$x_1,x_2\in\mathbb{R^3}$ are linearly dependent if and only if ${\rm Span}(x_1)={\rm Span}(x_2)$ 
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*$x_1,x_2\in\mathbb{R^3}$ are linearly dependent if and only if ${\rm Span}(x_1)={\rm Span}(x_2)$.


My proof trying.$\left( \Rightarrow \right)$. Assume $x_1,x_2$ are linearly dependent. Assume for $\alpha,\beta\in\mathbb{R}$, 
$$\alpha x_1+\beta x_2=0$$
Then, either $\alpha\neq 0$ or $\beta\neq 0$. Assume $\alpha\neq 0$. Then, we obtain,
$$x_1=-\dfrac {\beta }{\alpha }x_2$$
So, is ${\rm Span}(x_1)={\rm Span}\Big(-\dfrac {\beta }{\alpha }x_2\Big)={\rm Span}(x_2)?$
Let, $t=-\dfrac {\beta }{\alpha }$. Then, so, my question is ${\rm Span}(tx_2)={\rm Span}(x_2)$?
 A: By definition $\text{Span}[x_1, \dots ,x_n]=\left\{\sum_{i=1}^n\lambda_i x_i\mid  \lambda_i\in \mathbb{R}\right\}$. So $\text{Span}[x_1]=\left\{\lambda x_1\mid  \lambda\in \mathbb{R}\right\}$ (in words, $\text{Span}[x_1]$ are all multiples of $x_1$). 
Clearly if $x_1=\lambda x_2$, we have that $\text{Span}[x_1]=\text{Span}[x_2]$. You basically showed that if $x_1$ and $x_2$ are linearly dependent, then $x_1=\lambda x_2$. Conversely, if $\text{Span}[x_1]=\text{Span}[x_2]$, then $ x_1=\lambda x_2$ for a certain non-zero $\lambda$. Thus they are linearly dependent.
A: Yes. The span of a single vector $v$ are all vectors $\lambda v$, where $\lambda$ is an arbitrary scalar. So, if $v$ is replaced by $tv$ (with $t\neq0$), then the span will be the same
A: Yes. $$\rm{Span}(x_1)=\{ y \mid y= \lambda x_1\quad \lambda \in \mathbb{R} \} = \{ y \mid y= \lambda t x_2 \quad \lambda \in \mathbb{R} \}= \{ y \mid y= \bar \lambda  x_2 \quad \bar \lambda \in \mathbb{R} \},$$
that coincide with definition of $\rm{Span}(x_2)$
