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Matrix of a linear transformation

  1. Apr 7, 2006 #1
    This is how the question appears in my textbook
    Find the matrix of T corresponding to the bases B and D and use it to compute [itex] C_{D}[T(v)] [/itex] and hence T(v)

    T; P2 - > R2
    T(a + bx + cx^2) = (a+b,c)
    B={1,x,x^2}
    D={(1,-1),(1,1)}
    v = a + bx + cx^2

    ok i cna find Cd no problem it is
    [tex] C_{D}[T(v)] = \frac{1}{2} \left(\begin{array}{ccc} 1&1&-1 \\ 1&1&1 \end{array}\right) [/tex]

    now am i supposed to solve for X where
    [tex] C_{D}[T(v)] X = T(v) [/tex]

    but textbook doesnt do that...
    it does [tex] C_{D}[T(v)] \left(\begin{array}{c} a \\ b \\ c \end{array}\right) = X [/tex]

    It doesnt make sense... doesnt the question ask to compute Cd[T(v)] and T(v) from it??

    Next question
    Verify this theorem for the given transformation and uses the standard basis in Rn
    Theorem:
    [itex] M_{EB} (ST) = M_{ED} (S) \bullet M_{DB} (T) [/itex]

    T;R3 -> R4
    S: R4->R2
    T(a,b,c) = (a+b,b+c,c+a,b-a)
    S(a,b,c,d) = (a+b,c-d)

    [tex] M_{EB} (S) = \left[C_{D}\left(\begin{array}{c} 1 \\ 0 \end{array}\right) \ C_{D}\left(\begin{array}{c} 1 \\ 0 \end{array}\right) \ C_{D}\left(\begin{array}{c} 0 \\ 1 \end{array}\right) \
    C_{D}\left(\begin{array}{c} 0 \\ -1 \end{array}\right)\right] [/tex]
    [tex] M_{EB} (S) = \left[\begin{array}{cc} 1&0 \\ 1&0 \\ 0&1 \\ 0&-1 \end{array}\right] [/tex]

    [tex] M_{DB} (T) = \left[C_{D}\left(\begin{array}{c} 1 \\ 0 \\1 \\ -1 \end{array}\right) \ C_{D}\left(\begin{array}{c} 1 \\ 1 \\ 0 \\ 1 \end{array}\right) \ C_{D}\left(\begin{array}{c} 0 \\ 1 \\ 1 \\ 0 \end{array}\right)\right] [/tex]

    [tex] M_{DB} (T) = \left[\begin{array}{cccc} 1&0&1&-1 \\ 1&1&0&1 \\ 0&1&1&0 \end{array}\right] [/tex]

    the dimensions of the matrices are not correct though..

    is there something wrong in the way i am forming the matrices?
     
    Last edited: Apr 7, 2006
  2. jcsd
  3. Apr 7, 2006 #2

    Hurkyl

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    What does [itex]C_D[T(v)][/itex] mean?

    First off, did you notice that T(v) is a vector in R²?


    I think it might help you a great deal to write down explicitly exactly what type everything is in your equations. E.G. you might write down:

    v : element of P2
    T : transformation from P2 --> R²
    T(v) : element of R²
    [itex]C_D[T(v)][/itex] : 2x1 matrix.


    I'm going to assume [itex]C_D[w][/itex] means the coordinate representation of the vector w, with respect to the basis D.

    So you didn't compute [itex]C_D[T(v)][/itex] at all -- instead, you computed [itex]M_{B,D}[T][/itex] which is the coordinate representation of the linear transformation T, with respect to the bases B and D!

    You really ought to go reread the section on coordinates and bases. The whole point of coordinates is that they simply transform the equation into matrices. In other words, the equation:

    T(v) = w

    is true if and only if the equation

    [tex]M_{B,D}[T] \cdot C_B[v] = C_D[w][/tex]

    is true.
     
    Last edited: Apr 7, 2006
  4. Apr 7, 2006 #3

    Hurkyl

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    Well, you're certainly not internally consistent:

    First, you wrote down the matrix with three columns, each of which is supposed to be the coordinate representation of a vector in R^4.

    And then you wrote down a matrix whose rows are the transpose of those coordinate representations.


    Incidentally, based on how the problem's stated, E is the basis on R^4, and D is the basis on R³, so you meant [itex]M_{ED}(S)[/itex] and not [itex]M_{EB}(S)[/itex].
     
  5. Apr 9, 2006 #4
    well then what is the different between [itex] M_{ED}(s) [/itex] and [itex] M_{EB}(S) [/itex] is it even possible to form the former?
     
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