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I want to find out the intersection of two surfaces that exist in $n$-dimensions. These surfaces are defined by a system of $2$ linear equations.

$$A_1x+B_1y+C_1z+\cdots = D_1$$

$$A_2x+B_2y+C_2z+\cdots = D_2$$

But first, how would these surfaces look like? In $3$-D space, they would be planes, but in dimensions higher than $3$ are these not planes? Can we generalize a way to find the intersection of such linear expressions in some way?

Also can this problem be solved using some tools, preferably Matlab?

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    $\begingroup$ The surfaces are hyperplanes. Solve a linear system of the form $\rm A x = b$, where $\rm A$ is a $2 \times n$ matrix. If $n > 2$, the linear system is underdetermined. If $n=3$, the intersection could be a line. If $n=4$, the intersection could be a $2$-dimensional plane. $\endgroup$ – Rodrigo de Azevedo Jul 16 '17 at 15:27
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    $\begingroup$ You can use Gaussian elimination to find a parametrization of the affine space that is the intersection, assuming that the intersection is not empty. $\endgroup$ – Rodrigo de Azevedo Jul 16 '17 at 15:31
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    $\begingroup$ @RodrigodeAzevedo And how do I figure out t1 and t2? $\endgroup$ – Vidor Vistrom Jul 16 '17 at 15:44
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    $\begingroup$ Suppose we have $$\begin{bmatrix} 1 & 0 & -1 & -1\\ 0 & 1 & -1 & -1\end{bmatrix} \begin{bmatrix} x_1\\ x_2\\ x_3\\ x_4\end{bmatrix} = \begin{bmatrix} 1\\ 1\end{bmatrix}$$ The solution set is the plane parameterized as follows $$\left\{ \begin{bmatrix} 1\\ 1\\ 0\\ 0\end{bmatrix} + t_1 \begin{bmatrix} 1\\ 1\\ 1\\ 0\end{bmatrix} + t_2 \begin{bmatrix} 1\\ 1\\ 0\\ 1\end{bmatrix} : t_1, t_2 \in \mathbb R \right\}$$ $\endgroup$ – Rodrigo de Azevedo Jul 16 '17 at 15:48
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    $\begingroup$ @RodrigodeAzevedo: I suggest that you post your solution, perhaps mentioning the reduced row-echelon form of the augmented matrix for the $2\times n$ system, as an Answer. I will certainly upvote even if you mostly transcribe your Comments into the Answer box. $\endgroup$ – hardmath Jul 16 '17 at 21:49
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Suppose we would like to find the intersection of $2$ hyperplanes in $\mathbb R^n$

$$\begin{array}{rl} a_{11} x_1 + a_{12} x_2 + \cdots + a_{1n} x_n &= b_1\\ a_{21} x_1 + a_{22} x_2 + \cdots + a_{2n} x_n &= b_2\end{array}$$

In matrix form,

$$\begin{bmatrix} — \mathrm a_1^\top —\\ — \mathrm a_2^\top —\end{bmatrix} \mathrm x = \begin{bmatrix} b_1\\ b_2\end{bmatrix}$$

or, more economically, $\rm A x = b$, where $\rm A$ is a $2 \times n$ matrix. Without loss of generality, we assume that both hyperplanes are in Hessian normal form, i.e., $\| \mathrm a_1 \|_2 = \| \mathrm a_2 \|_2 = 1$.

If vectors $\rm a_1$ and $\rm a_2$ are collinear, then the $2$ given hyperplanes are either parallel or overlapping, neither of which are very interesting. Thus, we assume that vectors $\rm a_1$ and $\rm a_2$ are not collinear, i.e., matrix $\rm A$ has full row rank. Let $\theta := \arccos ( \mathrm a_1^\top \mathrm a_2 )$ be the angle formed by vectors $\rm a_1$ and $\rm a_2$.

