In ${\bf M}_n(\mathbb R)$, let us consider the usual operator norm
$$\|A\|=\sup\frac{\|Ax\|}{\|x\|},$$
where $\|x\|$ is the Euclidian norm.

The unit ball $B$ is the set of *contractions*. It is a convex compact subset of ${\bf M}_n(\mathbb R)$. By Krein-Milman (finite dimensional case), it is the convex hull of its subset ${\rm ext}(B)$ of extremal points. It turns out that ${\rm ext}(B)$ is the orthogonal group ${\bf O}_n(\mathbb R)$. 

Now, remember that  ${\bf O}_n(\mathbb R)$ has **two** connected components, a positive one  ${\bf SO}_n(\mathbb R)$ and a negative one ${\bf O}_n^-(\mathbb R)$. 

> What is the convex hull of  ${\bf SO}_n(\mathbb R)$ ?

Clearly, it is a compact convex subset, included in $B$. It is a strict subset of $B$, because it does not meet  ${\bf O}_n^-(\mathbb R)$. The title refers to the "positive part" of $B$, but this could be inappropriate, in the sense that it could meet the convex hull of ${\bf O}_n^-(\mathbb R)$ non-trivially.