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Linear algebra 1: cauchy schwarz problem

  1. Jan 28, 2014 #1
    1. The problem statement, all variables and given/known data

    If llull = 4, llvll = 5 and u dot v = 10, find llu+vll. u and v are vectors

    2. Relevant equations

    llu+vll = llull + llvll cauchy schwarz

    3. The attempt at a solution

    (1) llu+vll = llull + llvll

    (2) (llu+vll)^2 = (llull + llvll)^2

    (3) (llu+vll)^2 = llull^2 + 2uv +llvll^2

    (4) (llu+vll)^2 = 4^2 + 2(10) +5^2

    (5)(llu+vll)^2 = 16 + 20 + 25

    (6) sqr(llu+vll)^2 = sqr61

    (7) (llu+vll) = 7.81

    I think that I am doing it wrong for some reason. Any feed back would be greatly appreciated.
     
  2. jcsd
  3. Jan 28, 2014 #2
    Line (1) is ofcourse not correct.

    Hint: square ||u+v||.
     
    Last edited: Jan 28, 2014
  4. Jan 28, 2014 #3

    HallsofIvy

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    No, that is NOT correct! The Cauchy-Schwartz in-equality says that
    [tex]||u+ v||\le ||u||+ ||v||[/tex]

    again, no.

    No.

    That "2uv" (2 time the dot product of u and v) is NOT 0 tells you immediately that ||u+ v|| is NOT equal to ||u||+ ||v||!

    It looks right to me! What reason do you have to think this is wrong?
     
  5. Jan 28, 2014 #4
    aha! I see what you are saying now. (llu+vll)^2 = uu + uv +vu + vv, which becomes llu+vll^2 = llull^2 + 2uv + llvll^2

    llu+vll = llull +llvll, IFF 2uv = 0

    Which in this case it isn't. Because uv is equal to 10.
     
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