# Characterizing the dual cone of the squares of skew-symmetric matrices

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).

• If we're talking about real skew-symmetric matrices, then the negative eigenvalues of $B^2$ must come in duplicate pairs, which complicates things. Dec 25, 2019 at 5:28
• 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 $1$. Dec 25, 2019 at 5:30
• I assume that you're talking about real skew-symmetric matrices, by the way. Dec 25, 2019 at 5:40

## 2 Answers

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 $$U$$) $$B = U \pmatrix{0 & -\lambda_1 \\ \lambda_1 & 0 \\ && \ddots \\ &&& 0 & -\lambda_k \\ &&& \lambda_k & 0 \\ &&&&& 0 } U^T \\ = \lambda_1 \ [u_2u_1^T - u_1 u_2^T] + \cdots + \lambda_{k}\ [u_{2k}u_{2k-1}^T - u_{2k-1}u_{2k}^T]$$

• 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. Dec 25, 2019 at 16:00
• 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. Dec 25, 2019 at 16:02
• I’ll fix that when I get the chance. And you’re welcome! You always seem to find interesting questions to ask. Dec 25, 2019 at 17:21
• See my latest edit for a proof of the other direction Dec 25, 2019 at 17:51
• @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. Dec 26, 2019 at 15:47

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 $$\langle D, B^2\rangle = \sum_{i,j:i\neq j}-\tfrac12(d_i+d_j) b_{i,j}^2\leq 0.$$

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