Linear Transformation and spanning set I have the following question below:

Let $V$ and $W$ be real vector spaces and $T:V\to W$ be a linear transformation 
  such that $\ker(T) = {0} \subset V$. Let vectors $v_1,v_2,v_3,v_4$ belong to $V$, and $\{T(v_1), T(v_2)\}$ is a generating set for W.
  Is the set of vectors $v_1,v_2,v_3,v_4$ belonging to $V$ a generating set for $V$?

I found this question from an old book. The more current books I think use the word spanning set instead of generating set.
I am not sure how to start this proof off.
My initial guess is to assume 

$w1$ = $T(v1)$, $w2$ = $T(v2)$, so $\{w1,w2\}$ is the spanning set in $W$. Such that some vector $wi$ in $W$ can be represented as $wi$ = a$w1$ + b$w2$. Then by linearity I guess that we can write $wi$ = $T( a$v1$+ b $v2$)$ .

From this point I am stuck to how to show that these 2 vectors are a spanning set for V, and if I can do that, I guess I can then add v3 and v4 to the first two and claim this is a spanning set for V, because a spanning set can have redundant vectors, this is not asking for a basis set, but a spanning set.
Hope someone can help here.
Palu 
 A: First of all, mentioning the vectors $v_3$ and $v_4$ is pure smoke and mirrors, since we know nothing about them (they could even be zero). What is important is the fact that $\ker(T)=0$. This shows that $T$ is injective, so that the image $T(V)$ is isomorphic to $V$. Moreover the fact that $T(v_1)$ and $T(v_2)$ generate $W$ implies that $T$ is also surjective, so $T$ is an isomorphism, and $V$ and $W$ are isomorphic and thus $T(V)=W$. To show that $v_1$ and $v_2$ span $V$ take any vector $v \in V$, take it's image $T(v)$ in $W$ and decompose $T(v)=aw_1+bw_2$ and then take inverse images $v:=av1+bv_2$. This can only happen when $T$ is an isomorphism. An example where $T$ is not injective is given is T is the map $T:\mathbb{R}^2 \rightarrow \mathbb{R}^2$ given by the matrix $
\left(\begin{array}{cc}
1&-2\\
2&-4\\                
\end{array}\right)
$. The vectors $v_1=\left(\begin{array}{c}
1\\
0\\                
\end{array}\right)$ and The vectors $v_2=\left(\begin{array}{c}
0\\
1\\                
\end{array}\right)$ span the vectorspace $V=\mathbb{R}^2$ but their images by $T$ do not. 
A: Hints: Since $W$ have a spanning set with two elements this means
$\dim(W)\leq2$
Also note that $T$ is $1-1$ since it have a trivial kernel, this
means $\dim(V)\leq\dim(W)\leq2$
Now - what in general can we say about $\dim(T(V))$ ? 
Added in the comments: 
Look at the set of inequalities we got: $$\dim(W)=\dim(T(V))\leq\dim(V)\leq\dim(W)$$
and thus $\dim(V)=\dim(W)$ and since $T$ is $1-1$ it is also onto.
Now, since $\{T(v_{1}),T(v_{2})\}$ spans $W$ it have a subset that
is a basis of $W$ (if $\dim(W)=2$ then it is simply $\{T(v_{1}),T(v_{2})\}$
and if $\dim(W)=1$ it is $\{T(v_{1})\}$ or $\{T(v_{2}\}$) 
and $T^{-1}$
is also an isomorphism since $T$ is and it will map a basis to a
basis so $\{v_{1},v_{2}\}$ or $\{v_{1}\}$ or $\{v_{2}\}$ is a basis
for $V$
