Why is the affine hull of the unit circle $\mathbb R^2$?

Question

In Boyd's "Convex Optimization" it defines the affine hull of a subset $C$ of $\mathbb R^n$ as

$$\text{aff} C = \left\{\theta_1 x_1 + \ldots +\theta_k x_k \mid x_1, \ldots x_k \in C, \theta_1 + \ldots \theta_k = 1 \right\}.$$ Then, it claims $\text{aff } U = \mathbb R^2$ if $U$ is the unit circle. Why is this? Isn't any arc (or convex subset of the circle) entirely contained in the circle? I would think $\text{aff } U = U$ if $U$ is the unit circle.

Answer 1

To have an answer recorded as such, I'll add a few words. If $C$ contains the origin, then the affine hull is the same as linear span, since we can include $0$ with any coefficient we want. Also, translating $C$ by a vector translates its affine hull by the same vector. Thus, we can find the affine hull by moving the coordinate system so that the origin lies in $C$, and then taking the linear span. This shows at once that the affine hull of any three non-collinear points in the plane is the entire plane.

Answer 2

The unit circle $U$ is the set of all points (x,y) such that $x^2 + y^2 = 1$. So, if we take the affine hull of $U$, we will generate $\mathbb{R}^2$ since there exists at least 3 non-collinear points in $U$.

In particular, all we need to look at is a three element subset of the unit circle where the three points do not all lie along a line in order to generate $\mathbb{R}^2$ as a set of affine combinations.

If you are still confused by my answer and LVK's answer, you may want to review the definition of dimension and affine combination.

Answer 3

I guess I know your confusion. The point is that $\theta_i$ is not constrained to be greater than zero. So now you may understand why three non-collinear points will fill the whole $\Bbb R^2$.