Parametric form of hyperplanes The standard definition of a hyperplane in $\mathbb{R}^n$ is a set $\mathbb{H}$ of the form
$$\mathbb{H}=\left\{(x_1, x_2, \ldots, x_n) \in \mathbb{R}^n : \sum_{i=1}^n \alpha_ix_i = L \right\}$$
where $L$ is fixed, and the $\alpha_i$'s are fixed and not all $0$. 
We can also think of the hyperplane as a subset of $\mathbb{R}^n$ in "parametric" terms. In other words, for some linearly independent subset $l = \left\{\overline{y}_1, \overline{y}_2, \ldots, \overline{y}_{n-1}\right\} \subset \mathbb{R}^n$ and a "base" point $\overline{r} \in \mathbb{R}^n$ a hyperplane is the image of the function $g:\mathbb{R}^{n-1} \to \mathbb{R}^n$ given by $g(t_1, t_2, \ldots, t_{n-1}) = \left(\sum_{i=1}^{n-1} t_i\overline{y_i}\right) + \overline{r}$.
Given the "standard" definition of the hyperplane above, is it possible to explicitly describe the plane parametrically in terms of the second definition (ie., find the set $l$ and $r$ corresponding to $\mathbb{H}$)?
 A: Of course. You just need the algorithm for reading off a basis for the nullspace (kernel) of the matrix
$$\begin{bmatrix} \alpha_1 & \cdots & \alpha_n\end{bmatrix}.$$
Even easier, assuming $\alpha_1\ne 0$, just solve for $x_1$ in your equation and then set each free variable $x_j$ ($j\ge 2$) equal to $1$ while setting the others equal to $0$.
A: Of course.  First, fix a vector $v_0 = c_1x_1 + \cdots + c_nx_n$ which lies in $\mathbb H$, which is to say $c_1 + \cdots + c_n = 0$.
Then $ W := \mathbb H - v_0 = \{ h - v_0 : w \in \mathbb H \}$ is easily seen to be a subspace of $\mathbb{R}^n$.  It is exactly the set of $r_1x_1 + \cdots + r_nx_n$ such that $r_1 + \cdots + r_n = 0$.
Also, $W$ is homeomorphic to $\mathbb H$, so it must have the same topological dimension.  Hence $W$ is an $(n-1)$ dimensional vector space.  If $y_1, ... , y_{n-1}$ is a basis for this space, then 
$$\mathbb H = W + v_0 = \{ r_1y_1 + \cdots r_{n-1}y_{n-1} + v_0 : r_i \in \mathbb{R} \}$$
If you want to choose your basis for $W$ explicitly in terms of the basis $x_1, ... , x_n$ (and in turn, give a nontopological proof of the dimension of $W$), you can just do something like $x_1 - x_n, ... , x_{n-1} - x_n$.
A: Given that $\sum_{i=1}^n \alpha_ix_i= L$, with at least one of the $\alpha_i$ non-zero, rearrange the sum so that $\alpha_1= \beta\ne 0$.  Then we can write $\beta x_1= L- \sum_{i=2}^n \alpha_i x_i$ and then $x_1= \sum_{i= 2}^n \frac{\alpha_i}{\beta}x_i$.  
Now, using parameters $t_1$, $t_2$, ..., $t_n$, we can write $x_1= \sum_{i= 2}^n \frac{\alpha_i}{\beta}t_i$, $x_2= t_2$, ..., $x_n= t_n$.
