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Linear transformation (bases)

  1. Oct 16, 2008 #1
    Let T: R3 -> M(2,2) be the linear transformation given by

    T(x,y,z) = [ z .......-z ]
    ..............[ 0 ..... x-y]


    Fix bases B = {(1,0,0),(0,1,0),(0,0,1)} and C = { [1 0] , [0 1] , [0 0] , [0 0] }
    ..................................................................[0 0].....[0 0]...[1 0]...[0 1]


    for R3 and M(2,2) respectively


    a) Find the matrix [T]c,b of T with respect to the bases B and C.
    b) Use the matrix from part (a) to find the basis for Im(T)
    c) Use the matrix from part (a) to find the basis for Ker(T)

    For parts b and c I'm pretty sure thats just finding the column space (part b) and solution space (part c) of the matrix.

    As for a its more so to do with my own confusion when it comes to bases.

    I worked out the matrix to be [ 0 0 0 1 ]
    .........(3 by 4 matrix)...........[ 0 0 0 -1]
    ........................................[ 1 -1 0 0]

    The way i got this was by taking each vector from B applying the transformation then writing it in terms of the basis C and then writing the coefficients of each matrix in the rows of the above matrix. I hope that even makes sense >_>.

    Any help greatly appreciated :)

    P.S. sorry for the garbagety matrices :S


    EDIT: Maybe its the transpose of that matrix.... im lost >.<
     
    Last edited: Oct 16, 2008
  2. jcsd
  3. Oct 16, 2008 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Yes, the way you found the matrix certainly makes sense and, in fact, is the best way to do it. It is also true that "For parts b and c I'm pretty sure thats just finding the column space (part b) and solution space (part c) of the matrix."

    Now, since you have done part (a) and know how to do parts (b) and (c), what is your question?
     
  4. Oct 16, 2008 #3
    So the first matrix i got is correct?

    [ 0 0 0 1 ]
    [ 0 0 0 -1]
    [ 1 -1 0 0]


    or is is the transpose of it?
     
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