# Bi-monotone maps on the set of all variance-covariance matrices with respect
to minus partial order

Gregor Dolinar  University of Ljubljana, Faculty of Electrical Engineering,
Tržaška cesta 25, SI-1000 Ljubljana, Slovenia, and IMFM, Jadranska 19, SI-1000
Ljubljana, Slovenia<EMAIL_ADDRESS>,  Dijana Ilišević
University of Zagreb, Faculty of Science, Department of Mathematics, Bijenička
30, 10000 Zagreb, Croatia<EMAIL_ADDRESS>,  Bojan Kuzma  University of
Primorska, Glagoljaška 8, SI-6000 Koper, Slovenia, and IMFM, Jadranska 19,
SI-1000 Ljubljana, Slovenia<EMAIL_ADDRESS>and  Janko Marovt  University
of Maribor, Faculty of Economics and Business, Razlagova 14, SI-2000 Maribor,
Slovenia, and IMFM, Jadranska 19, SI-1000 Ljubljana, Slovenia
<EMAIL_ADDRESS>
###### Abstract.

Let $H_{n}^{+}(\mathbb{R})$ be the cone of all positive semidefinite $n\times
n$ real matrices. We describe the form of all surjective maps on
$H_{n}^{+}(\mathbb{R})$, $n\geq 3$, that preserve the minus partial order in
both directions.

###### Key words and phrases:

monotone map, minus partial order, variance-covariance matrix, linear model

###### 2020 Mathematics Subject Classification:

15A09, 15B48, 47B49, 47L07, 54F05, 62J99

The first, the third, and the fourth author acknowledge the financial support
from the Slovene Research Agency, ARRS (research programs No. P1-0285 and No.
P1-0288, and research projects No. N1-0210 and No. N1-0296). The authors also
acknowledge the bilateral project between Croatia and Slovenia (Generalized
inverses with emphasis on searching for applications in statistics) which was
financially supported by the Ministry of Science and Education, Republic of
Croatia, and by the Slovene Research Agency, ARRS (grant BI-HR/20-21-036).

##  1\. Introduction

Let $M_{m,n}(\mathbb{F})$ be the set of all $m\times n$ matrices over a field
$\mathbb{F}$. Unless stated otherwise,
$\mathbb{F}\in\\{\mathbb{R},\mathbb{C}\\}$. If $m=n$, that is, for the set of
all $n\times n$ square matrices over $\mathbb{F}$ we will write
$M_{n}(\mathbb{F)}$ instead of $M_{n,n}(\mathbb{F)}$. A matrix $A\in
M_{n}(\mathbb{F)}$ has a (unique) inverse $A^{-1}$ (a matrix satisfying
$AA^{-1}=A^{-1}A=I$, where $I$ is the identity matrix) if and only if it is a
square matrix with linearly independent columns (or rows). However, the need
for some kind of a partial inverse of a more general class of matrices (even
non-square ones) has arisen from applications. For given $A\in
M_{m,n}(\mathbb{F})$, any matrix $X\in M_{n,m}(\mathbb{F})$ which is a
solution to the equation $AXA=A$ is called an inner generalized inverse of $A$
and it is denoted by $A^{-}$. Every matrix $A\in M_{m,n}(\mathbb{F})$ has an
inner generalized inverse, but it is not unique in general. If $A\in
M_{n}(\mathbb{F})$ has inverse $A^{-1}$, then its inner generalized inverse is
unique and equal to $A^{-1}$. This justifies the use of the word
’generalized’. More on inner generalized inverses (and other generalized
inverses, such as the Moore-Penrose inverse $A^{\dagger}$ for example) can be
found in, e.g., [2, 13]. As usual, let $A^{T}\in M_{n,m}(\mathbb{F})$ denote
the transpose, $A^{\ast}\in M_{n,m}(\mathbb{C})$ the conjugate transpose,
$\text{Im}A$ the image (i.e. the column space), and $\text{Ker}A$ the kernel
(the nullspace) of $A\in M_{m,n}(\mathbb{F})$.

Generalized inverses are closely related to partial orders on matrices. There
are many partial orders which may be defined on various sets of matrices
(e.g., the Löwner, the minus, the sharp, the star, and also the corresponding
one-sided orders). The minus partial order may be defined on the full set
$M_{m,n}(\mathbb{F})$. For $A,B\in M_{m,n}(\mathbb{F})$ we say that $A$ is
below $B$ with respect to the minus partial order and write

$A\leq^{-}B\quad\text{when}\quad A^{-}A=A^{-}B\text{ and }AA^{-}=BA^{-}$

for some inner generalized inverse $A^{-}$ of $A$. The minus partial order is
also known as the rank subtractivity partial order since for $A,B\in
M_{m,n}(\mathbb{F})$ we have

(1) $A\leq^{-}B\quad\text{if and only
if}\quad\text{rank}(B-A)=\text{rank}(B)-\text{rank}(A).$

Let us mention that the minus partial order was originally introduced by
Hartwig in [7] and independently by Nambooripad in [14] on a general regular
semigroup however it was mostly studied on $M_{n}(\mathbb{F})$ (see [12] and
the references therein).

The minus partial order has some applications in statistics, especially in the
theory of linear statistical models (see [13, Sections 15.3, 15.4]). Let

$y=X\beta+\epsilon$

be the matrix form of a linear model. We call $y$ the response vector (also
known as the observation vector) and $X\in M_{m,n}(\mathbb{F})$ the model
matrix (also known as the design or regressor matrix). In order to complete
the description of the model, some assumptions about the nature of the errors
have to be made. It is assumed that $E(\epsilon)=0$ and
$V(\epsilon)=\sigma^{2}D$, i.e., the errors have the zero mathematical
expectation and covariances are known up to a scalar (real number). Here $V$
denotes the variance-covariance matrix (also known as dispersion or covariance
matrix). The nonnegative parameter $\sigma^{2}$ and the vector of parameters
(real numbers) $\beta\in\mathbb{R}^{n}$ are unspecified, and $D$ is a known
$m\times m$ real, positive semidefinite matrix. We denote this linear model
with the triplet $(y,X\beta,\sigma^{2}D)$. Let us now present an example of an
application of the minus partial order in the theory of linear models (see
[13, Theorem 15.3.6]).

######  Proposition 1.1.

Let $L_{1}=(y,X_{1}\theta,\sigma^{2}D)$ and $L_{2}=(y,X_{2}\beta,\sigma^{2}D)$
be any two linear models. Then $X_{1}\leq^{-}X_{2}$ if and only if there
exists a matrix $A$ with
$\mathop{\mathrm{I}m}A^{T}\subseteq\mathop{\mathrm{I}m}X_{2}^{T}$ and $L_{1}$
is the model $L_{2}$ constrained by $A\beta=0$.

