How to find the matrix of orthonormal base vectors of a hyperplane?

Question

How to find the matrix of orthonormal base vectors of a hyperplane translated such that it crosses the origin of parameter space? for example this hyperplane: $p1 + p2 = 1$ in a 3D space with this parameters: $(Theta, p1, p2)$.

Answer 1

Let’s say we have some hyperplane H cut out by the equation

$a_1 x_1 + a_2 x_2 + \cdots + a_n x_n = 0$

in some vector space $F_n$. We need to get a basis for this kernel, at which point we can just apply Gram-Schmidt. First, note that at least one of the $a_i$ must be nonzero. Let’s say, without loss of generality, that $a_1 \neq 0$. Then for each $i \neq 1$ the vector

$a_i e_1 - a_1 e_i = (a_i, 0, \ldots, 0, -a_1, 0, \ldots, 0)$

obviously lies on H. There are (n-1) such vectors, and since each has a nonzero coordinate that’s zero in all the other vectors they’re obviously linearly independent. So this gives us a basis of H. With this basis in hand we can finish by applying Gram-Schmidt.

In your example, you consider an equation $p_1 + p_2 = 1$ in $\mathbb{R}^3$ where the coordinates are $(\theta, p_1, p_2)$. Translated to go through the origin, this becomes $p_1 + p_2 = 0$. Since the coefficient of $p_2$ is one — in particular, since it’s not zero — we can use it to get a basis of H, namely

$\{(1, 0, 0), (0, 1, -1)\}$

Let’s orthonormalize. (1,0,0) already has length 1. And it’s already perpendicular to (0, 1, -1). So all we have to do is normalize (0, 1, -1). It has length$ \sqrt{2}$, so our orthonormal basis is

$\left\{(1, 0, 0), \left(0, \displaystyle\frac{1}{\sqrt{2}}, \displaystyle\frac{-1}{\sqrt{2}}\right)\right\}$