Expressing a hyperplane as the span of several vectors. How do I express the hyperplane $x+y=1$ as the span of two vectors or more?
P. S. We have a 3D space.
 A: Using only basic analytic geometry: find three different non-collinear points on the plane, for example
$$A=(1,0,1)\;,\;\;B=(1,0,0)\;,\;\;C=(0,1,0)$$
and now construct the directed vectors
$$\vec{AB}=B-A=(0,0,-1)\;,\;\;\vec{AC}=C-A=(-1,1,-1)$$
and then the plane is
$$\pi:\;A+r\vec{AB}+s\vec{AC}=(1,0,1)+r(0,0,-1)+s(-1,1,-1)\;,\;\;r,t\in\Bbb R$$
Check: take the vectorial product of the direction vectors to get a perpendicular vector to the plane:
$$\vec{AB}\times\vec{AC}=\det\begin{vmatrix}i&j&k\\0&0&-1\\-1&1&-1\end{vmatrix}=(1,1,0)$$
and thus our plane is $\;x+y+d=0\;$ , and to find $\;d\;$ we can substitute any  point on the plane here, say $\;A\;$ , to obtain
$$0+0+d=0\implies d=0$$
and the wanted plane is
$$x+y-1=0$$
which, of course, it is the same as you give. This way is just a standard form to check that what we got at the beginning is correct.
Anyway, you can look at your plane as the translation of a subspace, so;
$$\pi:\;\;\text{Span}\,\left\{\;(0,0,-1)\,,\,\,(-1,1,-1)\;\right\}+(1,0,1)$$
A: Note that it cannot be expressed as span of some vectors since the set of vectors satisfying the hyplerplane equation does not form a subspace. Nevertheless, let $u=\begin{bmatrix} x \\ y \end{bmatrix}$ be any vector satisfying the equation $x+y=1$. Rearranging gives $y=1-x$ and substituting this into $u$ gives
\begin{align*}
u = \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} x \\ 1-x \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \end{bmatrix} + x\begin{bmatrix} 1 \\ -1 \end{bmatrix}
\end{align*}
