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Linear Trans. & Bases

  1. May 3, 2007 #1
    Let T: P3 ----> P4
    Im attempting to find the matrix for T relative to the bases B and B'

    T maps P3 to P4

    [tex]

    \begin{array}{l}
    T(ax^3 + bx^2 + cx + d) = (3x + 2)(ax^3 + bx^2 + cx + d) \\
    B = \{ x^3 ,x^2 ,x,1\} \\
    B^' = \{ x^4 ,x^3 ,x^2 ,x,1\} \\
    \end{array}



    [/tex]



    Im able to do these porblems when say T:R3--->R3 and my bases are ordered vectors as a pair, triplets ect. but Im not seeing how to find the image of say X cubed under T. Do I just plug it in? then i get another polynomial, which I would need to write as a linear combination of the base B'?

    If someone could show me how to find the image of the first one, T(x^3) I could go on from there.
     
  2. jcsd
  3. May 3, 2007 #2
    correct.
    and just for the fun of it:
    T(x^3)=(3x+2)x^3=3x^4+2x^3
    T(x^2)=(3x+2)x^2=3x^3+2x^2
    T(x)=(3x+2)x=3x^2+2x
    T(1)=3x+2
    which gives you somethin like this:
    Code (Text):

    3000
    2300
    0230
    0023
    0002
     
     
  4. May 3, 2007 #3
    Thanks, for some reason I didnt see the where the substitutions went. I was trying to plug in my basis pieces into each x variable on the right hand side in the general polynomial.
    Seems kinda funny now.
     
  5. May 6, 2007 #4

    just to make sure that im seeing this correctly, the column entrys are also the scaler numbers in the linear combination (co efficients because im working with a polynomial) of B'


    So if base B' had a first vector of (2x^4) my matricie would look like this,

    6000
    2300
    0230
    0023
    0002


    Am I seeing this right?


    3000
    2300
    0230
    0023
    0002


    is my matrix for T relative to B and B'
     
    Last edited: May 6, 2007
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