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Math Help - Linear transformations

  1. #1
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    Linear transformations

    Consider the basis S=\{v1,v2\} where v1=(1,1) and v2=(1,0) and is T:R^2=>R^2 the linear operator such that T(v1)=(1,-2) and T(v2)=(-4,1). Find a formula for T(x,y) and use this formula to get T(5,-3)
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  2. #2
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    So we use linearity properties to know the range by T of the canonic basis of \mathbb{R}^{2} : (1,0) and (0,1).

    T((1,0))=(-4,1)
    T((0,1))=T((1,1)-(1,0))=T((1,1))-T((1,0))=(1,-2)-(-4,1)=(5,-3)

    So, using again the linearity of T, we can conclude :

    \forall (x,y) \in \mathbb{R}^{2},\ T((x,y))=(-4x+5y,x-3y)

    Therefore T((5,-3))=(-35,14)
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  3. #3
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    Thank you. How do you find: T(x,y)=(-4x+5y,x-3y)
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  4. #4
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    Just write (x,y) in the basis \{(1,0),(0,1)\}:

    T((x,y))=T((x,0)+(0,y))=T(x(1,0)+y(0,1))=xT((1,0))  +yT((0,1))

    But we know that T((1,0))=(-4,1) and T((0,1))=(5,-3)

    So we get:

    T((x,y))=xT((1,0))+yT((0,1))=x(-4,1)+y(5,-3)=(-4x,x)+(5y,-3y) =(-4x+5y,x-3y)
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  5. #5
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    ok thank you
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