linear algebra and vector spaces, proving w is an element of the span of a vector space

Question

Suppose that $(v_1,v_2,...,v_n) \subseteq V $ is linearly independent and let $w \in V$. If

$$(v_1+w,v_2+w,...,v_n+w)$$

is linearly dependent, prove that $w \in span(v_1,v_2,...,v_n)$

I'm not sure where I would start this one. I know that if $(v_1+w,v_2+w,...,v_n+w)$ is dependent then there are $s_1 \neq s_2$ where $0=s_1(v_1)+s_2(v_2)$. From here I'm not sure where to proceed.

Answer 1

$\def\vec#1{{\bf#1}}$I think the contrapositive will be the way to go here. So, let $\{\vec v_1,\vec v_2,\ldots,\vec v_n\}$ be linearly independent, and suppose $\vec w\notin span\{\vec v_1,\vec v_2,\ldots,\vec v_n\}$; we have to show that $\{\vec v_1+\vec w,\vec v_2+\vec w,\ldots,\vec v_n+\vec w\}$ is linearly independent.

Let $$\alpha_1(\vec v_1+\vec w)+\alpha_2(\vec v_2+\vec w)+\cdots+\alpha_n(\vec v_n+\vec w)=\vec 0\ .$$ Then $$(\alpha_1+\alpha_2+\cdots+\alpha_n)\vec w=-\alpha_1\vec v_1-\alpha_2\vec v_2-\cdots-\alpha_n\vec v_n$$ which is in $span\{\vec v_1,\vec v_2,\ldots,\vec v_n\}$; since $\vec w\notin span\{\vec v_1,\vec v_2,\ldots,\vec v_n\}$, the only way this can be true is if $$\alpha_1+\alpha_2+\cdots+\alpha_n=0\ .$$ This means $$\alpha_1\vec v_1+\alpha_2\vec v_2+\cdots+\alpha_n\vec v_n=\vec 0\ ;$$ since the vectors $\vec v_k$ are linearly independent, all the scalars $\alpha_k$ are zero, and this completes the proof.

Answer 2

Proof by contradiction.

Let us assume that $\{v_1, \dots, v_n\}$ are linearly independent, $\{v_1+w, \dots, v_n+w\}$ are linearly dependent and that $w \not\in \mathrm{span}\{v_1, \dots, v_n\}$. Because $\{v_1 + w, \dots, v_n + w\}$ are linearly dependent, we know that there are non-zero constants $\beta_1, \dots, \beta_n \in \mathbb{F}$ such that $$ \beta_1(v_1+w)+\dots+\beta_n(v_n+w)=\mathbf{0}, $$ in other words, $$ \beta_1 v_1 + \dots + \beta_n v_n + w(\beta_1 + \dots + \beta_n) = \mathbf{0}. $$

There are two situations: 1) $\beta_1 + \dots + \beta_n \not= 0$; and 2) $\beta_1 + \dots + \beta_n=0$.

If $\beta_1 + \dots + \beta_n \not= 0$, then we get $$ w = \frac{-\beta_1}{\beta_1+\dots+\beta_n}v_1 + \dots + \frac{-\beta_n}{\beta_1 + \dots + \beta_n} v_n $$ and thus $w \in \mathrm{span}\{v_1,\dots,v_n\}$ which contradicts the assumption that $w \not\in\mathrm{span}\{v_1,\dots,v_n\}$.

If $\beta_1 + \dots + \beta_n = 0$, then we have the equality $$ \beta_1 v_1 + \dots + \beta_n v_n = \mathbf{0}, $$ which contradicts the assumption that $\{v_1, \dots, v_n\}$ are linearly independent because $\beta_1, \dots, \beta_n$ are non-zero constants.

Comment: "non-zero constants" means that at least one of them is non-zero.

Answer 3

Suppose the vectors $$(v_1+w,v_2+w,...,v_n+w)$$ are linearly dependent.

Then for some constants $ c_1, c_2,...,c_n$, where at least one of them is non-zero, we have $$ c_1(v_1+w)+c_2(v_2+w)+...+c_n(v_n+w)=0$$ Solve for $w$ and we get $$ w=\frac {-(c_1v_1 +c_2 v_2 +...+c_nv_n)}{(c_1 +c_2 +...+c_n)}$$

Thus $$w \in span(v_1,v_2,...,v_n)$$