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Math Help - Operator Eigenvalues

  1. #1
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    Operator Eigenvalues

    I need find all the eigenvalues of the operator T:\ell^2(\mathbb{N},\mathbb{C})\rightarrow \ell^2(\mathbb{N},\mathbb{C}) defined by (Tx)(j)=\frac{1}{2^j}x(j) for all j and show that \lambda =0 is an approximate eigenvalue.

    So far I have deduced that since (Tx)(j)=\frac{1}{2^j}x(j) and \lambda must satisfy Tx=\lambda x for x\in\ell^2 with x\neq 0 that \lambda_j=\frac{1}{2^j}, is this correct?

    Then for the approximate value \lambda=0 for a sequence x_n(j) such that \|x_n\|=1 this must satisfy (Tx_n)(j)-\lambda x_n(j)\rightarrow0 as n\rightarrow\infty but I have no idea what x_n(j) could be?
    Any help would be great
    Last edited by ejgmath; April 30th 2010 at 09:11 AM.
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  2. #2
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    Quote Originally Posted by ejgmath View Post
    I need find all the eigenvalues of the operator T:\ell^2(\mathbb{N},\mathbb{C})\rightarrow \ell^2(\mathbb{N},\mathbb{C}) defined by (Tx)(j)=\frac{1}{2^j}x(j) for all j and show that \lambda =0 is an approximate eigenvalue.

    So far I have deduced that since (Tx)(j)=\frac{1}{2^j}x(j) and \lambda must satisfy Tx=\lambda x for x\in\ell^2 with x\neq 0 that \lambda_j=\frac{1}{2^j}, is this correct?
    That is correct as far as the eigenvalue is concerned, but you need to be a bit more precise about what the corresponding eigenvector is. let e_n be the vector with a 1 in the n'th coordinate and zeros everywhere else. Then Te_n = 2^{-n}e_n. So e_n is the eigenvector corresponding to the eigenvalue 2^{-n}.

    Quote Originally Posted by ejgmath View Post
    Then for the approximate value \lambda=0 for a sequence x_n(j) such that \|x_n\|=1 this must satisfy (Tx_n)(j)-\lambda x_n(j)\rightarrow0 as n\rightarrow\infty but I have no idea what x_n(j) could be?
    Take x_n to be the same vector e_n described above.
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