Given the model $M=(y,X\beta,\sigma^{2}I)$ one might rather work with the
transformed model $\hat{M}=(y,\hat{X}\beta,\sigma^{2}I)$ because the model
matrix $\hat{X}$ has more attractive properties then $X$ (e.g. elements of $X$
that are very close to zero are transformed to zero). It is natural to demand
that the transformed model still retains most of properties of the original
model (e.g. has similar relations to other transformed models) so in view of
the above application of the minus partial order it is interesting to know
what transformations $\Phi$ on $M_{m,n}(\mathbb{R})$, or on
$M_{n}(\mathbb{R})$, or on some subset of $M_{n}(\mathbb{R})$ preserve the
order $\leq^{-}$ in one or in both directions.

A generalization of the concept of the minus partial order from matrices to
operators (more precisely, to the algebra $B(\mathcal{H})$ of all bounded
linear operators on a Hilbert space $\mathcal{H}$) was studied by Šemrl in
[16]. Let $\mathcal{A}$ be some subset of $B(\mathcal{H})$. We say that a map
$\Phi:\mathcal{A}\rightarrow\mathcal{A}$ is a bi-monotone map with respect to
the minus partial order $\leq^{-}$ (or that it preserves the minus partial
order $\leq^{-}$ in both directions) when

$A\leq^{-}B\quad\text{if and only if}\quad\Phi(A)\leq^{-}\Phi(B)$

for every $A,B\in\mathcal{A}$. Linear bijective maps on $M_{n}(\mathbb{F})$,
where $\mathbb{F}$ is a field of more then $n$ elements, that preserve the
minus partial order in one direction (i.e., $A\leq^{-}B$ implies
$\Phi(A)\leq^{-}\Phi(B)$) were investigated in [6]. Bijective bi-monotone maps
with respect to the minus partial order on $M_{n}(\mathbb{F})$, $n\geq 3$,
were characterized in [9], and those on $B(\mathcal{H})$, where a Hilbert
space $\mathcal{H}$ is infinite-dimensional, in [16].

A set of matrices that deserves special attention is the set of all positive
semidefinite matrices. Such matrices are matrix analogues to nonnegative real
numbers. We may identify them with variance-covariance matrices since (see,
e.g., [3]) every variance-covariance matrix is positive semidefinite and
conversely, every (real) positive semidefinite matrix is a variance-covariance
matrix of some multivariate distribution. These matrices appear not only in
statistics but also as elements of the search space in convex and semidefinite
programming, as kernels in machine learning, as conductivity matrices in
thermodynamics, as diffusion tensors in medical imaging, as overlap matrices
in quantum chemistry, etc. More about the fundamental role played by positive
(semi)definite matrices can be found in [1].

Let $H_{n}(\mathbb{F})$ denote the set of all Hermitian (i.e., symmetric in
the real case) matrices in $M_{n}(\mathbb{F})$ and let $H_{n}^{+}(\mathbb{F})$
be the set of all positive semidefinite matrices in $H_{n}(\mathbb{F})$. Note
that $H_{n}^{+}(\mathbb{F})$ is not a linear space, but it is a convex cone
(it is closed under addition and scalar multiplication by non-negative scalars
only). Thus, linear preservers on $H_{n}^{+}(\mathbb{F})$ cannot be studied,
but at least additive ones can. Partial orders on $H_{n}^{+}(\mathbb{F})$ and
the corresponding bi-monotone maps were studied in [5]. One of the main
results from [5] follows.

######  Theorem 1.

Let $n\geq 3$ be an integer. Then $\Phi:H_{n}^{+}(\mathbb{R})\rightarrow
H_{n}^{+}(\mathbb{R})$ is a surjective, additive map that preserves the minus
order $\leq^{-}$ in both directions if and only if there exists an invertible
matrix $S\in M_{n}(\mathbb{R)}$ such that

$\Phi(A)=SAS^{T}$

for every $A\in H_{n}^{+}(\mathbb{R})$.

In this paper we study bi-monotone maps with respect to the minus partial
order on $H_{n}^{+}(\mathbb{R})$ without the additivity assumption. In Section
2, we give some preliminary results. It turns out that ellipses centered at
the origin in $\mathbb{R}^{2}$ can be identified with operators in
$H_{n}^{+}(\mathbb{R})$ whose images are equal to a fixed 2-dimensional
subspace of $\mathbb{R}^{n}$, while the points on these ellipses correspond to
rank-one positive semidefinite operators which are dominated by the
corresponding rank-two operators in $H_{n}^{+}(\mathbb{R})$ with respect to
the minus partial order. Thus, in Section 3 we study incidence relation among
concentric ellipses. The main result which describes the form of all
surjective bi-monotone maps with respect to the minus partial order on
$H_{n}^{+}(\mathbb{R})$, $n\geq 3$, is stated and proved in Section 4.

##  2\. Preliminaries

Let us present some tools that will be useful throughout the paper. As before,
let $\mathbb{F=R}$ or $\mathbb{F=C}$. Let $P_{n}(\mathbb{F})$ be the set of
all projectors in $H_{n}^{+}(\mathbb{F})$, i.e., the set of all orthogonal
projection matrices in $M_{n}(\mathbb{F)}$. Let $V$ be a subspace of
$\mathbb{F}^{n}$. By $P_{V}\in P_{n}(\mathbb{F})$ we will denote the projector
with $\mathop{\mathrm{I}m}P_{V}=V$. We will denote by $x\otimes y^{\ast}$ a
rank one linear operator on $B(\mathcal{H})$ defined by $(x\otimes
y^{\ast})z=\left\langle z,y\right\rangle x$ for every $z\in\mathcal{H}$. Note
that every rank-one linear operator on $\mathcal{H}$ may be written in this
form. Recall that when $\mathcal{H}$ is finite dimensional, say that the
dimension of $\mathcal{H}$ equals $n$, we may identify $B(\mathcal{H})$ with
$M_{n}(\mathbb{F})$, and that in $\mathbb{F}^{n}$, $\left\langle
z,y\right\rangle=y^{\ast}z$. It turns out that $P\in P_{n}(\mathbb{F})$ is of
rank-one if and only if $P=x\otimes x^{\ast}$ for some $x\in\mathbb{F}^{n}$
with $\left\|x\right\|=1$. Indeed, let first $x\in\mathbb{F}^{n}$ with
$\left\|x\right\|=1$. Then $(x\otimes x^{\ast})^{\ast}=x\otimes x^{\ast}$ and
for every $z\in\mathbb{F}^{n}$ we have $(x\otimes x^{\ast})^{2}z=\left\langle
z,x\right\rangle\left\langle x,x\right\rangle x=\left\langle z,x\right\rangle
x=(x\otimes x^{\ast})z$. Conversely, let $P$ be of rank-one and
$P^{2}=P=P^{\ast}$. Then $P=x\otimes y^{\ast}$ for some nonzero
$x,y\in\mathbb{F}^{n}$. By transferring the appropriate scalar to the second
factor we may assume without loss of generality that $\left\|x\right\|=1$.
From $x\otimes y^{\ast}=(x\otimes y^{\ast})^{\ast}=y\otimes x^{\ast}$ it
follows that $y=\mu x$ for some nonzero $\mu\in\mathbb{F}$ and $(x\otimes
y^{\ast})^{2}=x\otimes y^{\ast}$ then implies that $\mu=1$, i.e., $P=x\otimes
x^{\ast}$ with $\left\|x\right\|=1$. The following auxiliary result was proved
in [16].

