Determine the kernel of the linear map $T : \mathbb{R}^3\to\mathbb{R}^2$ with... Determine the kernel of the linear map $T : \mathbb{R}^3 \to \mathbb{R}^2$ with:

$$T\left(\begin{matrix} x \\ y \\ z \end{matrix}\right)=\left(\begin{matrix} x-y+z \\ x+z \end{matrix}\right)$$

Can anyone please explain.
 A: There are 2 ways to write this up, both are named by aston villa but one is explicit and one is not (he just says to pick a basis). I will show you how to pick the basis. Start with the generic $(x,y,z)^T$ and impose $y=0,x=-z$ to get
$$
\begin{pmatrix} x \\ y \\ z \end{pmatrix}
= \begin{pmatrix} -z \\ 0 \\ z \end{pmatrix}
= z \begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix},
$$
so the set of all solutions is 1-dimensional (i.e. a line) with the basis $(-1,0,1)^T$.
A: We need to find all $v=(x,y,z)$ for which $Tv=0$. Now $Tv=0$ if and only if $Av=0$ (check, it is very easy), where \begin{equation}A=\begin{bmatrix}1&-1 &1 \\1 & 0 &1\end{bmatrix}\sim\begin{bmatrix}1&0 &1\\ 0 & 1 &0\end{bmatrix}=B,\end{equation} where $B$ is a row echelon form of $A$. Again $Av=0$ if and only if $Bv=0$ and $B$ is 'simple' enough. It tells us that $y=0$ and $x+z=0$. Rank is 2, and we need a fee parameter ($3-2$), so we let $z=s\in R$ and use the previous two equations to write $x,y$ in terms of this $s$.
Note: All we needed to do here is to solve for the solution space of $A$.
The kernel is a subspace of domain space of $T $ so to be precise it should be given as three touples, or we can give a basis.
