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Thread: Linear maps

  1. #1
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    Linear maps

    Let S: $\displaystyle \mathbb{R}^2 \to \mathbb{R}^2$ be the function defined by S(x, y) = (x - y, y) for all (x,y) $\displaystyle \varepsilon \mathbb{R}^2$

    Show that S is a linear map

    Now I know that a linear map preserves the shape and takes lines to lines and points to points, but how would you show that S is a linear map using a formula such as
    $\displaystyle S(\alpha \mathbf{x} + \beta \mathbf{y})$?
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  2. #2
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    Quote Originally Posted by SyNtHeSiS View Post
    Let S: $\displaystyle \mathbb{R}^2 \to \mathbb{R}^2$ be the function defined by S(x, y) = (x - y, y) for all (x,y) $\displaystyle \varepsilon \mathbb{R}^2$

    Show that S is a linear map

    Now I know that a linear map preserves the shape and takes lines to lines and points to points, but how would you show that S is a linear map using a formula such as
    $\displaystyle S(\alpha \mathbf{x} + \beta \mathbf{y})$?
    Here, your x is a pair, $\displaystyle (x_1, y_1)$ and your y is a pair, $\displaystyle (x_2, y_2)$. $\displaystyle \alpha\mathbf{x}+ \beta\mathbf{y}= \alpha(x_1, y_1)+ \beta(x_2, y_2)= $$\displaystyle (\alpha x_1+ \beta x_2, \alpha y_1+ \beta y_2)$.

    What to you get when you apply S to that pair?

    (It might be less confusing to do the sum and scalar multiplication separately. That is, find $\displaystyle S(x_1+ x_2, y_1+ y_2)$ and $\displaystyle S(\alpha x_1, \alpha x_2)$.)
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  3. #3
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    You get $\displaystyle S((\alpha x1 + \beta x2) - (\alpha y1 + \beta y2 ), \alpha y1 + \beta y2)$, but I dont know what to to do next.
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  4. #4
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    Quote Originally Posted by SyNtHeSiS View Post
    You get $\displaystyle S((\alpha x1 + \beta x2) - (\alpha y1 + \beta y2 ), \alpha y1 + \beta y2)$, but I dont know what to to do next.
    No, you get $\displaystyle S(\alpha(x_1, y_1)+ \beta(x_2, y_2))= (\alpha x1 + \beta x2) - (\alpha y1 + \beta y2 ), \alpha y1 + \beta y2)$. What you have still has the "S" but S has already been performed.

    Now, separate those: $\displaystyle (\alpha x1 + \beta x2) - (\alpha y1 + \beta y2 ), \alpha y1 + \beta y2)= (\alpha x1-\alpha y1+ \beta x2- \beta y2, \alpha y1+ \beta y2)$$\displaystyle = (\alpha (x1- y1)+ \beta(x2- y2), \alpha y1+ \beta y2)= (\alpha (x1- y1), \alpha y1)+ (\beta(x2- y2), \beta y2)= \alpha(x1- y1, y1)+ \beta(x2- y2, y2)$.
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  5. #5
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    Quote Originally Posted by HallsofIvy View Post
    = \alpha(x1- y1, y1)+ \beta(x2- y2, y2)[/tex].
    How would you get from this step to $\displaystyle \alpha S(x1, y1) + \beta S(x2,y2)$? (I didnt understand this part while looking at the answer in my notes).
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  6. #6
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    Quote Originally Posted by SyNtHeSiS View Post
    How would you get from this step to $\displaystyle \alpha S(x1, y1) + \beta S(x2,y2)$? (I didnt understand this part while looking at the answer in my notes).
    $\displaystyle \alpha(x1- y1, y1)+ \beta(x2-y2, y2)= \alpha S(x1, y1)+ \beta S(x2, y2)$

    Well, you are told that S(x1, y1)= (x1- y1, y1) and S(x2, y2)= (x2- y2, y2), aren't you?
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