How is the affine hull of the unit circle the entire $\mathbb{R}^2$? From the definition in Convex Optimization by Boyd, I don't see how you can get the vector (2,0) for example.
The affine set doesn't even have (0,0) as one of the points.
 A: The affine hull of a set $S$ is the smallest affine set containing $S$ (equivalently, the intersection of all affine sets containing $S$.) The affine sets of $\Bbb R^2$ are:


*

*Points

*Straight lines

*$\Bbb R^2$


Which one of these contains the unit circle?
A: I don't know the definition given in that book, but if it's anything like the standard definition, a point $X$ is in the affine hull if there are points $P, Q$ in $S^1$ with $(1-t)P + t Q = X$, i.e., if $X$ lies on the line through $P$ and $Q$. (The entire affine hull is the result of repeating this process of "affine closure" infinitely often, but for this problem, you don't need that.)
Picking $P = (1,0)$ and $Q = (-1, 0)$ and $t = 0.5$, we see that $(0,0)$ is in the affine hull. 
Indeed, for any point $(x, y)$, let $ r = \sqrt{x^2 + y^2}$ and $\theta = arctan(y/x)$  (or better still, $\theta = atan2(x, y)$. Let $P = (x/r, y/r)$ and $ Q = -P$.
Then 
$$
X = (1-t)P + t Q = (1-t)P - t P = (1-2t)P
$$
when $t = \frac{1-r}{2}$. So every point in the plane is in the affine hull of the circle.  