######  Lemma 2.1.

Let $\mathcal{H}$ be a Hilbert space and let $x,y\in\mathcal{H}$ be nonzero
vectors and $A\in B(\mathcal{H})$. Then $x\otimes y^{\ast}\leq^{-}A$ if and
only if $x\in\mathop{\mathrm{I}m}A$ and there exists $z\in H$ such that
$\left\langle x,z\right\rangle=1$ and $y=A^{\ast}z$. In particular, rank-one
$x\otimes x^{\ast}\in H_{n}^{+}(\mathbb{F})$ satisfies $x\otimes
x^{\ast}\leq^{-}A$ if and only if $x\in\mathop{\mathrm{I}m}A$ and there exists
$z\in\mathbb{F}^{n}$ such that $\left\langle x,z\right\rangle=1$ and $x=Az$.

We denote by $e_{1},e_{2},\ldots,e_{n}\in\mathbb{F}^{n}$ the standard basis
vectors. In the next lemma we answer the question of when a rank-one positive
semidefinite matrix is dominated with respect to the minus partial order by a
given positive semidefinite matrix.

######  Lemma 2.2.

Let $A\in H_{n}^{+}(\mathbb{F})$ be nonzero with the eigenvalue decomposition

$A=\alpha_{1}e_{i_{1}}\otimes e_{i_{1}}^{\ast}+\alpha_{2}e_{i_{2}}\otimes
e_{i_{2}}^{\ast}+\ldots+\alpha_{j}e_{i_{j}}\otimes e_{i_{j}}^{\ast}$

where $i_{1},i_{2},\dots,i_{j}\in\\{1,2,\dots,n\\}$ are pairwise distinct,
$1\leq j\leq n$, and $\alpha_{k}>0$ for every
$k\in\left\\{1,2,\ldots,j\right\\}$. Let $x\in\mathbb{F}^{n}$ be a nonzero
vector. Then $x\otimes x^{\ast}\leq^{-}A$ if and only if there exist
$\beta_{1},\beta_{2},\ldots,\beta_{j}\in\mathbb{F}\ $such that

(2)
$x=\beta_{1}e_{i_{1}}+\beta_{2}e_{i_{2}}+\ldots+\beta_{j}e_{i_{j}}\quad\text{and\quad}\frac{\left|\beta_{1}\right|^{2}}{\alpha_{1}}+\frac{\left|\beta_{2}\right|^{2}}{\alpha_{2}}+\ldots+\frac{\left|\beta_{j}\right|^{2}}{\alpha_{j}}=1.$

###### Proof.

Suppose first that $x\otimes x^{\ast}\leq^{-}A$. By Lemma 2.1,
$x\in\mathop{\mathrm{I}m}A$ and thus there exist
$\beta_{1},\beta_{2},\ldots,\beta_{j}\in\mathbb{F}$ such that
$x=\beta_{1}e_{i_{1}}+\beta_{2}e_{i_{2}}+\ldots+\beta_{j}e_{i_{j}}$. Also (see
Lemma 2.1), there exists $z\in\mathbb{F}$, such that $\left\langle
x,z\right\rangle=1$ and $x=Az$. It follows that
$\left\langle\beta_{1}e_{i_{1}}+\beta_{2}e_{i_{2}}+\ldots+\beta_{j}e_{i_{j}},z\right\rangle=1$
and therefore

$\beta_{1}\left\langle e_{i_{1}},z\right\rangle+\beta_{2}\left\langle
e_{i_{2}},z\right\rangle+\ldots+\beta_{j}\left\langle
e_{i_{j}},z\right\rangle=1.$

From $Az=x$ we have $\alpha_{1}\left\langle z,e_{i_{1}}\right\rangle
e_{i_{1}}+\alpha_{2}\left\langle z,e_{i_{2}}\right\rangle
e_{i_{2}}+\ldots+\alpha_{j}\left\langle z,e_{i_{j}}\right\rangle e_{i_{j}}=$
$\beta_{1}e_{i_{1}}+\beta_{2}e_{i_{2}}+\ldots+\beta_{j}e_{i_{j}}$ and thus
$\alpha_{k}\left\langle z,e_{i_{k}}\right\rangle=\beta_{k}$ for every
$k\in\left\\{1,2,\ldots,j\right\\}$. So,

$\left\langle
e_{i_{k}},z\right\rangle=\frac{\overline{\beta_{k}}}{\alpha_{k}},\quad
k=1,2,\ldots,j.$

It follows that
$\beta_{1}\frac{\overline{\beta_{1}}}{\alpha_{1}}+\beta_{2}\frac{\overline{\beta_{2}}}{\alpha_{2}}+\ldots+\beta_{j}\frac{\overline{\beta_{j}}}{\alpha_{j}}=1$
and therefore

$\frac{\left|\beta_{1}\right|^{2}}{\alpha_{1}}+\frac{\left|\beta_{2}\right|^{2}}{\alpha_{2}}+\ldots+\frac{\left|\beta_{j}\right|^{2}}{\alpha_{j}}=1.$

Conversely, let $\beta_{1},\beta_{2},\ldots,\beta_{j}\in\mathbb{F}\ $be such
that $x=\beta_{1}e_{i_{1}}+\beta_{2}e_{i_{2}}+\ldots+\beta_{j}e_{i_{j}}$ and

$\frac{\left|\beta_{1}\right|^{2}}{\alpha_{1}}+\frac{\left|\beta_{2}\right|^{2}}{\alpha_{2}}+\ldots+\frac{\left|\beta_{j}\right|^{2}}{\alpha_{j}}=1.$

