Is it possible to find a vector $x$ such that for diagonal matrix $D$, $xx^T+D$ has a displacement structure?

Question

Suppose that $D \in \mathbb{R}_+^{n \times n}$ is a diagonal matrix with all positive elements. Is it possible to find a vector $x \in \mathbb{R}^n$ such that $xx^T + D$ has a displacement structure? For example, Vandermonde, Cauchy, Hankel, Toeplitz, etc.

Answer 1

Using the definition of displacement in that paper $\nabla_{F,A}(R) = FR-RA$, we can see that

$\nabla_{F,A}(xx^T+D) = \nabla_{F,A}(xx^T)+\nabla_{F,A}(D) = \underbrace{Fxx^T-xx^TA}_{\text{rank} \ \le \ 2}+\nabla_{F,A}(D)$

Thus, $\text{rank}\left[\nabla_{F,A}(D)\right] - 2 \le \text{rank}\left[\nabla_{F,A}(xx^T+D)\right] \le \text{rank}\left[\nabla_{F,A}(D)\right] + 2$.

This means that if $xx^T+D$ can only have low displacement rank when $D$ itself has low displacement rank.

Of course, whether or not $D$ has low displacement rank will depend on what matrices $F$ and $A$ you choose for your displacement operator.