Find kernel and image of linear transformation.

Question

I am given transformation : $f:R^3 \rightarrow R^2$
$ f(x,y,z)=(-x+y+z,x-y+z)$

I am requested to find kernel and image of this transformation. I am finding kernel: $ (-x+y+z,x-y+z)=(0,0 )$

$\begin{cases} -x+y+z=0 \\ x-y+z=0 \\ \end{cases} \Rightarrow \begin{cases} x=y\\ z=0 \\\end{cases}$

$ Ker f = \{ (x,x,0): x\in R^{3} \} = Lin\{(1,1,0) \} $ Am I doing this right? I have a problem with finding Im of this transformation. I do not really understand how to do this. I will be glad to be instructed how to do so.

Answer 1

You're right: the kernel is correct. Now by the Rank-Nullity theorem: $$\operatorname{rank}(f)=\dim \mathbb R^3-\dim\ker f=2=\dim \mathbb R^2$$ hence $f$ is surjective.