Linear Algebra, Orthonormal question

wtmoore

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

jbunniii

Correct, sin(x) and cos(x) are orthogonal [their inner product with each other is zero] but not orthonormal [their norms are not 1]. There's a very simple modification you can make to sin(x) and cos(x) which will make them orthonormal. Do you see what it is?

jbunniii

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?

genericusrnme

What I find helps when you're first working with the notion of 'functions as vectors' is to look back to your basic 3-vectors and think about what these properties mean for them.

Orthogonal 3-vectors 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 3-vectors. Think about how you orthogonalise 3-vectors.
Here's another little problem you can do which is kinda fun, build up a set of orthogonal polynomials using the gram-schmidt process (these polynomials are the legendre polynomials)

wtmoore

I think so, I sort of spotted it but wasn't sure. Do I define some sort of new function, lets say h(x), where h(x) = (π^-0.5)sin(x)? Then <h,h> = 1? This is normalization right?

wtmoore

I can explain it but don't know how to write this in a neat way. For example if I plug in values for x between 0 and π, lets take 0, then I know that this makes the sin term 0 and cos positive, and if we take pi/2 for example then it's the other way around. So for whatever value is taken between 0 and pi, one of them is a postive and one is 0, therefore a=b=0.

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

jbunniii

Right idea, but I think your scale factor is off. If your original function had <f,f> = pi/2, then you would want to define h = sqrt(2/pi)*f.

wtmoore

I mean <f,f> = pi sorry, then h=pi^(-0.5)*f would work?

jbunniii

Yes, you have the right idea. You want to show that if

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.

jbunniii

But is <f,f> = pi?

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).

wtmoore

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)?

wtmoore

Yeh I tried these,

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?

jbunniii

The square root of that.

jbunniii

That's enough. If the linear combination equals zero, it has to equal zero at every value of x, including the two you chose (0 and pi/2). Those two alone are enough to imply that a = b = 0. You could have picked almost any pair of values of x, as long as they resulted in a solvable system of two equations with two unknowns. 0 and pi/2 just make the math easier.

wtmoore

Thanks jbunniii,

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?

jbunniii

No, suppose you set [itex]h = \alpha f[/itex]. Then

[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]

wtmoore

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