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Given three vectors $\boldsymbol u,\,\boldsymbol v,\,\boldsymbol w$ in $\mathbb R^n$ that Span $\mathbb R^n$ then prove that the vectors $\boldsymbol u-2 \boldsymbol w,\, \boldsymbol v+\boldsymbol w$, and $\boldsymbol w$ span $\mathbb R^n$ as well. Hint: It helps to represent the original vectors $\boldsymbol u,\,\boldsymbol v,\,\boldsymbol w$ as linear combinations of these new vectors

Okay so I'm thinking this has to do with 'closure under addition'? I'm very confused by how to approach/solve this. Isn't w the same in both so obviously spans $\mathbb R^n$. All help is welcome and appreciated.

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Let $x$ be any vector of the vector space. Then according to what is given, we know that some scalars $a,b,c$ exist so that $au+bv+cw=x$ for any chosen $x$ Now choosing the new combinations, we know that some scalars $d,e,f$ exists so that $d(u-2w)+e(v+w)+fw=x$ , working out gives $du+ev+(-2d+e+f)w=x$ so that comparing scalars we have $a=d, b=e, c=-2d+e+f$ and so the "relation" is established, i.e. if $a,b,c$ are known to make some $x$, so are $d,e,f$ (and in a unique manner!)

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