Is the following Proof Correct?
Theorem. Given that $v_1,v_2,...v_m$ is a linearly independent list in $V$ and $w$ is a vector in $V$ such that $v_1+w,v_2+w,...,v_m+w$ is linearly dependent then $w\in\operatorname{span}(v_1,v_2,...,v_m)$.
Proof. Since the list $v_1+w,v_2+w,...,v_m+w$ is linearly dependent it follows that for some $j\in\{1,2,...,m\}$ it is the case that $$(v_j+w)\in\operatorname{span}(v_1+w,v_2+w,...,v_{j-2}+w,v_{j-1}+w)\tag{1}$$ consequently for some $\alpha_1,\alpha_2,...,\alpha_{j-1}\in\mathbf{F}$ we have $$(v_j +w)= \sum_{i=1}^{j-1}\alpha_i(v_i+w)\tag{2}$$ which implies that $$(1-\sum_{i=1}^{j-1}\alpha_i)w = \sum_{i=1}^{j-1}\alpha_iv_i+v_{j}\tag{3}$$
Assume now for the purpose of contradiction that $$\sum_{i=1}^{j-1}\alpha_i = 1\tag{4}$$ which implies that $$\sum_{i=1}^{j-1}\alpha_iv_i+v_{j} = 0\tag{5}$$ and by extension $$v_j = -\sum_{i=1}^{j-1}\alpha_iv_i = \sum_{i=1}^{j-1}(-\alpha_i)v_i\tag{6}$$ thus $v_j\in\operatorname{span}(v_1,v_2,...,v_{j-1})$ but the list $v_1,v_2,...,v_m$ is linearly independent and so is $v_1,v_2,...,v_{j-1},v_j$ implying that $v_j\not\in\operatorname{span}(v_1,v_2,...,v_{j-1},v_j)$ resulting in a contradiction and so by using $(3)$ we can see that $$w = \frac{\sum_{i=1}^{j-1}\alpha_iv_i+v_j}{(1-\sum_{i=1}^{j-1}\alpha_i)}\in\operatorname{span}(v_1,v_2,...,v_j)\subset\operatorname{span}(v_1,v_2,...,v_m)$$
$\blacksquare$
