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Math Help - linear transformations

  1. #1
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    linear transformations

    I have to show that transformation T defined by T(x1,x2)=(2x1-3x2,x1+4,5x2) is not linear?. I know by definition that T(u+v)= T(u)+T(v) and that T(cv)=cT(v). But how do I use this to prove the question?
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  2. #2
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    Re: linear transformations

    Quote Originally Posted by bonfire09 View Post
    I have to show that transformation T defined by T(x1,x2)=(2x1-3x2,x1+4,5x2) is not linear?. I know by definition that T(u+v)= T(u)+T(v) and that T(cv)=cT(v). But how do I use this to prove the question?
    You just need to show that either of the two equations does not hold for ALL vectors.

    Hint: Try the values

    \mathbf{u}=\mathbf{v}=\vec{0}

    and show that

    T(\mathbf{u}+\mathbf{v}) \ne T(\mathbf{u})+T(\mathbf{v})

    Remeber to disprove a universal statement (for all) you need to only produce one counter example.
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  3. #3
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    Re: linear transformations

    one simple property of ANY linear transformation T is that: T(0) = 0 (where these are 0-vectors, not 0-scalars).

    here is the proof:

    T(0) = T(0+0) (since 0 = 0+0)

    = T(0) + T(0) (since T is linear).

    subtracting T(0) from both sides, we get:

    0 = T(0) - T(0) = [T(0) + T(0)] - T(0) = T(0) + [T(0) - T(0)] = T(0) + 0 = T(0).

    now let's look at YOUR function:

    T(x,y) = (2x-3y,x+4,5y).

    the 0-vector of R2 is (0,0), and T(0,0) = (2*0-3*0,0+4,5*0) = (0,4,0) ≠ (0,0,0), so....
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  4. #4
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    Re: linear transformations

    I tried it with arbitrary values actually. I let u=(x1,x2) and v=(y1,y2). So T(u+v)=T(x1+y1,x2+y2). So it then becomes T=(2x1+2y1-3x2-3y2,x1+y1+4,5x2+5y2 which is the same as T(u)+T(v)=(2x1-3x2,x1,5x2)+(2y1-3y2,y1+4,5y2) but i dont know where to take it from here? As you can see T(u) /= T(v),
    Last edited by bonfire09; September 15th 2012 at 09:24 PM.
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  5. #5
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    Re: linear transformations

    Quote Originally Posted by bonfire09 View Post
    So T(u+v)=T(x1+y1,x2+y2). So it then becomes T=(2x1+2y1-3x2-3y2,x1+y1+4,5x2+5y2 which is the same as T(u)+T(v)=(2x1-3x2,x1,5x2)+(2y1-3y2,y1+4,5y2)
    You got the second coordinate of T(u) wrong.

    Quote Originally Posted by bonfire09 View Post
    but i dont know where to take it from here?
    Quote Originally Posted by TheEmptySet View Post
    Remeber to disprove a universal statement (for all) you need to only produce one counter example.
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  6. #6
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    Re: linear transformations

    My bad I think I figured it out. Since T(u+v)=T(u)+T(v), im supposed to evaluate each vector separately on the RHS of the equation. So vector u=[x1,x2] and v=[y1,y2]. T(u)=[2x1-3x2,x1+4,5x2] and T(v)=[2y1-3y2,y1+4,5y2] so T(u)+T(v)=[2x1-3x2+2y1-3y2,x1+y1+8.,5x2+5y2]. And that does not equal the LHS of the equation making this not a linear transformation. I know when writing vectors im not supposed to use comma's to separate the entries but I
    don't know latex yet.
    Last edited by bonfire09; September 16th 2012 at 08:57 PM.
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