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Math Help - Help with Linear Algebra , Orthonormal Basis

  1. #1
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    Help with Linear Algebra , Orthonormal Basis

    Greetings , I'm new to the site and well , I have the following question.

    In a polynominal space in the order smaller then 3 , a given standard basis is {1,x,x^2,x^3} , find the Orthonormal Basis.
    (I had to translate the question from another language so forgive me for any mistakes made in the translation)

    Basically I've been a real slacker on my algebra course so now I have study by myself :P .
    Anyways , I think the answer is {1,x+x^2+x^3} but im not sure , nor do I have a proper solution , so if possible I'd like to ask for a proper full solution.
    Help would be much appriciated , Thanks.
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  2. #2
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    Quote Originally Posted by Zipi View Post
    Greetings , I'm new to the site and well , I have the following question.

    In a polynominal space in the order smaller then 3 , a given standard basis is {1,x,x^2,x^3} , find the Orthonormal Basis.
    (I had to translate the question from another language so forgive me for any mistakes made in the translation)

    Basically I've been a real slacker on my algebra course so now I have study by myself :P .
    Anyways , I think the answer is {1,x+x^2+x^3} but im not sure , nor do I have a proper solution , so if possible I'd like to ask for a proper full solution.
    Help would be much appriciated , Thanks.
    Google for Gram-Schmidt process.

    RonL
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  3. #3
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    Sorry to bump an older thread , but figured it was better then opening a new one. Anyways I've looked around for some gram schmidt process , wikipedia , and some random find on google , they all use some proj function wich I'm unfamiliar with , most likely because the linear algebra course I'm taking is for physicists so I figure they removed part of the material . From the class notes our gram schmidt process is diffrent , but after trying to use it on the {1,x,x^2,x^3} basis all i get is U1=1 U2=0 U3=0 ... and so forth .

    i define U(1) = 1 U(2) = A*1 + B*x
    so 0=(U(2),U(1)) = (A,1) + (B*x,1) = A +B*x , i choose B=1 so i get A=-x and there goes the endless loop of U(n)=0 for n>1

    note : (x,y) - inner product space , just to clarify in case this forum signs it diffrently
    help would be much appriciated
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  4. #4
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    Quote Originally Posted by Zipi View Post
    Sorry to bump an older thread , but figured it was better then opening a new one. Anyways I've looked around for some gram schmidt process , wikipedia , and some random find on google , they all use some proj function wich I'm unfamiliar with , most likely because the linear algebra course I'm taking is for physicists so I figure they removed part of the material . From the class notes our gram schmidt process is diffrent , but after trying to use it on the {1,x,x^2,x^3} basis all i get is U1=1 U2=0 U3=0 ... and so forth .

    i define U(1) = 1 U(2) = A*1 + B*x
    so 0=(U(2),U(1)) = (A,1) + (B*x,1) = A +B*x , i choose B=1 so i get A=-x and there goes the endless loop of U(n)=0 for n>1

    note : (x,y) - inner product space , just to clarify in case this forum signs it diffrently
    help would be much appriciated

    First, we need an inner product space for an ortho-normal basis and you have not specified one. We could assume we are working with the inner product:

    \langle u, v \rangle = \int_0^1 u(x) v(x) \ dx

    where u and v are real polynomial of degree not more than 3. That is our inner product space is that of real polynomials of degree not more than 3 on (0,1) with the given inner product.

    The first component of our ortho-normal basis is just the normalised version of the first component of the given basis, which is the constant unit function f_0(x)=1. Now:

    \langle f_0, f_0 \rangle = \int_0^1 1 \ dx=1

    so it is already normalised.

    We construct the next component by removing the projection of x onto the space spanned by f_0:

    f_1^*=x-\langle x,f_0 \rangle f_0

    and then normalise to get:

    f_1=\frac{f^*_1}{\sqrt{\langle f^*_1, f^*_1\rangle}}

    and we continue in like manner for the other components of the ortho-normal basis, remove the projections of the next element of the original basis onto the subspaces spanned by each of the already found orthon-normal basis elements, and then normalise.


    RonL
    Last edited by CaptainBlack; September 14th 2008 at 02:16 PM.
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  5. #5
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    i see , i guess i wrongly assumed it was a simple action like (a(n),b(n)) = a1*b1 + a2*b2 +...
    the reason i didnt specifie an inner space product is because i simply quated the question and translated , so i had no idea about it. well makes more sense now , also i poked around wikipedia meanwhile and noticed that they tend to define the integration limits from a to b , rather then 0 to 1 , is it a certain norm or does each question usually defines it's own inner product space , and the one i bumped into simply was lacking it ?
    also thanks for explanation Captainblack , pretty much cleared stuff up for me on the subject , pretty much had to learn the whole course by myself in the last week.
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  6. #6
    Grand Panjandrum
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    Quote Originally Posted by Zipi View Post
    i see , i guess i wrongly assumed it was a simple action like (a(n),b(n)) = a1*b1 + a2*b2 +...
    the reason i didnt specifie an inner space product is because i simply quated the question and translated , so i had no idea about it. well makes more sense now , also i poked around wikipedia meanwhile and noticed that they tend to define the integration limits from a to b , rather then 0 to 1 , is it a certain norm or does each question usually defines it's own inner product space , and the one i bumped into simply was lacking it ?
    also thanks for explanation Captainblack , pretty much cleared stuff up for me on the subject , pretty much had to learn the whole course by myself in the last week.
    Each problem should either specify the inner product (the example that I gave was just that), the resuling ortho-normal basis will be different for each choice of inner product (usually), or limits of integration.

    RonL
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