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$(2.A.10)$ 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$(v_1, \dots , v_m)$.

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    $\begingroup$ Yes. Use the definition of linear dependence, and solve for $w$. $\endgroup$ – JDZ Sep 17 '19 at 1:00
  • $\begingroup$ @JDZ I updated my post, I think I got it $\endgroup$ – Evan Kim Sep 17 '19 at 1:22
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    $\begingroup$ There's an error in your update. The denominators should be $\sum_i a_i$. $\endgroup$ – Chris Custer Sep 17 '19 at 1:27
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    $\begingroup$ I thought I was just dividing each side by $(a_1 + a_2 + \dots + a_m)$. I didn't write that correctly? $\endgroup$ – Evan Kim Sep 17 '19 at 1:37
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    $\begingroup$ Thus $w = -\frac{a_1v_1}{k} -\frac{a_2v_2}{k} - \dots - -\frac{a_mv_m}{k}$ where $k = -\sum_{1}^{m} a_i $ is a linear combination of the vectors $v_1, \dots , v_m$ and is in the span$(v_1, \dots , v_m)$. $\endgroup$ – Evan Kim Sep 17 '19 at 1:46
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$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!

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