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Let $W=\lbrace( a, b, 0): a, b \in \Bbb R\rbrace$ be a sub space of a vector space $\Bbb R^3(\Bbb R)$. Then each vector of $W$ is generated by $\lbrace( 1, 0, 0), (0, 1, 0)\rbrace$.

Is it correct? Justify.

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Yes it is right, indeed just note that $\forall a,b$

$$(a,b,0)=a(1,0,0)+b(0,1,0)$$

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  • $\begingroup$ @gete The standard basis in $\mathbb R^n$ is the set of $n$ vectors $(1,0,...,),...,(0,0,...,1)$ also denoted by $e_1,...,e_n$. $\endgroup$ – user Oct 22 at 5:27
  • $\begingroup$ @gete In this case the subspace is generated by $e_1$ and $e_2$ but also by $(2,1,0)$ and $(1,1,0)$ for example. $\endgroup$ – user Oct 22 at 5:31
  • $\begingroup$ Ok. Also, keeping aside the above correct answers, my argument is that in this case, $W$ is a two dimensional vector space (because its basis has two elements), But the vectors of $W$ are of three tupples (vectors of three dimensional space). How is it so? $\endgroup$ – gete Oct 22 at 5:38
  • $\begingroup$ @gete The subspace is indeed a plane in $\mathbb R^3$ therefore it’s spanned by $2$ independent vectors in $\mathbb R^3$. $\endgroup$ – user Oct 22 at 5:41
  • $\begingroup$ Thanks....$...$ $\endgroup$ – gete Oct 22 at 5:44
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Indeed, these vectors are the vectors that the subspace $W$ is generated by As $a(1,0,0)+b(0,1,0)=(a,b,0)$

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