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Math Help - uniqueness of linear map

  1. #1
    Member
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    uniqueness of linear map

    I have a set of 4 vectors (a1,a2,a3,a4) in R3 and a set of 4 vectors (b1,b2,b3,b4)in R4. I need to show that there is precisely one linear map f: R3 --> R4 with f(ai) = bi


    a1 = (1,0,0)
    a2= (0,1,0)
    a3 = (0,0,1)
    a4 = (2,1,3)
    b1 = (1,2,4,1)
    b2 = (1,1,0,1)
    b3=(-1,0,4,-1)
    b4 = (0,5,20,0)

    I have found the linear map (x+y-z,2x+y,4x+4z,x+y-z)
    how would I show that this is unique? I then need to find the kernel and the image of f.. would the image just be (x+y-z,2x+y,4x+4z,x+y-z)

    and the kernel would be (1,-2,-1)^T ?

    many thanks
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  2. #2
    Senior Member JaneBennet's Avatar
    Joined
    Dec 2007
    Posts
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    Let g:\mathbb{R}^3\to\mathbb{R}^4 be any linear transformation satisfying g(a_i)=b_i for i=1,2,3,4.

    Given any vector \mathbf{u}=(x,y,z)\in\mathbb{R}^3, we have

    g(\mathbf{u})\ =\ g(xa_1+ya_2+za_3)

    ___... =\ xg(a_1)+yg(a_2)+zg(a_3)

    ___... =\ xf(a_1)+yf(a_2)+zf(a_3)

    ___... =\ f(xa_1+ya_2+za_3)

    ___... =\ f(\mathbf{u})

    Hence g(\mathbf{u})=f(\mathbf{u}) for all u\in\mathbb{R}^3, proving the uniqueness of f.

    and the kernel would be (1,-2,-1)^T ?
    You need to write it down properly.

    \ker{f}\ =\ \{t(1,-2,-1)\in\mathbb{R}^3:t\in\mathbb{R}\}

    would the image just be (x+y-z,2x+y,4x+4z,x+y-z)
    \mathrm{image}\,{f}\ =\ \{(x+y-z,2x+y,4x+4z,x+y-z)\in\mathbb{R}^4:x,y,z\in\mathbb{R}\}

    Write (x+y-z,2x+y,4x+4z,x+y-z)=(u,v,w,u) and express v and w in terms of u.
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