Equal principal minors of matrix plus rank-1 and inverse

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). 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 obtain $$b$$ and $$c$$ ?
3. As a general characterization of $$A$$ might be difficult, I am particularly interested in a solution of the form $$A = O G$$, with diagonal matrix $$G \neq I$$ and orthogonal $$O$$. Can you think of a class of matrices $$O$$ which simplifies this problem?

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 of $$B$$ and $$A^{-1}$$ are equivalent.

Attempt Following the work of Engel and Schneider and assuming fully connectedness of $$A$$: Let $$H = B \div A^{-1}$$, where $$\div$$ is element-wise. In Corrolary 3.11., it can be seen (from the fully connectedness) that $$H$$ is diagonally similar to $$\mathbf{1}$$, i.e., a matrix of 1s. Thus, for $$B$$ and $$A^{-1}$$ to be diagonally similar, there is a diagonal matrix $$X$$ such that $$X^{-1}HX = \mathbf{1}.$$ In particular, $$c_i = \frac{ A_{ii} - (A^{-1})_{ii} }{b_{i}}.$$ Remains to determine $$b$$. Please note that alternatively $$B$$ and $$A^{-T}$$ may be diagonally similar.

Possible Answer to Question 2 For indices $$J$$, the principal submatrix with rows and colomns $$J$$ is indicated by $$A_J$$. We want for all $$J$$

$$\det(B_J) = \det((A^{-1})_J).$$ With Sylvester's determinant identity this is $$b_J \textrm{adj}(A_J) c_J^T = \det((A^{-1})_J) - \det(A_J)$$ which is a system of bilinear equations and can be solved by vectorization.

Example Size=2

For

$$A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$$ and $$b = [3, 4.5]^T$$ and $$c = [1,1]^T$$. Then

$$B = \begin{bmatrix} -2 & -1 \\ -1.5 & -0.5 \end{bmatrix}$$ which is diagonally similar to $$A^{-1} = \begin{bmatrix} -2 & 1 \\ 1.5 & -0.5 \end{bmatrix}.$$

• What's the motivation to compare a rank-1 modification and a inverse? They have different "degrees". – Bullet51 Feb 14 at 12:05
• @Bullet51 I added the motivation – Sebastian Schlecht Feb 14 at 12:17