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Linear maps: finding matrix

  1. Nov 3, 2009 #1
    1. The problem statement, all variables and given/known data

    Let T:R[x]2->R[x]3 be defined by T(P(x))=xP(x). Compute the matrix of T with respect to bases {1,x,x^2} and {1,x,x^2,x^3}. Find the kernel and image of T.

    3. The attempt at a solution
    I genuinely have no idea where to start on this, any pointers you can give me would be greatly appreciated.
  2. jcsd
  3. Nov 3, 2009 #2


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    i would start be computing the action of T on each basis vector of R2 & write in terms of basis of R3

    then use that to make a matrix

    notice the polynomials are considered as vectors, so for example ax2 +b x + c in the basis of R2 could just be written (c,b,a)
    Last edited: Nov 4, 2009
  4. Nov 4, 2009 #3
    so would the basis in R^2 in vector form be: {1,0,0},{0,1,0},{0,0,1}? If so, how would you get this in terms of the R^3 basis?
  5. Nov 4, 2009 #4


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    What you give, since they are in R3, form a basis for R3, not R2. A basis for R2 is {(1, 0), (0,1)}. (Don't use "{" and "}" for individual vectors. Those are set delimiters.)
  6. Nov 4, 2009 #5


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    as halls points out its not R^2, I'm not too sure what the proper notation is, but as there are 3 independent basis vectors, it much more like R^3

    so for the space, like R^3 with the basis {1,x,x2}, i think you're correct that you identify
    (1,0,0) with 1
    (0,1,0) with x & so on

    simarlarly for the space where the image resides, with the basis {1,x,x2,x^3}, I would identify
    (1,0,0,0) with 1
    (0,1,0,0) with x & so on

    is what you wrote exactly how the question is written?
    Last edited: Nov 4, 2009
  7. Nov 4, 2009 #6
    okay, so i have used the basis for R^3 and got a diagonal matrix with the elements x. Given the equation T(P(x))=x P(x) it looks like it could work. Is this correct? cheers
  8. Nov 4, 2009 #7


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    what matrix do you get, the way I'm thinking there shoudn't be any x's in the matrix?

    if its from a 3 dimensional space to a 4 dimensional space, i think it should be a 4x3 matrix

    Also i think it should have constant entries... consider what multiplying by x does, it shifts you from one basis vector to another... (similar to 90degree rotation in normal R^3)
    Last edited: Nov 4, 2009
  9. Nov 4, 2009 #8
    well i have just changed direction a bit but here is what i have so:
    T(1,0,0)=(0,1,0,0); T(0,1,0)=(0,0,1,0); T(0,0,1)=(0,0,0,1)
    So the matrix would be:

    (Each sub-bracket is a row)
  10. Nov 4, 2009 #9


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    sounds reasonable to me & all lines up with the initial definition
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