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Linear Algebra: Finding the Standard Matrix from a Function

  1. Nov 6, 2010 #1

    Cod

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    1. The problem statement, all variables and given/known data

    Find the standard matrix of T(f(t)) = f(3t-2) from P2 to P2.

    2. Relevant equations

    n/a

    3. The attempt at a solution

    The overall question has to do with finding the determinants, so the matrix is provided; however, I want to know how the author came up with the standard matrix of T.

    Any help is greatly appreciated.
     
  2. jcsd
  3. Nov 6, 2010 #2

    vela

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    You can find the columns of the matrix representation by applying the transformation to the basis vectors. If you're using the basis {1, t, t2}, the first column of the matrix would correspond to T(1), the second column to T(t), and the third column to T(t2).
     
  4. Nov 7, 2010 #3

    Cod

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    Thanks for the guidance. What happens when the basis is Rn? I realize R2 is a 3x3 matrix, R3 is a 4x4, and so on.
     
  5. Nov 7, 2010 #4

    vela

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    Your question doesn't make sense. Rn is a vector space, not a basis.
     
  6. Nov 7, 2010 #5

    Cod

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    Sorry, I was looking at my homework when I typed the last post. I meant vector space.
     
  7. Nov 7, 2010 #6

    vela

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    Same thing. You choose a basis and apply the transformations to the basis vectors to get the columns of the matrix representing the transformation.
     
  8. Nov 8, 2010 #7

    Cod

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    So, I could choose a basis of 1, t, t2; 1, t, t2, t3; and so on (to tnth)?
     
  9. Nov 8, 2010 #8

    vela

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    Not for Rn because those aren't vectors in Rn. P2 consists of polynomials of degree less than or equal to 2, and each polynomial is a linear combination of 1, t, and t2. It turns out 1, t, and t2 are also independent, so they form a basis for P2. Rn, however, is a set of n-tuples, not polynomials. You need a collection of n linearly independent n-tuples to have a basis for Rn.
     
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