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Matrix of Transformation (non standard basis)

  1. Sep 19, 2009 #1
    1. The problem statement, all variables and given/known data

    Define T : R3x1 to R3x1 by T = (x1, x2,x3)T = (x1, x1+x2, x1+x2+x3)T

    1 Show that T is a linear transformation
    2 Find [T] the matrix of T relative to the standard basis.
    3 Find the matrix [T]' relative to the basis
    B' = {(1,0,0)t, (1,1,0)t, (1,1,1)t
    4 Find the transition matrix Q from B to B'
    5 Verify that [T]' = Q[T]Q-1



    2. Relevant equations



    3. The attempt at a solution
    1) I think this is done I showed that T(c(alpha + beta)) = c* T(alpha) + T(Beta) by substituting in vectors (a1, a2, a3) and (b1,b2,b3) and working throught the algebra.

    2) This is what I did, I think I understood this properly:

    I built a transformation Matrix [T] = T(e1) | T(e2) | T(e3)

    I got
    1 0 0
    1 1 0
    1 1 1

    3) Then I think is wrong but I'm not sure

    I built another matrix [T]' = T(v1) | T(v2) | t(v3)
    where v1, v2, and v3 are the the vectors given above

    I got
    1 1 1
    1 2 2
    2 2 3


    4) Then I built a transformation matrix by finding the coordinates of e1, e2, and e3 relative to v1, v2 and v3

    I got
    1 -1 0
    0 1 -1
    0 0 1

    I calculated an inverse

    1 1 0
    0 1 1
    0 0 1

    5) Then if I had done everything right
    [T]' should have been equal to Q[T]Q-1

    But it wasn't even close

    1 1 1
    0 1 1
    0 0 1

    Any help would be greatly appreciated.
    Margaret
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 21, 2009 #2

    One error is that T(v1) should be (1,1,1)t instead of (1,1,2)t. A more important error is that the resulting numbers "1,1,1" are coefficients with respect to the standard basis, not B'. You need to figure out x, y, and z such that
    (1,1,1)t=x v1 + y v2 + z v3. This is easy to do by inspection: x=0, y=0, z=1. The first column of [T]' is therefore (0,0,1)t. Do a similar computation for T(v2) and T(v3). The appropriate coefficients x, y, and z are easy to find by inspection, although not as trivial as for T(v1).

    Also, the inverse of

    1 -1 0
    0 1 -1
    0 0 1

    is almost but not quite

    1 1 0
    0 1 1
    0 0 1

    The row 1 column 3 entry is incorrect.
     
  4. Sep 21, 2009 #3
    Thank you very much. I think I've got it now.
     
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