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Math Help - [SOLVED] Linear transform - need a hand

  1. #1
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    [SOLVED] Linear transform - need a hand

    Let T:V->W be a linear transformation with NS(T)={0}. Suppose {u1, u2} is a linearly independent subset of V. Show that {T(u1), T(u2)} is linearly independent.

    My attempt:
    Suppose {T(u1), T(u2)} is linearly dependent, then there exsists c1, c2 =/ 0 (can't find the "is not equal" symbol, sorry) such that:
    c1T(u1) + c2T(u2) = 0
    c1u1 + c2u2 = 0 (as T(0)=0 ) **
    c1=c2=0, as {u1, u2} are linearly indepedent
    Then {T(u1), T(u2)} is also linearly independent as the only solution of c1T(u1) + c2T(u2) = 0 is the trivial solution.

    The part I'm wondering if I've done wrong is at the **. Given NS(T)={0}, instead of having c1u1 + c2u2 = 0, I could have c1u1 + c2u2 = x where x \in V : T(x) = 0 , couldn't I? Then I haven't proved anything at all...
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  2. #2
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    Quote Originally Posted by Cloud_pin View Post
    Let T:V->W be a linear transformation with NS(T)={0}. Suppose {u1, u2} is a linearly independent subset of V. Show that {T(u1), T(u2)} is linearly independent.

    My attempt:
    Suppose {T(u1), T(u2)} is linearly dependent, then there exsists c1, c2 =/ 0 (can't find the "is not equal" symbol, sorry) such that:
    c1T(u1) + c2T(u2) = 0
    c1u1 + c2u2 = 0 (as T(0)=0 ) ** your reasoning here is not correct!
    T(c_1u_1 + c_2u_2)=c_1T(u_1)+c_2T(u_2)=0, which means c_1u_1 + c_2u_2 \in NS(T)=\{0\}. thus c_1u_1 + c_2u_2 = 0.
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  3. #3
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    Quote Originally Posted by NonCommAlg View Post
    T(c_1u_1 + c_2u_2)=c_1T(u_1)+c_2T(u_2)=0, which means c_1u_1 + c_2u_2 \in NS(T)=\{0\}. thus c_1u_1 + c_2u_2 = 0.
    Aaaaah. Thanks oodles!!
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