Mapping of zero vector under a linear transformation Let $T$ be a linear transformation from a vector space $V$ to $W$ that are defined over a field $\mathbb F$. 
Now, $T(\mathbf0)=\mathbf0$. 
Which implies the zero vector in $V$ can't move or be transformed to a new vector in $W$. Why is it so? 
Does it also have a geometrical significance? 

Since you're asking for a geometrical interpretation:
First of all, if you have a linear transformation $T$ and use it on all vectors in a vector space $T$, then the mapped vectors will form a vector space again. What's important is that the zero vector is a part of every vector space. 
So intuitively, when you map all vectors from a space $V$ to another space $W$, then there must be some vector that gets mapped to the zero vector. (As it has already been pointed out, it follows from the linearity criterion that $T(0)=0$, which means that this vector we are looking for is indeed the zero vector from $V$)
Let's visualise what a transformation $T: \mathbb R^2 \rightarrow \mathbb R^2$ does to our space. In this case, we are not moving from one vector space to another, but rather just mapping $\mathbb R^2$ onto itself. This leads to a distortion of the space.
For example, if we take the matrix
$$B = \begin{pmatrix}1 & 0 \\ 1 & 1\end{pmatrix}$$
and apply it to all vectors on the unit circle, the result will look like this:
Basically, what is happening is that we stretch and turn our space in different directions. Think of drawing a circle on a balloon and then, depending on your transformation, you just stretch out the surface. 
It should be clear now, why the zero vector gets mapped to itself: If you stretch something that has $0$ length, then of course it will have no length afterwards either. You could again think of the balloon example and see for yourself that a small dot would not really change that much, no matter how much you stretch the balloon.