The solution set of the linear system $\rm A x = b$ is a $(n-2)$-dimensional affine space. To find this affine space, we must find a particular solution and the null space of $\rm A$. One particular solution would be the least-norm solution, i.e., the one closest to the origin

$$\begin{array}{rl} \mathrm x_{\text{LN}} := \mathrm A^\top \left( \mathrm A \mathrm A^\top \right)^{-1} \mathrm b &= \begin{bmatrix} | & | \\ \mathrm a_1 & \mathrm a_2\\ | & |\end{bmatrix} \begin{bmatrix} \| \mathrm a_1 \|_2^2 & \langle \mathrm a_1, \mathrm a_2 \rangle\\ \langle \mathrm a_2, \mathrm a_1 \rangle & \| \mathrm a_2 \|_2^2\end{bmatrix}^{-1} \begin{bmatrix} b_1\\ b_2\end{bmatrix}\\ &= \begin{bmatrix} | & | \\ \mathrm a_1 & \mathrm a_2\\ | & |\end{bmatrix} \begin{bmatrix} 1 & \cos (\theta)\\ \cos (\theta) & 1\end{bmatrix}^{-1} \begin{bmatrix} b_1\\ b_2\end{bmatrix}\\ &= \frac{1}{\sin^2 (\theta)} \begin{bmatrix} | & | \\ \mathrm a_1 & \mathrm a_2\\ | & |\end{bmatrix} \begin{bmatrix} 1 & -\cos (\theta)\\ -\cos (\theta) & 1\end{bmatrix} \begin{bmatrix} b_1\\ b_2\end{bmatrix}\\ &= \left(\frac{b_1 - b_2 \cos (\theta)}{\sin^2 (\theta)}\right) \begin{bmatrix} | \\ \mathrm a_1 \\ | \end{bmatrix} + \left(\frac{b_2 - b_1 \cos (\theta)}{\sin^2 (\theta)}\right) \begin{bmatrix} | \\ \mathrm a_2 \\ | \end{bmatrix}\end{array}$$

where $\sin (\theta) \neq 0$ is ensured by the non-collinearity of vectors $\rm a_1$ and $\rm a_2$.

To find the $(n-2)$-dimensional null space of $\rm A$, we solve the homogeneous linear system $\rm A x = 0_2$. Introducing an $n \times n$ permutation matrix $\rm P$ with certain desired properties, we reorder the columns of $\rm A$, i.e., $\rm A P P^\top x = 0_2$, and obtain a homogeneous linear system in $\rm y:= P^\top x$

$$\rm A P y = 0_2$$

If $\rm P$ is chosen wisely, then there is a $2 \times 2$ matrix $\rm E$, a product of elimination matrices, that puts $\rm A P$ in reduced row echelon form (RREF), i.e.,

$$\rm E A P = \begin{bmatrix} \mathrm I_2 & \mathrm F\end{bmatrix}$$

where $\rm F$ is a $2 \times (n-2)$ matrix. Hence, the null space of $\rm A P$ is given by

$$\left\{ \begin{bmatrix} -\mathrm F \\ \mathrm I_{n-2} \end{bmatrix} \eta : \eta \in \mathbb R^{n-2} \right\}$$

and, thus, the null space of $\rm A$ is given by

$$\left\{ \mathrm P \begin{bmatrix} -\mathrm F \\ \mathrm I_{n-2} \end{bmatrix} \eta : \eta \in \mathbb R^{n-2} \right\}$$

Lastly, the intersection of the $2$ given hyperplanes is given by

$$\left\{ \color{blue}{\mathrm x_{\text{LN}} + \mathrm P \begin{bmatrix} -\mathrm F \\ \mathrm I_{n-2} \end{bmatrix} \eta} : \eta \in \mathbb R^{n-2} \right\}$$

In MATLAB, we can use function rref to find permutation matrix $\rm P$ and matrix $\rm F$. Of course, we can also use function null to find an orthonormal basis for the null space of $\rm A$.

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