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Say there are two independent vectors in R3, to be specific we can assume the vectors to be [1 0 0] & [0 1 0]. I'm confused what will be the span of these vector and what will be its dimension? actually my doubt arises from the link https://www.youtube.com/watch?v=AqXOYgpbMBM in this video there are 4 vectors in R4 but only 3 of them are linearly independent therefore it spans R3 and not R4. So my question is, is it possible to span lower dimension when there are higher number of components in a vector? I'm not able to visualize it Please help!

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Yes, of course. A plane has two dimensions even if the coordinate are three. In a cartesian plane a straight line has 1 dimension but its points have 2 coordinates. A point alone has zero dimension whatever is the dimension of the space it is in.

Hope this helps

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  • $\begingroup$ I totally gt your point, but I have also learnt that Any basis in Fn must have exactly n vectors in it. And if the no. of vectors is less than n then the basis doesnot span the space. S, does both the things go hand in hand. I read it in the answer of this link -: math.stackexchange.com/questions/1992386/… $\endgroup$ – megha sonik Oct 3 '17 at 9:06
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I think you misunderstood the assertion. A set of vectors in $\mathbf R^n$ cannot possibly generate $\mathbf R^p$ if $p < n$, ut it can generate a subspace of dimension $p$, i.e. a subspace isomorphic to $\mathbf R^p$.

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