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$\operatorname{Span}(\{u, v\})$ is a set that contains all the linear combinations $au + bv$ where $a$ and $b \in \mathbb R$.

$\operatorname{Span}(\{v, w\})$ is a set that contains all the linear combinations $av + bw$ where $a$ and $b \in \mathbb R$.

Now am I going to have to find $u, v, w$ such that $u + v = v + w$ and $u \ne w$?

Thanks.

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  • $\begingroup$ You can define $w$ as $u+v$. $\endgroup$ – user130512 Apr 24 '14 at 19:19
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    $\begingroup$ You don't need $u + v = v + w$ for $\text{span}(\{u,v\}) = \text{span}(\{v,w\})$. $\endgroup$ – Michael Joyce Apr 24 '14 at 19:24
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Let $w=u+v$, then we immediatly have $\operatorname{span}(v, w)\subseteq\operatorname{span}(u, v)$ and since $u=v-w$ the other inclusion holds as well.

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