Find the dimension of the affine subspace of $\mathbb{R^5}$ generated by the points $$p=(-1,2,-1,0,4)$$ $$q=(0,-1,3,5,1)$$ $$r=(4,-2,0,0,3)$$ $$s=(3,-1,2,5,2)$$ Is it as trivial as simply finding $\vec{pq}, \vec{qr}, \vec{rs}, \vec{sp}$ and finding a basis?

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    $\begingroup$ It's that simple yes. Note that the greatest the dimension could be is $3$ though so you'll definitely have to throw out at least one vector. $\endgroup$ – user137731 Jan 13 '16 at 4:02
  • $\begingroup$ From top of my head, it should be $4$ or less than it. $\endgroup$ – kaka Jan 13 '16 at 4:03
  • $\begingroup$ Yeah, sp is useless when I have the other three. Thanks. $\endgroup$ – markoff Jan 13 '16 at 4:06

Technically the way that we define the affine space determined by those points is by taking all affine combinations of those points:

$$\mathcal A = \left\{a_1p + a_2q + a_3r + a_4 s \mid \sum a_i = 1\right\}$$

Notice though that this is equivalent to choosing (arbitrarily) any one of those points as our reference point, let's say we choose $p$, and then considering this set $$\big\{p + b_1(q-p) + b_2(r-p) + b_3(s-p) \mid b_i \in \Bbb R\big\}$$ Confirm for yourself that this set is equal to $\mathcal A$.

This tells us that $\dim\big(\operatorname{span}(q-p, r-p, s-p)\big) = \dim(\mathcal A)$.

It turns out to also be equivalent to find the dimension of the span of $\{q-p, r-q, s-r, p-s\}$ (which are exactly the vectors in your question), so feel free to do it that way as well. Notice though that not all of them are necessary. Can you see why?


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