based on a residual of 1%, while the MMR rank $r_{0}=6$ is based on the “knee”
criterion (see text). As with the models shown in Fig. 12, the resulting model
is stable and roughly matches the true dynamics, but is not as accurate as the
model reported in Sec. 6.4.

As a more principled example, Fig. 13 shows a model constructed by choosing a
kinematic reconstruction accuracy threshold of 99% for the POD-Galerkin rank
$r$. The MMR rank $r_{0}$ is then selected with the “knee” criterion, which
selects a truncation level based on the inflection point of the singular value
distribution. Again, the model is less accurate than the results shown in in
Sec. 6.4, but significantly more stable and accurate than any Galerkin system.
Thus, while a principled approach to rank selection is an open problem for
both POD-Galerkin modeling and the proposed multiscale reduction, the latter
appears generally less sensitive to this choice.

##  Appendix C Nonlinear eddy viscosity model

Multiscale model reduction is similar in spirit to the finite-time
thermodynamics (FTT) closure framework introduced by Noack et al. (2008), but
notably different in its particulars, as shown in Sec. 4.5. One drawback to
FTT closure modeling is that it increases the state dimension by a factor of
2, since the fluctuation energies $\\{E_{i}\\}$ have to be simulated along
with the state variables. Alternatively, Schlegel et al. (2009), Noack et al.
(2011), and Cordier et al. (2013) suggested a nonlinear eddy viscosity-type
closure model based on the structure of the FTT formalism without introducing
additional degrees of freedom. For a truncated Galerkin system

$\dot{x}_{i}=f_{i}(\bm{x})\equiv\sum_{j=1}^{r}L^{B}_{ij}x_{j}+\mbox{{Re}}^{-1}\sum_{j=1}^{r}L^{V}_{ij}x_{j}+\sum_{j,k=1}^{r}Q_{ijk}x_{j}x_{k},$
(78)

where $\bm{L}^{B}$ and $\bm{L}^{V}$ are the parts of the linear term
corresponding to base (or mean) flow advection and viscosity, respectively,
the FTT closure is based on the total mean fluctuation energy
$K_{\infty}=\sum_{i=1}^{r}\overline{x_{i}^{2}}/2$ and instantaneous
fluctuation energy $K(t)=\sum_{i=1}^{r}x_{i}^{2}(t)/2$. The nonlinear FTT
closure takes the form

$\dot{x}_{i}=f_{i}(\bm{x})+\nu_{T}\sqrt{\frac{K(t)}{K_{\infty}}}L^{V}_{ii}x_{i}.$
(79)

\begin{overpic}[width=390.25534pt]{figs/ftt_closure.pdf} \end{overpic} Figure
14:  Closure model based on the finite-time thermodynamics framework. The
plots show the fluctuating kinetic energy in the first 16 modes of Galerkin
(top) and FTT closure models (bottom) for the mixing layer on the longer
domain. Although the FTT closure model does stabilize the energy at a much
lower level than the unmodified Galerkin system, it does not reproduce the
natural dynamics of the flow over long time horizons.

The scalar eddy viscosity parameter $\nu_{T}$ is selected so that the mean
value $\overline{K(t)}$ predicted by the model (79) is consistent with the
computed value of $K_{\infty}$ by the least-squares solution of the energy
balance equation

$0=\mathop{\mathbb{E}}\left\\{\sum_{i=1}^{r}x_{i}f_{i}(\bm{x})\right\\}+\nu_{T}\mathop{\mathbb{E}}\left\\{\sum_{i=1}^{r}L_{ii}^{V}x_{i}^{2}\sqrt{\frac{K(t)}{K_{\infty}}}\right\\}.$
(80)

With this approach the solution of the closure model is guaranteed to be
bounded, with analytic estimates of the maximum kinetic energy growth (Cordier
et al., 2013).

Figure 14 compares the FTT-based nonlinear eddy viscosity model to the full
POD-Galerkin system modeling the mixing layer on the longest domain. While the
FTT model is significantly more stable than the truncated Galerkin system, it
neither matches the frequency content of the DNS over long time horizons nor
remains phase-coherent as for the MMR model (Figure 10). This is likely
because the nonlinearity introduced by the FTT model (79) models transfer of
energy to unresolved scales, but does not do so in a way that leads to the
synchronizing coupled-oscillator model in the same way as the MMR closure. The
truncation values $r$ shown here are typical of any values tested for $r<64$
for both the truncated POD-Galerkin system and the FTT closure.

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# The Fiber of the principal minor map

Abeer Al Ahmadieh  School of Mathematics, Georgia Institute of Technology,
Atlanta, USA 30332<EMAIL_ADDRESS>
###### Abstract.

This paper explores the fibers of the principal minor map over a general
field. The principal minor map is the map that assigns to each $n\times n$
matrix the $2^{n}$-vector of its principal minors. In $1984$, Hartfiel and
Loewy proposed a condition that was sufficient to ensure that the fiber of the
principal minor map is a single point up to diagonal equivalence. Loewy later
improved upon this condition in $1986$. In this paper, we provide a necessary
and sufficient condition for the fiber to be a point up to diagonal
equivalence. Additionally, we establish a connection between the reducibility
of a matrix and the reducibility of its determinantal representation. Using
this connection, we fully characterize the fiber of symmetric and Hermitian
matrices in the space of $n\times n$ matrices over any field $\mathbb{F}$. We
also use these techniques to answer a question of Borcea, Brändén, and Liggett
concerning real stable matrices.

##  1\. Introduction

The primary objective of this paper is to investigate the relationship between
two matrices with equal principal minors over a field $\mathbb{F}$. It is easy
to see that if two matrices are diagonally equivalent, then their principal
minors are equal. Two matrices $A$ and $B$ are diagonally equivalent if there
exists an invertible diagonal matrix $D$ such that $A=DBD^{-1}$ or
$A=DB^{T}D^{-1}$. The intriguing question is to determine the conditions under
which this remains the sole connection between the two matrices. Reducible
matrices, for instance, might have equal principal minors without being
diagonally equivalent. A matrix is reducible if it can be written as a block
upper triangular matrix after permuting some rows and the corresponding
columns. If we replace each nonzero entry with one and the diagonal entries
with zero, then a reducible matrix corresponds to the adjacency matrix of a
directed graph with at least two strongly connected components. The block
diagonal matrix $\tiny{\begin{pmatrix}E&0\\\ 0&G\end{pmatrix}}$, for instance,
has principal minors equal to those of all matrices of the form
$\tiny{\begin{pmatrix}E&F\\\ 0&G\end{pmatrix}}$.

Another motivating goal of this paper is to characterize real stable matrices
and answer a question of Borcea, Brändén, and Liggett [5, Question 3.4]. An
$n\times n$ complex matrix $A$ is real stable if its determinantal polynomial
$f_{A}=\det(A+\operatorname{diag}\\{x_{1},\ldots,x_{n}\\})$ is real stable. A
polynomial $f$ is real stable if it has no zeros with strictly positive
imaginary parts. Real stable polynomials have many applications in different
fields, see for instance [3, 4, 5, 10, 17]. In [2], the author and Vinzant
prove that if $f_{A}$ is real stable for some complex matrix $A$, then there
exists a Hermitian matrix $H$ such that
$f_{A}=\det(H+\operatorname{diag}\\{x_{1},\ldots,x_{n}\\})$. Since the
coefficients of $f_{A}$ are the principal minors of $A$, the matrices $H$ and
$A$ share the same principal minors. Therefore, to characterize real stable
matrices, we study the structure of matrices that have principal minors equal
to that of a Hermitian matrix.

