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][1], 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][2], 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][3] and assuming fully connectedness of $A$: Let $H = B \div A^{-1}$, where $\div$ is element-wise. In [Corrolary 3.11.][3], 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][4]. 

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


  [1]: https://www.sciencedirect.com/science/article/pii/0024379586900157
  [2]: https://ccrma.stanford.edu/~jos/cfdn/Feedback_Delay_Networks.html
  [3]: https://www.math.wisc.edu/hans/paper_archive/scanned_papers/hs066.pdf
  [4]: https://arxiv.org/abs/1303.4988