Problem: Let $V$ be a vector space over a field $\mathbb{F} $. Suppose that $\left\{ v_1,v_2,v_3,v_4 \right\} $ spans $V$, then prove that \[ \left\{ v_1-v_2,v_2-v_3,v_3-v_4,v_4 \right\} \] also spans $V$.
Solution: Let $v\in V$. We need to show that there exists $\alpha ,\beta ,\gamma ,\delta \in \mathbb{F}$ such that \begin{align*} & \alpha \left( v_1-v_2 \right) + \beta \left( (v_2-v_3) \right)+ \gamma \left(v_3-v_4\right) + \delta v_4 = v. \end{align*} The above implies \begin{equation}\label{eq:09Apr2023-1} \alpha v_1 + \left( \beta -\alpha \right) v_2 + \left( \gamma -\beta \right) v_3 + \left( \delta -\gamma \right) v_4 = v. \end{equation}
Since $\left\{ v_1,v_2,v_3,v_4 \right\} $ spans $V$, so there exits $a,b,c,d\in \mathbb{F} $ such that \begin{equation}\label{eq:09Apr2023-2} av_1 + bv_2 + cv_3 + dv_4 = v. \end{equation} Using \eqref{eq:09Apr2023-1} and \eqref{eq:09Apr2023-2}, we have \begin{align*} & \alpha = a, \\ & \beta -\alpha =b \implies \beta = \alpha +b = a+b \\ & \gamma - \beta = c \implies \gamma = c+\beta = a + b + c \\ & \delta -\gamma = d \implies \delta = d+ \gamma = a + b + c + d. \end{align*} Therefore, we proved that \[ v \in \text{span} \left\{ v_1-v_2,v_2-v_3,v_3-v_4,v_4 \right\} \] and hence $\left\{ v_1-v_2,v_2-v_3,v_3-v_4,v_4 \right\} $ spans $V$.