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I can describe an $n-1$ dimensional hyperplane in $R^n$ with a point and a single $n$ dimensional vector (namely, the normal vector). Similarly, I can describe a $1$ dimensional hyperplane (a line) in $R^n$ with a point and a single vector (the vector pointing in the direction of the line). However, in order to represent an $n-2$ dimensional hyperplane in $R^n$, I need TWO normal vectors and a point. This generalizes, and to represent an $m$ dimensional hyperplane in $R^{2m}$, I need fully $m$ vectors in $R^m$ and a point. What is the intuition behind this explosion in the number of free parameters necessary to represent a hyperplane? Is this merely an artifact of the way that I'm representing them (with normal vectors or parallel vectors), or is this an intrinsic property of embedding lower dimensional spaces in higher dimensional ones?

Him
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  • There is something intrinsic here, and it is not just your choice of "parameters." Actually the edge cases are exceptional here. Maybe you will be interested in projective duality. Or perhaps grassmanians & friends (in particular, their dimensions as manifolds.) – Andres Mejia Nov 06 '18 at 16:19
  • It is not convenient to "change" the definition of hyperplane. By internationally accepted definition, all hyperplane of a vector space must have a codimension equal to 1. In particular, the enunciation of this theorem would have to be changed "In a normed space, all hyperplane is closed or dense". – Piquito Nov 06 '18 at 16:41
  • @Piquito, This is fair. What would be a concise way to say "the subset of $R^n$ formed by all linear combinations of $m$ orthogonal vectors $\vec{v_i},\ i=1,...m$ plus some fixed point $p \in R^n$". A line is $1$-d, a plane is $2$-d and a hyperplane is $n-1$-d... Is there a word for the things in between? – Him Nov 06 '18 at 17:10
  • Also, @Piquito, for pedantry's sake, lines, planes and other $m$-dimensional linear-esque objects described in the above definition in $R^n$ are closed, so the theorem you quote would still hold. :) – Him Nov 06 '18 at 17:41
  • Obviously but the quoted theorem is for a normed space. – Piquito Nov 07 '18 at 18:29

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