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Math Help - Finding eigenvalues

  1. #1
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    Finding eigenvalues

    Let T: P_2 \mapsto P_2 be defined by T(p)=xp'. Find eigenvalues of T.

    I'm not sure what the notation of T(p)=xp' means and how to go about finding the eigenvalues.

    Thanks.
    Last edited by wattkow; May 17th 2010 at 08:23 AM.
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  2. #2
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    Quote Originally Posted by wattkow View Post
    Let T: P_2 \mapsto P_2 be defined by T(p)=xp'. Find eigenvalues of T.

    I'm not sure what the notation of T(p)=xp' means and how to go about finding the eigenvalues.

    Thanks.
    That is the linear transformation which is x times the derivative of the polynomial.
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  3. #3
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    Quote Originally Posted by dwsmith View Post
    That is the linear transformation which is x times the derivative of the polynomial.
    Ok, so p is a polynomial? How would I go about finding eigenvalues for said polynomial?
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    Quote Originally Posted by wattkow View Post
    Ok, so p is a polynomial? How would I go about finding eigenvalues for said polynomial?
    I am assuming P is a polynomial since we are mapping P_2 on P_2. Unless your book notates this differently, I am almost positive these are polynomials; however, different books notate P_2 differently. For instance, mine says P_2 are polynomial of degree 1 or less. Thus, the 2 in P_2 is the amount of elements {1, x} but some books refer to P_2 as degree two or less, { 1, x, x^2}. What does your book say?
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    Quote Originally Posted by dwsmith View Post
    I am assuming P is a polynomial since we are mapping P_2 on P_2. Unless your book notates this differently, I am almost positive these are polynomials; however, different books notate P_2 differently. For instance, mine says P_2 are polynomial of degree 1 or less. Thus, the 2 in P_2 is the amount of elements {1, x} but some books refer to P_2 as degree two or less, { 1, x, x^2}. What does your book say?
    I believe in my case it's the 2nd degree polynomial.
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    Quote Originally Posted by wattkow View Post
    I believe in my case it's the 2nd degree polynomial.
    Ok so your linear transformation goes from ax^2+bx+c\rightarrow xp'(x)=x(2ax+b)=2ax^2+bx.

    Now you need to find the eigenvalues of the linear transformation.
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  7. #7
    Super Member Random Variable's Avatar
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    I would do the following:

    A basis for  P_{2} is  \{1,x,x^{2} \}

     T(1) = 0

     T(x) = x

    and  T(x^{2}) = 2x^{2}

    so the transformation matrix is  \begin{bmatrix}<br />
0 & 0 & 0 \\<br />
0 & 1 & 0\\<br />
0 & 0 & 2<br />
 \end{bmatrix}

    which has eigenvalues 0,1, and 2
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  8. #8
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    Quote Originally Posted by wattkow View Post
    I believe in my case it's the 2nd degree polynomial.
    The "set of all 2nd degree polynomials" does not form a vector space. As dwsmith said, P_2 is the set of all polynomials of degree 2 or less.

    Random Variable showed how to represent this transformation as a matrix- worth knowing for itself.

    But for this problem, you don't have to do that. As dwsmith said, if p= ax^2+ bx+ c, then T(p)= xp'= x(2ax+ b)= 2ax^2+ bx. If \lambda is an eigenvalue for T, we must have some a, b, c, not all 0, such that T(p)= 2ax^2+ bx= \lambda(ax^2+ bx+ c). Since that must be true for all x, we can set corresponding coefficients equal: 2a= \lambda a,  b= \lambda b, and 0= \lambda c.

    If a is not 0, then \lambda= 2, if b is not 0, then \lambda= 1, and if c is not 0, then \lambda= 0.

    That is, 0 is an eigenvalue with eigenvectors any multiple of 1, 1 is an eigenvalue with eigenvectors any multiple of x, and 2 is an eigenvalue with eigenvectors any multiple of x^2, exactly what Random Variable got.
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  9. #9
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    Thanks fellas.
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