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Problem text:
Suppose ##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##
Using the definition of the span, we obtain $$\mathrm{span}\left(v_1,v_2,v_3,v_4\right) = \left \{ a_1v_1 + a_2v_2 + a_3v_3 + a_4v_4 : a_1, a_2, a_3, a_4 \in \mathbb{F}\right \}$$
Suppose $$\mathrm{span}\left ( v_1v_2,v_2v_3,v_3v_4,v_4 \right ) = \left \{a_1(v_1v_2) + a_2(v_2v_3) + a_3(v_3v_4) + a_4v_4 : a_1, a_2, a_3, a_4 \in \mathbb{F}\right \}$$
The terms can be rearranged as follows: $$\mathrm{span}\left ( v_1v_2,v_2v_3,v_3v_4,v_4 \right ) = \left \{a_1v_1a_1v_2 + a_2v_2a_2v_3 + a_3v_3a_3v_4 + a_4v_4 : a_1, a_2, a_3, a_4 \in \mathbb{F}\right \}$$ $$\Updownarrow$$ $$\mathrm{span}\left ( v_1v_2,v_2v_3,v_3v_4,v_4 \right ) = \left \{a_1v_1(a_1+a_2)v_2(a_2+a_3)v_3(a_3+a_4)v_4 : a_1, a_2, a_3, a_4 \in \mathbb{F}\right \}$$
Which shows that the list indeed spans ##V## since the coefficients are arbitrary elements of ##\mathbb{F}##.
Is this correct?
Suppose $$\mathrm{span}\left ( v_1v_2,v_2v_3,v_3v_4,v_4 \right ) = \left \{a_1(v_1v_2) + a_2(v_2v_3) + a_3(v_3v_4) + a_4v_4 : a_1, a_2, a_3, a_4 \in \mathbb{F}\right \}$$
The terms can be rearranged as follows: $$\mathrm{span}\left ( v_1v_2,v_2v_3,v_3v_4,v_4 \right ) = \left \{a_1v_1a_1v_2 + a_2v_2a_2v_3 + a_3v_3a_3v_4 + a_4v_4 : a_1, a_2, a_3, a_4 \in \mathbb{F}\right \}$$ $$\Updownarrow$$ $$\mathrm{span}\left ( v_1v_2,v_2v_3,v_3v_4,v_4 \right ) = \left \{a_1v_1(a_1+a_2)v_2(a_2+a_3)v_3(a_3+a_4)v_4 : a_1, a_2, a_3, a_4 \in \mathbb{F}\right \}$$
Which shows that the list indeed spans ##V## since the coefficients are arbitrary elements of ##\mathbb{F}##.
Is this correct?