Calculating the intersection of two affine subspaces

Question

Question

Let $A$ and $B$ be affine subspaces of a Euclidean vector space $E$. The intersection of two affine subspaces is an affine subspace. But, how do I calculate the affine subspace $A\cap B$?

What I got so far

As far as I understand I can say that an affine space is the image of the function

$$f(x) = M \cdot (x - v)$$

for all $x\in E$ with $v\in E$ and $M$ being a matrix of size $(m\ \times\ n)$, where $m$ is the dimension of the affine space and $n$ the dimension of $E$.

So I was thinking, that I could write

$$ M_A \cdot (x - v_A) \overset!= M_B \cdot (x - v_B) \\ M_A \cdot x - M_A \cdot v_A = M_B \cdot x - M_B \cdot v_B \\ (M_A - M_B) \cdot x = M_A \cdot v_A - M_B \cdot v_B \\ (M_A - M_B)^T \cdot (M_A - M_B) \cdot x = (M_A - M_B)^T \cdot (M_A \cdot v_A - M_B \cdot v_B) \\ x = ((M_A - M_B)^T \cdot (M_A - M_B) )^{-1} \cdot (M_A - M_B)^T \cdot (M_A \cdot v_A - M_B \cdot v_B) \\ $$

and this would give me all (?) the elements in $E$ that belong to $A\cap B$. But wouldn't that be only one element? How do I calculate the other ones? Where did my train of thought go wrong?

Answer 1

1. Every affine space can be written as the image of $f(x)=M(x-v)$. So, if $y\in A$, then there exists some $x_A$ such that $$y=f(x_A)=M_A(x_A-v_A)$$ Likewise, if $y\in B$, then there exists some $x_B$ such that $$y=f(x_B)=M_B(x_B-v_B)$$
2. But $x_A$ and $x_B$ need not be the same! Indeed, if $A$ and $B$ have different dimensionalities, then $x_A$ and $x_B$ certainly cannot coincide. This is why your strategy failed.
3. Given a set $S$, there are two ways to represent it: as the image of a function $\alpha$ (i.e. $S=\{\alpha(x):x\in P\}$), or the fiber of some function $\beta$ (i.e. $S=\{y:\beta(y)=0\in Q\}$, where $Q$ is some high-dimensional space). The first corresponds to a giant "or" over $P=\{x_1,x_2,\dots\}$: $$y\in S\Leftrightarrow (y=\alpha(x_1))\vee(y=\alpha(x_2))\vee\dots$$ The second corresponds to a giant "and": each coordinate of $\beta$ must be zero.
4. If $A$ and $B$ are represented as the images of functions, then $$y\in A\cap B\Leftrightarrow\left(\bigvee_{x\in P_A}{y=\alpha_A(x)}\right)\wedge\left(\bigvee_{x\in P_B}{y=\alpha_B(x)}\right)$$ — "ors" nested within "and". If $A$ and $B$ are represented as the fibers of functions, then $$y\in A\cap B\Leftrightarrow(\beta_A(y)=0)\wedge(\beta_B(y)=0)$$ — just a single flat "and". Suffice it to say, mixed logical operators are harder to work with than unmixed ones.
5. This suggests that representation from (1) is suboptimal for your task. Instead, note that any basis of $A$ can be extended to a basis of $E$ (and likewise for $B$). By sufficient cleverness, you can make the added basis elements perpendicular to $A$ (respectively $B$). Let the matrix of those added basis elements be $N$. Then $x\in A$ iff $g_A(x)=0\in\mathbb{R}^{n-m}$, where $g_A(x)=N_A^{\mathsf{T}}x-c$.
6. Using part (5), find a function $h$ such that $h(x)=0\in(\text{some high-dimensional space})$ iff $x\in A\cap B$. Then use part (5) again to achieve an affine basis, etc.