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Let $X$ be the set of all real $n \times n$ diagonal matrices $D$ satisfying $\langle D,B^2 \rangle \le 0$ for any (real) skew-symmetric matrix $B$. (I am using the Frobenius Euclidean product here).

$X$ is a convex cone.

Can we give an explicit characterization of $X$?

Comment:

If we denote by $C$ the space of all squares of skew-symmetric matrices, we can characterize its dual cone as follows:

Since every square of a skew-symmetric matrix is symmetric, and the symmetric and the skew-symmetric matrices are orthogonal, we know that every skew-symmetric matrix belongs to the dual cone of $C$. So, the question whether a given matrix $A$ belongs to the dual cone of $C$ depends solely on the symmetric part of $A$. Since $C$ is invariant under orthogonal conjugation, we can orthogonally diagonalize $\text{sym}(A)$ and deduce that $A$ lies in $C^*$ if and only if the diagonal matrix whose entries are the eigenvalues of $\text{sym}(A)$ is in $C^*$. Thus, the question reduces to determining the case of diagonal matrices.

Edit:

Omnomnomnom proved in this answer that every $D$ in $X$ has at most one negative entry, and the absolute value of the negative entry is less than or equal to the next-smallest entry.

I have a strangely complicated proof for the converse, namely I can prove that every diagonal matrix satisfying the condition above is in $X$.

I would like to find a "direct" proof based on linear algebra\matrix analysis. (my proof is based on rather convoluted variational considerations).

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  • $\begingroup$ If we're talking about real skew-symmetric matrices, then the negative eigenvalues of $B^2$ must come in duplicate pairs, which complicates things. $\endgroup$ Dec 25, 2019 at 5:28
  • $\begingroup$ Note that if a matrix $D$ satisfies $\langle D, S \rangle \leq 0$ for any symmetric $S$ with non-positive eigenvalues, then we would be able to conclude that $D$ must be positive definite. This is can be seen quickly by testing matrices $S$ of rank $1$. $\endgroup$ Dec 25, 2019 at 5:30
  • $\begingroup$ I assume that you're talking about real skew-symmetric matrices, by the way. $\endgroup$ Dec 25, 2019 at 5:40

2 Answers 2

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Claim: $D$ has at most one negative eigenvalue, and the absolute value of the negative eigenvalue is less than or equal to the next-smallest eigenvalue.

Proof: Let $E_{ij}$ denote the matrix with a $1$ in the $i,j$ entry and zeros elsewhere.

It suffices to show that if the $i$th and $j$th diagonal entries of $D$ have a negative sum, then $D$ cannot satisfy the criterion. To that end, it suffices to note that there exists a skew-symmetric matrix with $B^2 = -(E_{ii} + E_{jj})$ (take $B = E_{ij} - E_{ji}$ for instance). $\square$

I am not sure whether this condition is equivalent to your inequality.


We can also prove that the condition above is sufficient as follows. Suppose that $D$ has at most one negative eigenvalue, and the absolute value of the negative eigenvalue is less than or equal to the next-smallest eigenvalue.

We first note that every matrix of the form $M = B^2$ for a skew-symmetric $B$ can be written in the form $$ M = -[a_1 \, (x_1x_1^T + y_1y_1^T) + \cdots + a_k \, (x_kx_k^T + y_ky_k^T)]. $$ where the coefficients $a_i$ are non-negative and $x_i,y_i$ are a pair of orthonormal unit vectors for all $i$. So, it suffices to show that $\langle D,M\rangle \leq 0$ where $M = -(xx^T + yy^T)$ for some orthonormal $x,y$.

Now, let $v_1,\dots,v_n$ be an orthonormal basis for $\Bbb R^n$ such that $x = v_1$ and $y = v_2$. Let $V$ be the orthogonal matrix whose columns are $v_1,\dots,v_n$, and let $A = V^TDV$. We now note that $$ \langle D, xx^T + yy^T \rangle = x^TDx + y^TDy = a_{11} + a_{22}. $$ From here, it suffices to apply the $(\implies)$ direction of the Schur-Horn theorem to $-A$ in order to conclude that $a_{11} + a_{22} \geq \lambda_{n}(D) + \lambda_{n-1}(D)$.


