Why is any subspace affine?

Question

I am studying 'Convex Optimization' written by Stephen Boyd. I am confused by an assertion in the book(page 27). Any one can tell me why and give an explanation ?

Any subspace is affine, and a convex cone (hence convex).
--Convex Optimization

Answer 1

If $u, v \in U$, then $c_1u+ c_2v \in U$ where $c_1,c_2 \in \Bbb R$ . This is by the closure of subspace.

Thus if we replace them with familiar constants this is simply, $c_1u+ (1-c_1)v \in U$ where we have chosen $c_2 = 1 - c_2.$

Answer 2

Definition : A set $C ⊆ \mathbb R^n$ n is affine if the line through any two distinct points in $C$ lies in $C$, i.e., if for any $x_1$ , $x_2 ∈ C$ and $θ ∈ \mathbb R$, we have $θx_1 +(1−θ)x_2 ∈ C$

This is the definition given in the book of S. Boyd and L. Vandenberghe, the answer of your question is as @Nameless clarified before. all vector spaces satisfies this definition.

But I want to add an intuition useful to have, it is that the affine space is any set of the form $\{x_0\} + V$, where $V $ is a (nonzero)vector subspace and $x_0 \in \mathbb R^n$. So, a special case when $x_0 = 0$, gives $V$ which is thus an affine space. ( you can find this in page 22)

To set an example in $\mathbb R ^ 2$. (nontrivial)Affine spaces are all straight lines of the plane $\mathbb R ^ 2$, while (nontrivial) vector subspaces are the straight lines passing through the origin. (ie a special case of affine space)