Let $z~{}=~{}\left(z_{1},z_{2},\ldots,z_{n}\right)^{T}\in\mathbb{F}^{n}$ be
such that $z_{i_{k}}=\frac{\beta_{k}}{\alpha_{k}}$ for every
$k\in\left\\{1,2,\ldots,j\right\\}$ while other (possible) components can be
arbitrary. Then

$\left\langle
x,z\right\rangle=\frac{\left|\beta_{1}\right|^{2}}{\alpha_{1}}+\frac{\left|\beta_{2}\right|^{2}}{\alpha_{2}}+\ldots+\frac{\left|\beta_{j}\right|^{2}}{\alpha_{j}}=1$

and

$Az=\alpha_{1}\left\langle z,e_{i_{1}}\right\rangle
e_{i_{1}}+\alpha_{2}\left\langle z,e_{i_{2}}\right\rangle
e_{i_{2}}+\ldots+\alpha_{j}\left\langle z,e_{i_{j}}\right\rangle
e_{i_{j}}=\alpha_{1}\frac{\beta_{1}}{\alpha_{1}}e_{i_{1}}+\alpha_{2}\frac{\beta_{2}}{\alpha_{2}}e_{i_{2}}+\ldots+\alpha_{j}\frac{\beta_{j}}{\alpha_{j}}e_{i_{j}}=x$

and thus by Lemma 2.1, $x\otimes x^{\ast}\leq^{-}A$.

######  Remark 2.3.

Notice that equation (2) is equivalent to $x^{\ast}A^{\dagger}x=1$ where
$A^{\dagger}$ is the Moore-Penrose inverse of the matrix $A$. Therefore, every
rank-one operator $x\otimes x^{\ast}\in H_{n}^{+}(\mathbb{F})$ which is
majorized by a nonzero operator $A\in H_{n}^{+}(\mathbb{F})$ is induced by a
vector $x\in\mathop{\mathrm{I}m}A$ which determines a point that lies on an
ellipsoid $\left\\{x\in\mathop{\mathrm{I}m}A:x^{\ast}A^{\dagger}x=1\right\\}$.
Clearly, every nonzero $x=Az\in\mathop{\mathrm{I}m}A$ satisfies
$x^{\ast}A^{\dagger}x=z^{\ast}AA^{\dagger}Az=z^{\ast}Az=z^{\ast}\sqrt{A}\sqrt{A}z=\|\sqrt{A}z\|^{2}\neq
0$ so, when appropriately scaled belongs to ellipsoid induced by $A$. It
follows that if $B\in H_{n}^{+}(\mathbb{F})\backslash\\{0\\}$ induces the same
ellipsoid $\mathfrak{E}$ as $A\in H_{n}^{+}(\mathbb{F})$, then each nonzero
$x\in\mathop{\mathrm{I}m}A$, when appropriately scaled, lies on $\mathfrak{E}$
so satisfies $x\in\mathop{\mathrm{I}m}B$. Thus,
$\mathop{\mathrm{I}m}A=\mathop{\mathrm{I}m}B$. Moreover,
$x^{\ast}A^{\dagger}x=1$ if and only if $x^{\ast}B^{\dagger}x=1$ for each
$x\in\mathop{\mathrm{I}m}A=\mathop{\mathrm{I}m}B$. Next, $AA^{\dagger}$ is a
projector onto $\mathop{\mathrm{I}m}A$, so
$A^{\dagger}x=(A^{\dagger}AA^{\dagger})x=A^{\dagger}(AA^{\dagger})x=0$ for
each $x\in\left(\mathop{\mathrm{I}m}A\right)^{\bot}$, and likewise
$B^{\dagger}x=0$ for each
$x\in\left(\mathop{\mathrm{I}m}B\right)^{\bot}=\left(\mathop{\mathrm{I}m}A\right)^{\bot}$.
Therefore, $x^{\ast}\left(A^{\dagger}-B^{\dagger}\right)x=0$ for every
$x\in\mathbb{F}^{n}$, and since $A^{\dagger}$ and $B^{\dagger}$ are Hermitian
matrices, $A=B$. This shows that $A=B$ is equivalent to $C\leq^{-}A$ if and
only if $C\leq^{-}B$ for every rank-one $C\in H_{n}^{+}(\mathbb{F})$.

In particular, ellipses centered at the origin in $\mathbb{R}^{2}$ can be
identified with operators in $H_{n}^{+}(\mathbb{R})$ whose images are equal to
a fixed 2-dimensional subspace of $\mathbb{R}^{n}$ while points on these
ellipses correspond to rank-one positive semidefinite operators majorized by
the corresponding rank-two operators in $H_{n}^{+}(\mathbb{R})$.

Given a rank-two $A\in H_{n}^{+}(\mathbb{R})$, let

(3)
$\mathfrak{E}_{A}=\left\\{x\in\mathop{\mathrm{I}m}A:x^{\ast}A^{\dagger}x=1\right\\}$

be the corresponding ellipse induced by $A$.

##  3\. Incidence relation among concentric ellipses

######  Lemma 3.1.

Let $\Pi_{1},\Pi_{2}\subseteq\mathbb{R}^{n}$ be two two-dimensional subspaces,
and let $E_{1}\subseteq\Pi_{1}$ and $E_{2}\subseteq\Pi_{2}$ be the ellipses,
centered at the origin. The following are equivalent.

  * (i)

Ellipses $E_{1}$ and $E_{2}$ lie in the same space (i.e. $\Pi_{1}=\Pi_{2}$).

  * (ii)

There exists an ellipse $E$, which may coincide with $E_{1}$ or with $E_{2}$,
centered at the origin and lying is some two-dimensional subspace $\Pi$
(perhaps $E=E_{1}$) such that $\\#(E\cap E_{1})=\\#(E\cap E_{2})\geq 4$.

###### Proof.

If $\\#(E_{1}\cap E)\geq 4$, then the two-dimensional spaces $\Pi_{1}$ and
$\Pi$ contain at least four nonzero points which cannot belong to the same
line. Hence, $\Pi=\Pi_{1}$.

Conversely, if $E_{1},E_{2}$ belong to the same plane then it is easy to
construct another ellipse $E$ with the required properties.

Throughout the rest of this section the ambient space is the plane
$\mathbb{R}^{2}$. We assume that all ellipses are concentric (i.e., their axes
intersect at the same point, called the center of ellipse) with the center at
the origin. Thus, they are given by the equation $x^{T}Qx=1$ and denoted by
$E_{Q}$, where $Q\in H_{2}^{+}(\mathbb{R})$ is a positive-definite matrix.
Note that an ellipse $E_{Q}$ lies inside the unit circle if and only if both
eigenvalues of $Q$ are greater or equal to $1$. To shorten the narrative, a
concentric ellipse whose axes lie on the coordinate axes will be called a
_standard ellipse_.

