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? 

You are being confused. There are two statements going on here.
The first statement is a property of linear transformations. Specifically, it is true that $G(\lambda \mathbf{u}) = \lambda G(\mathbf{u})$ for any $\lambda$ and any $\mathbf{u}$.
A consequence of this is that every linear map must map $\mathbf{0}$ to $\mathbf{0}$, because $\mathbf{0} = 0\mathbf{v}$ for any vector $v$, so $$G(\mathbf{0}) = G(0\mathbf{v}) = 0G(\mathbf{v}) = \mathbf{0}.$$
This property holds true for any vector $\mathbf{v}$ we choose, and certainly it holds if we wish to choose $\mathbf{v} = \mathbf{0}$.
So, this is a necessary condition for a mapping $G$ to be linear.
Now, let's see what happens under your mapping $G_1$ when applied to $\mathbf{0}$:
$$G_1(\mathbf{0}) = (1,0,0)^T.$$
Since $G_1$ doesn't map $\mathbf{0}$ to $\mathbf{0}$, then it cannot be linear!
Therefore, it doesn't make sense to try to pull zero out of the function, because the function is not linear. Hence, $G_1$ does not satisfy the property that $G_1(\lambda \mathbf{u}) = \lambda G_1(\mathbf{u})$.
