In an n-dimensional space, how many vectors construct a hyperplane?

My guess is n, because a line in R^2 is a hyperplane and a plane in R^3 is a hyperplane. But how can I prove this?

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- Nov 25th 2011, 07:34 AMnorickaHyperplane
In an n-dimensional space, how many vectors construct a hyperplane?

My guess is n, because a line in R^2 is a hyperplane and a plane in R^3 is a hyperplane. But how can I prove this? - Nov 25th 2011, 08:16 AMDevenoRe: Hyperplane
n-1, not n.

a hyperplane is defined by a single equation of the form:

$\displaystyle a_1x_1 + a_2x_2 + \dots + a_nx_n = b$, where not all $\displaystyle a_i$ are 0.

the solution set consists of all elements of the solution space of the associated homogeneous system:

$\displaystyle a_1x_1 + a_2x_2 + \dots + a_nx_n = 0$, plus any (particular) vector $\displaystyle (b_1,b_2,\dots,b_n)$ lying on the hyperplane

(in other words we have a coset, or translate, of the nullspace of the homogeneous system).

since the homogeneous system has rank 1, its solution space has dimension n-1 (by the rank-nullity theorem), therefore.... - Nov 28th 2011, 04:20 AMnorickaRe: Hyperplane
Thank you very much!

- Nov 28th 2011, 06:13 AMHallsofIvyRe: Hyperplane
What, exactly, do you mean by "construct a plane"? In n dimensional space, there will be, pretty much by definition, n-1 independent vectors lying on a hyperplane

**through the origin**. But there will be**no**vectors lying on a hyper plane [b]not[/b ] through the origin. There is, however, always**one**vector**perpendicular**to the given hyperplane and that can be used to write the equation of the plane.

As Deveno said, any hyperplane can be written as $\displaystyle a_1(x_1- p_1)+ a_2(x_2- p_2)+ \cdot\cdot\cdot+ a_n(x_n- p_n)= 0$, where $\displaystyle <a_1, a_2, \cdot\cdot\cdot, x_n>$ is a vector normal to the hyperplane and $\displaystyle (p_1, p_2, \cdot\cdot\cdot, p_n)$ is a point in the hyperplane.