Linear Algebra, Orthonormal question

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  • #1
wtmoore
21
0

Homework Statement



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.


Homework Equations





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
 
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  • #2
wtmoore said:

Homework Statement



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.


Homework Equations





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

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?
 
  • #3
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
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)
 
  • #5
jbunniii said:
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?

I think so, I sort of spotted it but wasn't sure. Do I define some sort of new function, let's say h(x), where h(x) = (π^-0.5)sin(x)? Then <h,h> = 1? This is normalization right?
 
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  • #6
jbunniii said:
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?

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, let's 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.
 
  • #7
wtmoore said:
I think so, I sort of spotted it but wasn't sure. Do I define some sort of new function, let's say h(x), where h(x) = (π^-0.5)sin(x)? Then <h,h> = 1? This is normalization right?

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.
 
  • #8
jbunniii said:
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.

I mean <f,f> = pi sorry, then h=pi^(-0.5)*f would work?
 
  • #9
wtmoore said:
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, let's 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.

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.
 
  • #10
wtmoore said:
I mean <f,f> = pi sorry, then h=pi^(-0.5)*f would work?

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).
 
  • #11
jbunniii said:
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).
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
jbunniii said:
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.

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?
 
  • #13
wtmoore said:
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)?

The square root of that.
 
  • #14
wtmoore said:
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?

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.
 
  • #15
jbunniii said:
The square root of that.

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?
 
  • #16
wtmoore said:
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?

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]
 
  • #17
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
 
Last edited:

1. What is Linear Algebra?

Linear Algebra is a branch of mathematics that deals with linear equations and their representations in vector spaces. It involves the study of linear transformations and their properties, as well as the use of matrices and determinants to solve systems of linear equations.

2. What are the basic concepts of Linear Algebra?

The basic concepts of Linear Algebra include vectors, matrices, linear transformations, determinants, and eigenvalues and eigenvectors. These concepts are used to represent and solve systems of linear equations and to understand the properties of linear transformations.

3. What does "orthonormal" mean in Linear Algebra?

Orthonormal refers to a set of vectors that are both orthogonal (perpendicular) and normalized (unit length). In other words, the vectors are at right angles to each other and have a magnitude of 1. Orthonormal vectors are important in Linear Algebra because they form a basis for vector spaces and are used for many mathematical operations.

4. How is Linear Algebra used in real life?

Linear Algebra has a wide range of applications in fields such as physics, engineering, computer science, economics, and statistics. It is used to solve systems of equations, analyze data, create computer graphics and animations, compress and encrypt data, and much more.

5. What are some common applications of Orthonormal matrices?

Orthonormal matrices have many practical applications, including image and signal processing, data compression, and cryptography. They are also used in machine learning algorithms, such as principal component analysis, which is used for dimensionality reduction and feature extraction.

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