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Determining the matrix of a linear transformation

  1. May 17, 2009 #1
    1. The problem statement, all variables and given/known data
    Let D : P3--> P2 be differentiation of polyonimals. Determine the matrix of D with
    respect to the standard basis of P3.



    2. Relevant equations
    None


    3. The attempt at a solution
    I think D=[1 0 0; 0 1 0; 0 0 0]. This is from inspection though because I know that the resulting matrix must exclude the numbers in the last row of the matrix.
    Could someone give me a hint on how to start this question?
     
  2. jcsd
  3. May 17, 2009 #2

    quasar987

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    Well first, do you know what is means by "the standard basis of P3"?
     
  4. May 17, 2009 #3
    The standard basis is [1 0 0; 0 1 0 ; 0 0 1]
     
  5. May 17, 2009 #4

    Hurkyl

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    How can that be? None of those are even elements of P3.
     
  6. May 17, 2009 #5
    ohh wait wait , I meant [ 1 0 0 0; 0 1 0 0 ; 0 0 1 0; 0 0 0 1]
     
  7. May 17, 2009 #6

    Hurkyl

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    You have the dimension right, but those still aren't elements of P3.
     
  8. May 17, 2009 #7
    um....then is it the set of different polynomials? {1, x, x^2, x^3}...i think
     
  9. May 17, 2009 #8

    Hurkyl

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    Yes, that's the standard basis for P3.
     
  10. May 17, 2009 #9
    It's either that or the other way around. I think that P3, however, means polynomials of degree less than 3; so x^3 is excluded.

    Now, what you do is you do the linear transformation on each of the basis vectors. This will produce a vector which is an element of P2. Say that U = P3 and V = P2, and D goes from U to V. Then L(u) = v, where u is in U and v is in V.

    So you need to find: v1 = L(u1), v2 = L(u2), v3 = L(u3) (and v4 = L(u4), if you include x^3)

    Then the matrix for D will have columns v1, v2, and v3 (, and v4 if you include x^3), in that order. Then if you take M, the matrix for D, and try

    M*u = v, you'll find that v is a vector in P2 which represents the derivative of the polynomial u.
     
  11. May 17, 2009 #10
    So
    v1 = L(u1) = 0
    v2 = L(u2) = 1
    v3 = L(u3) = x
    v4 = L(u4) = x^2

    How am i suppose to rewrite this to represent the matrix D?.....because this is now the standard basis for P2
     
  12. May 17, 2009 #11
    Well, the derivative of x^3 isn't really x^2... it's 3x^2.

    And, like I said, I believe that P2, P3, etc. means "the set of all polynomials of degree less than 2, 3, etc.". I might be wrong. That's how it was presented to me, however.

    I'll do one part of it for you - so here we go.

    U = P3 = [1, x, x^2], V = P2 = [1, x]

    D(x^2) = 2x = transform(0, 2) wrt V

    So the first column of M will be this vector, namely,

    Code (Text):

    M = 0   ?   ?
        2   ?   ?
     
    To fill in the other question marks, do the same thing with the other basis vectors and put them in. Then you'll see that things like

    M * transform(3, 7, 2) = transform(7, 4).

    Does that make more sense?
     
  13. May 17, 2009 #12

    Hurkyl

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    What is L? You were supposed to be finding the coordinate matrix for D.


    Do you know how to find the coordinates of a vector with respect to a basis?
     
  14. May 17, 2009 #13
    To find the coordinates of a vector with respect to a basis, row reduce the vector with the new basis.

    To:AUMathTutor
    I think D works out to be [0 1 0 0 ; 0 0 2 0; 0 0 0 3] if we include polynomials of degree 3 in P3
     
  15. May 17, 2009 #14
    I think that's right. I see I made a mistake in putting (0, 2) in the first column if x^2 comes last in my scheme.

    I think that what you have is correct, if of course you allow polynomials of degree 3 in P3 and polynomials of degree 2 in P2.
     
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