yasusar0

Answered

2022-07-15

Can two n-dimensional vectors that are not scalar multiples of each other be added to make any point on an n-dimensional coordinate plan?

If we have $u\phantom{\rule{1px}{0ex}}u,v\phantom{\rule{1px}{0ex}}v\in {\mathbb{R}}^{n}$ and $cu\phantom{\rule{1px}{0ex}}u\ne v\phantom{\rule{1px}{0ex}}v$ where $c\in \mathbb{R}$, then is

${\mathrm{\forall}}_{{a}_{1},\dots ,{a}_{n}\in \mathbb{R}}\left({\mathrm{\exists}}_{m,n\in \mathbb{R}}(mu\phantom{\rule{1px}{0ex}}u+nv\phantom{\rule{1px}{0ex}}v=\left[\begin{array}{c}{a}_{1}\\ \vdots \\ {a}_{n}\end{array}\right])\right)\text{true?}$

Note: Bolded letters represent vectors.

Answer & Explanation

Selden1f

Expert

2022-07-16Added 14 answers

We know that ${\mathbb{R}}^{\mathbb{n}}$ has a basis of n vectors (such as the standard unit basis {${e}_{1},{e}_{2},....{e}_{n}$}). By a known theorem from linear algebra, we then have that any set of less than n vectors in ${\mathbb{R}}^{\mathbb{n}}$ does not span ${\mathbb{R}}^{\mathbb{n}}$. Hence, as long as n>2, vectors u and v could not span ${\mathbb{R}}^{\mathbb{n}}$. That is, $\mathrm{\exists}$ ${a}_{1}...{a}_{n}$ such that $\mathrm{\forall}m,n\in \mathbb{R}$, $mu+nv\ne ({a}_{1}...{a}_{n})$, which means the given statement cannot be true.

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