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Linear Algebra - Minimize the Norm

  1. Apr 9, 2008 #1
    1. The problem statement, all variables and given/known data
    In R4, let U = span((1, 1, 0, 0), (1, 1, 1, 2)). Find u in U such that ||u - (1, 2, 3, 4)|| is as small as possible.

    2. Relevant equations

    3. The attempt at a solution
    I came up with a vector u = (-.5, -.5, 0, 0) + (2, 2, 2, 4) = (1.5, 1.5, 2, 4). Then u - (1, 2, 3, 4) = (0.5, -0.5, -1, 0). Using the standard dot product as the norm, I get ||(0.5, -0.5, -1, 0)|| = sqrt(.25 + .25 + 1) = sqrt(1.5).

    This was probably a bad approach, but I can't think of any other real way to do this. Is this the right answer at least? If not, how could I approach this in a better manner other than just randomly picking vectors? Thanks.
     
  2. jcsd
  3. Apr 9, 2008 #2
    Try playing around with orthogonal projections onto U.
     
  4. Apr 9, 2008 #3
    Duh... :rolleyes: Thanks. So now I want to find a unit vector which spans U, but how do I do that? Once I find the unit vector, I can find the orthogonal projection P(u), and then the minimal distance will be ||u - P(u)||.
     
  5. Apr 9, 2008 #4
    Since U is two dimensional you will not find a single vector spanning U. If you are talking about "the" normal vector to U, then I have to tell you that the orthogonal complement of U is two-dimensional as well, so there will be no unique normal vector (even after normalization.)

    The rest is right, ||u-Pu|| will be the distance you are looking for.
     
  6. Apr 9, 2008 #5
    Ok, I did what you said, and ended up with P(u) = (3/2)(1, 1, 0, 0) + 2(1, 1, 1, 2), which was what I originally had, so I guess I was right. Thanks for your help!
     
  7. Apr 10, 2008 #6
    I was just writing up my work today, and I found that I must have messed up somewhere. Here's how I went about the problem:

    P(u) = <u, e1>e1 + <u, e2>e2
    ||e1|| = sqrt(12 + 12) = sqrt(2) => e1 = (1/sqrt(2))*(1, 1, 0, 0).
    ||e2|| = sqrt(1 + 1 + 1 + 4) = sqrt(7) => e2 = (1/sqrt(7))*(1, 1, 1, 2).
    P(u) = (1/2)[(1*1) + (1*2) + (0*3) + (0*4)]e1 + (1/7)[(1*1) + (1*2) + (1*3) + (1*4)]e2
    = (3/2)*(1, 1, 0, 0) + 2*(1, 1, 1, 2) = (3.5, 3.5, 2, 4)

    But when I calculate ||P(u) - (1, 2, 3, 4)|| I get sqrt(10), which is higher than my random guess of sqrt(1.5). Where did I go wrong?
     
  8. Apr 10, 2008 #7
    the coefficient of e2 in P(u) -- shouldn't it be 10/7 instead of 2 ..?

    Oh no I see you made a type, it should read 2*4, so the coefficient is right.

    As you may have noticed, (Pu-u) as you calculated is orthogonal to neither of the vectors spanning U, so you made a mistake in calculating Pu.

    Why do you consider P(u) = <u, e1>e1 + <u, e2>e2 to be the right formula for the projection?
     
    Last edited: Apr 10, 2008
  9. Apr 10, 2008 #8
    I have a formula that says Pv = v, e1>e1 + ... + <v, em>em, where {e1, ..., em} is an orthonormal basis for U. Apparently {e1, e2} is not an orthonormal basis. Do I just need to find an orthonormal basis and then calculate P(u)?
     
    Last edited: Apr 10, 2008
  10. Apr 10, 2008 #9
    Sounds like a good idea. :smile:
     
  11. Apr 10, 2008 #10
    Ok, to be orthonormal, <v_i, v_j> = 0 for i != j. So I want <(a, a, 0, 0), (b, b, b, 2b)> = 0. This means sqrt(ab + ab) = sqrt(2ab) = 0. But the only way this can happen is if a and/or b are zero. But if I solve <(a, a, 0, 0), (a, a, 0, 0)> = 1 I get a = 1/sqrt(2) and if I solve <(b, b, b, 2b), (b, b, b, 2b)> = 1 I get a = 1/sqrt(7)>.

    This doesn't seem to make sense. What am I doing wrong? Thanks for your continued help.
     
  12. Apr 10, 2008 #11
    What you were trying to do was taking v_1 to be a multiple of (1,1,0,0) and v_2 a multiple of (1,1,1,2). However, (1,1,0,0) and (1,1,1,2) are not orthogonal, so any multiples of them will not be orthogonal either. That's why the method didn't work.

    Try taking v_1 to be equal to (1,1,0,0), that;s fine. Then you need to find another vector in U which is orthogonal to this v_1. A general vector in U is a linear combination of (1,1,0,0) and (1,1,1,2) that is

    [tex]
    v_2 = a (1,1,0,0) + b (1,1,1,2)
    [/tex]

    Try to find a and b such that <v_1, v_2> = 0. Then you can normalize v_1 and v_2 and have an orthonormal basis.

    In general you might want to look up Gram-Schmidt orthogonalization which is a method to find an orthogonal basis if you start from any given basis in a finite-dimensional Euclidean vector space. It's not necessary for this problem, though.
     
    Last edited: Apr 10, 2008
  13. Apr 10, 2008 #12
    Ok, so I take u1 = (1, 1, 0, 0) and then take u2 = a(1, 1, 0, 0) + b(1, 1, 1, 2) = (a + b, a + b, b, 2b). Then <u1, u2> = a+b + a+b = 2(a+b). So the only way for this to be zero is if a = -b. So I take a = 1, b = -1 to get u2 = (0, 0, -1, -2). Then ||u1|| = sqrt(2) and ||u2|| = sqrt(5), so e1 = (1/sqrt(2))*u1, and e2 = (1/sqrt(5))*u2.

    So now we have P(u) = 1/2[(1*1) + (1*2) + (0*3) + (0*4)](1, 1, 0, 0) + 1/5[(0*1) + (0*2) + (-1*3) + (-2*4)](0, 0, -1, -2) = (3/2)*(1, 1, 0, 0) - (11/5)*(0, 0, -1, -2) = (3/2, 3/2, 11/5, 11/5).

    Then P(u) - u = (0.5, 0.5, -0.8, 0.4) and ||P(u) - u|| = sqrt((0.25) + (0.25) + (0.64) + (0.16)) = sqrt(1.3).

    I think I've FINALLY got this problem solved. Please let me know if I did this right. Thanks a lot for your help, Pere! :smile:
     
  14. Apr 10, 2008 #13
    Seems right. So your first guess was pretty close :smile:
     
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