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Linear algebra question

  1. Dec 28, 2008 #1
    the question is in the link:
    http://img517.imageshack.us/img517/3830/70738563la6.gif

    i know thats how i find coordinated(x,y,z) U=x*v1 +y*v2 +z+v3

    but i dont know how to build this structure here?
     
    Last edited: Dec 28, 2008
  2. jcsd
  3. Dec 28, 2008 #2
    In imageshack you see a link below which is named "hotlink for forums", copy THAT link here and the picture will be directly shown in your post.
     
  4. Dec 28, 2008 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    You have a groupm, B, containing three members, [itex]{b_1, b_2, b_3}[/itex] and the vector space of all functions from B to R. (you say only "group of functions" but it must be a vector space for this to make sense.) You are also given a "basis" for that vector space defined by [itex]g_i(b_j)[/itex] equals 1 if i= j, 0 otherwise. You are asked to write g, defined by [itex]g(b_1)= 1[/itex], [itex]g(b_2)= 4[/itex], [itex]g(b_3)= 5.

    Okay, you must have [itex]g(x)= a_1g_1(x)+ a_2g_2(x)+ a_3g_3(x)[/itex]. Set x= [itex]b_1, b_2, b_3[/itex] to get three very simple equations to solve for [itex]a_1, a_2, a_3[/itex].
     
  5. Dec 28, 2008 #4
    [itex]
    g(b_1)= 1
    g(b_2)= 4
    g(b_3)= 5
    g(x)= a_1g_1(x)+ a_2g_2(x)+ a_3g_3(x)
    [/itex]
    [itex]
    x=b1,b2,b3

    [/itex]

    [itex]
    g(b1,b2,b3)= a_1g_1(b1,b2,b3)+ a_2g_2(b1,b2,b3)+ a_3g_3(b1,b2,b3)
    [/itex]

    what is the next step
    for constracting the equations
     
  6. Dec 28, 2008 #5

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    There is NO "[itex]g(b2, b3,b3)[/itex]! Set x equal to b1, b2, and b3 separately to get three equations.
     
  7. Dec 28, 2008 #6
    like this?

    X(1,2,0)+y(0,1,2)+z(0,0,1)=(1,2,5)
     
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