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 closed unit ball $B$ is the set of *contractions* (in the terminology used by operator theorists). 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.

Remark also that this convex hull is invariant under multiplication at right or left by an element of ${\bf SO}_n(\mathbb R)$. Therefore it would be enough to decide which diagonal matrices ${\rm diag}(a_1,\ldots,a_n)$ with $|a_1|\le a_2\le\cdots\le a_n$ it contains.

When $n=2$, ${\bf SO}_n(\mathbb R)$ is a circle and its convex hull is a disk, obviously a much smaller set (even from the dimensional point of view) than $B$.