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Let $V$ be a vector space and $v_1,v_2,v_3,v_4,w in V$

If $w in text{span}{v_1,v_2,v_3}$, then $w in text{span}{v_1,v_2}$ - I believe this to be false, however I cannot come up with an appropriate counter-example, what would be a suitable counter-example?

If $w in text{span}{v_1,v_2,v_3}$, then $w in text{span}{v_1,v_2,v_3,v_4}$ - I believe this to be true but what is an appropriate proof to show this is true?

Yes you are right, we have that

$$ w ∈ operatorname{span}{v_1,v_2,v_3} implies w ∈ operatorname{span}{v_1,v_2}$$

only if $v_3=av_1+bv_2$.

1.- $win {rm span}{v_1,v_2,v_3} Rightarrow win {rm span}{v_1,v_2}$ is FALSE.

Take, for example, $v_1=(1,0,0)$, $v_2=(0,1,0)$ and $w=v_3=(0,0,1)$. Or just take any two vectors $v_1,v_2$ and $w=v_3$ independent to them.

2.- $win {rm span}{v_1,v_2,v_3} Rightarrow win {rm span}{v_1,v_2,v_3,v_4}$ is TRUE.

If $win {rm span}{v_1,v_2,v_3} $,then $w=alpha_1v_1+alpha_2v_2+alpha_3v_3 =
alpha_1v_1+alpha_2v_2+alpha_3v_3 +0cdot v_4$, so $win{rm span}{v_1,v_2,v_3,v_4}$.

1. What about $v_3$? (Given ofcourse that $v_1,v_2,v_3$ are linearly independent)




2. If $wintext{span}{v_1,v_2,v_3}$, then $$w=a_1v_1+a_2v_2+a_3v_3tag{1}$$ You can also write $(1)$ as $$w=a_1v_1+a_2v_2+a_3v_3tag{2}+0v_4$$ so $wintext{span}{v_1,v_2,v_3,v_4}$