plane from 2 vectors in N dimensions With reference to this post:
Equation of plane containing two vectors
the answer given by one user: https://math.stackexchange.com/q/1016717 made me think, in particular whether that method could be applied in higher dimensions, and how it differs from the other methods mentioned in the post.
The linear combination of the two vectors $u = \left(1,\ 0,\ \sqrt3 \right)$ and $v = (1,\ \sqrt3,\ 0)$ results in a sort of parametric expression of the plane defined by the vectors themselves:
$$(a + b, \sqrt{3}b, \sqrt{3}a) = (x, y, z)$$
The user then writes "and observe that: 
$$z - \sqrt{3}x + y = 0$$
"
So my first question is how to go from the parametric expression of the plane (3 equations) to the single equation that is "observed" here.
I can see it is just an elimination of variables $a, b$ from the system of equations in $x, y, z, a, b$, but how can one do that using linear algebra? Meaning, what would be the system to solve? I'm pretty sure it's obvious, unfortunately not to me.
As for the higher dimensional case, suppose you have two vectors in 4D: $u = (1,\ 0,\ \sqrt3 ,\ 1/2 )$ and $v = (1,\ \sqrt3,\ 0,\ -1)$.
Applying the same concept of linear combination, any point $P$ on the plane defined by $u$ and $v$ will have to be:
$$P = \{a \cdot u + b \cdot v \}$$
$a, b \in \mathbb{R}$
Thus:
$$(a + b, \sqrt{3}b, \sqrt{3}a, \frac 1 2 a - b) = (x, y, z, w)$$
Eliminating $a,b$ this yields:
$$\{-\sqrt3 \ y + x - 2 \ w = 0, \ \sqrt3 \ z - 2 \ x - 2 \ w = 0\}$$
As expected, there are 2 equations left, not only because a plane in 4D requires that, but because 2 variables were eliminated from a system of 4 equations.
So my second question is whether you think this is a correct and general conclusion (that one can find the equation of a 'two-dimensional' plane in N dimensions by the above linear combination and elimination).
I think this post may be quite closely related, but the last answer has not been commented on by anyone, so I don't know what to make of it.
Thanks!
 A: As I’ve explained here and here, there are two basic ways to specify a subspace $V$ of $\mathbb R^n$ (or of any vector space, for that matter): One is to list a set of vectors $\{\mathbf v_1,\dots,\mathbf v_m\}$ that span $V$. That is, $V$ is the set of all linear combinations of the $\mathbf v_k$. If the set of spanning vectors is linearly independent, it is a basis of $V$, and the number of basis vectors is the dimension of $W$. You will sometimes see $V$ called the join of the vectors (or points). The space $V$ is built “bottom-up,” as it were.  
The other basic way to specify $V$ is “top-down” as the intersection of a number of $(n-1)$-dimensional subspaces $W_1\cap\cdots\cap W_m$. Each of these hyperspaces can be defined by a single homogeneous linear equation, so an equivalent specification is as the solution to a system of homogeneous linear equations. If $\dim V=m$, then the Rank-Nullity theorem tells us that we need $n-m$ independent equations to fully specify the space. If you write the system in matrix form, the solution set is the null space of the coefficient matrix, so we can call this the “null space representation” of the subspace. Or, just as the spanning set representation is called the join of the defining points in some sources, this null space representation is also called the meet of the subspaces $W_i$.  
If $m\lt n$, the system is underdetermined. Its solution can be expressed in parametric form as $t_1\mathbf v_1+\cdots+t_m\mathbf v_m$, but this is exactly the span of $\{\mathbf v_1,\dots,\mathbf v_m\}$, so converting from the intersection representation to a span representation is just a matter of solving the system of equations. One can also convert from a span to an intersection by solving a system of homogeneous linear equations. Note that in the equation $a_1x_1+\cdot+a_nx_n=0$ the roles of the constant coefficients and variables are symmetric: we can just as well substitute the coordinates of some fixed vector for the $x_i$ and treat the $a_i$ as the unknowns to be found. In this way a spanning set generates a system of homogeneous linear equations in the $a_i$. Any basis for the solution set provides the coefficients for a minimal system of linear equations with solution set equal to $V$.  
We can make this completely symmetric by setting $\mathbf a=(a_1,\dots,a_n)^T$ so that we can write the linear equation $a_1x_1+\cdots+a_nx_n=0$ as $\mathbf a\cdot\mathbf x=0$. This equation says that $\mathbf a$ and $\mathbf x$ are orthogonal. In matrix form, the system of equations that define $V$ is then $$\begin{bmatrix}\mathbf a_1^T\\\vdots\\\mathbf a_{n-m}^T\end{bmatrix}\mathbf x = 0.$$ If $\mathbf x$ is orthogonal to a set of vectors, then it’s orthogonal to every linear combination of them, too, so we have that the null space of a matrix—our subspace $V$—is the orthogonal complement of its row space—the span of the $\mathbf a$’s. Given the defining equations, finding a span representation amounts to computing a basis for the null space of the coefficient matrix, and given a spanning set $\{\mathbf v_1,\dots,\mathbf v_k\}$, finding the corresponding set of equations amounts to computing a basis for the null space of $\begin{bmatrix}\mathbf v_1&\cdots&\mathbf v_k\end{bmatrix}^T$.  
