Given an invertible real matrix $A$ and real column vectors $b$ and $c$.

For which $A$,$b$ and $c$ are all corresponding principal minors of $B = A-bc^T$ and $A^{-1}$ equal?

According to a result by Loewy, this is true if $B$ and $A^{-1}$ are diagonally similar with transpose (plus some extra conditions). This work by Engel and Schneider seems more promising, but still I'm stuck. We can assume that both $A$ and $A^{-1}$ are adjacency matrices of fully connected graphs, i.e., all entries are non-zero.

My main interests are:

1. For which matrices $A$ is the problem solvable?
2. Given a matrix $A$, how attain $b$ and $c$ numerically (best possible if no exact solution exist)?
3. As a general characterization of $A$ might be difficult, I am particularly interested in a solution of the form $A = G O G$, with diagonal matrix $G \neq I$ and orthogonal $O$. Can you think of a class of matrices $O$ which simplifies this problem?

Attempt Following the work of Engel and Schneider and assuming fully connectedness: Let $H = A \div B$, where $\div$ is element-wise. Then for $A$ and $B$ to be diagonally similar, we need $H_{ii} = 1$ for all $i$. Hence, $$ b_i = \frac{ A_{ii} - (A^{-1})_{ii} }{c_{i}}. $$ Note, $(A^{-1})_{ii}$ can be expressed via Jacobi's identity in terms of $A$. Remains, to determine $c$. In Corrolary 3.11., it can be easily seen (from the fully connectedness) that $H$ is diagonally similar to a matrix of only 1s.

A simple case is when assuming that $A$ and $bc^T$ are symmetric. Then $A-dd^T$ and $A^{-1}$ are in canonical form and therefore $dd^T = A - A^{-1}$. This is only true if all eigenvalues of $A$ are 1 except one.

Motivation: The problem arises in control theory, where transfer function from a state-space formulation is:

$$ H(z) = \frac{\det(A) \det(D(z) - (A - bc^T))}{\det(D(z) - A)}, $$ where $D(z) = diag([z^{m_1},\dots,z^{m_n}])$ for integer $m_i$. The goal is now to choose $A$, $b$ and $c$ such that $|H(z)|=1$ for all $z$. This is true if the numerator and denominator of $H(z)$ are "flipped", i.e.,

$$ flip(\det(D(z) - A)) = \det(A) \det(D(z) - A^{-1}). $$

Thus, for any $m_i$, we need: $$ \det(D(z) - A^{-1}) = \det(D(z) - (A - bc^T)), $$ which is true if all principal minors are equivalent.