How many vectors are needed to define a plane/hyperplane in n-dimensional space? In 3 dimensions, if there are 2 vectors with tails at the origin and the heads in differing locations (and the vectors aren't parallel), that information is sufficient to define a plane. In higher dimensions, how many vectors with their tails at the origin are needed to define a plane/hyperplane?

  • $\begingroup$ Well, what's your definition of hyperplane? :D $\endgroup$ – dani_s Jan 10 '14 at 16:29
  • $\begingroup$ If the space has n dimensions, the hyperplane has n-1 dimensions. $\endgroup$ – Lee Jan 10 '14 at 16:31
  • $\begingroup$ Do you know how to define the dimension of a space? $\endgroup$ – dani_s Jan 10 '14 at 16:34

If by plane you mean a 2-dimensional subspace, then the answer is $2$. Since you also asked for a hyperplane, which is a subspace of codimension $1$, meaning an $(n-1)$-dimensional subspace of a vector space of dimension $n$, you need $n-1$ linearly independent vectors to span a hyperplane.

  • $\begingroup$ This is what I know from my Manifold course +1 $\endgroup$ – mrs Jan 10 '14 at 16:39

If you happen to have an inner product on your $n$-dimensional space, then you can specify a $(n-1)$-dimensional hyperplane using a single normal vector for it (plus a point in the hyperplane if you don't want it to go through the origin).

  • $\begingroup$ I understand inner product in terms of 2 vectors, but what do you mean by inner product on n-dimensional space? Can I extend what you're saying to mean that 2 vectors with tails at the origin could define a plane? Thanks. $\endgroup$ – Lee Jan 10 '14 at 16:38
  • $\begingroup$ @Lee: I mean this only works if you have an inner product operation defined for the $n$-dimensional space you're working with. $\mathbb R^n$ comes with a standard inner product you can use, but if your space is, say "the vector space of all polynomials of degree at most 36", then it's not clear how to take inner products, and so just saying such-and-such is a normal vector to your hyperplane does not in itself tell which plane you're talking about. $\endgroup$ – Henning Makholm Jan 10 '14 at 18:12

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