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


$$ A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} $$ 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 $$ A^{-1} = \begin{bmatrix} -2 & 1 \\ 1.5 & -0.5 \end{bmatrix}. $$

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

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