Let $K_{\rho}$ be a circle with the center at the origin and radius $\rho$. In
particular, ${\mathbf{K}}=K_{1}$ is the unit circle. We will consider
bijection $\Phi\colon\mathbb{R}^{2}\to\mathbb{R}^{2}$ which map concentric
ellipses into concentric ellipses. Hence, if $E$ is an ellipse, then $\Phi(E)$
is either an ellipse or a part of a uniquely determined ellipse (uniqueness
follows because $\Phi$ is injective, so the set $\Phi(E)$ is infinite). We
denote by $\mathrm{int}\Phi(E)$ the interior of this unique ellipse.

Let us start with the following.

######  Lemma 3.2.

Let $\Phi\colon\mathbb{R}^{2}\to\mathbb{R}^{2}$ be a bijection which maps
concentric ellipses into concentric ellipses. Then, $\Phi$ uniformly preserves
containment of ellipses inside each other or uniformly reverses it.

###### Proof.

Let $E_{1},E_{2},E_{3}$ be concentric ellipses. It is easy to see the
equivalence of the following statements:

  * (i)

$E_{1}\subseteq\mathrm{int}E_{2}\subseteq\mathrm{int}E_{3}$ or
$E_{1}\supseteq\mathrm{int}E_{2}\supseteq\mathrm{int}E_{3}$.

  * (ii)

    * (a)

$E_{1},E_{2},E_{3}$ are pairwise disjoint.

    * (b)

If some ellipse $E$ touches $E_{1}$ and $E_{3}$, then $E$ intersects $E_{2}$
in at least four points.

Suppose $E_{1},E_{2},E_{3}$ satisfy the assumptions of (i). Then one can
always find $E$ which touches $E_{1}$ and $E_{3}$, and hence intersects
$E_{2}$ in at least four points. Applying the above equivalence to
$\Phi$-images and using injectiveness of $\Phi$ we see that either
$\Phi(E_{1})\subseteq\mathrm{int}\Phi(E_{2})\subseteq\mathrm{int}\Phi(E_{3})$
or else
$\mathrm{int}\Phi(E_{1})\supseteq\mathrm{int}\Phi(E_{2})\supseteq\Phi(E_{3})$.
Arguing recursively, the same conclusion holds for any finite collection of
nested ellipses. Since each interior point of $E_{3}$ belongs to some
concentric ellipse contained in $\mathrm{int}E_{3}$ we deduce that either
$\Phi$ maps $\mathrm{int}E_{3}$ into the interior of the ellipse $\Phi(E_{3})$
or into its exterior.

We claim that $\Phi$ does this uniformly for each ellipse. Assume otherwise,
and let $\hat{E}_{3}$ be a concentric ellipse such that $\Phi$ maps
$\mathrm{int}E_{3}$ into interior of $\Phi(E_{3})$ and
$\mathrm{int}\hat{E}_{3}$ into exterior of $\Phi(\hat{E}_{3})$. If $E_{3}$ and
$\hat{E}_{3}$ do not intersect, then one is contained in another, say
$E_{3}\subseteq\mathrm{int}\hat{E}_{3}$. We may enlarge this inclusion by
adding additional ellipse $E_{2}$ contained in $\mathrm{int}E_{3}$; then on
the one hand, $\Phi(E_{2})\subseteq\mathrm{int}\Phi(E_{3})$, so by the above,
$\Phi(E_{2})\subseteq\mathrm{int}\Phi(E_{3})\subseteq\mathrm{int}\Phi(\hat{E}_{3})$
but on the other hand, $\Phi(\hat{E}_{3})\subseteq\mathrm{int}\Phi(E_{3})$, so
again, by the previous arguments,
$\Phi(\hat{E}_{3})\subseteq\mathrm{int}\Phi(E_{3})\subseteq\mathrm{int}\Phi(E_{2})$,
which implies that $\Phi(E_{2})\subseteq\mathrm{int}\Phi(E_{2})$ a
contradiction.

Assume $E_{3}$ and $\hat{E}_{3}$ intersect or touch. Then we choose an ellipse
$E_{4}$ which contains both $E_{3}$ and $\hat{E}_{3}$ in its interior and we
choose ellipse $E_{2}\subseteq\mathrm{int}E_{3}\cap\mathrm{int}\hat{E}_{3}$.
The inclusion $E_{2}\subseteq\mathrm{int}E_{3}\subseteq\mathrm{int}E_{4}$ is
preserved, while the inclusion
$E_{2}\subseteq\mathrm{int}\hat{E}_{3}\subseteq\mathrm{int}E_{4}$ is reversed,
so $\Phi(E_{2})$ belongs to interior and to exterior of ellipse $\Phi(E_{4})$,
a contradiction.

######  Lemma 3.3.

Let $\Phi\colon\mathbb{R}^{2}\to\mathbb{R}^{2}$ be a bijection which maps
concentric ellipses into concentric ellipses and its restriction to the unit
circle ${\mathbf{K}}$ equals $x\mapsto Tx/\|Tx\|$ for some invertible diagonal
matrix $T={\mathrm{d}iag\,}(\mu,\lambda)$. Then, $\Phi$ fixes ${\mathbf{K}}$
set-wise and fixes the points $(\pm 1,0)$ and $(0,\pm 1)$. Moreover, it leaves
the interior of the unit circle invariant.

###### Proof.

The first part is clear. For the second part, the previous lemma implies that
$\Phi$ either maps $\mathrm{int}{\mathbf{K}}$ into itself or into exterior of
the unit circle. Assume the former. Then the previous lemma implies that
$\Phi$ also maps its exterior onto its interior. Let us choose a circle
$K_{\rho}$ of radius $\rho<1$ and let $E_{1}$ and $E_{2}$ be two standard
ellipses, touching the unit circle ${\mathbf{K}}$ at $(\pm 1,0)$ and $(0,\pm
1)$ respectively and intersecting at $\frac{1}{\sqrt{2}}(\pm\rho,\pm\rho)$;
note that $E_{1}$ and $E_{2}$ intersect on the circle $K_{\rho}$.

