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Math Help - Linear Mapping Image

  1. #1
    Senior Member slevvio's Avatar
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    Linear Mapping Image

     f: \mathbb{R}^3 \rightarrow \mathbb{R}^2 given by  f(x,y,z) = (x+ 2y + 3z, x-y) is a linear mapping. Determine its image and give a basis for the image.

    Is this correct?

     Im f = \{f(\bold{x}) : \bold{x} \in \mathbb{R}^3 \} = \{f(x,y,z): x,y,z \in \mathbb{R}\}
    = \{(x+2y+3z)(1,0) + (x-y)(0,1): x,y,z \in \mathbb{R} \}
     = sp((1,0),(0,1))

    So the basis is (1,0),(0,1). Any help would be appreciated.
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  2. #2
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    Quote Originally Posted by slevvio View Post
     f: \mathbb{R}^3 \rightarrow \mathbb{R}^2 given by  f(x,y,z) = (x+ 2y + 3z, x-y) is a linear mapping. Determine its image and give a basis for the image.

    Is this correct?

     Im f = \{f(\bold{x}) : \bold{x} \in \mathbb{R}^3 \} = \{f(x,y,z): x,y,z \in \mathbb{R}\}
    = \{(x+2y+3z)(1,0) + (x-y)(0,1): x,y,z \in \mathbb{R} \}
     = sp((1,0),(0,1))

    So the basis is (1,0),(0,1). Any help would be appreciated.
    The image is the set, \{ (x+2y+3z,x-y)|x,y,z\in \mathbb{R} \}.
    Now, (x+2y+3z,x-y) = (x,x) + (2y,-y) + (3z,0) = x(1,1)+y(1,-1)+z(1,0).

    Thus, \{ (1,1),(1,-1),(1,0) \} spams the the image. But it is not a basis we just need to remove the unnecessary vectors that can be duplicated from other ones, such as, (1,0) = \tfrac{1}{2} \left[ (1,1)+(1,-1) \right].
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