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Math Help - A prove on linear mapping and basis and range

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    A prove on linear mapping and basis and range

    Hi everyone , can anyone of you help me with this question ?
    My background in Algebra is weak
    Thanks a lot guys

    Let T : V -> W be a linear mapping and let {v1, v2, ..., vn} be a basis for V such that {vr+1,...,vn} is a basis for Null(T). Prove that {T(v1),...,T(vr) is a basis for the range of T.

    clear explanation is appreciated.
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    Quote Originally Posted by dull1234 View Post
    Hi everyone , can anyone of you help me with this question ?
    My background in Algebra is weak
    Thanks a lot guys

    Let T : V -> W be a linear mapping and let {v1, v2, ..., vn} be a basis for V such that {vr+1,...,vn} is a basis for Null(T). Prove that {T(v1),...,T(vr) is a basis for the range of T.

    clear explanation is appreciated.
    First, let y be in the range of T. Then y= Tx for some x in V. Since {v1, v2, ..., vn} is a basis for V, x= a1v1+ a2v2+ ...+ anvn for some numbers a1, a2, ..., an. Then Tx= a1Tv1+ a2Tv2+ ...+ an Tvn. But T(vr+1)= 0, T(vr+2)= 0, ... T(vn)= 0 so y= Tx= a1T(v1)+ a2Tv2+ ...+ arTvr. That shows that T(vr) spans the range of T.

    To show independence, assume that b1T(v1)+ b2T(v2)+ ...+ brT(vr)= 0.

    Then T(b1v1+ b2v2+ ...+ brvr)= 0. Since {vr, vr+1, ..., vn} is a basis for Null(T), and b1v1+ b2v2+ ...+ brvr is in Null(T), we must have b1v1+ b2v2+ brvr= cr+1 vr+1+ ...+ cnvn. Show, using the fact that {v1, v2, ..., vn} is a basis, that all of b1, b2, ..., br, cr+1, ...cn must be 0.
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