How can I prove that an arbitrary set of vectors spans some space V?

Question

Supposed $v_{1}, v_{2}, v_{3}, v_{4}$ spans $V$. Prove that the list $v_{1}-v_{2}, v_{2}- v_{3}, v_{3}-v_{4}, v_{4}$ also spans $V$.

I started out with:

If $v_{1}-v_{2}, v_{2}- v_{3}, v_{3}-v_{4}, v_{4}$ spans $V$, then:

$\alpha(v_{1}-v_{2}), \alpha(v_{2}- v_{3}), \alpha(v_{3}-v_{4}), \alpha(v_{4})$ = $V$

= $\alpha v_{1} - \alpha v_{2} + \alpha v_{2} - \alpha v_{3} + \alpha v_3 - \alpha v_4 + \alpha v_4$

= $\alpha v_1$ = $V$

I'm not sure where to go from here or if this was even valid.

Answer 1

Spans mean that we can write any $v \in V$ as $v = \sum_{i=1}^4 \alpha_i v_i $ where $\alpha_i$ are constants. Now, we wanna find $\beta_i$ constants so that

$$ \beta_1 (v_1-v_2) + \beta_2 (v_2-v_3) + \beta_3 (v_3-v_4) +\beta_4 v_4 = v $$

Rearranging gives

$$ \beta_1 v_1 + (\beta_2-\beta_1) v_2 + (\beta_3-\beta_2 ) v_3 + \beta_4 v_4 = v $$

can you take it from there?

Answer 2

HINT

Note that in the basis $\{v_1,v_2,v_3,v_4\}$ the set of vectors are

• $u_1=v_1-v_2=(1,-1,0,0)$
• $u_2=v_2-v_3=(0,1,-1,0)$
• $u_3=v_3-v_4=(0,0,1,-1)$
• $u_4=v_4=(0,0,0,1)$

then we can consider the associated matrix

\begin{bmatrix} 1&-1&0&0\\0&1&-1&0\\0&0&1&-1\\0&0&0&1\end{bmatrix}

Answer 3

Note that \begin{align*} v_1 &= (v_1 - v_2) + (v_2 - v_3) + (v_3 - v_4) + v_4 \\ v_2 &= (v_2 - v_3) + (v_3 - v_4) + v_4 \\ v_3 &= (v_3 - v_4) + v_4. \end{align*} Thus, $v_1, v_2, v_3, v_4 \in \operatorname{span}(v_1 - v_2, v_2 - v_3, v_3 - v_4, v_4)$, and so $$V = \operatorname{span}(v_1, v_2, v_3, v_4) \subseteq \operatorname{span}(v_1 - v_2, v_2 - v_3, v_3 - v_4, v_4) \subseteq V,$$ hence $$\operatorname{span}(v_1 - v_2, v_2 - v_3, v_3 - v_4, v_4) = V.$$