Linear Algebra, Orthonormal questionby wtmoore Tags: algebra, linear, orthogonal, orthonormal, vector 

#1
Mar112, 02:50 PM

P: 21

1. The problem statement, all variables and given/known data
I have this question that I am trying to figure out about orthonormality,I have tried to take a picture of it and put it on here but I can't figure out the url. Anyway I will try and write it out. Show that the vector {sin(x),cos(x)} is a basis for the vector space defined by: V={asin(x) + bcos(x) l a,b ε ℝ, 0≤ x ≤ pi} using the inner product : <f,g>=∫(0 to pi)fgdx, fgεV and determine an orthonormal basis. 2. Relevant equations 3. The attempt at a solution I found the integral of sin(x)cos(x)dx between 0 and pi to be 0. This make it orthonormal right as it's the same as the dot product. Now I think I have to find out whether <f,f> is = 1 (of unit length) so I did integral of sin^2(x)dx between 0 and pi but found pi/2. I also found the same for cos^2(x). Does this mean they are not orthonormal? I don't know if it makes a difference that <f,f>=<g,g>. Thanks 



#2
Mar112, 03:06 PM

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PF Gold
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#3
Mar112, 03:09 PM

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P.S. It's clear that V is spanned by sin(x) and cos(x), but the definition of a basis also requires that these functions must be linearly independent. Did you show that?




#4
Mar112, 03:13 PM

P: 615

Linear Algebra, Orthonormal question
What I find helps when you're first working with the notion of 'functions as vectors' is to look back to your basic 3vectors and think about what these properties mean for them.
Orthogonal 3vectors have a.b = 0, orthogonal functions have f.g = 0 Normalised vectors have a.a = 1, normalised functions have f.f = 1 Think about how you normalise 3vectors. Think about how you orthogonalise 3vectors. Here's another little problem you can do which is kinda fun, build up a set of orthogonal polynomials using the gramschmidt process (these polynomials are the legendre polynomials) 



#5
Mar112, 03:20 PM

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#6
Mar112, 03:23 PM

P: 21

I however don't know how to express this in proper notation. 



#7
Mar112, 04:36 PM

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#8
Mar112, 04:37 PM

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#9
Mar112, 04:40 PM

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a cos(x) + b sin(x) = 0 for all x, then a and b must be 0. So pick some specific values of x that are easy to work with. I suggest x = 0 and x = pi/2. 



#10
Mar112, 04:41 PM

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Isn't it [tex]\int_{0}^{\pi} sin^2(x) dx[/tex]? This should work out to pi/2, not pi. And similarly for cos(x). 



#11
Mar112, 04:45 PM

P: 21

Yeh originally I did it from 0 to pi but when I double checked for some reason I made a mistake and did it from pi to pi. Ok so I just need (pi/2)^(1)? 



#12
Mar112, 05:15 PM

P: 21

acos(0)+bsin(0)=0 Then a must be 0. acos(pi/2)+bsin(pi/2)=0 then b must be 0. Is it enough to write this, or is there some sort of other form I can generalize it for? 



#14
Mar112, 07:04 PM

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#15
Mar112, 07:47 PM

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I understand all of it, and have managed to complete 2 similar questions now. My only question is, why is it the square root? Surely if <f,f> = pi/2 then we need 2/pi to make this 1? 



#16
Mar212, 12:05 AM

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[tex]<h,h> = <\alpha f, \alpha f> = \alpha^2 <f,f>[/tex] You want this equal to 1, so [tex]\alpha^2 = 1/<f,f>[/tex] and hence [tex]\alpha = \sqrt{1/<f,f>}[/tex] 



#17
Mar212, 03:37 AM

P: 21

Yeh I was thinking just at the end not the fact that they are going to multiply each other. Thanks for all your help Jbunnii



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