Results 1 to 4 of 4

Math Help - Invertible Matrix

  1. #1
    Junior Member
    Joined
    Jan 2010
    Posts
    32

    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?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    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.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jan 2010
    Posts
    32
    Thats great, thanks for the help. One question though, is my original arguement for the invertability of T-\lambda I correct?

    Thanks again
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    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.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Invertible matrix.
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: March 1st 2011, 07:35 PM
  2. invertible matrix
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: November 19th 2010, 04:47 AM
  3. Invertible matrix
    Posted in the Math Challenge Problems Forum
    Replies: 9
    Last Post: July 1st 2010, 03:49 AM
  4. Nilpotent matrix Invertible matrix
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: January 6th 2010, 12:08 PM
  5. Invertible Matrix
    Posted in the Advanced Algebra Forum
    Replies: 5
    Last Post: September 17th 2009, 07:05 PM

Search Tags


/mathhelpforum @mathhelpforum