By the assumptions, the ellipses $\Phi(E_{1})$ and $\Phi(E_{2})$ lie outside
the unit circle and touch it at $(\pm 1,0)$ and $(0,\pm 1)$ respectively
because these four points are fixed by $\Phi$. Thus, $\Phi(E_{1})$ and
$\Phi(E_{2})$ are also standard ellipses. We note that $\Phi(E_{1})$ lies in
the vertical strip $[-1,1]\times\mathbb{R}$ and $\Phi(E_{2})$ lies in the
horizontal strip $\mathbb{R}\times[-1,1]$. Therefore, $\Phi(E_{1})$ and
$\Phi(E_{2})$ intersect outside the unit circle but inside the square spanned
by $(\pm 1,0)$ and $(0,\pm 1)$. Since $\Phi(E_{1})$ and $\Phi(E_{2})$ are
standard ellipses, they intersect at points that are placed symmetrically
relatively to the coordinate axes.

Figure 1. $E_{1},E_{2}$ touch the unit circle and intersect at a concentric
circle $K_{\rho}$. The image of $K_{\rho}$ is an ellipse passing through the
symmetrically placed points of intersection of $\Phi(E_{1})$ and
$\Phi(E_{2})$. These points belong to the interior of the shaded rectangle
$[-1,1]\times[-1,1]$.

The ellipses $E_{1}$ and $E_{2}$ intersect on $K_{\rho}$, so $\Phi(E_{1})$ and
$\Phi(E_{2})$ intersect on the ellipse $\Phi(K_{\rho})$. Thus, the ellipse
$\Phi(K_{\rho})$ contains four symmetric points inside the square
$[-1,1]\times[-1,1]$ and therefore its axes lie on the coordinate axes.

On the other hand, let us choose ellipses $E_{1}^{{}^{\prime}}$ and
$E_{2}^{{}^{\prime}}$, again with axes lying on the coordinate axes and
touching the unit circle at $(\pm 1,0)$ and $(0,\pm 1)$ respectively, but
touching the circle $K_{\rho}$ at $(0,\pm\rho)$ and $(\pm\rho,0)$
respectively. Then $\Phi(E_{1}^{{}^{\prime}})$ and $\Phi(E_{2}^{{}^{\prime}})$
are ellipses touching the unit circle and lying in its exterior, so one of
their axes is 1 and the other ($b_{\rho}$, respectively $a_{\rho}$) is greater
than $1$. Since $\Phi(K_{\rho})$ lies outside the unit circle, its axes lie on
the coordinate axes and it touches both $\Phi(E_{1}^{{}^{\prime}})$ and
$\Phi(E_{2}^{{}^{\prime}})$, its equation is

$\frac{x^{2}}{a_{\rho}^{2}}+\frac{y^{2}}{b_{\rho}^{2}}=1.$

Furthermore, since $\Phi(K_{\rho})$ contains four symmetric points inside the
square $[-1,1]\times[-1,1]$, the vertices of that square, i.e., the points
$(\pm 1,\pm 1)$ lie outside $\Phi(K_{\rho})$. Thus,

$\frac{1}{a_{\rho}^{2}}+\frac{1}{b_{\rho}^{2}}>1.$

Then at least one of $a_{\rho}$, $b_{\rho}$ is less than $\sqrt{2}$ and this
is true for every radius $\rho<1$.

If for $\Phi(K_{\rho_{1}})$ we have $b_{\rho_{1}}\geq\sqrt{2}$ and for
$\Phi(K_{\rho_{2}})$ we have $a_{\rho_{2}}\geq\sqrt{2}$ then
$a_{\rho_{1}}<\sqrt{2}$ and $b_{\rho_{2}}<\sqrt{2}$, so $\Phi(K_{\rho_{1}})$
and $\Phi(K_{\rho_{2}})$ intersect; a contradiction. Therefore, either for
every radius $\rho<1$ we have $a_{\rho}<\sqrt{2}$, or for every radius
$\rho<1$ we have $b_{\rho}<\sqrt{2}$. In the first case $\Phi(K_{\rho})$ lies
in the vertical strip $(-\sqrt{2},\sqrt{2})\times\mathbb{R}$ for each
concentric circle $K_{\rho}$ contained inside the unit circle. Each point in
the interior of ${\mathbf{K}}$ lies on some concentric circle and hence it is
mapped into that vertical strip. Then, in particular, the point $(4,0)$ cannot
be the image of any point inside the unit circle. It cannot be the image of
any point outside the unit circle either since $\Phi$ maps exterior of the
unit circle onto its interior. The case when $b_{\rho}<\sqrt{2}$ for every
radius $\rho<1$ analogously contradicts surjectivity of $\Phi$.

We continue with a technical lemma which will help us in calculating the
touching ellipses.

Let $R_{\phi}=\left(\begin{smallmatrix}\cos\phi&-\sin\phi\\\
\sin\phi&\cos\phi\\\ \end{smallmatrix}\right)$ be a rotation matrix. Observe
that the angle between one of the axis of the ellipse induced by the matrix

$R_{\phi}\left(\begin{smallmatrix}1&0\\\ 0&r\\\
\end{smallmatrix}\right)R_{\phi}^{T}$

and the $x$-axis is equal to $\phi$. We say that such ellipse is at angle
$\phi$ relative to $x$-axis.

######  Lemma 3.4.

Let an ellipse $E_{Q}$, induced by a matrix $Q$, lie in the interior of the
unit circle and let $\phi\in(0,\frac{\pi}{2})$ be a given angle. Then there
exists a unique ellipse $E$, at angle $\phi$ relative to $x$-axis,
simultaneously touching the unit circle and the ellipse $E_{Q}$. It is induced
by the matrix $R_{\phi}\left(\begin{smallmatrix}1&0\\\ 0&r\\\
\end{smallmatrix}\right)R_{\phi}^{T}$, where $r$ satisfies the equation

(4) $\det\left(Q-R_{\phi}\left(\begin{smallmatrix}1&0\\\ 0&r\\\
\end{smallmatrix}\right)R_{\phi}^{T}\right)=0.$

###### Proof.

The ellipse $E$ touches the unit circle, hence one of its axis has length $1$,
and the other has length $\rho$ which is yet to be determined. Since the
ellipse $E$ is at an angle $\phi$ relative to $x$-axis, $E$ is obviously
induced by $R_{\phi}\left(\begin{smallmatrix}1&0\\\ 0&r\\\
\end{smallmatrix}\right)R_{\phi}^{T}$, where $r=1/\rho^{2}$. We need to
determine $r$ so that $E_{Q}$ and $E$ touch at some vector $x$. Then,

(5) $x^{T}Qx=1=x^{T}R_{\phi}\left(\begin{smallmatrix}1&0\\\ 0&r\\\
\end{smallmatrix}\right)R_{\phi}^{T}x$

which implies

$x^{T}\left(Q-R_{\phi}\left(\begin{smallmatrix}1&0\\\ 0&r\\\
\end{smallmatrix}\right)R_{\phi}^{T}\right)x=0.$

