Finding cartesian equation for a plane in $\mathbb R^4$

Question

Let's assume that I have previously found an orthonormal basis for the plane (dim 2) whose cartesian equation I want to find. Is it as simple as using Gram-Schmidt a third time to find a vector that's orthogonal to both basis vectors, and then plugging in the coordinates in the equation $ax+by+cz+dw=0$ (substituting $a$, $b$, $c$ and $d$)?

Answer 1

There are a couple of complementary ways to specify a subspace of $\mathbb R^n$ (or any vector space $V$, for that matter).

One is by listing a set of vectors that generate the space. If these vectors are linearly independent, then they are a basis for that space, of course.

The complementary method is via a system of homogeneous linear equations. These equations can be written in the form $\mathbf n\cdot\mathbf x=0$, so what you’re really doing is listing a set of vectors orthogonal to the space. More generally, you can specify a subspace $W$ of $V$ by listing a set of vectors $\mathbf\alpha_i$ in the dual space $V^*$ that annihilate the subspace, i.e., such that for all $\mathbf w\in W$, $\mathbf\alpha[\mathbf v]=0$. For a finite-dimensional vector space, you will need $\dim V-\dim W$ such dual vectors/equations.

So, you can find a set of linear equations that elements of your subspace satisfy by finding a basis $\mathbf n_i$ for the orthogonal complement of that space. You have a two-dimensional subspace of $\mathbb R^4$, so you’ll need two equations. Since $W^{\perp\perp}=W$ this amounts to finding the kernel (null space) of the matrix that has your two basis vectors as rows.