# Thread: Linear Mapping Image

1. ## 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.

2. Originally Posted by slevvio
$f: \mathbb{R}^3 \rightarrow \mathbb{R}^2$ 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, $(1,0) = \tfrac{1}{2} \left[ (1,1)+(1,-1) \right]$.