How do I find the Kernel and Range of composition linear transformations? 
Let $G(x,y,z)=(0,x+y+z)$ and $H(x,y)=(x−y,0,x)$.
Find the kernel and range of $G\circ H : \Re^2 → \Re^2$

My Approach
Finding $G\circ H$:
$$G\circ H= G(H(x,y))$$
$$=G(x-y,0,x)$$
$$=(0,x-y+x)=(0,2x-y)$$
How does one find the Kernel and Range of this? I can't find it in my book.
 A: To find the kernel, you're looking for $(x,y)$ such that $(0,2x-y) = (0,0)$. So $(x,y)$ will be in the kernel if and only if $y=2x$. So the elements of the kernel are the elements of the form $(x,2x) = x(1,2)$. In other words, this is the span of the vector $(1,2)$ Or the line y=2x, if you like.
To find the range, you need to determine for what vectors $(a,b)$ is there a vector $(x,y)$ with $(0,2x-y) = (a,b).$ Clearly we must have $a=0$. Now if $b$ is any real number, then $G(H(0,-b)) = (0,b)$. The vectors of the form $(0,b)$ are just the vectors in the span of $(0,1)$. So the range is just the $y$-axis.
A: Since $G \circ H(x,y) = (0,2x-y)$, the kernel of $G \circ H$ is given by
\begin{equation}
\begin{split}
\text{ker}(G \circ H) & = \{(x,y) \in \mathbb{R}^2: (0,2x-y) = (0,0)\}\\
& = \{(x,y)\in \mathbb{R}^2: y = 2x\}\\
& = \{t(2,1) \in \mathbb{R}^2: t \in \mathbb{R}\}\\
& = \langle (2,1) \rangle.
\end{split}
\end{equation}
For $t \in \mathbb{R}$, we have
$$(G \circ H)(0,-t) = (0,t).$$
Hence the range of $G \circ H$ is given by
\begin{equation}
\begin{split}
\text{Im}(G \circ H) & = \{(0,2x-y) \in \mathbb{R}^2: x,y \in \mathbb{R}\}\\
& = \{(0,t)\in \mathbb{R}^2: t \in \mathbb{R}\}\\
& = \langle (0,1) \rangle.
\end{split}
\end{equation}
