Prove that set of linear transformations is infinite-dimensional if domain of linear maps is finite-dimensional and codomain is infinite-dimensional

Question

Suppose $V$ is finite-dimensional with $dim\ V > 0$, and suppose $W$ is infinite-dimensional. Prove that $\mathcal{L}(V,W)$ is infinite-dimensional.

Notation: V, W are vector spaces over complex or real numbers. L(V,W) stands for set of linear maps/linear transformations from V to W.

My question is regarding this solution(source:https://linearalgebras.com/3a.html): why is it correct to look at the particular value of the functions $T_1,...,T_m$ on $v=v_1$ to prove that the list of functions(in particular linear transformations) $T_1,...,T_m$ is linearly independent? Shouldn't the implication of the statement:

$a_1T_1(v_1)+a_2T_2(v_1)...+a_mT_m(v_1)=0$ implies $a_1=...=a_m=0$

be

"$T_1(v_1),T_2(v_1),...,T_m(v_1)$ is a list of elements of $W$ that is linearly independent."

instead of:

"$T_1,T_2,...,T_m$ is a list of elements of $\mathcal{L}(V,M)$ that is linearly independent."?

Why isn't this proof as erroneous as saying that "the list of polynomials(reals to reals) $f,g$ in which $f(x)=x$ and $g(x)=x^2$ is linearly dependent because $a_1f(0)+a_2g(0)=0a_1+0a_2=0$ is true for multiple real values of $a_1,a_2$"?

Answer 1

So the fact that $T_1(v_1), T_2(v_2), ...$ are linearly independent is trivial because they were defined to be that way. The implication is that we assumed $$a_1T_1 + a_2T_2 + ... + a_mT_m = 0 $$ and now we have shown that $a_1, ..., a_m = 0$. Notice that there are no $v_i$'s in that line. Thats because this is a linear combination of linear maps, not vectors. So we have shown this linear combination of functions is the 0 function.