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Thread: Linear transformations

  1. #1
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    Linear transformations

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

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

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

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

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

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

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

    So we get:

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