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Math Help - Using the Dimension Theorem...

  1. #1
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    Using the Dimension Theorem...

    Let L:R^{3}\rightarrow R^{2} be the linear mapping defined by L\left(x,y,z\right)=\left(x-y,y-z\right).
    Find the standard 2x3 matrix representation of L and compute L(1,1,1).
    Show that the kernel of L is a subspace of R^{3}. What is the dimension of the kernel?
    State the Dimension Theorem. Hence or otherwise find the image of R^{3} under the transformation L.

    So I have L = \left( \begin{array}{ccc}1 & -1 & 0 \\ 0 & 1 & -1 \end{array} \right). L(1,1,1)=(0,0).
    The kernel of L has basis (1,1,1) and has dimension 1, and I've proven it is a subspace of R^{3}.

    My problem is with the final piece of the question.
    The Dimension Theorem states that given a vector space, any two bases have the same cardinality...

    How am I supposed to use that to find the image of R^{3} under L?


    Many thanks...
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  2. #2
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    Quote Originally Posted by Unenlightened View Post
    Let L:R^{3}\rightarrow R^{2} be the linear mapping defined by L\left(x,y,z\right)=\left(x-y,y-z\right).
    Find the standard 2x3 matrix representation of L and compute L(1,1,1).
    Show that the kernel of L is a subspace of R^{3}. What is the dimension of the kernel?
    State the Dimension Theorem. Hence or otherwise find the image of R^{3} under the transformation L.

    So I have L = \left( \begin{array}{ccc}1 & -1 & 0 \\ 0 & 1 & -1 \end{array} \right). L(1,1,1)=(0,0).
    The kernel of L has basis (1,1,1) and has dimension 1, and I've proven it is a subspace of R^{3}.

    My problem is with the final piece of the question.
    The Dimension Theorem states that given a vector space, any two bases have the same cardinality...

    How am I supposed to use that to find the image of R^{3} under L?


    Many thanks...
    1. Convince yourself that Img(L) is a sub-space of R^2.
    2. Now note (1,0) and (0,1) belong to Img(L). This Img(L) contains the sub-space spanned by (1,0) and (0,1)
    3. (1) and (2) above give Img(L) = R^2

    Other wise use rank-nullity theorem to get dim(Img(L)) = 2. As dimension of R^2 = 2 and Img(L) is a sub-space of R^2. Img(L)=R^2 is an immediate consequence.
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