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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. My guess (have to check all details, but I have a proof) is that if $n=2$, then the extremal points of the (complex) unit ball are the nilpotent matrices of Frobenius norm $2$, that is the matrices $ab^T$ with $b^Ta=0$ and $\|a\|^2\|b\|^2=4$.
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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. One could wander whether these are

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. My guess (have to check all details, but I have a proof) is that if $n=2$, then the extremal points of the (complex) unit ball are the nilpotent matrices of Frobenius norm $B_{\rm nr}$.2$, that is the matrices $ab^T$ with $b^Ta=0$ and $\|a\|^2\|b\|^2=4$.
show/hide this revision's text 2 added 490 characters in body

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 teh 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 of 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. One could wander whether these are all the extremal points of $B_{\rm nr}$.
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