Prove there is a homogeneous system of equations where solution is $M \subseteq \mathbb{R}^n$ So, I've been thinking of how to prove this.
For example.
Let $\{(1, 1, 0, -1), (1, 0, 1, 2)\}$ be the base vectors of subspace $M \subseteq \mathbb{R}^4$.
One needs to show there is a system of linear equations where its solution would be equal to $M$.
The system solution to the example is
$$x_1 - x_2 - x_3 = 0$$
$$x_2 - 2x_3 + x_4 = 0 .$$
The process of the solution is next.
Space $M$ can be described as a linear combination of its base vectors.
$$\alpha(1, 1, 0, -1) + \beta(1, 0, 1, 2)$$
This means that 
$$x_1 = \alpha + \beta,~ x_2 = \alpha,~ x_3 = \beta,~ x_4 = -\alpha + 2\beta$$
From this one can derive the system whose coefficients can be put into a matrix.
$$\left( \begin{smallmatrix} 1&-1&-1&0&|&0\\ 0&1&-2&1&|&0 \end{smallmatrix} \right)$$
The solution is definitely equal to $M$.
How can one form a generalized proof?
 A: Let $B=\{v_1,.., v_n \}$ be a basis for $M$. Complete $B$ to a basis $B \cup B'=\{v_1,.., v_n,v_{n-1},..,v_m \}$ of $\mathbb R^d$. use Gram Schmith to get a basis $ \{ w_1,.,w_m \}$ of $\mathbb R^d$.
Then $M$ is the null space of the matrix whose colums are $w_{n+1},..,w_m$.
A: You need to find a basis for $M^\bot$. One way would be to apply Gram Schmidt to the vectors $(1, 1, 0, -1)^T, (1, 0, 1, 2)^T, e_1,...,e_4$, and use the last two non-zero vectors (which will span $M^\bot$).
Then if $M^\bot = \text{sp} \{ v_1,...,v_k \}$, you can write $M = \{x | v_i^T x = 0, i=1,...,k \}$.
In this case, $v_1 = (6,-7,-4,-1)^T$, $v_2 = (0 , 1, -2, 1)^T$ work (I modified the results to have nice integer coefficients).
A: Represent $M$ as a blade in the clifford algebra.  Here, you have two vectors $u_1, u_2$ that span $M$.  The bivector $u_1 \wedge u_2$ therefore represents this planar subspace.
One can then find the dual blade $M^*$, which is equal to $iM$, where $i$ is the pseudoscalar of the space.
Finally, one can construct a basis for $M^*$ through an orthonormalization procedure.
Example: let's take your two vectors $u_1 = e_1 + e_2 - e_4$ and $u_2 = e_1 + e_3 + 2 e_4$.  The bivector $M$ is then
$$\begin{align*} M &= u_1 \wedge u_2 \\ &= e_1 e_3 + 2 e_1 e_4 + e_2 e_1 + e_2 e_3 + 2 e_2 e_4 - e_4 e_1 - e_4 e_3 \\ &=-e_1e_2 +  e_1 e_3 + 3 e_1 e_4 + e_2 e_3 + 2 e_2 e_4 + e_3 e_4\end{align*}$$
Multiply by $i = e_1 e_2 e_3 e_4$, the pseudoscalar, to find the dual.
$$M^* = iM = e_3 e_4 + e_2 e_4 - 3 e_2 e_3 - e_1 e_4 + 2 e_1 e_3 - e_1 e_2$$
Now we need two vectors $b_1, b_2$ that will span the dual space.  Try $b_1 = e_1 \cdot M^*$.
$$b_1 = e_1 \cdot M^* = -e_4 + 2 e_3 - e_2$$
Now find $b_2 = b_1 \cdot M^*$:
$$\begin{align*} b_2 &= (-e_4 + 2 e_3 - e_2) \cdot M^* \\ &= -e_4 + 3 e_3 - e_1 + 2 e_4 + 6 e_2 - 4 e_1 + e_3 + e_2 - e_1 \\ &= -6 e_1 + 7 e_2 + 4 e_3 + e_4\end{align*}$$
These vectors yield two equations:
$$\begin{align*} -x^2 + 2 x^3 - x^4 &= 0 \\ -6 x^1 + 7 x^2 + 4 x^3 + x^4 &= 0 \end{align*}$$
And as you can check yourself, these equations hold for any vector in $M$.
So, what we're doing is finding the dual space $M^*$ and just finding a basis for it.  These basis vectors must be orthogonal to any vector in $M$.
