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Affine transformation proof

  1. Feb 13, 2010 #1
    1. The problem statement, all variables and given/known data

    An affine transformation T : R^n --> R^m has the form T(x) = Ax + y, where x,y are vectors
    and A is an m x n matrix and y is in R^m. Show that T is not a linear transformation when
    b != 0.


    2. Relevant equations

    There are couple of useful definitions, but I think this one will suffice :

    Definition :

    If T is a linear transformation, then T(0) = 0

    and T(cU + dV) = cT(U) + dT(V), for all vectors U,V in the domain of T and all scalars c,d.


    3. The attempt at a solution

    I figure I would show that the affine transformation does not satisfy the definition.

    Attempt :

    Let U,V be vectors, and c,d be scalers.

    Then the affine transformation has to satisfy T(cU + dV) ?= cT(U) + dT(V).

    since T(x) = Ax + y then

    T(cU + dV) = A(cU + dV) + y

    = cAU + dAV + y

    = cAU + dT(V)

    thus cAU + dT(V) != cT(U) + dT(V), which means that affine transformation is not a
    linear transformation.

    Would this suffice or Do I make some assumption that are not plausible. BTW, there
    was a theorem that said that If A is an m x n matrix, then the transformation x -> Ax has
    the properties A(U + V) = AU + aV and A(sU) = sA(U), where U,V are vectors in R^n
    and s is a scaler.
     
  2. jcsd
  3. Feb 14, 2010 #2
    Yes, it suffices to show that T(cU + dV) =!= cT(U) + dT(V). However, there's an easier way:

    If T(x) = Ax + y, then T(0) = y =!= 0 unless y=0.
     
  4. Feb 14, 2010 #3
    Oh, I should have read the question carefully. Thanks a lot.
     
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