I have to find the basis for the kernel and image of the following linear map.
$ \phi: R^3 → R^2, ϕ \begin{pmatrix} \begin{pmatrix} x \\y \\z \end{pmatrix}\end{pmatrix}= \begin{pmatrix} x -y \\z \end{pmatrix} $
For the range, I think we can express any arbitrary linear transformation as:
$ x\begin{pmatrix} 1 \\ 0 \end{pmatrix} - y\begin{pmatrix} 1 \\ 0 \end{pmatrix} + z\begin{pmatrix} 0 \\ 1 \end{pmatrix} $. So I think that a basis for the range would be $\left\{{{\begin{pmatrix} 1 \\ 0 \end{pmatrix}},{\begin{pmatrix} 0 \\ 1 \end{pmatrix}}}\right\}$
As for the kernel, we set
$\begin{pmatrix} x -y \\z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$
We get $x=y$ and $z=0$. Therefore a basis for the kernel would be
$\left\{\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}\right\}$
I am doing this right?
Thanks in advance