Another way of rephrasing the same problem is by considering the principal
minor map. Given an $n\times n$ matrix $A$ with entries in a field
$\mathbb{F}$, let $A_{S}$ denote the principal minor indexed by $S$, that is
the determinant of the submatrix of $A$ indexed by the set $S$ on the rows and
columns. The principal minor map is the map that assigns to each matrix the
vector of its principal minors, namely

$\varphi:\mathbb{F}^{n\times n}\longrightarrow\mathbb{F}^{2^{n}}\ \ \text{
given by }\ \ A\longrightarrow\left(A_{S}\right)_{S\subseteq[n]}.$

where we set $A_{\emptyset}=1$. In this paper we are interested in studying
the preimage of this map. This problem dates back to $1980$ when Engel and
Schneider [9] proved that a completely reducible matrix $A$ that belongs to
the fiber of a symmetric matrix $B$ must be diagonally equivalent to it. In
$1984$, Hartfiel and Loewy [11] considered the fibers of general matrices over
any field $\mathbb{F}$. They proved that if the fiber contains an irreducible
matrix $A$ with the property that the submatrices $A[X,X^{c}]$ and
$A[X^{c},X]$ have rank at least two for any subset $X$ of $[n]$ with
$2\leq|X|\leq n-2$, then the fiber is a point up to diagonal equivalence. Here
$X^{c}$ denotes the complement of $X$ in $[n]$. This result was improved in
$1986$ by Loewy [15], who proved that if only one of the submatrices
$A[X,X^{c}]$ or $A[X^{c},X]$ of an irreducible matrix $A$ has rank at least
two, then the fiber of $A$ is still a point up to diagonal equivalence. In
both papers, the authors consider the inverse of the matrix
$\left(\operatorname{diag}\\{x_{1},\ldots,x_{n}\\}+A\right)$. We say that a
matrix $A$ has a cut $X\subset[n]$ if $2\leq|X|\leq n-2$ and the ranks of
$A[X,X^{c}]$ and $A[X^{c},X]$ are at most one.

In this paper we consider the fiber of any $n\times n$ matrix over a field
$\mathbb{F}$. While Loewy [15] proves that an irreducible matrix with no cuts
have only one preimage up to diagonal equivalence, we show that such matrices
and irreducible symmetric matrices are the only matrices that satisfy this
property.

######  Main Result 1 (Theorem 4.1).

The fiber of a matrix over any field $\mathbb{F}$ under the principal minor
map consists of only one point up to diagonal equivalence if and only if the
matrix is irreducible and either it has no cuts or it is diagonally equivalent
to a symmetric matrix.

Using this result, it is straightforward to check on a computer whether the
fiber of a matrix is a point up to diagonal equivalence. If that is not the
case, then using the proof of the theorem, one can generate an algorithm to
produce another point in the fiber.

By relating the structure of the matrix $A$ to the factorization of its
determinantal polynomial $f_{A}$

$f_{A}=\det(\operatorname{diag}\\{x_{1},\ldots,x_{n}\\}+A),$

we characterize the fibers of symmetric and Hermitian matrices in the space of
$n\times n$ matrices. In particular, we prove that a matrix $A$ can be written
as a block upper triangular matrix with $s$ diagonal blocks if and only if its
determinantal polynomial $f_{A}$ factors into $s$ irreducible factors. A
direct consequence is that the fiber of an irreducible matrix under the
principal minor map consists only of irreducible matrices. Using this relation
we characterize the fiber of a symmetric matrix under the principal minor map
in the space of $n\times n$ matrices over any field $\mathbb{F}$. Notice that
a symmetric matrix is either irreducible or completely reducible. A completely
reducible matrix is a matrix that can be written as a block diagonal matrix by
permuting some rows and the corresponding columns, with each diagonal block
being irreducible.

######  Main Result 2 (Theorem 6.2).

Let $A$ be a symmetric matrix with entries in a field $\mathbb{F}$. Then
either

  1. (1)

The matrix $A$ is irreducible and the fiber of $A$ consists of all matrices
that are diagonally equivalent to $A$, or

  2. (2)

The matrix $A$ can be written as a completely reducible matrix with $s$
diagonal blocks $A_{1},\ldots,A_{s}$ that are irreducible, and the fiber of
$A$ in $\mathbb{F}^{n\times n}$ consists of all matrices that can be written
as block upper triangle matrices with $s$ irreducible diagonal blocks that are
diagonally equivalent to the diagonal blocks of $A$.

In particular, this theorem implies that the fiber of an irreducible symmetric
matrix with entries in a field $\mathbb{F}$ consists of a single point up to
diagonal equivalence. Notice also that the result of Engel and Schneider [9,
Theorem $3.5$] follows from this theorem. Moreover, one can conclude that the
fiber of a symmetric matrix $A$ in the space of symmetric matrices,
$\rm{Sym}_{n}(\mathbb{F})$, consists of matrices that are diagonally
equivalent to $A$ via a diagonal matrix with entries equal $\pm 1$ (see for
instance [18]) from the above theorem.

We also investigate the fibers of Hermitian matrices over the complex field.
While the fiber of an irreducible symmetric matrix consists of a single point,
the fiber of an irreducible Hermitian matrix can be bigger. An example
illustrating this is provided in [2, Example 4.8]. However, we establish that
the fiber consists solely of irreducible Hermitian matrices, up to diagonal
equivalence. This characterization, combined with the previous argument
regarding the structure of a matrix based on its determinantal polynomial
factorization, grants us a comprehensive understanding of the fiber of a
Hermitian matrix. Similar to the case of symmetric matrices, a reducible
Hermitian matrix can be expressed as a completely reducible matrix.

######  Main Result 3 (Theorem 7.2).

Let $A$ be a Hermitian matrix. Then either

  1. (1)

The matrix $A$ is irreducible and the fiber of $A$ consists of Hermitian
irreducible matrices up to diagonal equivalence, or

  2. (2)

The matrix $A$ can be written as a completely reducible matrix with $s$
diagonal blocks $A_{1},\ldots,A_{s}$ that are irreducible and the fiber of $A$
consists exactly of matrices that can be written as block upper triangular
matrices with $s$ diagonal blocks that are irreducible and Hermitian up to
diagonal equivalence and that have principal minors equal to those of one of
the $A_{i}$s.

This characterization along with the author’s work with Vinzant [2, Theorem
$6.4$] answer a question of Borcea, Brändén, and Liggett:

######  Question.

[5, Question $3.4$] Characterize the class of all real stable $n\times n$
matrices, that is, $n\times n$ matrices $A$ such that
$\det(A+\operatorname{diag}\\{x_{1},\ldots,x_{n}\\})$ is real stable.

######  Main Result 4 (Corollary 7.3).

A matrix $A$ is real stable if it can be written as a block upper triangular
matrix with diagonal blocks that are irreducible and Hermitian, up to diagonal
equivalence, and that have determinantal polynomials equal to the irreducible
factors of the polynomial
$f_{A}=\det(\operatorname{diag}\\{x_{1},\ldots,x_{n}\\}+A)$.

The fibers of skew symmetric matrices were addressed by Boussaïria and
Cherguia [6]. They prove that if two skew symmetric matrices have equal
principal minors of size at most $4$, and one of the matrices is dense and
does not contain any cut, then the matrices are diagonally equivalent by a
diagonal matrix with diagonal entries equal $\pm 1$.

While in this paper we focus on the preimage of the principal minor map, the
image of this map has been studied in various contexts. Holtz and Sturmfels
[12] explored the image of the space of real and complex symmetric matrices
under the principal minor map. They demonstrated that the image is both closed
and invariant under the action of the group
$\operatorname{SL}_{2}(\mathbb{R})^{n}\rtimes S_{n}$. They conjectured that
the vanishing of polynomials in the orbit of the hyperdeterminant, under the
action of this group, defines the image of the principal minor map over the
field of complex numbers. This conjecture was resolved by Oeding [16]. In a
recent work by the author and Vinzant [1], techniques from a previous study by
Kummer, Plaumann, and Vinzant [13] were utilized to generalize this result,
making it applicable to arbitrary unique factorization domains. Moreover in
[2], the author and Vinzant prove that the image of the space of Hermitian
matrices, under the principal minor map, is cut out by the orbit of finitely
many equations and inequalities under the action of
$\operatorname{SL}_{2}(\mathbb{R})^{n}\rtimes S_{n}$. We also examined such
representations over more general fields with quadratic extensions. However,
in contrast to the Hermitian case, it was shown that for any field
$\mathbb{F}$, there is no finite set of equations whose orbit under
$\operatorname{SL}_{2}(\mathbb{R})^{n}\rtimes S_{n}$ defines the image of
$n\times n$ matrices over $\mathbb{F}$ under the principal minor map for all
values of $n$.

This paper is organized as follows. In Section 2, we introduce terminology and
the basic notations that we use throughout the paper. In Section 3 we state
and prove the generalized Laplace theorem, and we use it in Section 4 to
construct a matrix in the fiber of an irreducible matrix with a cut and
consequently characterize matrices with a single point fiber. In Section 5, we
give a description of the structure of a matrix $A$ according to its
determinantal polynomial. This description gives a characterization of the
fibers of symmetric matrices in Section 6 and of Hermitian matrices in Section
7, where a description of stable matrices is also provided.

Acknowledgements. We are grateful to Cynthia Vinzant for the helpful comments
and discussions. We also thank Jonathan Leake and Josephine Yu for their
helpful comments on the paper.

