I am having a bit of trouble trying to understand the last little bit of this solution: Prove that there exist $w_1, \dots, w_m \in W$ such that no $T \in \mathcal L(V,W)$ satisfies $Tv_k=w_k$ for each $k=1, \dots ,m$.
The question: Suppose $v_1,…,v_m$ is a linearly dependent list of vectors in V. Suppose also that $W≠{0}$. Prove that there exist $w_1,…,w_m∈W$ such that no $T∈L(V,W)$ satisfies $Tv_k=w_k$ for each $k=1,…,m$.
My solution went as follows: Because $v_1, ... , v_m$ is linearly dependent, for some $a_i \neq 0$, $a_1v_1 + ... + a_mv_m = 0$. Then assume $Tv_k = w_k$ for $k = 1, ... , m$. Then this implies $a_1w_1 + ... + a_mw_m = 0$ where again the same $a_i \neq 0$. The difficulty arises here because for each $a_i \neq 0$, what if the respective $w_i = 0$? Or can I just define the linear map such that each of the necessary $w_i \neq 0$ to get the desired contradiction of $a_iw_i \neq 0$ because $W \neq {0}$ itself?
The linked solution suggests "Now choose $w_k=0$ if $k≠i$, and $w_i≠0$: this is possible since $W≠{0}$."