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Math Help - Image of linear operator on finite vector space

  1. #1
    Senior Member vincisonfire's Avatar
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    Image of linear operator on finite vector space

    Suppose that T is a linear operator on the vector space V (over the field
    F ). Then for any natural number m,  Im(T^{m+1}) \leq Im(T^m) .
    If V is finite-dimensional, then  Im(T^{m+1}) = Im(T^m) must happen for some natural number m.
    Can you give me a counterexample for the case where V is infinite-dimensional? (that is the image is decreasing in size for ever)
    Thanks.
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    Senior Member vincisonfire's Avatar
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    I just think of one ... polynomials with T = d/dx.
    Any more inventive ideas?
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    Quote Originally Posted by vincisonfire View Post
    I just think of one ... polynomials with T = d/dx.
    Why does this work?

    Any more inventive ideas?
    I came up with a one that is a little strange.
    Consider, \mathbb{R}^{\infty} = \mathbb{R}\times \mathbb{R}\times ... - an infinite tuple.
    Okay, I am abusing notation but I think you have an idea of what I mean.
    Now, \mathbb{R}^{\infty} is a vector space over \mathbb{R} in a natural way.

    Define, T: \mathbb{R}^{\infty} \to \mathbb{R}^{\infty} by T(a_1,a_2,a_3,...) = (0,a_1,a_2,a_3,...).
    Notice that T(\bold{a}+\bold{b}) = T(\bold{a}) + T(\bold{b}) and T(a\bold{a}) = aT(\bold{a}).
    Thus, T is a linear transformation.

    Define, \mathbb{R}^{\infty}_1 = \{ (0,r_1,r_2,r_3,...)| r_j \in \mathbb{R}\}.
    Define, \mathbb{R}^{\infty}_2 = \{(0,0,r_1,r_2,r_3,...)|r_j \in \mathbb{R}\}.
    Define, \mathbb{R}^{\infty}_3 = \{(0,0,0,r_1,r_2,...)|r_j \in \mathbb{R}\}.
    And in general, \mathbb{R}^{\infty}_n for n\geq 1.

    Notice that \mathbb{R}^{\infty}_{n+1} \subset \mathbb{R}^{\infty}_n.
    However, \text{im}(T^k) = \mathbb{R}^{\infty}_k.
    Thus, \text{im}(T^{k+1}) \subset \text{im}(T^k).
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    Senior Member vincisonfire's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    Why does this work?
    No sorry, I was thinking of the increasing Kernel.
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    Quote Originally Posted by ThePerfectHacker View Post
    Consider, \mathbb{R}^{\infty} = \mathbb{R}\times \mathbb{R}\times ... - an infinite tuple.
    Okay, I am abusing notation but I think you have an idea of what I mean.
    Now, \mathbb{R}^{\infty} is a vector space over \mathbb{R} in a natural way.

    Define, T: \mathbb{R}^{\infty} \to \mathbb{R}^{\infty} by T(a_1,a_2,a_3,...) = (0,a_1,a_2,a_3,...).
    Notice that T(\bold{a}+\bold{b}) = T(\bold{a}) + T(\bold{b}) and T(a\bold{a}) = aT(\bold{a}).
    Thus, T is a linear transformation.
    In operator theory, this is called the unilateral shift. It is the device used by the manager of the Hilbert Hotel to ensure that an empty room can always be guaranteed, no matter how many guests arrive.
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