$(2.A.10)$ Suppose $v_1, \dots , v_m$ is a linearly independent list in $V$ and $w \in V$. Prove that if $v_1 + w, \dots v_m + w$ is a linearly dependent list, then $w \in $ span$(v_1, \dots , v_m)$.
Is $w$ being added to each vector in the linearly independent list? I don't see how that implies $w \in $ span$(v_1, \dots , v_m)$.
edit:
\begin{align*}
a_1(v_1 + w) + a_2(v_2 + w) + \dots + a_m(v_m + w) &= 0 \\
a_1v_1 + a_2v_2 + \dots + a_mv_m + w(a_1 + a_2 + \dots + a_m) &= 0 \\
\end{align*}
Thus $w = -\frac{a_1v_1}{a_1} -\frac{a_2v_2}{a_2} – \dots – -\frac{a_mv_m}{a_m}$ is a linear combination of the vectors $v_1, \dots , v_m$ and is in the span$(v_1, \dots , v_m)$.
Best Answer
$v_i+w's $ are dependent means there are scalars $a_1,a_2, \cdots, a_m$ not all zero such that $$a_1(v_1+w)+a_2(v_2+w)+\cdots+a_m(v_m+w)=0$$ That is $$a_1v_1+a_2v_2+\cdots+a_mv_m+w(a_1+a_2+\cdots+a_m)=0 \tag1$$ which implies $$w=\frac{a_1}{k}v_1+\frac{a_2}{k}v_2+\cdots+\frac{a_m}{k}v_m \in \text{span}(v_1,v_2,\cdots,v_m)$$ where $k=-(\sum a_i)$
Note that $\sum a_i \neq 0$. Otherwise, $(1)$ implies, using independence of $v_i$, all $a_i's$ are zero. Which contradict our assumption!