##  2\. Background and notation

Through out this paper we use $\mathbb{F}^{n\times n}$ to denote the set of
$n\times n$ matrices with entries in the field $\mathbb{F}$. Let
$\mathscr{D}(n)$ denote that set of all matrices $A$ such that the fiber of
$A$ consists of matrices that are diagonally equivalent to $A$ or $A^{T}$,
where $A^{T}$ denotes the transpose of the matrix $A$. For simplicity, we use
the expression diagonally equivalent by which we mean diagonally equivalent to
the matrix or to its transpose. One of the goals of this paper is to
characterize the set $\mathscr{D}(n)$.

We denote by $A^{\rm adj}$ the adjugate of the matrix $A$ and by ${\rm rk}(A)$
the rank of the matrix $A$. The determinantal polynomial of a matrix $A$ is
the polynomial $f_{A}=\det(\operatorname{diag}\\{x_{1},\ldots,x_{n}\\}+A)$.
Notice that a determinantal polynomial is multiaffine. A polynomial is
multiaffine if it has degree at most one in each variable.

###  2.1. The Rayleigh differece of a polynomial

The Rayleigh difference of a polynomial $f$ with respect to two variables
$x_{i}$ and $x_{j}$ is given by

$\Delta_{ij}(f)=\frac{\partial f}{\partial x_{i}}\frac{\partial f}{\partial
x_{j}}-f\frac{\partial^{2}f}{\partial x_{i}\partial x_{j}}.$

In [1] and [2], the author and Vinzant use the Rayleigh difference of a
polynomial in order to characterize multiaffine determinantal polynomials of
symmetric and Hermitian matrices. We also use them to characterize
determinantal polynomials of $n\times n$ matrices as the following theorem
shows.

######  Theorem 2.1 (Theorem 3.1 in [2]).

Let $f\in\mathbb{F}[x_{1},\ldots,x_{n}]$ be multiaffine in the variables
$x_{1},\ldots,x_{n}$ with its coefficient of $x_{1}\cdots x_{n}$ equals one.
Then $f=\det({\rm diag}(x_{1},\ldots,x_{n})+A)$ for some
$A\in\mathbb{F}^{n\times n}$ if and only if for every $i\neq j\in[n]$, the
polynomials $\Delta_{ij}(f)$ factor as the product $g_{ij}\cdot g_{ji}$ where

  * (a)

$g_{ij}\in\mathbb{F}[x_{k}:k\neq i,j]$ is multiaffine in $x_{1},\ldots,x_{n}$
and

  * (b)

for every $k\in[n]\backslash\\{i,j\\}$, ${\rm
res}_{x_{k}}(g_{ij},f)=g_{ik}g_{kj}$.

In this case, we can take $g_{ij}$ to be the $(i,j)$th entry of $({\rm
diag}(x_{1},\ldots,x_{n})+A)^{\rm adj}$, with $M^{\rm adj}$ represents the
adjugate matrix of $M$.

We use ${\rm res}_{x_{k}}(g,h)$ to denote the resultant of two polynomials $g$
and $h$ with respect to a variable $x_{k}$:

${\rm res}_{x_{k}}(g,h)=(g|_{x_{k}=0})\cdot\frac{\partial}{\partial
x_{k}}h-(h|_{x_{k}=0})\cdot\frac{\partial}{\partial x_{k}}g.$

Moreover, we use the Rayleigh differences to study the irreducibility of a
multiaffine polynomial as shown in the following lemma from [2]. We include
the proof here for completeness.

######  Lemma 2.2 (Proposition 2.2 in [2]).

If $f\in\mathbb{F}[x_{1},\ldots,x_{n}]$ has degree one in each of $x_{i}$ and
$x_{j}$, then $\Delta_{ij}(f)=0$ if and only if $f$ factors into polynomial
$g\cdot h$ with $g\in\mathbb{F}[x_{k}:k\neq i]$ and
$h\in\mathbb{F}[x_{k}:k\neq j]$.

###### Proof.

By assumption we can write $f=ax_{i}x_{j}+bx_{i}+cx_{j}+d$ for
$a,b,c,d\in\mathbb{F}[x_{k}:k\neq i,j]$. Then $\Delta_{ij}(f)=bc-ad$. If
$\Delta_{ij}(f)=0$, then there is some factorization $b=b_{1}b_{2}$ and
$c=c_{1}c_{2}$ for which $a=b_{1}c_{1}$ and $d=b_{2}c_{2}$. Hence
$f=(b_{1}x_{i}+c_{2})(c_{1}x_{j}+b_{2})$. Similarly, if
$f=(b_{1}x_{i}+c_{2})(c_{1}x_{j}+b_{2})$ for some
$b_{1},b_{2},c_{1},c_{2}\in\mathbb{F}[x_{k}:k\neq i,j]$, then
$\Delta_{ij}(f)=bc-ad=0$. ∎

######  Theorem 2.3.

Let $A$ be an $n\times n$ matrix in $\mathbb{F}^{n\times n}$. Let
$f=\det(\operatorname{diag}\\{x_{1},\ldots,x_{n}\\}+A)$ and
$G=(\operatorname{diag}\\{x_{1},\ldots,x_{n}\\}+A)^{\rm adj}$. Then

$\Delta_{ij}(f)=G_{ij}G_{ji},$

where $G_{ij}$ denotes the $(i,j)^{\rm th}$ entry of $G$.

###### Proof.

Let $S,T\subset[n]$ be of equal cardinality. Let $M(S,T)$ denote the submatrix
of $M$ obtained by dropping rows $S$ and columns $T$ from $M$. Then for any
$i\neq j\in[n]$,

$\det(M(i,i))\cdot\det(M(j,j))-\det(M)\det(M(\\{i,j\\},\\{i,j\\}))=\det(M(i,j))\cdot\det(M(j,i)).$

This follows from a classical equality used by Dodgson [8] as a method for
computing determinants. Since $M={\rm diag}(x_{1},\ldots,x_{n})+A$, then for
any subset $S\subseteq[n]$, the principal minor $\det(M(S,S))$ equals the
derivative of $f$ with respect to the variables in $S$, $\left(\prod_{i\in
S}\frac{\partial}{\partial x_{i}}\right)f$. Since $G_{ij}=\det(M(i,j))$, the
equation above then gives that $\Delta_{ij}(f)$ equals $G_{ij}G_{ji}$. ∎

##  3\. The Generalized Laplace Expansion

In this section we give a complete proof of the generalized Laplace theorem.
While the Laplace expansion theorem states that the determinant of an $n\times
n$ matrix $A$ over any field $\mathbb{F}$ can be computed by summing the
products of the elements of some row or column with their corresponding
cofactors, or $n-1\times n-1$ minors, the generalized Laplace expansion [14]
computes the determinant by summing up minors of size $k\times k$ and their
corresponding $n-k\times n-k$ minors. In this section we state the theorem and
we provide a proof based on exterior algebra for completeness.

######  Theorem 3.1.

[The Generalized Laplace Theorem] Let $A$ be an $n\times n$ matrix with
entries in a field $\mathbb{F}$. Let $S\subset[n]$ such that $|S|=k$ with
$1\leq k\leq n-1$. The determinant of $A$ is given by:

${\rm det}(A)=\sum_{\begin{subarray}{c}T\subset[n]\\\
|T|=k\end{subarray}}(-1)^{\sum T+\sum S}A_{S,T}A_{S^{c},T^{c}}$

where $\sum T=\sum_{t\in T}t$.

To prove the theorem we use the following lemma.

