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Polynomials and isomorphism

  1. Feb 27, 2013 #1
    Ok for the longest while I've been at war with polynomials and isomorphisms in linear algebra, for the death of me I always have a brain freeze when dealing with them. With that said here is my question:

    Is this pair of vector spaces isomorphic? If so, find an isomorphism T: V-->W.

    V= R4 , W = {p[itex]\in[/itex]P4(R) | p(0) = 0}

    Here is the issue. What kind of polynomial am I examining?

    It says it is the set of polynomials of degree 4 s.t p(0) = 0.

    What does the p(0) = 0 mean? For example if I used the set of standard basis vectors of P4(R), what would the set of p(0) = 0 look like? All I could picture is

    P(1) = 1 , p(x) = 0 , p(x2) = 0,.....p(x4) = 0.

    Is that the right way to look at it?

    There was no specified transformation given either.

    Thanks
     
  2. jcsd
  3. Feb 27, 2013 #2

    jbunniii

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    A general polynomial of degree 4 looks like ##p(x) = a_4 x^4 + a_3 x^3 + a_2 x^2 + a_1 x^1 + a_0##. So plug in ##x = 0## to see what ##p(0) = 0## means.

    By the way, are you sure ##P_4(\mathbb{R})## means the set of polynomials of degree 4, and not the set of polynomials with degree less than or equal to 4?
     
  4. Feb 27, 2013 #3


    It would be the set of degree less than or equal to 4. Ok but doing that would then mean all the polynomials get reduced down to their constant term so the dimension is 1 which is not the same as the dimension of R4 i.e not isomorphic.
     
  5. Feb 27, 2013 #4

    jbunniii

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    No, that's not true. What do you get when plug ##x = 0## into ##p(x) = a_4 x^4 + a_3 x^3 + a_2 x^2 + a_1 x^1 + a_0##? Which coefficient(s) does the condition ##p(0) = 0## force to zero?
     
  6. Feb 27, 2013 #5
    it forces all of the coefficients except for a0 to be 0's. So it would be the set of polynomials represented only by their constants. But how can I show its dimension in order to prove it's an isomorphism.
     
  7. Feb 27, 2013 #6

    jbunniii

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    No, that's not right. If ##p(x) = a_4 x^4 + a_3 x^3 + a_2 x^2 + a_1 x^1 + a_0##, then ##p(0) = a_0##. So ##p(0) = 0## forces ##a_0 = 0##. All the other coefficients can be anything.
     
  8. Feb 28, 2013 #7
    unfortunately you had to give me the answer to finally see it, but I get your logic behind it now. So the set of polynomials where p(0)=0 implies that the constant is going to be 0. Now since any of the other terms can be anything, which in this case reduces the dimension of the polynomial by 1 making it the same dimension as R4 therefore it is isomorphic.

    Now I have to try and find an isomprphism........which is the other thing I suck at. So I'm going to have the general form of the polynomial:

    ##p(x) = a_4 x^4 + a_3 x^3 + a_2 x^2 + a_1 x^1 + a_0## but a0 = 0 so it is now of the form ##p(x) = a_4 x^4 + a_3 x^3 + a_2 x^2 + a_1 x^1##

    now for it to be an isomorphism I have to get this polynomial to turn into a vector of 4 elements by a transformation. How do I create that transformation?
     
  9. Feb 28, 2013 #8

    pasmith

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    What are the first four derivatives of [itex]p(x)[/itex] at [itex]x = 0[/itex]?
     
  10. Feb 28, 2013 #9

    hmmm..... I suppose each one of those derivatives could be considered a component of the vector then. Interesting.
     
  11. Feb 28, 2013 #10

    jbunniii

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    Well, you have four coefficients, not counting the one that was forced to zero. The natural thing to try is to group those four coefficients into a vector, ##(a_1, a_2, a_3, a_4)## [or in whichever order you prefer]. Then check whether this is indeed an isomoprhism.
     
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