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Standard basis math

  1. Mar 17, 2005 #1
    I think I sort of understand this but need confirmation/correction:

    We have this formula in the book: X = P(b)X' (point = point-matrix * coord vector).

    So the follow-up problem is this:

    given basis X1 = [1, 0]t X2 = [1, 1]t what point X has b-coord vector X' = [-3, 2]t.
    Well, it is straight-forward, by formula one gets X = [-1, 2]t.

    So it looks to me that this point is with respect to the standard basis, am I correct?
     
  2. jcsd
  3. Mar 17, 2005 #2

    HallsofIvy

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    Assuming that by "b-coord" means with respect to this given basis, then you are looking for X'= -3[1, 0]+ 2[1, 1]= [-3+2, 2]= [-1, 2] in the standard basis.
     
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