# Basis of an affine subspace

Consider an affine subspace $$D$$ of an affine space or affine plane $$\mathcal{A}$$.

• Every set of points that are not elements of a proper affine subspace of $$D$$ is called a generating set of $$D$$.
• If every point $$x$$ of a set (of points) $$S \subseteq D$$ has the property that there exists an affine subspace of $$D$$ that contains $$S \backslash \{x\}$$, then we call $$S$$ an independent set of $$D$$.
• A basis for $$D$$ is a generating and independent set.

I'm having difficulties understanding the first and second definitions. I can't really think of a good visual representation or an intuitive explanation. Maybe somebody has an equivalent or clearer formulation for both definitions. Are there easy examples to illustrate them?

Your second bullet is not quite correct. It should be the following:

• If $$S \subseteq D$$ is a generating set of the affine subspace $$C\subseteq D$$, and if for every $$x\in S$$ there is a proper affine subspace of $$C$$ containing $$S\setminus\{x\}$$, then $$S$$ is called an independent set of $$D$$.

It's almost the exact same as with vector spaces and subspaces. If you had correct definitions of those, you could say the following:

Consider a vector subspace $$W$$ of a vector space $$V$$.

• A set of vectors $$S\subseteq W$$ which is not contained in a proper subspace of $$W$$ is called a spanning set of $$W$$.

• A set $$S\subseteq W$$ which is a spanning set of the subspace $$X\subseteq W$$ and which has the property that for any $$v\in S$$ there is a proper subspace of $$X$$ containing $$S\setminus\{v\}$$ is called a linearly independent subset of $$W$$.

• A basis for $$W$$ is a linearly independent spanning set.

These definitions are very geometric in the vector space setting, and it's much the same in the affine setting. In fact, if you pick an origin $$O\in S$$ and replace $$S$$ with $$S' = \{x-O : x\in S\setminus\{O\}\}$$, you get back the vector space definitions. So the affine definitions are geometrically really like spanning sets with an origin and linearly independent sets with an origin, only that the origin is not fixed in the affine setting.

For example, $$S=\{(1,1),(1,0),(1,-1)\} \subseteq \mathbb{R}^2$$ is a generating set for the affine line $$x = 1$$, since any an affine subspace which is more than a point must contain a line (and there's only one line between two points). But it is not an independent set, since you also need the full line $$x=1$$ to capture $$(1,0)$$ and $$(1,1)$$ in an affine subspace. However, $$S = \{(1,1),(1,0)\}$$ is independent: it generates the line $$x=1$$, and if you remove either point, the other point is contained in the affine subspace which just consists of that point.

If you pick $$(1,0)\in S$$ to think of as your origin in $$\mathbb{R}^2$$ so that you now have a vector space structure, this is like saying that $$(1,1)-(1,0) = (0,1)$$ and $$(1,-1)-(1,0) = (0,-1)$$ span the line $$\{x=1\}-(1,0) = \{x=0\}$$ but are not linearly independent vectors. However, $$(0,1)$$ also spans that line, and $$\{(0,1)\}$$ is a linearly independent set of vectors.

• Thanks for correcting the definition! I don't understand why $S = \{(1,1),(1,0),(1,-1)\}$ is a generating set for $x=1$. By definition the points $(1,1),(1,0)$ and $(1,-1)$ cannot be elements of a proper subspace of the affine line (which has dimension 1). However $(1,0) \in \{(1,0)\}$ (dimension 0). Or does the definition imply that all three points cannot be elements of the same subspace? – Zachary Feb 20 at 19:44
• Yes, it means all of them. – csprun Feb 20 at 23:26