Moreover, gradients of these two ellipses at the point where they touch are
parallel. Now, the gradient of a quadratic form $x\mapsto x^{T}Ax$ is given by

$\partial x^{T}Ax=x^{T}(A^{T}+A)$

and since in our case the matrices $Q$ and
$R_{\phi}\left(\begin{smallmatrix}1&0\\\ 0&r\\\
\end{smallmatrix}\right)R_{\phi}^{T}$ are symmetric, we get that for some
scalar $\lambda$,

$x^{T}Q=\lambda x^{T}R_{\phi}\left(\begin{array}[]{cc}1&0\\\ 0&r\\\
\end{array}\right)R_{\phi}^{T}.$

Multiplying this by $x$ and using (5) we obtain $\lambda=1$, so
$x^{T}(Q-R_{\phi}\left(\begin{smallmatrix}1&0\\\ 0&r\\\
\end{smallmatrix}\right)R_{\phi}^{T})=0$. Thus, $x$ is an eigenvector of the
symmetric matrix $Q-R_{\phi}\left(\begin{smallmatrix}1&0\\\ 0&r\\\
\end{smallmatrix}\right)R_{\phi}^{T}$ corresponding to the eigenvalue $0$.
Therefore, we are looking for $r$ such that

$\det\left(Q-R_{\phi}\left(\begin{smallmatrix}1&0\\\ 0&r\\\
\end{smallmatrix}\right)R_{\phi}^{T}\right)=0.$

This is equivalent to

(6) $\det(R_{\phi}^{T}QR_{\phi}-\left(\begin{smallmatrix}1&0\\\ 0&r\\\
\end{smallmatrix}\right))=0.$

Since the ellipse $E_{Q}$ is inside the unit circle, $x^{T}Qx>1$ for each unit
vector $x$, so the $(1,1)$ entry of
$R_{\phi}^{T}QR_{\phi}-\left(\begin{smallmatrix}1&0\\\ 0&r\\\
\end{smallmatrix}\right)$ is nonzero. Thus, the left hand side of (6) is a
linear polynomial in $r$ and hence there does exist a unique $r$ satisfying
(6).

Moreover, with such $r$, the matrix $Q-R_{\phi}\left(\begin{smallmatrix}1&0\\\
0&r\\\ \end{smallmatrix}\right)R_{\phi}^{T}$ is singular but nonzero, and
hence it has exactly one nonzero eigenvalue. Therefore, it is a semi-definite
$2$-by-$2$ matrix, and it follows easily after its diagonalization that,
modulo a scalar multiple, there exists at most one nonzero $x$ with

$x^{T}(Q-R_{\phi}\left(\begin{smallmatrix}1&0\\\ 0&r\\\
\end{smallmatrix}\right)R_{\phi}^{T})x=0.$

It remains to prove that at least one such $x$ does exist. In fact, if $x$ is
a unit vector in the kernel of $Q-R_{\phi}\left(\begin{smallmatrix}1&0\\\
0&r\\\ \end{smallmatrix}\right)R_{\phi}^{T}$ , then
$Qx=R_{\phi}\left(\begin{smallmatrix}1&0\\\ 0&r\\\
\end{smallmatrix}\right)R_{\phi}^{T}x$. We only need to scale $x$ to conclude
that it belongs to ellipses induced by $Q$ and induced by
$R_{\phi}\left(\begin{smallmatrix}1&0\\\ 0&r\\\
\end{smallmatrix}\right)R_{\phi}^{T}$.

######  Remark 3.5.

As a computational complement to Lemma 3.4 let us note that

(7) $R_{\phi}\left(\begin{matrix}1&0\\\ 0&r\\\
\end{matrix}\right)R_{\phi}^{T}=\left(\begin{matrix}(r-1)\sin^{2}{\phi}+1&-(r-1)\sin{\phi}\cos{\phi}\\\
-(r-1)\sin{\phi}\cos{\phi}&(r-1)\cos^{2}{\phi}+1\\\ \end{matrix}\right).$

If $Q$ in Lemma 3.4 is given by $Q=\left(\begin{smallmatrix}a&0\\\ 0&b\\\
\end{smallmatrix}\right)$ then

$Q-R_{\phi}\left(\begin{matrix}1&0\\\ 0&r\\\
\end{matrix}\right)R_{\phi}^{T}=\left(\begin{matrix}a-r\sin^{2}{\phi}-\cos^{2}{\phi}&(r-1)\sin{\phi}\cos{\phi}\\\
(r-1)\sin{\phi}\cos{\phi}&b-\sin^{2}{\phi}-r\cos^{2}{\phi}\\\
\end{matrix}\right),$

so (4) first implies

$r-1=\frac{(a-1)(b-1)}{(a-1)\cos^{2}{\phi}+(b-1)\sin^{2}{\phi}}$

and then

$r=\frac{(a-1)b+a(b-1)\tan^{2}{\phi}}{(a-1)+(b-1)\tan^{2}{\phi}}.$

For such $r$, the matrix (7) becomes

(8)
$\displaystyle\frac{1}{(a-1)+(b-1)\tan^{2}{\phi}}\left(\begin{matrix}(a-1)+a(b-1)\tan^{2}{\phi}&-(a-1)(b-1)\tan{\phi}\\\
-(a-1)(b-1)\tan{\phi}&(a-1)b+(b-1)\tan^{2}{\phi}\\\ \end{matrix}\right).$

From now on, $\Phi\colon\mathbb{R}^{2}\to\mathbb{R}^{2}$ is a surjection,
which maps concentric ellipses into concentric ellipses, maps interior of
${\mathbf{K}}$ onto itself, and maps a point $(c,s)\in\mathbb{R}^{2}$ on the
unit circle to the point

$\frac{(\mu c,\lambda s)}{\|(\mu c,\lambda s)\|}$

again on the unit circle, i.e., if $x=(c,s)\in{\mathbf{K}}$, then
$\Phi(x)=Tx/\|Tx\|$ where $T={\mathrm{d}iag\,}(\mu,\lambda)$ is a diagonal
matrix. Without loss of generality, due to polar decomposition, we may assume
$\lambda$ and $\mu$ are both positive. Within a sequence of steps we will show
that $\lambda=\mu$ and that $\Phi$ fixes all the points from $\mathbb{R}^{2}$.

Step 1. The points $(\pm 1,0)$ and $(0,\pm 1)$ are fixed.

Proof. Obvious.

Step 2. Each concentric circle $K_{\rho}$ of radius $\rho<1$ is mapped into a
standard ellipse.

