1 closed subspace of H_1 and M_2 closed subspace of H_2 then show that T(M_1) \subseteq M_2 if and only if T^*(M_2) \subseteq M_1.
    That is not correct. It should be: T(M_1) \subseteq M_2 if and only if T^*(M_2^\perp) \subseteq M_1^\perp.

    The reason is that M_2 = M_2^{\perp\perp} and so

    . . . . . . . . . . . \begin{aligned}T(M_1) \subseteq M_2\ &\Longleftrightarrow\ \langle Tx,y\rangle = 0 \quad(x\in M_1,\ y\in M_2^\perp) \\ &\Longleftrightarrow\ \langle x,T^*y\rangle = 0 \quad(x\in M_1,\ y\in M_2^\perp) \\ &\Longleftrightarrow\ T^*(M_2^\perp) \subseteq M_1^\perp .\end{aligned}
    Last edited by Opalg; January 15th 2011 at 09:08 AM. Reason: left out a couple of "perp"s
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  3. #3
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    Hey thank you for your reply..Why is the question wrong?
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  4. #4
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    Quote Originally Posted by rishirich View Post
    Hey thank you for your reply..Why is the question wrong?
    The simplest way to see that it is wrong is to think about what happens if H_2=H_1 and M_2=M_1. Then the question is saying that if a subspace is invariant under  T then it is also invariant under T^*. That is certainly not true. For example, in a 2-dimensional space, the subspace spanned by the second basis vector is invariant under the transformation with matrix \begin{bmatrix}0&1\\0&0\end{bmatrix}, but not under its adjoint.
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