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Thread: Invertible Matrix

  1. #1
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    Invertible Matrix

    I need to show that if T:\ell^2(\mathbb{N},\mathbb{C})\rightarrow\ell^2(\  mathbb{N},\mathbb{C}) is an operator defined by (Tx)(j)=\frac{1}{2^j}x(j) and \lambda\neq0 is NOT in the set of eigenvalues of T then T-\lambda I is invertible. I also need to find the set of approximate eigenvalues of T and the spectrum of T.

    So far I have argued that since \lambda is not a eigenvalue of T and is not zero, there is no vector x such that Tx=\lambda x which implies that Tx-\lambda x\neq0 and therefore Tx-\lambda Ix=(T-\lambda I)x\neq0 and therefore T-\lambda I\neq0 hence T-\lambda I is non singular and therefore invertible. Is this correct?
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  2. #2
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    Quote Originally Posted by ejgmath View Post
    I need to show that if T:\ell^2(\mathbb{N},\mathbb{C})\rightarrow\ell^2(\  mathbb{N},\mathbb{C}) is an operator defined by (Tx)(j)=\frac{1}{2^j}x(j) and \lambda\neq0 is NOT in the set of eigenvalues of T then T-\lambda I is invertible. I also need to find the set of approximate eigenvalues of T and the spectrum of T.

    So far I have argued that since \lambda is not a eigenvalue of T and is not zero, there is no vector x such that Tx=\lambda x which implies that Tx-\lambda x\neq0 and therefore Tx-\lambda Ix=(T-\lambda I)x\neq0 and therefore T-\lambda I\neq0 hence T-\lambda I is non singular and therefore invertible. Is this correct?
    The set S = \{2^{-j}: j\in\mathbb{N}\}\cup{0} is closed. So if \lambda\notin S then d, the distance from \lambda to S, is greater than 0. The inverse of T-\lambda I is given by ((T-\lambda I)^{-1}(x))(j) = (2^{-j}-\lambda)^{-1}x(j). This is a bounded operator, with \|(T-\lambda I)^{-1}\|\leqslant d. So \lambda is not in the spectrum of T.

    Therefore the spectrum of T consists of the eigenvalues 2^{-j}\  (j\in\mathbb{N}) and the approximate eigenvalue 0.
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  3. #3
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    Thats great, thanks for the help. One question though, is my original arguement for the invertability of T-\lambda I correct?

    Thanks again
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    Quote Originally Posted by ejgmath View Post
    Thats great, thanks for the help. One question though, is my original arguement for the invertability of T-\lambda I correct?
    No, it's not. There are two things wrong with it. First, on an infinite-dimensional space like \ell^2(\mathbb{N},\mathbb{C}), an operator can be injective without being surjective. So the fact that (T-\lambda I)x\neq0 for all x\ne0 does not imply that T-\lambda I is invertible. Second, I'm assuming that you are dealing with bounded operators here. So it's not enough to show that T-\lambda I has an algebraic inverse, you need to show that the inverse is a bounded operator.
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