The boundary of the convex hull of squares of skew-symmetric matrices

Question

Let $n \ge 3$, and let $C$ be the convex cone generated by the squares of all real $n \times n$ skew-symmetric matrices. Is $C$ closed in $\mathbb{R}^{n^2}$? What is its boundary?

$C$ is a strictly contained in the cone of negative semi-definite matrices.

I know that the set of the squares of all real $n \times n$ skew-symmetric matrices is closed, but in general, a convex cone which is generated by a closed linear cone does not have to be closed.

Edit:

Let's focus on the case where $n=3$, and try to describe $C$ more explicitly.

Let $T:\mathbb{R}^3 \to \mathbb{R}^3$ be a skew-symmetric operator on $\mathbb{R}^3$. $T$ has the form of $T(x) = v \times x$ for some $v \in \mathbb{R}^3$, where $\times$ is denotes the cross product.

The triple vector product implies that

$$ T^2(x)=v \times (v \times x)=\langle v,x \rangle v-\langle v,v \rangle x. $$

Thus, $C$ is the set of all operators $\mathbb{R}^3 \to \mathbb{R}^3$ of the form

$$ x \to \sum_i \langle v_i,x \rangle v_i-|v_i|^2 x.$$

I am not sure if this observation really advance us...

Answer 1

I will prove in the below that $C$ is identical to $$ L=\left\{S: S\preceq0,\ 2\lambda_\min(S)\ge\operatorname{tr}(S)\right\}. $$ Note that the conditions $S\preceq0$ and $2\lambda_\min(S)\ge\operatorname{tr}(S)$ together imply that any $S\in L$ cannot be a rank one matrix, and when $S$ has rank two, its two nonzero eigenvalues must be the same. It follows that if $S\in L$ has rank $\le2$, it must be the square of some skew-symmetric matrix.

If $C$ is indeed equal to $L$, then $C$ is closed in the cone of negative semidefinite matrices, in the set $\mathcal S$ of all symmetric matrices, or in $M_n(\mathbb R)$, and

• $\partial C=C$ in $M_n(\mathbb R)$, as we can always perturb every $S\in C$ to some non-symmetric matrix, so that $C$ has an empty interior,
• $\partial C=\{S\in C: 2\lambda_\min(S)=\operatorname{tr}(S)\text{ or } S \text{ is singular}\}$ in the set of all symmetric matrices, and
• $\partial C=\{S\in C: 2\lambda_\min(S)=\operatorname{tr}(S)\}$ in the cone of negative semidefinite matrices.

To prove the inclusion $C\subseteq L$ is easy. The square of every (possibly zero) skew-symmetric matrix is clearly a member of $L$. It follows that every sum of squares of skew-symmetric matrices $S=K_1^2+\cdots+K_m^2$ is a member of $L$ too, because $$ 2\lambda_\min(S) =2\lambda_\min\left(\sum_iK_i^2\right) \ge2\sum_i\lambda_\min\left(K_i^2\right) \ge\sum_i\operatorname{tr}\left(K^2\right) =\operatorname{tr}(S). $$

To prove $L\subseteq C$, we prove by mathematical induction that $$ L^{(r)}=\left\{S\in L: \operatorname{rank}(S)\le r\right\}\subseteq C $$ for $r=2,3,\ldots$. The base case $r=2$ is true because every $S\in L^{(2)}$ is necessarily the square of some (possibly zero) skew-symmetric matrix. In the induction step, suppose $r\ge3$ and $L^{(r-1)}\subseteq C$. Pick any $S\in L^{(r)}$. By orthogonal diagonalisation, we may assume that $S=\operatorname{diag}(s_1,s_2,\ldots,s_r,0,\ldots,0)$ where $s_1\le s_2\le\cdots\le s_r<0$. The condition $2\lambda_\min(S)\ge\operatorname{tr}(S)$ is equivalent to that $s_1\ge\sum_{i>1}s_i$. There are three cases and we will prove in each case that $S\in C$:

1. $s_1=s_2$. Then $$ \begin{aligned} S&=(s_2-s_3)(I_2\oplus0)+(s_3-s_4)(I_3\oplus0)+\cdots+(s_{r-1}-s_r)(I_{r-1}\oplus0)+s_r(I_r\oplus0)\\ &=|s_2-s_3|(-I_2\oplus0)+|s_3-s_4|(-I_3\oplus0)+\cdots+|s_{r-1}-s_r|(-I_{r-1}\oplus0)+|s_r|(-I_r\oplus0). \end{aligned} $$ Since $-I_2\oplus0=(E_{12}-E_{21})^2$ and $$ -I_k\oplus0 =\frac12\left[(E_{12}-E_{21})^2+(E_{23}-E_{32})^2 +\cdots+(E_{k-1,k}-E_{k,k-1,k})^2+(E_{k1}-E_{1k})^2\right] $$ for each $k\ge3$, we have $S\in C$.
2. $s_1-s_2\ge s_3$. Let $A=\operatorname{diag}\left(s_3,\,s_2+s_3-s_1,\,s_3,\,s_4,\ldots,s_r,0,\ldots,0\right)$. Then $A\preceq0$ and the two most negative eigenvalues of $A$ are both equal to $s_3$. Therefore $A\in C$ by case 1. But $S-A=(s_1-s_3)\operatorname{diag}\left(1,1,0,\ldots,0\right)=|s_1-s_3|(E_{12}-E_{21})^2$ also belongs to $C$. Hence $S\in C$.
3. $s_1-s_2<s_3$. Let $B=\operatorname{diag}(s_1-s_2,0,s_3,\ldots,s_r,0,\ldots,0)$. Then $B\preceq0$ and $$ 2\lambda_\min(B)=2(s_1-s_2)\ge(s_1-s_2)+s_3+\cdots+s_r=\operatorname{tr}(B). $$ Hence $B\in L^{(r-1)}$ and by induction hypothesis, $B\in C$. But $S-B=\operatorname{diag}(s_2,s_2,0,\ldots,0)$ also belongs to $C$. Therefore $S\in C$.