# Show that a vector $w \in V$ is in the span $(v_1, \dots , v_m)$

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

• Yes. Use the definition of linear dependence, and solve for $w$. – JDZ Sep 17 '19 at 1:00
• @JDZ I updated my post, I think I got it – Evan Kim Sep 17 '19 at 1:22
• There's an error in your update. The denominators should be $\sum_i a_i$. – Chris Custer Sep 17 '19 at 1:27
• I thought I was just dividing each side by $(a_1 + a_2 + \dots + a_m)$. I didn't write that correctly? – Evan Kim Sep 17 '19 at 1:37
• 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$(v_1, \dots , v_m)$. – Evan Kim Sep 17 '19 at 1:46

$$v_i+w's$$ 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!