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Thread: Operator Eigenvalues

  1. #1
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    Operator Eigenvalues

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

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

    Then for the approximate value $\displaystyle \lambda=0$ for a sequence $\displaystyle x_n(j)$ such that $\displaystyle \|x_n\|=1$ this must satisfy $\displaystyle (Tx_n)(j)-\lambda x_n(j)\rightarrow0$ as $\displaystyle n\rightarrow\infty$ but I have no idea what $\displaystyle x_n(j)$ could be?
    Any help would be great
    Last edited by ejgmath; Apr 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 $\displaystyle T:\ell^2(\mathbb{N},\mathbb{C})\rightarrow \ell^2(\mathbb{N},\mathbb{C})$ defined by $\displaystyle (Tx)(j)=\frac{1}{2^j}x(j)$ for all $\displaystyle j$ and show that $\displaystyle \lambda =0$ is an approximate eigenvalue.

    So far I have deduced that since $\displaystyle (Tx)(j)=\frac{1}{2^j}x(j)$ and $\displaystyle \lambda$ must satisfy $\displaystyle Tx=\lambda x$ for $\displaystyle x\in\ell^2$ with $\displaystyle x\neq 0$ that $\displaystyle \lambda_j=\frac{1}{2^j}$, 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 $\displaystyle e_n$ be the vector with a 1 in the n'th coordinate and zeros everywhere else. Then $\displaystyle Te_n = 2^{-n}e_n$. So $\displaystyle e_n$ is the eigenvector corresponding to the eigenvalue $\displaystyle 2^{-n}$.

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