linear polynomial of linear transformations. I got stuck at the following problem 

Let $V,W$ two finite dimensional (m and n respectively) vector spaces over $\mathbb{C}$, and $T,S: V \to W$ linear applications, suppose that $T$ is onto. Prove that $f(x)=xS+T$ is onto for all $x\in \mathbb{C}$ except for a finite number of values. 

Now the thing is that I don't know how to get further than the following: Given $Z \in L(V,W)$, consider 
$$ Z= xS+ T $$
and find an $ x \in \mathbb{C}$ such that the equation holds, if I solve for $x$ I'll get $x=S^{-1}(Z-T)$ however I have no priori information on the invertibility  of $S$, and there's no use of $T$ being onto, another is to find the values of $x$ such that $0=xS+(T-Z)$ which only happens when $Z=T$ and $x=0$.
Any ideas?
Thanks a lot. 
 A: I have this idea. Like $n\leq m$ (because $T$ is onto), then in some bases $B$ and $B'$ of $V$ and $W$ respectively the matrix of $|f(x)|_{BB'}\in \mathbb{C}^{n\times m}$ and $f(x)$ will be onto if and only if all its columns are linearly independent. Or wich is the same if and only if all its minor determinants of $n \times n$ (which are a finite number, exactly ${m\choose n}$) are not zero. 
But a minor determinant of $n \times n$ of $|f(x)|_{BB'}\in \mathbb{C}^{m\times n}$ is a polinomial of degree $n$ or less in $x$, so only finite $x$ has the property that $f(x)$ is not onto.
Then there are at most $n{m \choose n}$ values of $x$ for wich $f(x)$ is not onto.
A: $V$ must have an $n$-dimensional subspace, call it $U$, such that the restriction of $T$ to $U$, call it $T'$, is onto (in fact, is an isomorphism). Let $S$ restricted to $U$ be called $S'$. Then the matrix representing $xS'+T'$ has, as its determinant, a polynomial of degree $n$ in $x$, hence, at most $n$ values of $x$ for which $xS'+T'$ is not onto. Then it follows that except for those values of $x$, $xS+T$ is onto. 