######  Lemma 3.2.

Let $S$ and $T$ be two subsets of $[n]$ of size $k$. Let $\sigma$ be the
permutation of $[n]$ given in two line notation by

$\sigma=\begin{pmatrix}S&S^{c}\\\ T&T^{c}\end{pmatrix}$

where the elements of $S,T,S^{c},$ and $T^{c}$ are arranged in increasing
order. Then

$\rm{sgn}(\sigma)=(-1)^{\sum S+\sum T}.$

###### Proof.

First assume that $S=\\{1,2,\ldots,k\\}$. We will proceed by induction on $k$.
Assume that $k=1$ and the smallest element of $T$ is $t_{1}$. Since the
elements of $T^{c}$ are listed in increasing order, then the sign of $\sigma$
is $(-1)^{t_{1}-1}=(-1)^{t_{1}+1}$. For the inductive step, $t_{1}$ should be
transposed $t_{1}-1+(k-1)$ times, and so ${\rm
sgn}(\sigma)=(-1)^{t_{1}+k-2}{\rm sgn}(\tau)$ where

$\tau=\tiny{\begin{pmatrix}1,\ldots,k-1&k,\ldots,n\\\
T_{1}&(T_{1})^{c}\end{pmatrix}}.$

Here $T_{1}$ denotes the set $T\setminus\\{t_{1}\\}$, and the elements of
$T_{1}$ and $T_{1}^{c}$ are arranged in increasing order. Applying the
inductive hypothesis to $\tau$, we infer that ${\rm sgn}(\tau)=(-1)^{\sum
T_{1}+\sum_{i=1}^{k-1}i}$. Therefore,

${\rm sgn}(\sigma)=(-1)^{t_{1}+k-2}(-1)^{T_{1}+\sum_{i=1}^{k-1}i}=(-1)^{\sum
T+\sum_{i=1}^{k}i}.$

For general $S$, $\sigma$ can be written as the composite of two permutations
$\sigma_{1}$ and $\sigma_{2}$ given in two line notation by
$\sigma_{1}=\tiny{\begin{pmatrix}S&S^{c}\\\
1,\ldots,k&k+1,\ldots,n\end{pmatrix}}$ and
$\sigma_{2}=\tiny{\begin{pmatrix}1,\ldots,k&k+1,\ldots,n\\\
T&T^{c}\end{pmatrix}}$. Thus, the sign of $\sigma$ is the product of the signs
of $\sigma_{1}$ and $\sigma_{2}$. Since a permutation has the same sign as its
inverse, we conclude that ${\rm sgn}(\sigma)=(-1)^{\sum
T+\sum_{i=1}^{k}i}(-1)^{\sum S+\sum_{i=1}^{k}i}=(-1)^{\sum T+\sum S}$ as
desired. ∎

###### Proof of Theorem 3.1.

For simplicity, we assume $S=[k]$. We denote by $c_{i}$ the $i^{{\rm th}}$
column of the matrix $A$ and by $e_{i}$ the vector in $\mathbb{F}^{n}$ with
$i^{\rm th}$ entry equal to one and zero otherwise. Then,

$c_{1}\wedge c_{2}\wedge\cdots\wedge c_{n}={\rm det}(A)e_{1}\wedge\cdots\wedge
e_{n}$

Let $c_{i}=\sum_{j=1}^{n}a_{ji}e_{j}$. Then the wedge product of the first $k$
columns can be written as

$c_{1}\wedge\cdots\wedge
c_{k}=\sum_{j=1}^{n}a_{j1}e_{j}\wedge\cdots\wedge\sum_{j=1}^{n}a_{jn}e_{j}=\sum_{\begin{subarray}{c}T\subset[n]\\\
|T|=k\end{subarray}}(\sum_{j\in T}a_{j1}e_{j}\wedge\cdots\wedge\sum_{j\in
T}a_{jn}e_{k})=\sum_{\begin{subarray}{c}T\subset[n]\\\
|T|=k\end{subarray}}A_{T,S}e_{T}$

where $e_{T}=e_{t_{1}}\wedge\cdots\wedge e_{t_{k}}$ and $t_{1},\ldots,t_{k}$
are the elements in $T$ arranged in increasing order. Therefore

$(c_{1}\wedge\cdots\wedge c_{k})\wedge(c_{k+1}\wedge\cdots\wedge
c_{n})=\sum_{\begin{subarray}{c}T\subset[n]\\\
|T|=k\end{subarray}}A_{T,S}e_{T}\wedge\sum_{\begin{subarray}{c}T\subset[n]\\\
|T|=k\end{subarray}}A_{T^{c},S^{c}}e_{T^{c}}.$

We denote by $\sigma_{S,T}$ the permutation written in two line notation as
$\tiny\begin{pmatrix}S&S^{c}\\\ T&T^{c}\end{pmatrix}$. Using Lemma 3.2, ${\rm
sgn}(\sigma_{T,S})=(-1)^{\sum T+\sum S}$. Thus

$\det(A)e_{[n]}=c_{1}\wedge\cdots\wedge
c_{n}=\sum_{\begin{subarray}{c}T\subset[n]\\\ |T|=k\end{subarray}}(-1)^{\sum
T+\sum S}A_{T,S}A_{T^{c},S^{c}}e_{[n]}.$

Therefore $\det(A)=\sum_{\begin{subarray}{c}T\subset[n]\\\
|T|=k\end{subarray}}(-1)^{\sum T+\sum S}A_{T,S}A_{T^{c},S^{c}}$ as desired. ∎

##  4\. Matrices with a single point fibers

In this section we characterize $\mathscr{D}(n)$, the set of all matrices with
a single point fiber up to diagonal equivalence.

######  Theorem 4.1.

Let $A$ be an $n\times n$ matrix with entries in a field $\mathbb{F}$ with
$n\geq 4$. The matrix $A$ belongs to the set $\mathscr{D}(n)$ if and only if
$A$ is irreducible and either

  1. (a)

$A$ is diagonally equivalent to a symmetric matrix or

  2. (b)

$A$ does not have any cut, that is ${\rm rk}\left(A[X,X^{c}]\right)\geq 2$ or
${\rm rk}\left(A[X^{c},X]\right)\geq 2$ for all $X\subset[n]$, with
$2\leq|X|\leq n-2$.

###### Proof of ($\Leftarrow$).

Follows from [15, Theorem 1] and Lemma 6.1. ∎

To prove the other directions, we will use the following lemma repeatedly. The
reader is referred to [11] for the proof.

######  Lemma 4.2.

[Theorem 1 in [11]] Let $A$ be an $n\times n$ irreducible matrix with entries
in a field $\mathbb{F}$. Let
$M=(\operatorname{diag}\\{x_{1},\ldots,x_{n}\\}+A)$ and $G=M^{\rm adj}$, the
adjugate matrix of $M$. Then the entries of $G$ are nonzero.

######  Lemma 4.3.

Let $A$ be an $n\times n$ irreducible matrix with entries in a field
$\mathbb{F}$ with $n\geq 4$. Suppose that $A$ has a cut $X\subset[n]$. Let
$G=(\operatorname{diag}\\{x_{1},\ldots,x_{n}\\}+A)^{\rm adj}$. Then for all
$i\in X$ and $j\in X^{c}$ we have

$G_{ij}=(-1)^{i}a_{i}b_{j}$

where $a_{i}\in\mathbb{F}[x_{k}:k\in X\setminus\\{i\\}]$,
$b_{j}\in\mathbb{F}[x_{k}:k\in X^{c}\setminus\\{j\\}]$ and both are nonzero
polynomials.

###### Proof.

Since $A$ is irreducible and has a cut $X$, then ${\rm rk}(A[X,X^{c}])={\rm
rk}(A[X^{c},X])=1$. Let $M=(A+\operatorname{diag}\\{x_{1},\ldots,x_{n}\\})$.
We denote by $X_{i}$ the set $X\setminus\\{i\\}$. For $S,T\subset[n]$, let
$M_{S,T}$ denote the determinant of the submatrix $M[S,T]$ indexed by $S$ on
the rows and $T$ on the columns. Since ${\rm rk}(A[X,X^{c}])=1$, then for each
$i\in X$ there exists $a_{i}\in\mathbb{F}[x_{k}:k\in X_{i}]$ such that
$M_{X,X_{i}\cup\\{j\\}}=c_{j}a_{i}$ for all $j\in X^{c}$ and for some
$c_{j}\in\mathbb{F}$. Now using Theorem 3.1 we have:

$G_{ij}=\sum_{\begin{subarray}{c}T\subset[n]_{i}\\\
|T|=k\end{subarray}}(-1)^{\sum X+\sum T}M_{X,T}M_{(X^{c})_{j},(T^{c})_{i}}$

for $i\in X$ and $j\in X^{c}$. Notice that for all $T\subset[n]_{i}$ such that
$|T\cap X^{c}|\geq 2$, the submatrix $M[X,T]$ is not full rank, since ${\rm
rk}(A[X,X^{c}])=1$. Thus for all such $T$, $M_{X,T}=0$. Therefore, $G_{ij}$
can be written as:

$\displaystyle G_{ij}$ $\displaystyle=\sum_{\ell\in X^{c}}(-1)^{\sum X+\sum
X-i+\ell}M_{X,\ell\cup X_{i}}M_{(X^{c})_{j},(X^{c})_{\ell}}$
$\displaystyle=(-1)^{i}a_{i}\sum_{\ell\in
X^{c}}(-1)^{\ell}c_{\ell}M_{(X^{c})_{j},(X^{c})_{\ell}}$

using $M_{X,\ell\cup X_{i}}=c_{\ell}a_{i}$. Letting
$b_{j}=\displaystyle\sum_{\ell\in
X^{c}}(-1)^{\ell}c_{\ell}M_{(X^{c})_{j},(X^{c})_{\ell},}$, we obtain the
desired result. By Lemma 4.2, $G_{ij}\neq 0$. Therefore, $a_{i}$ and $b_{j}$
are both nonzero. ∎

######  Lemma 4.4.

Let $A$ be an $n\times n$ irreducible matrix with entries in a field
$\mathbb{F}$ with $n\geq 4$. Consider
$G=(\operatorname{diag}\\{x_{1},\ldots,x_{n}\\}+A)^{\rm adj}$. If $A$ has a
cut $X\subset[n]$ such that for each $i\in X$ and $j\in X^{c}$ there exists
$\alpha_{ij}\in\mathbb{F}$ satisfying $G_{ij}=\alpha_{ij}G_{ji}$, then $A$ is
diagonally similar to a symmetric matrix.