About the squares of skew-symmetric matrices: by the spectral theorem, there exists a unitary $U$ with columns $u_1,u_2,\dots,u_n$ such that $$ B = U \pmatrix{i \lambda_1 \\ & - i\lambda_1 \\ && \ddots \\ &&& i \lambda_k \\ &&&& - i \lambda_k \\ &&&&& 0 } U^* \\ = \lambda_1 i \ [u_1u_1^* - u_2 u_2^*] + \cdots + i\lambda_{k}\ [u_{2k-1}u_{2k-1}^* - u_{2k}u_{2k}^*] $$ where each $\lambda_i$ is positive. Thus, squaring $B$ yields $$ B^2 = -(\lambda_1^2 \ [u_1u_1^* + u_2 u_2^*] + \cdots + \lambda_{k}^2\ [u_{2k-1}u_{2k-1}^* + u_{2k}u_{2k}^*]). $$ We could equivalently have used the canonical form (with a real, orthogonal $U$) $$ B = U \pmatrix{0 & -\lambda_1 \\ \lambda_1 & 0 \\ && \ddots \\ &&& 0 & -\lambda_k \\ &&& \lambda_k & 0 \\ &&&&& 0 } U^T \\ = \lambda_1 \ [u_2u_1^T - u_1 u_2^T] + \cdots + \lambda_{k}\ [u_{2k}u_{2k-1}^T - u_{2k-1}u_{2k}^T] $$

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  • $\begingroup$ Thanks! This is a very cute observation, really. Actually, it happens that I know that the converse direction holds, i.e. that any diagonal matrix satisfying your condition projects non-positively on the squares. (So, this condition is equivalent to the inequality). However, my proof is indirect, and I would like to find a more straightforward proof. You may see my edit to the question. $\endgroup$ Dec 25, 2019 at 16:00
  • $\begingroup$ Also, there is a minor typo in your question: It should be $B^2=-(E_{ii} + E_{jj})$ (you forgot a minus sign). Finally, this is a chance to thank you for all your many great answers on this site. You have given me valuable answers many times, and I appreciate it. $\endgroup$ Dec 25, 2019 at 16:02
  • $\begingroup$ I’ll fix that when I get the chance. And you’re welcome! You always seem to find interesting questions to ask. $\endgroup$ Dec 25, 2019 at 17:21
  • $\begingroup$ See my latest edit for a proof of the other direction $\endgroup$ Dec 25, 2019 at 17:51
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    $\begingroup$ @Asaf not every such convex-combination is the square of a skew symmetric matrix, but every square of a skew-symmetric matrix can be expressed as such a linear combination. The set of matrices expressible as $-[a_1 \, (x_1x_1^T + y_1y_1^T) + \cdots a_k \, (x_kx_k^T + y_ky_k^T)]$ is the convex cone generated by the squares of skew-symmetric matrices. Your statement on $D$ amounts to saying that $D$ is an element of the dual cone. $\endgroup$ Dec 26, 2019 at 15:47
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Here is a slightly different proof of the sufficiency of the condition $d_i+d_j\geq 0$ for all $i\neq j,$ which is the same as the condition in Omnomnomnom's answer.

Note that

\begin{align*} (B^2)_{ii} &=\sum_{j} b_{i,j}b_{j,i}\\ &=-\sum_{j:i\neq j} b_{i,j}^2 \end{align*}

So

$$\langle D, B^2\rangle = \sum_{i,j:i\neq j}-d_i b_{i,j}^2.\tag{1}$$ Swapping the roles of $i$ and $j,$ $$\langle D, B^2\rangle = \sum_{i,j:i\neq j}-d_j b_{i,j}^2.\tag{2}$$ Averaging (1) and (2) gives $$\langle D, B^2\rangle = \sum_{i,j:i\neq j}-\tfrac12(d_i+d_j) b_{i,j}^2\leq 0.$$

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  • $\begingroup$ Hi. Can you ask a question on this site some time this academic year? I'd be appreciative. $\endgroup$ Dec 27, 2019 at 10:40
  • $\begingroup$ @Dap Thanks, this is a great and self-contained answer. (which does not rely on heavier guns like the Schur-Horn theorem). $\endgroup$ Dec 30, 2019 at 19:43

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