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Math Help - Linear Transformation

  1. #1
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    Linear Transformation

    Let T: V -> W be a linear transformation, and let the dimension of V be n, Let { v_1, v_2, ..., v_m} be a basis for the kernel of T.

    1) Let V_(m+1), v_(m+2), ..., v_n be vectors in V such that { v_1, v_2, ..., v_m, v_(m+1), ...,  v_n} in a basis for V. Prove that {Tv_(m+1), Tv_(m+2), ..., Tv_n} is a basis for the image of T.

    2) Using 1), prove that the rank of T plus the nullity of T is n.

    Thanks!
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  2. #2
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    Quote Originally Posted by cindy View Post
    Let T: V -> W be a linear transformation, and let the dimension of V be n, Let { v_1, v_2, ..., v_m} be a basis for the kernel of T.
    It is safe to assume m\geq 1.

    1) Let V_(m+1), v_(m+2), ..., v_n be vectors in V such that { v_1, v_2, ..., v_m, v_(m+1), ...,  v_n} in a basis for V. Prove that {Tv_(m+1), Tv_(m+2), ..., Tv_n} is a basis for the image of T.
    Let \bold{a} be in the image of T.
    Then \bold{a} = k_1 \bold{v}_1 + ... + k_m \bold{v}_m + k_{m+1} \bold{v}_{m+1} + ... + k_n \bold{v}_n .
    Thus, T\bold{a} = k_1 T\bold{v}_1 + ... + k_m T\bold{v}_m + k_{m+1}T \bold{v}_{m+1} + ... + k_n T\bold{v}_n = k_{m+1}T \bold{v}_{m+1} + ... + k_n T\bold{v}_n.
    Thus, T\bold{v}_{m+1}, ... T\bold{v}_n spams the image of T.

    You need to show T\bold{v}_{m+1}, ... T\bold{v}_n is linearly independent.
    If a_{m+1}T\bold{v}_{m+1} + ... + a_n T\bold{v}_n = \bold{0}
    Then T(a_{m+1}\bold{v}_{m+1} + ... + a_n \bold{v}_n) = \bold{0}.
    Thus, a_{m+1}\bold{v}_{m+1} + ... + a_n \bold{v}_n  \in \ker T.

    We can therefore write, a_{m+1}\bold{v}_{m+1} + ... + a_n \bold{v}_n = a_1\bold{v}_1 + ... + a_2 \bold{v}_m.
    This implies a_1=a_2= ... 0 because those vectors form a basis.
    ---
    Then #2 follows because m + (n-m) = n.
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