Flatness directions of the operator norm

Question

It is known that the standard operator norm $\|\cdot\|_2$ over ${\bf M}_n({\mathbb R})$ is very flat, as is any operator norm (= subordinated norm) actually. The set of extremal points of the unit ball is the very small set ${\bf O}_n({\mathbb R})$ and the unit sphere contains faces (convex subsets) of dimension $(n-1)^2$.

I discovered that the directions of flatness actually are singular matrices: if a segment $[M,N]$ is included in the unit sphere, then $\det(M-N)=0$. I hardly pretend that this is an original result.

Is there any reference in the literature for the above result ?

I should be also interested in any related statement for other operator norms.

Notation. I use to write $\|\cdot\|_p$ for the norm over ${\bf M}_n({\mathbb R})$ induced by the $\ell^p$ norm over ${\mathbb R}^n$.

Answer 1

(You write $\|\cdot\|_2$ but say "operator norm", and I assume that is what you mean.)

I haven't seen this exact statement, but it follows easily from known facts about the facial structure of operator unit balls. The basic reference is Akemann and Pedersen, Facial structure in operator algebra theory, Proc. London Math. Soc. 64 (1992), 418–448.

Akemann and Pedersen show that the weakly closed faces of the unit ball of a von Neumann algebra are precisely the sets of the form $v + (1 - vv^*)B(1 - v^*v)$ where $B$ is the full unit ball and $v$ is a partial isometry. In finite dimensions all faces are weakly closed, of course. The difference of two elements of such a face would have the form $(1 - vv^*)x(1 - v^*v) = pxq$ where $p$ and $q$ are orthogonal projections. That can be nonsingular only if $p = q = 1$, which would imply $v = 0$, making the "face" be all of $B$ (if you consider that to be a face).

Your result is stated for real matrices, but any face of the unit ball of $M_n(\mathbb{R})$ is contained in a face of the unit ball of $M_n(\mathbb{C})$, so nothing more needs to be said.