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Homework Help: Linear Combinatio

  1. Nov 29, 2007 #1
    1. The problem statement, all variables and given/known data
    We know that B = { (1, -1, 2, 5) , (2,-3,-1,6) } is a basis of H.

    Express v as a linear combination of b(one) and b(two)

    Where
    v = (0,2,-6,8)
    b(one) = (1,-1,2,5)
    b(two) = (2,-3,-1,6)


    2. Relevant equations



    3. The attempt at a solution

    is it just

    v = tb(one) + pb(two)

    where t , p are parameters
    ?
    Do i need to solve for t and p?
     
    Last edited: Nov 29, 2007
  2. jcsd
  3. Nov 29, 2007 #2
    The second part of the question is:
    Fine the co-ordinate vector [v] of v relative to the basis B

    and I don't have a clue how to do this. Do i need to kno what's the linear transformation?
     
  4. Nov 29, 2007 #3

    HallsofIvy

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    For the first part of the equation, you do the algebra!
    You want t(1, -1, 2, 5)+ p(2, -3, -1, 6)= (0,2,-6,8) . That is,
    (2p- t, -t-3p, 2t- p, 5t+ 6p)= (0, 2, -6, 8). Solve for t and p.

    That is, of course, 4 equations for 2 values, p and t. The given basis vectors span a 2 dimensional subspace of R4. IF the given (0, 2, -6, 8) is in that subspace then this problem will have an answer.

    The "coordinate vector of v relative to the basis B" is just the vector with those two numbers (p, t).
     
  5. Nov 29, 2007 #4
    Yea i've tried solving for the variables p and t, and it doesnt work out. When I try solving it by augmented coefficient matrix and reducing it to row echelon form, I get a row of zeros and then a number ( [ 0 0 | # ] )
    Then i tried solving it using the TI-83+, and I get an "Invalid Dim" error
     
  6. Nov 30, 2007 #5

    HallsofIvy

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    Science Advisor

    Yes, if you have copied the numbers correctly, then the problem is that v is not in the subspace, H, spanned by the two given basis vectors! The four equations you get trying to "fit" it in are t+ 2p= 0, -t- 3p= 2, 2t-p= -6, 5t+ 6p= 8.
    From the first, t= -2p. Then the second becomes 2p- 3p= -p= 2 so p= -2 and then t= 4. Putting that into the third equation, 2t-p= 8+ 2= 10, not -6. Are you sure you haven't lost a sign somewhere? If that third equation were 2t+ p= 6 or -2t- p= -6, then it would work.
     
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