Linear map from zero vector to zero vector. I am reading an introduction on linear maps in my text book on linear algebra. The following statements are made:
Suppose $G_1 (\vec{u}) = (x_1 + 2x_2 + 3x_3 + 1, 4x_1, 9x_3)$
Then we can use the following property of linear maps. Let $\lambda = 0$ and $\vec{u} = \vec{0}$
$$G(\lambda\vec{u}) = \lambda G(\vec{u})$$
And specifically:
$$G(\vec{0}) = 0 \cdot G(\vec{0}) = \vec{0}$$
This means that a linear map maps the zero vector to the zero vector. It also means that $G_1$ cannot be a linear map, this is because $G_1(0,0,0) = (1,0,0) \neq (0,0,0)$. The constant term $1$ is breaking the linearity.
My analysis
I don't understands the above statements completely. For example this statement: $G(\vec{0}) = 0 \cdot G(\vec{0}) = 0$ should be true for any function $G(\vec{u})$, since whatever result of the  map $G(\vec{u})$ will be it will be multiplied by $0$ and result in $\vec{0}$. In the case above it would be $0 \cdot (1,0,0) = \vec{0}$. This would map the zero vector to the zero vector and hence be a correct linear map?
Can anyone please explain this to me? 

The vector $\vec{0}$ can be written as $0$ . $\vec{u}$ for any $\vec{u}$, therefore 
$G(\vec{0})= G(0\vec{u}) =0 \cdot G(\vec{u}) = \vec{0}$
You can also write $\vec{0}$ as $\vec{0}+\vec{0}$ and argue that 
$G(\vec{0}) = G(\vec{0}+\vec{0}) = G(\vec{0}) + G(\vec{0})$ $\Rightarrow$ $G(\vec{0})=\vec{0}$