The two representations are complementary. If $A$ is the coefficient matrix of the defining system of equations and $B$ the matrix $\begin{bmatrix}\mathbf v_1&\cdots&\mathbf v_k\end{bmatrix}$, i.e., the matrix with the spanning vectors as its columns, the fact that every $\mathbf v_i$ satisfies the system means that $AB=0$. The columns of $B$ lie in the null space of $A$ and the rows of $A$ lie in the null space of $B^T$.  
To make these ideas more concrete, we can use your two examples. We have a plane in $\mathbb R^3$ defined as the span of two. We can write an arbitrary linear combination of them as the matrix product $$\begin{bmatrix}1&1\\0&\sqrt3\\\sqrt3&0\end{bmatrix} \begin{bmatrix}a\\b\end{bmatrix}.$$ Since the two vectors are linearly independent, we need only one equation of the form $ax+by+cz=0$ to define this plane. To find it, we solve the system $$\begin{align} a+\sqrt3c&=0\\a+\sqrt3b&=0 \end{align}$$ from which we get $b=c$ and $a=-\sqrt3b$. Setting $b=1$ produces the equation in your example. Note that the solution isn’t unique, but this is to be expected. After all, you can multiply both sides of an equation by the same nonzero number without changing the solution set. This corresponds to the fact that the orthogonal complement of this plane is a line—a one-dimensional subspace of $\mathbb R^3$.  
Incidentally, in $\mathbb R^3$ we have a short-cut available to do this calculation. We’re looking for a nonzero vector that’s orthogonal to $(1,\sqrt3,0)^T$ and $(1,0,\sqrt3)^T$, which we can find by computing their cross product. You’ll see this used a lot in various answers here. Unfortunately, this doesn’t easily generalize to spaces of other dimensions. There is a generalized cross product of $n-1$ vectors in $\mathbb R^n$ (which is what that answer you referenced describes), but that’s only going to help find the single equation that defines the $(n-1)$-dimensional subspace spanned by those vectors. It doesn’t help us find the system of equations for a two-dimensional plane in $\mathbb R^n$.
For your second example, we have a plane in $\mathbb R^4$ spanned by $(1,0,\sqrt3,1/2)^T$ and $(1,\sqrt3,0,-1)^T$. Now we need a system of two equations, which we can find by solving the corresponding system of unknowns or by finding a pair of linearly-independent vectors that are orthogonal to these. Not seeing anything obvious off the bat, we find their orthogonal complement by row-reducing $$\begin{bmatrix}1&0&\sqrt3&\frac12\\1&\sqrt3&0&-1\end{bmatrix} \to \begin{bmatrix}1&0&\sqrt3&\frac12 \\ 0&1&-1&-\frac{\sqrt3}2\end{bmatrix}.$$ Using the method described here, we can read from this that $(\sqrt3,-1,-1,0)^T$ and $(1,-\sqrt3,0,-2)^T$ span the null space, which corresponds to the system $$\begin{align} \sqrt3x-y-z&=0 \\ x-\sqrt3y-2w&=0.\end{align}$$ These aren’t the same equations that you got, but the two systems are equivalent, which I’ll leave for you to verify.
The above can be generalized in two interesting ways. First, we can do the same thing for flats, i.e., affine subsets of $\mathbb R^n$. These can be defined by systems of inhomogeneous systems of linear equations, or as cosets of subspaces of $\mathbb R^n$, i.e., as the set of vectors $\mathbf w+\mathbf v$, where $\mathbf w$ is some fixed vector and $\mathbf v$ ranges over all elements of some subspace $V$. Passing to homogeneous coordinates, we can identity elements of $\mathbb R^n$ with lines through the origin in $\mathbb R^{n+1}$†, so that affine sets in $\mathbb R^n$ map to distinct subspaces of $\mathbb R^{n+1}$, and the above discussion can be applied.  
Generalizing a different way, we can work in any vector space $\mathbf V$ over a scalar field $\mathbb K$. The spanning-set representation is unchanged, but we need to do something different for the intersection representation. Since we’re not working with tuples of scalars any more, and haven’t even mentioned any particular a basis of $V$, so we can’t just write a system of linear equations that define the subspace $W$. We can, however, use the dual space $V^*$ of linear functionals $\varphi:V\to\mathbb K$. We can then choose a set of elements $\mathbf\alpha_i$ of $V^*$ such that $W$ is the solution set of the system of equations $\mathbf\alpha_i[\mathbf x]=0$. With an appropriate choice of bases for $V$ and $V^*$, these become the homogeneous linear equations in the coordinates of $\mathbf x$ with which you’re familiar.  
Finally, there’s no particular reason to restrict join to operate only on vectors (points) and meet to operate only on hyperspaces. These operations can be generalized in obvious ways to operate on any collection of affine subsets of $\mathbb R^n$.

† Strictly speaking, it’s really $\mathbb R^{n+1}\setminus\{0\}$, but that’s not an important detail here.
