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Math Help - Rank-nullity theorem

  1. #1
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    Rank-nullity theorem

    This may be a stupid question, but when I was studying the rank-nullity theorem, I was wondering about the next problem:

    The rank-nullity theorem states, for a linear map T: V --> W:
    "dim(Im T) + dim(Ker T) = dim V"
    But what happens if you, for example, have a linear map from R to R. Then the dimension of the image is 3 and the dimension of V is 2. But that means that the dimension of the kernel has to be -1, and that's not possible, is it?

    I hope someone can help me out with this.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by Wottie View Post
    But what happens if you, for example, have a linear map from R to R. Then the dimension of the image is 3 and the dimension of V is 2.
    Impossible. If B=\left\{{e_1,e_2}\right\} is a basis of \mathbb{R}^2 , then

    S=\left\{{f(e_1),f(e_2)}\right\}

    is a generator system of \textrm{Im}f so,

    \dim \textrm{Im}f\leq 2 .


    Fernando Revilla
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  3. #3
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    Quote Originally Posted by Wottie View Post
    This may be a stupid question, but when I was studying the rank-nullity theorem, I was wondering about the next problem:

    The rank-nullity theorem states, for a linear map T: V --> W:
    "dim(Im T) + dim(Ker T) = dim V"
    But what happens if you, for example, have a linear map from R to R. Then the dimension of the image is 3
    No, that's impossible. Are you clear on what the "image" of a linear map is? For T:V-> W, the image of T is a subspace of W but not necessarily all of W.

    and the dimension of V is 2. But that means that the dimension of the kernel has to be -1, and that's not possible, is it?
    No, it is not possible- and so you have proved that the dimension of the image of T cannot be larger than the dimension of V (but can be less).

    I hope someone can help me out with this.
    Last edited by HallsofIvy; January 16th 2011 at 07:09 AM.
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