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Homework Help: Linear Algebra - Linear transformations

  1. Aug 27, 2010 #1
    1. The problem statement, all variables and given/known data

    which of the following are linear transformations.

    a) L(x,y,z) = (0,0)
    b) L(x,y,z) = (1 ,2, -1)
    c) L(x,y,z) = (x^2 + y, y - z)

    3. The attempt at a solution

    I know that L is a linear transformation if L(u + v) = L(u) + L(v) and L(ku) = kL(u).

    I am not sure how to apply that to those equations.

    Please help.
     
  2. jcsd
  3. Aug 27, 2010 #2

    Mark44

    Staff: Mentor

    Start with a couple of vectors, say u = <x1, y1, z1> and v = <x2, y2, z2>.

    For a), see if L(u + v) = L(u) + L(v) and L(ku) = kL(u). Do the same for b) and c).
     
  4. Aug 27, 2010 #3
    I guess my problem is, I don't know how to apply that to L(x,y,z) = (0,0). If it was L(v) = Av I would understand. I just don't understand the transformations themselves and how to apply them.
     
  5. Aug 27, 2010 #4

    Office_Shredder

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Well if it was L(v)=Av it wouldn't be interesting to ask if the transformation was linear!

    For example for the first one if (x,y,z)=(1,2,4)

    L(1,2,4)=(0,0).

    And if (x,y,z)=(1,1,1)

    L(1,1,1)=(0,0)

    So L(1,2,4)+L(1,1,1)=(0,0)+(0,0)=(0,0).

    Is this equal to L(2,3,5)? ((1,2,4)+(1,1,1))
     
  6. Aug 27, 2010 #5
    Ok, I think I understand. so that is linear because L(2,3,5) = (0,0). And L(ku) = kL(u) = (0,0).

    But for b), the second part doesn't hold up because L(ku) = (1,2,-1) but kL(u) = k(1,2,-1) and they are not equal for all k.

    and c probably shows something similar. I will have to write it out in a bit in Maple.

    Thanks!
     
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