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    • Can someone explain why (yx) -> (y+12x) is not a linear transformation and why (yx) -> (3y-x) is linear transformation?thank you
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    (Original post by alesha98)
    • Can someone explain why (yx) -> (y+12x) is not a linear transformation and why (yx) -> (3y-x) is linear transformation?thank you
    When is something a linear transformation?
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    (Original post by alesha98)
    • Can someone explain why (yx) -> (y+12x) is not a linear transformation and why (yx) -> (3y-x) is linear transformation?thank you
    Does the first one satisfy the property that \mathbf{T}(ax) = a\mathbf{T}(x)?
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    (Original post by Zacken)
    Does the first one satisfy the property that \mathbf{T}(ax) = a\mathbf{T}(x)?
    Am i on the right line?
    For x:
    T(kx) = Tk(x)
    1(1x) = 1(2)(x)
    (x) = 2(x)
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    (Original post by alesha98)
    Am i on the right line?
    For x:
    T(kx) = Tk(x)
    1(1x) = 1(2)(x)
    (x) = 2(x)
    Yes, so the 'x' component works.

    What about the "y+1" bit? Does T(ay) = aT(y)?
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    (Original post by Zacken)
    Yes, so the 'x' component works.

    What about the "y+1" bit? Does T(ay) = aT(y)?
    T(ky) = Tk(y)
    (y) = (y+1)
    In this case, i cant move out +1 from the bracket , so it doesnt satisfy?
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    (Original post by alesha98)
    T(ky) = Tk(y)
    (y) = (y+1)
    In this case, i cant move out +1 from the bracket , so it doesnt satisfy?
    pretty much
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    (Original post by Zacken)
    pretty much
    so when i am doing this kind of questions, i need to solve x and y separately?
    Can u show me a working example?
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    (Original post by alesha98)
    so when i am doing this kind of questions, i need to solve x and y separately?
    Can u show me a working example?
    Uh, you should just be able to spot whether they're a linear transform or not? :confused: not sure what you want
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    (Original post by Zacken)
    Uh, you should just be able to spot whether they're a linear transform or not? :confused: not sure what you want
    ah ok, so i dont need to prove it.
    for linear transformation, it cant be
    1. ie (x,y) -> (xy,y). So no xy element in the transformation
    2. ie (x,y)->(x+1,y). So no constant in the transformation?
 
 
 
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