Calculate a linear transformation with a specific kernel I just want to make sure that what I'm doing is correct. Here's the question:

Determine a linear transformation $T$: $\mathbb{R^3} \rightarrow \mathbb{R^2}$ with kernel $W$:
$W$ = {$(x,y,z)$ | $x-2y+z=0$}

So what I did to define a transformation considering that
$T (1,0,0) = (0,0)$
$T (0,1,0) = (0,1)$
$T (0,0,1) = (1,0)$
I concluded that
$T(x(1,0,0) + y(0,1,0) + z(0,0,1))=(z,y)$
My doubt is in my next step...
So the kernel of the transformation is $z=0 \wedge y=0$
But because W should be the kernel and by W $z=-x+2y$ we have that our kernel is $(x,0,-x)$ so it's the vector (1,0,-1)...
Can someone please help me verify if this is correct?
 A: wrong part of your answer is defining  $ T (1,0,0) = (0,0), T (0,1,0) = (0,1), T (0,0,1) = (1,0)$. using these definitions, the constraint of null space wont be satisfied. a simple way i know for solving this category is as below. 
assume $(x\; y\; z)^T$ exists in null space of our transformation. then it can be written as:
$$\begin{pmatrix} x\\y\\z \end{pmatrix} = t \begin{pmatrix} 1\\-2\\1 \end{pmatrix}$$ this gives us two equations $x=z$ and $y=-2x$. these two equations are spanning row space of our transformation. writing them in following form and we get: 
$$\begin{matrix}
        (1)x & (0)y & (-1)z & = & 0 \\
        (2)x & (1)y & (0)z  & = & 0\\
\end{matrix}$$
which gives us the transformation matrix as: $$T = \begin{pmatrix}
        1 & 0 & -1\\
        2 & 1 & 0\\
\end{pmatrix}$$
A: Firstly, we can find a basis for the nullspace. We have that:
$$W \begin{array}[t]{l}= \{ (x,y,z)^T: x-2y+z = 0, x,y,z \in \mathbb R \}\\
= \{ (x,y,z)^T: z = -x+2y, x, y, z\in \mathbb R\}\\
= \{ (x,y,-x+2y)^T: x,y \in \mathbb R\}\\
= \{x\cdot (1,0,-1)^T + y\cdot(0,1,2)^T:x,y \in \mathbb R\}.
\end{array}$$
Thus, $$W = \langle \underbrace{(1,0,-1)^T}_{\vec e_1},\,  \underbrace{(0,1,2)^T}_{\vec e_2}\rangle.$$
From rank-nullity theorem we have that:
$$\dim (\text{Im} T) + \dim (\ker T) = 3 \implies \dim (\text{Im} T)=1.$$
This means that our $2\times 3$ transformation matrix will have rank $1.$
Thus,
$$A = \begin{bmatrix} a & b & c \\ \lambda a & \lambda b &\lambda c \end{bmatrix}.$$
So we need to define $a,b,c$.
From $\begin{bmatrix} a & b & c \\ \lambda a & \lambda b &\lambda c \end{bmatrix} \cdot \begin{bmatrix} 1 \\ 0 \\ -1\end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \implies a = c$
From $\begin{bmatrix} a & b & c \\ \lambda a & \lambda b &\lambda c \end{bmatrix} \cdot \begin{bmatrix} 0 \\ 1 \\ 2\end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \implies b = -2c.$
Thus, our matrix $A$ is of the form:
$$A = \begin{bmatrix} a & -2a & a\\ \lambda a & - 2\lambda a& \lambda  a\end{bmatrix}.$$
Choosing $a  = 1$ and $\lambda = 3$ , we have that:
$$T\begin{bmatrix} x \\ y \\ z\end{bmatrix} = \begin{bmatrix} 1 & - 2 & 1\\3 & -6 & 3\end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix}= \begin{bmatrix} x-2y+z \\ 3x-6y+ 3z \end{bmatrix}$$
