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Math Help - Linear Transformation problem

  1. #1
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    Linear Transformation problem

    Determine if the transformation T: R^{2}\rightarrow R^{2} is linear if T(x, y)= (x+1, 2y)



    1. T(u + v) = T(u) + T(v)
    2. T(c*u) = cT(u)
    3. T(0) = 0

    I believe I have to use the above provided equations to determine whether T(x, y)= (x+1, 2y) is linear or not. If I use the third equation above, I get T(0,0) = T(0 + 1, 0) = (1, 0). Therefore the transformation is not linear. Am I right?
    Last edited by temaire; February 5th 2010 at 01:27 AM.
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  2. #2
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    Quote Originally Posted by temaire View Post
    Determine if the transformation T: R^{2}\rightarrow R^{2} is linear if T(x, y)= (x+1, 2y)



    1. T(u + v) = T(u) + T(v)
    2. T(c*u) = cT(u)
    3. T(0) = 0

    I believe I have to use the above provided equations to determine whether T(x, y)= (x+1, 2y) is linear or not. If I use the third equation above, I get T(0,0) = T(0 + 1, 0) = (1, 0). Therefore the transformation is not linear. Am I right?
    You have to prove the homogeneity condition T(\lambda x,\lambda y) = (\lambda x+1, \lambda 2y) = \lambda(x+1, 2y) = \lambda T(x,y), for some scalar \lambda. And you must show the additivity condition as well.

    (To prove your third condition you must set \lambda = 0, so you will have T(0) = 0)
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  3. #3
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    So I couldn't have used the condition set by T(0) = 0? I believe it is a result of the two definitions you mentioned.
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  4. #4
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    Quote Originally Posted by temaire View Post
    So I couldn't have used the condition set by T(0) = 0? I believe it is a result of the two definitions you mentioned.
    Yeh, I went back and edited my post just before I saw this.
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  5. #5
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    Ok, thanks for the help.
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  6. #6
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    Quote Originally Posted by Roam View Post
    You have to prove the homogeneity condition T(\lambda x,\lambda y) = (\lambda x+1, \lambda 2y) = \lambda(x+1, 2y) = \lambda T(x,y), for some scalar \lambda. And you must show the additivity condition as well.

    (To prove your third condition you must set \lambda = 0, so you will have T(0) = 0)
    Since he is showing that this T is NOT a linear transformation, he only has to show that one of the conditions is not true. What temaire did in his first post is sufficient.
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  7. #7
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    Yes, what he needs to do is to verify that they are all true, or show that one or more are not true, since the violations of either conditions is sufficient for nonlinearity. The additivity condition is violated because, if we let u = (u_1, u_2) and v = (v_1, v_2) then T(u + v) = (u_1 + v_1+1, 2 u_2+v_2) \neq (u_1 +1, 2u_2) + (v_1+1, 2u_1) = T(u)+T(v).
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  8. #8
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    Quote Originally Posted by Roam View Post
    Yes, what he needs to do is to verify that they are all true, or show that one or more are not true, since the violations of either conditions is sufficient for nonlinearity. The additivity condition is violated because, if we let u = (u_1, u_2) and v = (v_1, v_2) then T(u + v) = (u_1 + v_1+1, 2 u_2+v_2) \neq (u_1 +1, 2u_2) + (v_1+1, 2u_1) = T(u)+T(v).
    Yes, that would be sufficient but my point is that showing that T(0,0)\ne 0 which is what he did, is also sufficient. And since it was his problem, his method is the one he should use.
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