###### Proof.

Lemma 4.3 implies that for all $i,j\in X$ and $k,\ell\in X^{c}$ we have
$G_{ik}/G_{jk}=G_{i\ell}/G_{j\ell}$ and $G_{ki}/G_{kj}=G_{\ell i}/G_{\ell j}$.
Therefore, $\alpha_{ik}/\alpha_{jk}=\alpha_{i\ell}/\alpha_{j\ell}$. Let
$i,j\in X$. We claim that $G_{ij}=\alpha_{ik}/\alpha_{jk}G_{ji}$ for any $k\in
X^{c}$. Using Theorem 2.3, $\Delta_{ij}(f)=G_{ij}G_{ji}$. Then

$\displaystyle{\rm
res}_{x_{j}}(G_{ik},f)=G_{ij}G_{jk}=\alpha_{jk}G_{ij}G_{kj},$

by Theorem 2.1. Since $G_{ik}=\alpha_{ik}G_{ki}$, then

${\rm res}_{x_{j}}(G_{ik},f)={\rm
res}_{x_{j}}(\alpha_{ik}G_{ki},f)=\alpha_{ik}{\rm
res}_{x_{j}}(G_{ki},f)=\alpha_{ik}G_{kj}G_{ji}.$

The two equations above now give $G_{ij}=\alpha_{ik}/\alpha_{jk}G_{ji}$. Using
similar argument, one can prove that $G_{ij}=\alpha_{kj}/\alpha_{ki}G_{ji}$
for $i,j\in X^{c}$. Fix $i\in X$ and $k\in X^{c}$ and let $D$ be the diagonal
matrix such that $D_{kk}=1$, $D_{jj}=\sqrt{\alpha_{jk}}$ if $j\in X$, and
$D_{jj}=\sqrt{\alpha_{ik}}/\sqrt{\alpha_{ij}}$ if $j\in X^{c}$. Since for all
$i,j\in X$ and $k,\ell\in X^{c}$,
${\alpha_{ik}}/{\alpha_{i\ell}}={\alpha_{jk}}/{\alpha_{j\ell}}$, then
$D^{-1}GD$ is a symmetric matrix and so is $DAD^{-1}$ as desired. ∎

If an irreducible matrix $A$ has a cut $X=[k]$, then Lemma 4.3 states that the
matrix $G=(\operatorname{diag}\\{x_{1},\ldots,x_{n}\\}+A)^{\rm adj}$ can be
written as

$G=\begin{pmatrix}P&R\\\ T&Q\end{pmatrix}$

where $P$ is a $k\times k$ submatrix, $R_{ij}=(-1)^{i}a_{i}b_{j}$ with
$a_{i}\in\mathbb{F}[x_{k}:k\in X]$ and $b_{j}\in\mathbb{F}[x_{k}:k\in X^{c}]$,
and $T_{ij}=(-1)^{j}c_{i}d_{j}$ with $c_{i}\in\mathbb{F}[x_{k}:x_{k}\in
X^{c}]$ and $d_{j}\in\mathbb{F}[x_{k}:k\in X]$. By switching suitable factors
between the entries $R_{ij}$ and $T_{ij}$, we get a new matrix $H$. This
matrix is the adjugate of the matrix
$(\operatorname{diag}\\{x_{1},\ldots,x_{n}\\}+B)$ where $B$ is a point in the
fiber of $A$ as the following lemma shows. We will use this operation later to
find a point in the fiber of $A$ that is not diagonally equivalent to $A$.

######  Lemma 4.5.

Let $A$ be an $n\times n$ irreducible matrix with entries in a field
$\mathbb{F}$ with $n\geq 4$. Assume that $A$ has a cut $X=[k]$. Let

$G=(\operatorname{diag}\\{x_{1},\ldots,x_{n}\\}+A)^{\rm
adj}=\begin{pmatrix}P&R\\\ T&Q\end{pmatrix}$

where $P$ is a $k\times k$ submatrix, $R_{ij}=(-1)^{i}a_{i}b_{j}$ with
$a_{i}\in\mathbb{F}[x_{k}:k\in X]$ and $b_{j}\in\mathbb{F}[x_{k}:k\in X^{c}]$,
and $T_{ij}=(-1)^{j}c_{i}d_{j}$ with $c_{i}\in\mathbb{F}[x_{k}:x_{k}\in
X^{c}]$ and $d_{j}\in\mathbb{F}[x_{k}:k\in X]$. Let $H$ be the matrix defined
as

$H=\begin{pmatrix}P&U\\\ V&Q^{T}\end{pmatrix}$

with $U_{ij}=(-1)^{i}a_{i}c_{j}$ and $V_{ij}=(-1)^{j}b_{i}d_{j}$. Then
$H=(\operatorname{diag}\\{x_{1},\ldots,x_{n}\\}+B)^{\rm adj}$ for some
$n\times n$ matrix $B$ such that
$\det(\operatorname{diag}\\{x_{1},\ldots,x_{n}\\}+A)=\det(\operatorname{diag}\\{x_{1},\ldots,x_{n}\\}+B)$.

###### Proof.

Let $f=\det(\operatorname{diag}\\{x_{1},\ldots,x_{n}\\}+A)$. Using Theorem 2.3
we have

$\Delta_{ij}(f)=G_{ij}G_{ji}=(-1)^{i+j}a_{i}b_{j}c_{j}d_{i}=H_{ij}H_{ji},$

for $i\in X$ and $j\in X^{c}$. Thus, $\Delta_{ij}(f)=H_{ij}H_{ji}$, where
$H_{ij}$ and $H_{ji}$ are multiaffine for all $i,j\in[n]$. Using Theorem 2.1,
it remains to show that for all $k$, ${\rm
res}_{x_{k}}(H_{ij},f)=H_{ik}H_{kj}$ for all distinct $i,j\in[n]$. We only
consider the case $i\in X$, $j\in X^{c}$, and $k\in X$, as the other cases
follow similarly. Since $b_{j}\in\mathbb{F}[x_{\ell}:\ell\in X^{c}]$ is
independent of the variable $x_{k}$, the resultant ${\rm
res}_{x_{k}}(G_{ij},f)={\rm
res}_{x_{k}}((-1)^{i}a_{i}b_{j},f)=(-1)^{i}b_{j}{\rm res}_{x_{k}}(a_{i},f)$.
Also, using Theorem 2.1 it holds that

${\rm res}_{x_{k}}(G_{ij},f)=G_{ik}G_{kj}=(-1)^{k}G_{ik}a_{k}b_{j}.$

This yields ${\rm res}_{x_{k}}(a_{i},f)=(-1)^{i+k}a_{k}G_{ik}$. Since
$c_{j}\in\mathbb{F}[x_{\ell}:\ell\in X^{c}]$ is independent of the variable
$x_{k}$, we obtain

${\rm res}_{x_{k}}(H_{ij},f)={\rm
res}_{x_{k}}((-1)^{i}a_{i}c_{j},f)=(-1)^{i}c_{j}{\rm
res}_{x_{k}}(a_{i},f)=(-1)^{k}c_{j}a_{k}G_{ik}=H_{kj}G_{ik}.$

Since $i,k\in X$, then $G_{ik}=P_{ik}=H_{ik}$. Therefore ${\rm
res}_{x_{k}}(H_{ij},f)=H_{kj}H_{ik}$. We apply Theorem 2.1 and obtain the
existence of a matrix $B$ with $\det(\\{x_{1},\ldots,x_{n}\\}+B)=f$. ∎

###### Proof of Theorem 4.1 ($\Leftarrow$).

First suppose that $A$ is reducible. Then, after permuting some rows and the
corresponding columns, $A$ can be written as $\tiny{\begin{pmatrix}E&F\\\
0&G\end{pmatrix}}$, with $E$ is a $k\times k$ submatrix. Let $H$ be any
$k\times n-k$ matrix such that $H_{ij}=0$ if and only if $F_{ij}\neq 0$ and
$H_{ij}=1$ otherwise. Let $B=\tiny{\begin{pmatrix}E&H\\\ 0&G\end{pmatrix}}$.
If $S\subseteq[k]$, then $A_{S}=E_{S}=B_{S}$, and similarly if
$S\subseteq[k]^{c}$. Otherwise, if $S=S_{1}\cup S_{2}$ with $S_{1}\subset[k]$
and $S_{2}\subset[k]^{c}$, then $A_{S}=E_{S_{1}}G_{S_{2}}=B_{S}$. Therefore,
$A$ and $B$ have the same principal minors. Notice that $B$ is not diagonally
similar to $A$. Therefore, $A\notin\mathscr{D}(n)$.

