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Simple Eigenvector Question

  1. Feb 20, 2012 #1
    Hi,

    So for some reason I have the hardest time trying to work with polynomials in linear algebra. I can't explain it, but whenever I see a question I draw a complete blank.

    Question: i) Find all the eigenvalues. ii) for each eigenvalue λ, find a basis of the eigenspace Eλ.

    T: P3(R) --> P3(R) defined by T(p)(x) = p'(x) + 2p(x)

    So this is all I'm given. My question is what polynomials do I use to find the eigenvalues, and once I find those eigenvalues how do I find the eigenvectors? I'm inclined to try and solve it like eigenvector problems with matrices, problem is I don't know how to put this into a matrix.
     
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  3. Feb 20, 2012 #2

    Ray Vickson

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    You need to solve the problem [itex] p'(x) + 2p(x) = \lambda p(x),[/itex] and the solution must be a polynomial.

    RGV
     
  4. Feb 20, 2012 #3
    But what am I solving for? it can't be for lambda. In simple algebra I would have the polynomial and solve for "x". Do I use the standard basis vectors?
     
  5. Feb 20, 2012 #4

    Ray Vickson

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    In this case you are solving a first-order linear differential equation that happens to have a parameter, λ, in it. In other words, you need to find the function p(x).

    RGV
     
  6. Feb 20, 2012 #5
    Oh, I should mention that I haven't done differential equations yet at my Uni. Only have Calc courses and Linear Algebra part I.

    What I wanted to do was solve it as if it was searching for an eigenvector but with matrices. But an issue I always have is I don't know how to transform the polynomial into a matrix, expecially when I'm not given a specific polynomial.
     
  7. Feb 20, 2012 #6

    kai_sikorski

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    What is the basis for your space? HINT: you need to find 4 polynomials, such that any cubic polynomial can be expressed as their sum. There are many possible choices, but probably only a few natural ones.

    Okay assuming you can answer that question here is how you would find a matrix representation for the operator. Say the basis is [itex] e_i [/itex] for [itex] i = 1 ... 4 [/itex].

    Define the vectors [itex] v_j [/itex] as

    [tex] v_j = T (e_j) [/tex]

    So above line means you take your polynomials that you chose as your basis, and you apply the operator T to them which amounts to adding twice the polynomial to its own derivative.

    Now since [itex] v_j [/itex] must be cubic polynomials, you must be able to express them uniquely in the basis you chose. Say this representation is
    [tex] v_j = b_{1j}e_1 + b_{2j}e_2 + b_{3j}e_3 + b_{4j}e_4 [/tex]
    Your matrix is now given by [itex] T_{ij} = b_{ij} [/itex]
     
    Last edited: Feb 20, 2012
  8. Feb 20, 2012 #7
    Ahhh. Thank you. So to build on that question, because it always seems to pop up in some fashion or another. If I'm not given any vectors that I have to specifically apply my linear transformation to, should I just assume that I can use the standard basis vectors? Because my troubles always occur when I don't know what vectors should be chosen.

    The reason I ask is because there is a similar question where i have to verify that the given vector is an eigenvector:

    p = x3 ....under the same P3 conditions, but defined by: T(x) =
    xp' - 4p. Now the eigenvector condition is T(x) = λx. So do I use the standard basis vectors of P3 in that transformation and try to obtain the λ that would prove this?
     
  9. Feb 20, 2012 #8

    kai_sikorski

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    It shouldn't matter what intermediate basis you work in. When you re-express the eigen-vectors as polynomials they should be the same no matter what basis you chose. If an eigen-space for a particular eigen-value had dimension greater than 1, then the individual polynomials you get might be different but collectively they should span the same space.
     
  10. Feb 20, 2012 #9
    Thanks. I'm going to give this all a try, hopefully there won't be any problems
     
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