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Math Help - Finding a basis for the subspace of P_2

  1. #1
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    Finding a basis for the subspace of P_2

    Problem Statement:
    Let p(x), q(x) \in P_2. You may assume that

    < p(x), q(x)> = \int_{-1}^1 p(x)q(x)dx

    defines an inner product on P_2

    Find a basis for the subspace of P_2:

    V = \{ a + b + ax + bx^2 | a,b \in R\}

    -----------------------

    I'm sure this is a very easy question, just not sure how to go about it. Would we not just use as the basis the standard basis of P_2, i.e.,
    \{1, x, x^2\}?

    Note: I'm not sure if the inner product is important for answering this specific question, as there is another question that follows to which it may apply instead.
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  2. #2
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    Quote Originally Posted by Ares_D1 View Post
    Problem Statement:
    Let p(x), q(x) \in P_2. You may assume that

    < p(x), q(x)> = \int_{-1}^1 p(x)q(x)dx

    defines an inner product on P_2

    Find a basis for the subspace of P_2:

    V = \{ a + b + ax + bx^2 | a,b \in R\}

    -----------------------

    I'm sure this is a very easy question, just not sure how to go about it. Would we not just use as the basis the standard basis of P_2, i.e.,
    \{1, x, x^2\}?

    Note: I'm not sure if the inner product is important for answering this specific question, as there is another question that follows to which it may apply instead.
    For some reason, I am having a mental block but I believe all you need to do is solve A\mathbf{x}=\mathbf{0}.

    \begin{bmatrix}<br />
0 & 1 & 0\\ <br />
1 & 0 & 0\\ <br />
1 & 1 & 0<br />
\end{bmatrix} so if I am correct we have an overdetermined system.

    rref=\begin{bmatrix}<br />
1 & 0 & 0\\ <br />
0 & 1 & 0\\ <br />
0 & 0 & 0<br />
\end{bmatrix}\to a=0, \ b=0

    Basis:
    { 0} which is also equal to the null space and ker(V); therefore, the range of V is { 1,x,x^2}

    Exercise extreme caution with what I have since I can't think right now.
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  3. #3
    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by Ares_D1 View Post
    Problem Statement:
    Let p(x), q(x) \in P_2. You may assume that

    < p(x), q(x)> = \int_{-1}^1 p(x)q(x)dx

    defines an inner product on P_2

    Find a basis for the subspace of P_2:

    V = \{ a + b + ax + bx^2 | a,b \in R\}

    -----------------------

    I'm sure this is a very easy question, just not sure how to go about it. Would we not just use as the basis the standard basis of P_2, i.e.,
    \{1, x, x^2\}?

    Note: I'm not sure if the inner product is important for answering this specific question, as there is another question that follows to which it may apply instead.
    a+b+ax+bx^2=a(1+x)+b(1+x^2) and therefore V = \mbox{span }(1+x,1+x^2). I'll let you show that 1+x, 1+x^2 are in fact linearly independent.

    The inner product part is irrelevant to this problem.
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  4. #4
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    Quote Originally Posted by Bruno J. View Post
    a+b+ax+bx^2=a(1+x)+b(1+x^2) and therefore V = \mbox{span }(1+x,1+x^2). I'll let you show that 1+x, 1+x^2 are in fact linearly independent.

    The inner product part is irrelevant to this problem.
    Of course. Thanks for clearing that up.

    Out of curiosity though; obviously \{1, x, x^2\} is still a basis for this subspace, no? So then, in finding an orthonormal basis for V (via Gram-Schmidt procedure), would there not be multiple, equally valid solutions which arise from starting with either basis?
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  5. #5
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    If the space is P2, why would you expect to have 3 basis functions?
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  6. #6
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    Quote Originally Posted by Ares_D1 View Post
    Of course. Thanks for clearing that up.

    Out of curiosity though; obviously \{1, x, x^2\} is still a basis for this subspace, no?


    If you write "obviously" then why the question "no?" But the answer is no: that is not a basis for that subspace, whose elements are characterized for having

    their free coefficient equal to the sum of the linear and the quadratic coefficients. The element x does not share this characteristic.

    Tonio

    So then, in finding an orthonormal basis for V (via Gram-Schmidt procedure), would there not be multiple, equally valid solutions which arise from starting with either basis?
    .
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