Now assume that $A$ is irreducible, $A$ is not diagonally equivalent to a
symmetric matrix, and $A$ has a cut $X=[k]\subset[n]$. Notice that ${\rm
rk}(A[X,X^{c}])$ and ${\rm rk}(A[X^{c},X])$ are nonzero since $A$ is
irreducible. Thus ${\rm rk}\left(A[X,X^{c}]\right)={\rm
rk}\left(A[X^{c},X]\right)=1$. Let $M={\rm diag}\\{x_{1},\ldots,x_{n}\\}+A$,
$f=\det\left(M\right)$, and $G=M^{\rm adj}$. Using Lemma 4.3, for all $i\in X$
and $j\in X^{c}$ the entry $G_{ij}=(-1)^{i}a_{i}b_{j}$ for some
$a_{i}\in\mathbb{F}[x_{k}:k\in X_{i}]$ and
$b_{j}\in\mathbb{F}[x_{k}:k\in(X^{c})_{j}]$. By Lemma 4.2, $G_{ij}\neq 0$.
therefore, neither $a_{i}$ nor $b_{j}$ can be zero and so the matrix $G$ can
be written as a block matrix

$G=\begin{pmatrix}P&R\\\ T&Q\end{pmatrix}$

where $P$ is a $k\times k$ submatrix, $R_{ij}=(-1)^{i}a_{i}b_{j}$ and
$T_{ij}=(-1)^{j}c_{i}d_{j}$ with $a_{i}\in\mathbb{F}[x_{k}:k\in X_{i}]$,
$b_{j}\in\mathbb{F}[x_{k}:k\in(X^{c})_{j}]$,
$c_{i}\in\mathbb{F}[x_{k}:x_{k}\in(X^{c})_{i}]$, and
$d_{j}\in\mathbb{F}[x_{k}:k\in X_{j}]$. Consider the matrix $H$ defined by

$H=\begin{pmatrix}P&U\\\ V&Q^{T}\end{pmatrix}$

with $U_{ij}=(-1)^{i}a_{i}c_{j}$ and $V_{ij}=(-1)^{j}b_{i}d_{j}$. We infer
that $H=(\operatorname{diag}\\{x_{1},\ldots,x_{n}\\}+B)^{\rm adj}$ where $B$
has the same principal minors as $A$, by Lemma 4.3.
We still need to check that $B$ is not diagonally similar to $A$. We present a
proof by contradiction. Indeed, assume there is a diagonal matrix $D$ such
that $B=DAD^{-1}$. Then $H=DGD^{-1}$. Let
$D=\operatorname{diag}\\{\lambda_{1},\ldots,\lambda_{n}\\}$ with
$\lambda_{i}\in\mathbb{F}$. Then $H_{ij}=\lambda_{j}/\lambda_{i}G_{ij}$ for
all $i,j\in[n]$. In particular,
$H_{ij}=(-1)^{i}a_{i}c_{j}=\lambda_{j}/\lambda_{i}(-1)^{i}a_{i}b_{j}$ for all
$i\in X$ and $j\in X^{c}$. Hence, $b_{j}=\alpha_{j}c_{j}$ for some
$\alpha_{j}\in\mathbb{F}$ and for $j\in X^{c}$. If this is the case, then we
switch $a_{i}$ and $d_{i}$ in $G$ instead of $b_{j}$ and $c_{j}$. That is we
let $H=\tiny\begin{pmatrix}P^{T}&U\\\ V&Q\end{pmatrix}$ such that
$U_{ij}=(-1)^{i}d_{i}b_{j}$ and $V_{ij}=(-1)^{i}a_{i}c_{j}$. Using the same
argument as the one above, we obtain a matrix $C$ with the same principal
minors as those of $A$. If $C$ is also diagonally congruent to $A$, then we
get $a_{i}=\beta_{i}d_{i}$ for some $\beta_{i}\in\mathbb{F}$ and for all $i\in
X$. In this case, by Lemma 4.4, $A$ is diagonally equivalent to a symmetric
matrix and this gives the desired contradiction. ∎

######  Example 4.6.

Consider the $4\times 4$ matrix $A$

$A=\tiny{\begin{pmatrix}2&-1&1&-2\\\ 1&1&-3&6\\\ 1&2&1&1\\\ -1&-2&2&-1\\\
\end{pmatrix}}.$

Notice that $A$ has a cut at $X=\\{1,2\\}$ since ${\rm rk}(A[X,X^{c}])={\rm
rk}(A[X^{c},X])=1$. We compute
$G=(A+\operatorname{diag}\\{x_{1},\ldots,x_{4}\\})^{\rm adj}$ and obtain

$G=\tiny{\begin{pmatrix}p_{1}&({x_{3}}+3)({x_{4}}+3)&-({x_{2}}-2)({x_{4}}+3)&({x_{2}}-2)(2{x_{3}}+3)\\\
-15-5x_{3}-4x_{4}-x_{3}x_{4}&p_{2}&(3{x_{1}}+7)({x_{4}}+3)&-(3{x_{1}}+7)(2{x_{3}}+3)\\\
-({x_{2}}-1){x_{4}}&-(2{x_{1}}+5){x_{4}}&p_{3}&-{x_{1}}{x_{2}}+11{x_{1}}-4{x_{2}}+29\\\
({x_{2}}-1)({x_{3}}+3)&(2{x_{1}}+5)({x_{3}}+3)&-(2{x_{1}}+5)({x_{2}}-2)&p_{4}\\\
\end{pmatrix}}$

with $p_{i}=G_{ii}$. If we switch the factors $(x_{2}-2)$ and $-(3x_{1}+7)$ in
$G[X,X^{c}]$ with the factors $(x_{2}-1)$ and $(2x_{1}+5)$ in $G[X^{c},X]$
respectively and we transpose the submatrix $A[X,X]$, we get

$H=\tiny{\begin{pmatrix}p_{1}&-{x_{3}}{x_{4}}-5{x_{3}}-4{x_{4}}-15&-({x_{2}}-1)({x_{4}}+3)&({x_{2}}-1)(2{x_{3}}+3)\\\
({x_{3}}+3)({x_{4}}+3)&p_{2}&-(2{x_{1}}+5)({x_{4}}+3)&(2{x_{1}}+5)(2{x_{3}}+3)\\\
-({x_{2}}-2){x_{4}}&(3{x_{1}}+7){x_{4}}&p_{3}&-{x_{1}}{x_{2}}+11{x_{1}}-4{x_{2}}+29\\\
({x_{2}}-2)({x_{3}}+3)&-(3{x_{1}}+7)({x_{3}}+3)&-(2{x_{1}}+5)({x_{2}}-2)&p_{4}\\\
\end{pmatrix}}$

which is the adjugate of the matrix
$(B+\operatorname{diag}\\{x_{1},\ldots,x_{n}\\})$ with

$B=\tiny{\begin{pmatrix}2&1&1&-2\\\ -1&1&2&-4\\\ 1&-3&1&1\\\
-1&3&2&-1\end{pmatrix}}.$

By computing the determinantal polynomial of $A$,
$f_{A}=\det(\operatorname{diag}\\{x_{1},\ldots,x_{n}\\}+A)$, and that of $B$,
$f_{B}=\det(\operatorname{diag}\\{x_{1},\ldots,x_{n}\\}+B)$, we see that $A$
and $B$ have the same principal minors. Simple computations show that $B$ is
not diagonally equivalent to $A$. $\diamond$

##  5\. The Structure of a matrix from the factorization of its determinantal
polynomial

In this section, we relate the irreducibility of a matrix $A$ to the
irreducibility of its determinantal polynomial
$f_{A}=\det(\operatorname{diag}\\{x_{1},\ldots,x_{n}\\}+A)$. Moreover, we use
the factors of this determinantal polynomial in order to study the structure
of the matrix.

