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?
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....
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.