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Help with making an orthonormal basis

  1. Jun 22, 2004 #1


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    I'm wanting to form an orthonormal basis from two non-parallel vectors.

    [tex]a = \left(\begin{array}{cc}3 & 4\end{array}\right)[/tex]

    [tex]b = \left(\begin{array}{cc}2 & -6\end{array}\right)[/tex]

    Could someone please walk me through the calculations needed? Much appreciated.
  2. jcsd
  3. Jun 22, 2004 #2


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    I think I've finally figured out. I had it explained to me visually, which didn't make sense. By visually, I am referring to the whole vector "arrow" idea which always troubles me.
  4. Jun 22, 2004 #3
    Look up the Gram-Schmidt process.

    Or consider the image I've attached. The vectors are probably pointing in the wrong directions, but it doesn't really matter. The red vector (call it [tex]a_o[/tex]) is orthogonal to b, and the green vector (call it [tex]a_p[/tex]) is parallel to b. Now, a_p is the projection of a onto b, and there's a nice formula for finding it:

    [tex]a_p = \frac{a.b}{b.b} * b[/tex]

    (It's easy to prove, see any linear algebra text). After some calculations, we find that [tex]a_p = \left(\begin{array}{cc}-9/10, & 27/10\end{array}\right)[/tex]. But since [tex]a = a_p + a_o[/tex], we have that [tex]a_o = a - a_p = \left(\begin{array}{cc}3, & 4\end{array}\right) - \left(\begin{array}{cc}-9/10, & 27/10\end{array}\right) = \left(\begin{array}{cc}39/10, & 13/10\end{array}\right)[/tex]. And so, we have "created" a vector that's orthogonal to b, so you can chose a_o and b as the basis vectors. (Of course, you've got to normalize them first, but that's trivial).

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    Last edited: Jun 22, 2004
  5. Jun 23, 2004 #4


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    Thanks very much for the explanation. Orthogonalization makes quite a bit more sense now. I've done some visual graphs with Mathematica, and that's helped quite a bit, also.
  6. Aug 16, 2004 #5


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    the reason the projection of a onto b is ghiven by that formula is because of the familiar formula a.b = |a| |b| cos(C) where C is the angle between a and b. I.e. from triangle trig, the projection of a onto b has length |a| cos(C). so the vector of that length in that direction is obtained by multiplying a unit vector by that length. Now of course b/|b| is a unit vector in the direction of b,

    so the vector parallel to b, with length |a| cos(C), is the product

    |a|cos(C) b/|b| = |a||b| cos(C)/|b|^2 b = (a.b/b.b) b.
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