######  Lemma 5.1.

Let $A$ be an $n\times n$ matrix with entries in a field $\mathbb{F}$. Let $f$
be its determinantal polynomial

$f=\det\left({\rm diag}\\{x_{1},x_{2},\ldots,x_{n}\\}+A\right).$

Then $A$ is irreducible if and only if $f$ is irreducible.

###### Proof.

$(\Longleftarrow)$ Suppose $A$ is reducible. Then $A$ can be written as a
block upper triangular matrix, possibly after permuting some rows and the
corresponding columns. Consequently $f$ factors into at least two factors.
$(\Longrightarrow)$ Suppose $A$ is irreducible. Let
$M=(A+\operatorname{diag}\\{x_{1},\ldots,x_{n}\\})$ and $G=M^{\rm adj}$. Then
using Lemma 4.2, we have $G_{ij}\neq 0$ for all $i,j\in[n]$. Thus, Theorem 2.3
implies that $\Delta_{ij}(f)=G_{ij}G_{ji}\neq 0$. In view of Lemma 2.2, we
conclude that $f$ is irreducible. ∎

######  Theorem 5.2.

Let $A$ be an $n\times n$ matrix with entries in a field $\mathbb{F}$. Let $f$
be the polynomial defined by

$f=\det\left({\rm diag}\\{x_{1},x_{2},\ldots,x_{n}\\}+A\right).$

Then $f$ factors as $f_{1}\cdots f_{s}$ such that each $f_{i}$ is irreducible
and $f_{i}\in\mathbb{F}[x_{k}:k\in T_{i}]$ if and only if $A$ can be written,
after permuting some rows and the corresponding columns, as a block upper
triangular matrix with $s$ diagonal blocks $A_{1},\ldots,A_{s}$ such that each
$A_{i}$ is irreducible and $f_{i}=\det(A_{i}+\operatorname{diag}\\{x_{k}:k\in
T_{i}\\})$.

###### Proof.

$\Longleftarrow$ Suppose that $A$ can be written as a block upper triangular
matrix with $s$ irreducible diagonal blocks. Then
$f=\prod_{i=1}^{s}\det(A_{i}+\operatorname{diag}\\{x_{k}:k\in
T_{i}\\})=\prod_{i=1}^{s}f_{i}$. Each $f_{i}$ is irreducible in view of Lemma
5.1.
$\Longrightarrow$ Suppose that $f$ is reducible and factors as $f_{1}\cdots
f_{s}$. We proceed by induction on $s$. The case $s=1$ follows from Lemma 5.1.
For $s>1$, Lemma 5.1 implies that $A$ is reducible and so it can be written as
a block upper triangular matrix with two diagonal blocks $A_{1}$ and $A_{2}$.
Without loss of generality, $\det(A_{1})=f_{1}\cdots f_{t}$ for $t<s$. Thus,
$A_{1}$ satisfies all the inductive hypothesis and it can be written as a
block upper triangular matrix with $t$ diagonal irreducible blocks, after
permuting some rows and the corresponding columns of $A_{1}$. This permutation
will not affect $A_{2}$. The same holds for $A_{2}$ and we write it as a block
upper triangular matrix with $s-t+1$ diagonal blocks. This proves that $A$ can
be written as a block upper triangular matrix with $s$ diagonal blocks with
associated determinantal polynomials that equal the factors of $f$. ∎

######  Remark 5.3.

Notice that Theorem 5.2 implies that any matrix can be rearranged by permuting
some rows and the corresponding columns to form a block upper triangular
matrix with irreducible diagonal blocks. This fact was proved by Brualdi and
Ryser [7, Theorem 3.2.4]. They gave a purely combinatorial proof. From a graph
theory perspective, the theorem can be stated as follows:
To each $n\times n$ matrix $A=(a_{ij})_{1\leq 1i,j\leq n}$, we associate a
direct graph $G$ with $n$ vertices $v_{1},\ldots,v_{n}$. There is a directed
edge $(i,j)$ from a vertex $v_{i}$ to a vertex $v_{j}$ if and only if
$a_{ij}\neq 0$. Suppose that $G$ is composed of strongly connected components
$C_{1},\ldots,C_{s}$. By contracting each component $C_{i}$ into a single
vertex $V_{i}$, we obtain a new graph $H$ with $s$ vertices $V_{1},...,V_{s}$.
Directed edges are added to $H$ such that there is a directed edge from
$V_{i}$ to $V_{j}$ if and only if there exists a directed edge from a vertex
in $C_{i}$ to a vertex in $C_{j}$. Thus, $H$ is a directed acyclic graph,
allowing for a topological ordering of its vertices. This topological ordering
corresponds to the diagonal irreducible blocks.
This gives a very nice connection between the strongly connected components of
the graph $G$ that corresponds to a matrix $A$ and the factors of
$f_{A}=\det(\operatorname{diag}\\{x_{1},\ldots,x_{n}\\}+A)$.

######  Corollary 5.4.

Let $A$ be an $n\times n$ matrix with entries in the field $\mathbb{F}$. Let
$f$ be the polynomial defined by

$f=\det\left({\rm diag}\\{x_{1},x_{2},\ldots,x_{n}\\}+A\right).$

Suppose that $f$ factors as $f_{1}\cdots f_{s}$ such that each $f_{i}$ is
irreducible and $f_{i}\in\mathbb{F}[x_{k}:k\in T_{i}]$. Then the fiber of $A$
consists of all matrices that can be written as block upper triangular
matrices with $s$ diagonal irreducible blocks $A_{1},\ldots,A_{s}$ with
$\det(A_{i}+\operatorname{diag}\\{x_{k}:k\in[T_{i}]\\})=f_{i}$ for each $i$.

######  Example 5.5.

Consider the $6\times 6$ matrix $A$

$A=\tiny{\left(\begin{array}[]{cccccc}1&-3&3&-2&-1&2\\\ 0&-3&5&1&0&2\\\
0&0&4&0&0&-4\\\ 0&1&2&1&0&5\\\ 1&0&-1&6&2&4\\\ 0&0&2&0&0&3\\\
\end{array}\right)}.$

Then the determinantal polynomial of $A$ is given by

$f_{A}=\det(\operatorname{diag}\\{x_{1},\ldots,x_{n}\\}+A)=(x_{1}x_{5}+2x_{1}+x_{5}+3)(x_{2}x_{4}+x_{2}-3x_{4}-4)(x_{3}x_{6}+3x_{3}+4x_{6}+20).$

According to Theorem 5.2, this matrix can be written as a block upper
triangular matrix with three irreducible diagonal blocks. Since the fiber of a
$2\times 2$ irreducible matrix is a point up to diagonal equivalence, then
Corollary 5.4 tells us that the fiber of $A$ consists of matrices that can be
written in the following form up to permuting the diagonal blocks and up to
diagonal equivalence:

$\tiny{\begin{pmatrix}\begin{matrix}2&1\\\
-1&1\end{matrix}&\mbox{\large$\star$ }&\mbox{\large$\star$ }\\\ \mbox{\large
0}&\begin{matrix}1&1\\\ 1&-3\end{matrix}&\mbox{\large$\star$}\\\ \mbox{\large
0}&\mbox{\large 0}&\begin{matrix}3&2\\\ -4&4\end{matrix}\end{pmatrix}}$

where $\star$ can be replaced with any $2\times 2$ matrix. $\diamond$

######  Corollary 5.6.

Let $A$ be an $n\times n$ irreducible matrix with entries in a field
$\mathbb{F}$ with $n\geq 4$. The fiber of $A$ consists of irreducible
matrices.

###### Proof.

Let $f=\det(A+\operatorname{diag}\\{x_{1},\ldots,x_{n}\\})$. By Lemma 5.1, $f$
is irreducible. Let $B$ be any matrix in the fiber of $A$, then
$\det(B+\operatorname{diag}\\{x_{1},\ldots,x_{n}\\})=f$. Hence this later
determinant is irreducible. By Lemma 5.1, $B$ is irreducible. ∎

##  6\. The fibers of Symmetric Matrices

######  Lemma 6.1.

Let $A$ be an $n\times n$ symmetric matrix with $n\geq 4$ and with entries in
the field $\mathbb{F}$. Then $A$ belongs to the set $\mathscr{D}(n)$ if and
only if $A$ is irreducible.

