In a book I am reading the author states that "from Linear Algebra we know that the intersection of two hyperplanes of the form $H_i =\{x: a_i^Tx = b_i \}$ for $a_i \in \mathbb{R}^n$ and $b \in \mathbb{R}^m$ is an affine subspace".
If the $b_i$ were $0$ this would be clear to me since then the intersection would be just the kernel of a matrix. But why should this be true for $b_i \ne 0$? Could you help me?
Best Answer
Well, the claim is not that the intersection is a vector space. So claim: If $X$ is a vector space (or just a general affine space) and $A,B\subseteq X$ are affine, then $A\cap B$ is affine.
This follows, since, if $x,y\in A\cap B$ and $t\in \mathbb{R},$ we know that $tx+(1-t)y\in A$ and $tx+(1-t)y\in B$ by assumption, so $tx+(1-t)y\in A\cap B$.
Now, why is a hyperplane affine?
Well, take $x$ and $y$ such that $a^T_ix=a^T_iy=b_i$ and take $t\in \mathbb{R}$. Then, by linearity, $$ a_i^T(tx+(1-t)y)=ta_i^Tx+(1-t)a_i^Ty=tb_i+(1-t)b_i=b_i, $$ proving that a hyperplane is, indeed, affine, even though it is certainly not a subspace of the larger vector space in general.