If $T$ is surjective, so is $T+tS$ except at finite values of $t$ I am trying to work out this problem

Let $V$ and $W$ be complex finite dimensional vector spaces. Let $T,S:V\to W$ be linear transformations with $T$ surjective. Show that $T+tS$ is onto for all but finitely many complex values of $t$.

I tried the following:
Suppose $V$ and $W$ have dimensions $n$ and $m$ respectively, clearly $n\geq m$. Suppose that $A=\{v_i\}$ be a basis for $V$. Since $T$ is surjective $\{Tv_i\}$ span $W$. Let us assume that $T+tS$ is not surjective, then $B_t:=\{(T+tS)v_i\}$ has cardinality $n\geq m$ and thus must be linearly dependent. Then there exist a non-trivial sequence $\{c_i(t)\}_i^n$ such that $T(u_t)+tS(u_t)=0$, where $u_t:=\sum_i c_i(t)v_i\neq 0$.
Now let $t,t'\in \Bbb C$ such that $T+tS$ and $T+t'S$ are not onto. Then
$$T(u_{t'})+t'S(u_{t'})=0\tag{1}.$$ Note that if $u_t=\lambda u_{t'}$, then we could multiply eq (1) by $\lambda$ and get $T(u_t)+t'S(u_{t})=0$, which in turn implies
$$(t'-t)S(u_t)=0.$$
So I was thinking that if we could prove that $S(u_t)\neq 0$, then we would have $t=t'$ and thus we could only choose up to $m$ possible complex values of $t$ for which $T+tS$ is not onto. So, am I in the right track? How could I continue?
 A: Let $B_W=\{w_1,\ldots,w_m\}$ be a basis of $W$. For each $j\in\{1,2,\ldots,m\}$, let $v_j\in V$ be such that $T(v_j)=w_j$. Since $B_W$ is linearly independent, then so is $\{v_1,\ldots,v_m\}$. Let $v_{m+1},\ldots,v_n\in V$ be such that $\{v_1,\ldots,v_n\}$ is a basis $B_V$ of $V$. Then if you consider the matrix $[T]_{B_V}^{B_W}$, its first $m$ columns will be$$\begin{array}{c}1\\0\\0\\\vdots\\0\end{array},\ \begin{array}{c}0\\1\\0\\\vdots\\0\end{array},\ \ldots,\ \begin{array}{c}0\\0\\0\\\vdots\\1\end{array},$$whose determinant is $1$. Now, consider the determinant of the matrix which consists of the first $m$ columns of $[T+tS]_{B_V}^{B_W}$. It's a polynomial in $t$ which takes the values $1$ at $0$. So, it is not the null polynomial, and therefore it can take the value $0$ only finitely many $t$'s. And for those $t$'s such that the determinant is not $0$, $T+tS$ is surjective, since$$\operatorname{span}\bigl(\{(T+tS)(v_1),\ldots,(T+tS)(v_m)\}\bigr)=W.$$
A: Let $m=\dim W$. Since $T$ is surjective, its restriction on some $m$-dimensional subspace $U\subseteq V$ is bijective. If $T+tS$ is not surjective, $(T+tS)|_U$ is also not surjective. Therefore the linear map $f:U\to U$ defined by
$$
f=T|_U^{-1}\circ(T+tS)|_U
=\operatorname{Id}_U\,+\,t\,T|_U^{-1}\circ S|_U
$$
is not surjective. This means $f$ is singular and $-\frac{1}{t}$ is an eigenvalue of $T|_U^{-1}\circ S|_U$. However, as $U$ is finite-dimensional, $T|_U^{-1}\circ S|_U$ has only finitely many eigenvalues. Hence the result follows.
