# Thread: Orthogonal Basis

1. ## Orthogonal Basis

After finding an orthogonal basis for a given set of vectors, I know it's okay to multiply any of the orthogonal vectors by a scalar, because this just changes the length of the vector and not it's direction (and hence not it's angle between any of the other vectors in the set).

However, is it okay to do this for any of the vectors in the equations used to find each orthognal vector in the first place? For example, the equation for finding the third orthogonal vector has vectors in it. Could I multiply any of these vectors by a scalar (to simplify calculations) and still get the correct answer?

Thanks!

2. ## Re: Orthogonal Basis

which equation are you referring to?

3. ## Re: Orthogonal Basis

1) I'd have to agree with Deveno. Speak you of a Gramm-Schmidt orthogonalization process?

2) Since you really need an orthoNORMAL basis, are you actually doing anything?

4. ## Re: Orthogonal Basis

if you are asking: if <u,v> = 0, can i replace v by av, the answer is yes: <u,av> = a<u,v> = a0 = 0 (in a REAL inner-product space).

so u is orthogonal to any scalar multiple of v iff u is orthogonal to v.

in fact, if {v1,v2,v3} is a basis, and if a,b,c are all non-zero, then {av1,bv2,cv3} is a basis, too.

however, if you are applying the Gram-Schmidt orthogonalization process to {v1,v2,v3} you won't necessarily get the same orthogonal basis {u1,u2,u3}

if you replace vj by avj. suppose our given basis is {(1,1,1),(1,0,1),(1,0,0)}.

we take u1 = v1.

to find u2, we take v2 - (<u1,v2> /<u1,u1>)u1 = (1,0,1) - (2/3)(1,1,1) = (1/3,-2/3,1/3)

to find u3, we take v3 - (<u1,v3>/<u1,u1>)u1 - (<u2,v3>/<u2,u2>)u2 = (1,0,0) - (1/3)(1,1,1) - (1/2)(1/3,-2/3,1/3) = (1/2,0,-1/2).

now, suppose we replaced v2 by 3v2, to "simplify the calculations":

again, u1 = v1 = (1,1,1) as before.

u2 = v2 - (<u1,v2> /<u1,u1>)u1 = (3,0,3) - (2)(1,1,1) = (1,-2,1) note this is a different u2 than we got before.

u3 = v3 - (<u1,v3>/<u1,u1>)u1 - (<u2,v3>/<u2,u2>)u2 = (1,0,0) - (1/3)(1,1,1) - (1/6)(1,-2,1) = (1/2,0,-1/2).

(it may seem surprising that although u2 is different, u3 is the same. but in our calculations for u3:

the only place the "new u2" occurs is: (<u2,v3>/<u2,u2>)u2, which in terms of our "original" u2 is (<3u2,v3>/<3u2,3u2>)3u2

= [3<u2,v3>/(9<u2,u2>)]3u2, so the factors of 3 cancel).

all that said, asTKHunny pointed out, you'll have to normalize at the end, anyway, so it's not like you're really saving any computational effort.

5. ## Re: Orthogonal Basis

Yes, I was talking about the grahm schmidt process for finding an orthonormal set, and wonder if, during that process, you can multiply any of the vectors by a scalar to simplify calculations. For example, multiply ,<.5,.5,.5> by 2 to get <1,1,1>. Mainly to take out fractions

6. ## Re: Orthogonal Basis

Truthfully, if a student submitted such work, I would consider it childish. If you are up to linear algebra, you should long have gotten over your aversion to mere fractions.

Just my opinion.