$\newcommand{\S}{\mathcal{S}}$
Question: Let $A$ be a symmetric matrix with zero diagonals. I was curious if there exist a diagonal matrix with positive diagonals $D$ and a skew-symmetric matrix $\Omega$ such that $A = \Omega D – D \Omega$.
Context: Lately I've been researching the manifold of positive-definite matrices, which I'll denote $\S_+$. It is written the the tangent space of $\S_+$ at any point is the symmetric matrices, which I'll denote $\S$. Let $P(t) \in \S_+$ be a smooth curve of PD matrices. Then we can write the eigendecomposition $P(t)=V(t)D(t)V(t)^T$ such that $D(t),V(t)$ are smooth curves of positive diagonal and orthogonal matrices, respectively. We can then write $$\dot{P}=\dot{V}DV^T+V\dot{D}V^T+VD\dot{V^T}=V\Omega DV^T+V\dot{D}V^T-VD\Omega V^T,$$ where $\dot{V}=V\Omega$, and $\Omega$ is skew-symmetric. Then $$\dot{P}=V(\Omega D-D\Omega+\dot{D})V^T.$$
As you can see, since $\Omega D – D\Omega + \dot{D}$ is symmetric, it follows $\dot{P}$ is symmetric. So, we can write $T_P \S_+ \subset \S$. However, I wanted to prove the converse. This implies that for any $S \in \S$, there exists a diagonal matrix with positive entries $D$, a skew-symmetric matrix $\Omega$, and a diagonal matrix with arbitrary diagonals $\dot{D}$ such that $S=\Omega D – D \Omega + \dot{D}$.
Now, I know $\Omega D – D \Omega$ has zero diagonals and symmetric. So, we can express $S=B + A$, where $B=diag(S)$ and $A=S-B$. Thus, we can set $\dot{D}=B$, and simply solve $A=\Omega D – D \Omega$. This leads me to my question written above: Can we express any zero-diagonal symmetric matrix as the commutator of a diagonal matrix with positive diagonals and a skew-symmetric matrix?
Best Answer
Yes. You may take any positive diagonal matrix $D=\operatorname{diag}(d_1,d_2,\ldots,d_n)$ with distinct diagonal elements and define $\Omega$ with $\omega_{ij}=\dfrac{a_{ij}}{d_j-d_i}$ for all $i\ne j$.