# Standard Basis of a vector space

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.

## 2 Answers

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

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

• @gete The standard basis in $\mathbb R^n$ is the set of $n$ vectors $(1,0,...,),...,(0,0,...,1)$ also denoted by $e_1,...,e_n$. – user Oct 22 at 5:27
• @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. – user Oct 22 at 5:31
• 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? – gete Oct 22 at 5:38
• @gete The subspace is indeed a plane in $\mathbb R^3$ therefore it’s spanned by $2$ independent vectors in $\mathbb R^3$. – user Oct 22 at 5:41
• Thanks....$...$ – gete Oct 22 at 5:44

Indeed, these vectors are the vectors that the subspace $$W$$ is generated by As $$a(1,0,0)+b(0,1,0)=(a,b,0)$$