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Finding a linear transformation.

  1. Dec 30, 2012 #1
    Hi,
    1. The problem statement, all variables and given/known data
    How may I find (or prove that there isn't) a linear transformation which satisfies T: R3->R1[x], ker T = Sp{(1,0,1), (2,-1,1)}?


    2. Relevant equations



    3. The attempt at a solution
    I am not sure how to approach this. I understand that kerT is the group of all vectors (x,y,z) in R3 so that T(x,y,z) = 0 = Sp{(1,0,1), (2,-1,1)}. So x = alpha, y = -beta, z = alpha + beta? Hence, x,y,z=0? Hence, T has to be one-to-one? Since dim(R1[x]) is 2, does that mean that there is no such linear transformation?
     
  2. jcsd
  3. Dec 30, 2012 #2

    mfb

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    I don't understand what you try to do with your alpha and beta.
    Every vector which can be written in that way is part of ker T - and now? What about vectors which cannot be written like that?

    You can use a dimensional analysis to prove that there is no surjective linear transformation (with the given ker T), but that is not relevant here.
     
  4. Dec 30, 2012 #3
    Won't dimensional analysis show that dimIm(T) must be equal to 1? How does that help me disprove that no such linear transformation exists?
     
  5. Dec 30, 2012 #4

    mfb

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    Why do you think you can disprove it at all?

    dim Im(T)=1, I agree
     
  6. Dec 30, 2012 #5
    I don't know, wrong inference from your previous post I presume. In any case, how shall I proceed?
     
  7. Dec 30, 2012 #6

    HallsofIvy

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    [itex]R_1[x][/itex] is the set of all polynomials of the form ax+ b for real numbers a and b (unless I am misunderstanding- I would call that R[x] without the "1"). We can represent that by the column vector [itex]\begin{bmatrix}a \\ b \end{bmatrix}[/itex] and so we can represent a linear transformation from [itex]R^3[/itex] to [itex]R_1[x][/itex] by a 3 by 2 matrix:
    [tex]\begin{bmatrix}a & b & c \\ d & e & f \end{bmatrix}\begin{bmatrix}x \\ y \\ z \end{bmatrix}= \begin{bmatrix}p \\ q\end{bmatrix}[/tex]

    Saying that the kernel is spanned by (1, 0, 1) and (2, -1, 1) means that
    [tex]\begin{bmatrix}a & b & c \\ d & e & f \end{bmatrix}\begin{bmatrix}1 \\ 0 \\ 1 \end{bmatrix}= \begin{bmatrix}a+ c \\ d+ e\end{bmatrix}= \begin{bmatrix}0 \\ 0 \end{bmatrix}[/tex]
    and
    [tex]\begin{bmatrix}a & b & c \\ d & e & f \end{bmatrix}\begin{bmatrix}2 \\ -1\\ 1\end{bmatrix}= \begin{bmatrix}2a- b+ c \\ 2d- e+ f\end{bmatrix}= \begin{bmatrix}0 \\ 0 \end{bmatrix}[/tex]

    So we must have a+ c= 0, d+ e= 0, 2a- b+ c= 0, and 2d- e+ f= 0. That gives four equations with which we can solve for four of a, b, c, d, e, and f in terms of the other two.
     
  8. Dec 30, 2012 #7
    Would you care to explain the logic behind such approach? For instance, why did you choose a matrix whose first two elements in the first row are the elements of the column vector [a,b]?
     
  9. Dec 30, 2012 #8

    mfb

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    a to f are just some constants (with arbitrary names), as this 2x3-matrix represents an arbitrary linear transformation from R^3 to R^2 (or R[x] in this special case).
     
  10. Dec 30, 2012 #9
    Right, but it is for a reason that a and b appear in both cases, namely first two elements in the first row of the matrix AND the elements of the column vector [a,b]. I am asking why is that.
    Moreover, I have a general question - are all different bases linearly dependent?
     
  11. Dec 30, 2012 #10

    mfb

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    No, they are not related. As you can see, HallsofIvy replaced the names by p and q in the first equation to avoid collisions.
     
  12. Dec 30, 2012 #11
    Okay. The result I obtained was a=b=-c and d=-e=-f/3.
    How do I derive my answer from these values?
     
  13. Dec 31, 2012 #12

    mfb

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    As you have no additional restrictions, you can choose two values for those parts to find a linear transformation.
    a=b=-c=5, d=e=-f/3=7 for example.
     
  14. Dec 31, 2012 #13
    Hence, how should the final answer be presented? Shouldn't it be presented in the form of a polynomial ax+b?
     
  15. Dec 31, 2012 #14

    mfb

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    The final answer should give some unique way to map all vectors (g,h,i) to vectors px+q.
     
  16. Dec 31, 2012 #15
    In that case, I don't think I understand. I thought I was to provide a general term for a transformation answering the criteria as presented in the question. Would you clarify, please?
     
  17. Dec 31, 2012 #16

    mfb

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    You have to give a single linear transformation - this is a specific way to map every vector of R^3 to a polynomial: Like an equation T(g,h,i)=... which can be used for all values of g,h,i and gives a result which depends on g,h,i.
     
  18. Dec 31, 2012 #17
    Let me see if I got it correctly. Supposing I take the values for a,b,c,d,e and f, and substitute them into the original matrix as suggested by HallsofIvy, I get a single column vector whose elements are 5x+5y-5z and 7x+7y-21z. To what end does that serve me?
     
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