###### Proof.

$\Longrightarrow$ If $A$ is reducible and symmetric, then $A$ can be written
as a block diagonal matrix with diagonal blocks that are symmetric. Let $B$ be
any block upper triangular matrix with the same diagonal blocks where we
replace all the upper blocks of the matrix with nonzero entries. Direct
computations show that $A$ and $B$ have equal principal minors, but $B$ is not
diagonally equivalent to $A$. Therefore $A\notin\mathscr{D}(n)$.
$\Longleftarrow$ Let $B$ be a matrix with the same principal minors as $A$.
Then

$f_{A}=\det\left({\rm
diag}\\{x_{1},x_{2},\ldots,x_{n}\\}+A\right)=\det\left({\rm
diag}\\{x_{1},x_{2},\ldots,x_{n}\\}+B\right)=f_{B}.$

Then $\Delta_{ij}(f_{A})=\Delta_{ij}(f_{B})$. Notice that
$G=(\operatorname{diag}\\{x_{1},\ldots,x_{n}\\}+A)^{\rm adj}$ is also
symmetric. Using Theorem 2.3, $\Delta_{ij}(f_{A})=G_{ij}^{2}$. Let
$H=(\operatorname{diag}\\{x_{1},\ldots,x_{n}\\}+B)^{\rm adj}$. Then
$\Delta_{ij}(f)$ equals $G_{ij}^{2}=H_{ij}H_{ji}$. Since $G_{ij}$ and $H_{ij}$
are multiaffine, there exists $\alpha_{ij}\in\mathbb{F}$ such that
$H_{ij}=\alpha_{ij}G_{ij}$ and $H_{ji}=\frac{1}{\alpha_{ij}}G_{ij}$ for $1\leq
i<j\leq n$. Since $A$ is irreducible, then so is $f$. By Lemma 2.2,
$\Delta_{ij}(f)\neq 0$ and so $\alpha_{ij}\neq 0$. Using Theorem 2.1, ${\rm
res}_{x_{k}}(H_{ij},f)=H_{ik}H_{kj}$. Then
$\alpha_{ij}=\alpha_{ik}\frac{1}{\alpha_{jk}}$ for $1\leq i<j<k\leq n$. Let
$D={\rm diag}\\{1,\alpha_{12},\ldots,\alpha_{1n}\\}$, then $B=DAD^{-1}$ and
$A\in\mathscr{D}(n)$. ∎

######  Theorem 6.2.

Let $A$ be an $n\times n$ symmetric matrix with entries in a field
$\mathbb{F}$ and with $n\geq 4$. Then either

  1. (a)

The matrix $A$ is irreducible and the fiber of $A$ consists of a point up to
diagonal equivalence or

  2. (b)

the matrix $A$ can be written as a completely reducible matrix with $s$
irreducible diagonal blocks in which case the fiber of $A$ consists of all
matrices that can be written as block upper triangle matrices with $s$
irreducible diagonal blocks that are diagonally equivalent to the diagonal
blocks of $A$, after permuting some rows and the corresponding columns.

###### Proof.

Part $(a)$ of the theorem is Lemma 6.1. For $(b)$, assume that $A$ is
completely reducible with $s$ irreducible diagonal blocks $A_{1},\ldots,A_{s}$
indexed on the rows and columns by $T_{1},\ldots,T_{s}$ respectively with
$T_{i}\subset[n]$ and let $B$ be a matrix in the fiber of $A$ under the
principal minor map. Using Theorem 5.2,
$f_{A}=\det(\operatorname{diag}{x_{1},\ldots,x_{n}}+A)$ is reducible and can
be written as $f_{1}\cdots f_{s}$ where each
$f_{i}=\det(A_{i}+\operatorname{diag}\\{x_{k}:k\in T_{i}\\})$ is irreducible.
Since $A$ and $B$ have equal determinantal polynomials
$f_{A}=f_{B}=\det(\operatorname{diag}{x_{1},\ldots,x_{n}}+B)$, then , in view
of Theorem5.2, $B$ can be written as a block upper triangular matrix with $s$
irreducible diagonal blocks $B_{i}$ with $1\leq i\leq s$ and such that
$f_{i}=\det(B_{i}+\operatorname{diag}\\{x_{k}:k\in T_{i}\\})$. Thus $B_{i}$
and $A_{i}$ have the same principal minors for each $i$. Since $B_{i}$ is
symmetric and irreducible, then by Lemma 6.1 $A_{i}$ is diagonally equivalent
to $B_{i}$ as desired. ∎

##  7\. The fibers of Hermitian matrices and Determinantal Stable Polynomials

In this section we restrict to the field of complex numbers $\mathbb{C}$.

######  Lemma 7.1.

Let $A$ be an $n\times n$ complex matrix with $n\geq 4$. The fiber of $A$
consists of irreducible Hermitian matrices up to diagonal equivalence if and
only if $A$ is diagonally equivalent to an irreducible Hermitian matrix.

###### Proof.

$\Longleftarrow$ Suppose that $A$ is Hermitian. Let
$f=\det(\operatorname{diag}\\{x_{1},\ldots,x_{n}\\}+A)$. In view of [5,
Theorem 3.2], $f$ is real stable. Moreover, by Lemma 5.1, $f$ is irreducible
since $A$ is irreducible. Let $B$ be a matrix in the fiber of $A$. Then
$\det(B+\operatorname{diag}\\{x_{1},\ldots,x_{n}\\})=f$. By [2, Theorem
$6.4$], $B$ is diagonally equivalent to a Hermitian irreducible matrix.
$\Longrightarrow$ Let $B$ be a matrix in the fiber of $A$. Then $B$ is
diagonally equivalent to an irreducible Hermitian matrix. By the above
argument, $A$ is diagonally equivalent to an irreducible Hermitian matrix. ∎

######  Theorem 7.2.

Let $A$ be an $n\times n$ Hermitian matrix with $n\geq 4$. Then either

  1. (a)

The matrix $A$ is irreducible and the fiber of $A$ consists of Hermitian
irreducible matrices up to diagonal equivalence or

  2. (b)

The matrix $A$ can be written as a completely reducible matrix with $s$
irreducible diagonal blocks, after permuting some rows and the corresponding
columns, in which case the fiber of $A$ consists exactly of matrices that can
be written as block upper triangular matrices with $s$ diagonal blocks that
are irreducible and diagonally equivalent to Hermitian irreducible matrices
and that have determinantal polynomials equal to the factors of $f$.

###### Proof.

Similar proof as that of Theorem 6.2 ∎

Notice that, unlike the symmetric case, the diagonal blocks of a matrix that
lies in the fiber of a Hermitian matrix $A$ need not be diagonally equivalent
to the diagonal blocks of $A$. See for instance [2, Example 4.8].

Now we are ready to answer Question Question by Borcea, Brändén, and Liggett
[5, Question 3.4] about real stable matrices. A real polynomial
$f\in\mathbb{R}[x_{1},\ldots,x_{n}]$ is _stable_ if it has no zeros with
strictly positive imaginary parts. A real stable matrix $A$ is an $n\times n$
complex matrix with real stable determinantal polynomial i.e
$f_{A}=\det(\operatorname{diag}\\{x_{1},\ldots,x_{n}\\}+A)$ is real stable.

######  Corollary 7.3.

Let $A$ be an $n\times n$ complex matrix such that
$f=\det(\\{x_{1},\ldots,x_{n}\\}+A)$ is real stable. Then

  1. (a)

If $f$ is irreducible, then $A$ is diagonally equivalent to an irreducible
Hermitian matrix with equal principal minors.

  2. (b)

If $f$ factors as $f_{1}\cdots f_{s}$ where each factor $f_{k}$ is
irreducible, then $A$ can be written, after permuting some rows and the
corresponding columns, as a block upper triangular matrix with $s$ diagonal
blocks that are irreducible and diagonally equivalent to Hermitian irreducible
matrices that have determinantal polynomials equal to the irreducible factors
of $f$.

###### Proof.

Part $(a)$ of the corollary follows from [2, Theorem $6.6$] and Lemma 5.1. For
part $(b)$, Theorem 5.2 implies that $A$ can be written as a block upper
triangular matrix with $s$ diagonal irreducible blocks that have determinantal
polynomials equal to the factors of $f$. Since each $f_{i}$ is stable and
irreducible, then by [2, Theorem 6.6] and Lemma 5.1 each diagonal block is
diagonally equivalent to an irreducible Hermitian matrix as desired. ∎

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