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Math Help - Linear transformation problem

  1. #1
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    Linear transformation problem

    Hi! Have some troubles with solving one problem in my assignment, would be great if somebody could help me. The problem is following:
    Let n>0 and let Ln,k , whear 1<=k<=n be a subspace spanned by vectors ej for j=1...k. Show that it exist a linear transformation T:Rn=>R(n-k) such as Ln,k=Ker(T)
    Last edited by bogdano; September 18th 2013 at 09:22 AM.
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  2. #2
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    Re: Linear transformation problem

    You can always find a basis for L_{n,k} contained in \{e_j\}_{j=1}^k then extend it to a basis for R^n. Define T(e_i)= 0 for e^i\in \{e^j\}_{j=1}^k, T(v)= 0 for v in the extended basis. T(v) is defined "by linearity" for all other v in R^n. That is, since \{e_1, e_2,..., e_k, f_1, f_2, ..., f_{n-k}\}, we can write v= a_1e_1+ a_2e_2+ ...+ a_ke_k+ b_1f_1+ b_2f_2+ ...+ b_{n-k}f_{n-k} and then Tv= a_1T(e_1)+ a_2T(e_2)+ ...+ a_kT(e_k)+ b_1T(f_1)+ b_2T(f_2)+ ...+ b_{n-k}T(f_{n-k})= a_1+ a_2+ ...+ a_k. Then Tv= 0 if and only if v is in the subspace.
    Last edited by HallsofIvy; September 18th 2013 at 09:56 AM.
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  3. #3
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    Re: Linear transformation problem

    Thanx for answe! Could you please spicify what do you mean by saying that Tv=0, is it a transformation of v? What is v if it can be represented as v=a(1)e(1)+....b(n-k)*f(n-k). And if you said that T(e_j)=0, how can Tv be equal to a_1+a_2+....+a_k if the all supposed to be transformed into null!?
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