The unit ball of ${\bf M}_n(\mathbb R)$ is a compact convex subset. As such, it is (Krein-Milman theorem) the convex envelop of its extremal points. So far, so good; but the unit ball depends of the choice of a norm. There are a few natural choices:
- the Schur-Frobenius (Hilbert-Schmidt) norm $\|X\|_F=\sqrt{{\rm Tr}X^TX}$. This is the standard Euclidian norm over ${\bf M}_n(\mathbb R)\sim\mathbb R^{n^2}$. The extremal points form the unit sphere.
- the operator norm $\|X\|_2$ associated with the Euclidian norm over $\mathbb R^n$. The extremal points form the orthogonal group ${\bf O}_n(\mathbb R)$. See a related question.
- the operator norm $\|X\|_1$ associated with the $\ell_1$-norm over $\mathbb R^n$. The extremal points are the sign-permutation matrices. There are only $2^nn!$ of them.
- the numerical radius $r(X)=\sup|y^*Xy|$, where the supremum is taken over all unit complex vectors (real vectors are not enough). It is the smallest radius of a disk $D(0;r)$ containing the numerical range (Hausdorffian) of $X$. So, this my question:
Let $B_{\rm nr}$ be the unit ball when we endow the $n\times n$-matrices with the norm $r$. What are the extremal points of $B_{\rm nr}$ ?
Edit. To answer Geoff's comment. If $A$ is a normal matrix, then $r(A)=\rho(A)$ (the numerical range is the convex envelop of the spectrum in this case). But if $A$ is not normal, one usually have (not always) $r(A)>\rho(A)$. Thus not all matrices with spectral radius equal to $1$ belong to $B_{\rm nr}$. And even if $A$ is normal and $\rho(A)=1$, it is not extremal unless all eigenvalues have unit modulus.
Re-edit. After thinking to the case $n=2$, I have the opinion that considering the unit ball for $r$ in ${\bf M}_n(\mathbb R)$ is unnecessarily complicated. It would be more reasonable to work in ${\bf M}_n(\mathbb C)$ instead.
$\|A\|\leq 1$
for all$A \in B_{nr}$
(resp.$\begin{pmatrix} 0 & 1\\1 &0\end{pmatrix} = 1/2 \begin{pmatrix} 0 & 2\\0 &0\end{pmatrix} + 1/2 \begin{pmatrix} 0 & 0\\2 &0\end{pmatrix}$
). $\endgroup$