linear-transformations
Let T be a linear transformation from a vector space U(F) into a vector space V(F). Then
T(0) = 0 where 0 on the left hand side is zero vector of U and 0 on the right hand side is zero vector of V.
Why is there a explicit mention about different zero vectors ? How can two vector spaces can have two different definitions for zero vector ?
We are given that $f$ is linear; we just need to prove that it is both injective and surjective.
Injectivity: we need to show that if $f(x) = f(y)$, then $x = y$. That is, if $$ f(a_1 v_1 + \cdots + a_n v_n) = f(b_1 v_1 + \cdots + b_n v_n) $$ then $$ a_1 v_1 + \cdots + a_n v_n = b_1 v_1 + \cdots + b_n v_n $$
Surjectivity: If $y \in W$, then we can write $$ y = a_1 f(v_1) + \cdots + a_n f(v_n) $$ find an $x \in V$ such that $f(x) = y$.
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.
Best Answer
Because a vector space is an abstract object, the elements of which can be anything. Just to give an example: the zero of $\mathbb{R}$ is not an element of $\mathbb{R}^{\mathbb{N}}$.