Help with making an orthonormal basis

In summary, to form an orthonormal basis from two non-parallel vectors, you can use the Gram-Schmidt process or find the projections of one vector onto the other using the formula a_p = a.b/b.b * b. This will create a new vector that is orthogonal to the original vector, which can be used as part of the orthonormal basis.
  • #1
graphic7
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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.
 
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  • #2
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.
 
  • #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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  • #4
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.
 
  • #5
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.
 

What is an orthonormal basis?

An orthonormal basis is a set of vectors in a vector space that are both orthogonal (perpendicular) and normalized (unit length). This means that the vectors are all at right angles to each other and have a magnitude of 1.

Why is an orthonormal basis important?

An orthonormal basis is important because it allows us to easily describe and manipulate vectors in a vector space. It also simplifies calculations, such as finding the magnitude or projection of a vector.

How do you create an orthonormal basis?

To create an orthonormal basis, you can use the Gram-Schmidt process. This involves taking a set of linearly independent vectors and applying a series of orthogonalization and normalization steps to create an orthonormal set.

What is the difference between an orthonormal basis and a regular basis?

An orthonormal basis is a special type of basis where the vectors are both orthogonal and normalized. In a regular basis, the vectors may not be orthogonal or normalized, but they still span the same vector space.

Can an orthonormal basis exist in any vector space?

No, an orthonormal basis can only exist in certain types of vector spaces, such as Euclidean spaces. In other types of vector spaces, such as complex vector spaces, an orthonormal basis may not be possible.

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