Proof. Let us consider a concentric ellipse $E_{1}\colon\frac{1}{\rho^{\prime
2}}x^{2}+y^{2}=1$ which touches the unit circle at points $(0,\pm 1)$ and
touches a concentric circle $K_{\rho^{\prime}}$ with radius
$\rho^{\prime}=\frac{\rho}{\sqrt{2-\rho^{2}}}<1$. Since points $(0,\pm 1)$ are
fixed, the image of $E_{1}$ is an ellipse which touches the unit circle again
at points $(0,\pm 1)$, therefore its axes lie on the coordinate axes.
Similarly for a concentric ellipse $E_{2}\colon x^{2}+\frac{1}{\rho^{\prime
2}}y^{2}=1$ which touches the unit circle at points $(\pm 1,0)$ and touches
the concentric circle $K_{\rho^{\prime}}$. Hence, the axes of ellipses
$\Phi(E_{1}),\Phi(E_{2})$ lie on the coordinate axes and both ellipses touch
the unit circle. Therefore, their intersection consists of four points,
symmetrically placed relative to the $x$ and $y$ axes.

Let us note that ellipses $E_{1}$ and $E_{2}$ intersect at points
$\frac{1}{\sqrt{2}}(\pm\rho,\pm\rho)$ on $K_{\rho}$. Therefore, the image of
$K_{\rho}$ must be a concentric ellipse which contains the four symmetrical
points of intersection of $\Phi(E_{1})$ and $\Phi(E_{2})$. The only such
ellipse again has axes lying on the coordinate axes.

Step 3. The points inside the unit circle belonging to $x$-axis (respectively,
$y$-axis), are mapped into $x$-axis (respectively, $y$-axis).

Proof. The concentric circle $K_{\rho^{\prime}}$ touches the above two
ellipses $E_{1}$ and $E_{2}$ respectively at points $(\pm\rho^{\prime},0)$ and
$(0,\pm\rho^{\prime})$. These four points are mapped into intersection of the
image of $E_{1}$ and $E_{2}$ with the image of $K_{\rho^{\prime}}$. The images
of $E_{1}$, $E_{2}$ and $K_{\rho^{\prime}}$ are standard ellipses, moreover,
the images of $E_{1},E_{2}$ also touch the unit circle at points $(0,\pm 1)$
and $(\pm 1,0)$, respectively. Hence, $\Phi(E_{1})$ and
$\Phi(K_{\rho^{\prime}})$ intersect on $x$-axis, while $\Phi(E_{2})$ and
$\Phi(K_{\rho^{\prime}})$ intersect on $y$-axis.

Step 4. Let us consider again two concentric ellipses $E_{1},E_{2}$, which
touch the unit circle at symmetrically placed points $(\cos\phi,\pm\sin\phi)$
relative to $x$-axis and touch a given concentric circle $K_{\rho}$ of radius
$\rho=1/\sqrt{r}$. These two ellipses are induced by the matrices

$R_{\pm\phi}\left(\begin{smallmatrix}1&0\\\ 0&r\\\
\end{smallmatrix}\right)R_{\pm\phi}^{T}=\left(\begin{matrix}(r-1)\sin^{2}{(\pm\phi)}+1&-(r-1)\sin{(\pm\phi)}\cos{(\pm\phi)}\\\
-(r-1)\sin{(\pm\phi)}\cos{(\pm\phi)}&(r-1)\cos^{2}{(\pm\phi)}+1\\\
\end{matrix}\right).$

The points $\pm(\cos\phi,\pm\sin\phi)$ of intersection of ${\mathbf{K}}$ with
$E_{1}$ and $E_{2}$, respectively, are mapped by $\Phi$ into points

$\pm\frac{(\mu\cos\phi,\pm\lambda\sin\phi)}{\sqrt{\lambda^{2}\sin^{2}\phi+\mu^{2}\cos^{2}\phi}}$

whose argument of polar coordinates satisfies
$\tan\psi=\pm\frac{\lambda\sin\phi}{\mu\cos\phi}=\pm\frac{\lambda}{\mu}\tan\phi$,
that is,

$\psi=\pm\arctan\left(\frac{\lambda}{\mu}\tan\phi\right)=\pm\arctan(\gamma\tan\phi)$

where $\gamma:=\lambda/\mu>0$.

Step 5. Since $\Phi(E_{1})$ and $\Phi(E_{2})$ are at angle $\pm\psi$,
respectively, relative to $x$-axis and $\tan{\psi}=\pm\gamma\tan{\phi}$, it is
sufficient to replace $\tan{\phi}$ in (8) with $\gamma\tan{\phi}$ to conclude
that $\Phi(E_{1})$ and $\Phi(E_{2})$ are given by the matrices

$\frac{1}{(a-1)+(b-1)\gamma^{2}\tan^{2}{\phi}}\left(\begin{matrix}(a-1)+a(b-1)\gamma^{2}\tan^{2}{\phi}&\mp(a-1)(b-1)\gamma\tan{\phi}\\\
\mp(a-1)(b-1)\gamma\tan{\phi}&(a-1)b+(b-1)\gamma^{2}\tan^{2}{\phi}\\\
\end{matrix}\right).$

Step 6. Let us now consider ellipses $E_{3},E_{4}$ which touch ${\mathbf{K}}$
at points $(\sin\phi,\pm\cos\phi)$ and again touch $K_{\rho}$. By repeating
the previous arguments (with $\pm\phi$ replaced by $\pm\phi+\frac{\pi}{2}$)
one gets that $\psi$ changes to
$\psi=\pm\arctan\frac{\lambda\cot\phi}{\mu}=\pm\arctan(\gamma\cot\phi)$ and
therefore it remains to replace $\tan{\phi}$ in (8) with $\gamma\cot{\phi}$ to
get that $\Phi(E_{1})$ and $\Phi(E_{2})$ are induced by the matrices

$\frac{1}{(a-1)+(b-1)\gamma^{2}\cot^{2}{\phi}}\left(\begin{matrix}(a-1)+a(b-1)\gamma^{2}\cot^{2}{\phi}&\mp(a-1)(b-1)\gamma\cot{\phi}\\\
\mp(a-1)(b-1)\gamma\cot{\phi}&(a-1)b+(b-1)\gamma^{2}\cot^{2}{\phi}\\\
\end{matrix}\right).$

Step 7. The original ellipses intersect at vertical points

$E_{1}\cap
E_{2}\ni\left(0,\pm\frac{1}{\sqrt{r\cos^{2}\phi+\sin^{2}\phi}}\right).$

Proof. To see this, one merely inserts $x=0$ and
$y=\pm\frac{1}{\sqrt{r\cos^{2}\phi+\sin^{2}\phi}}$ into the equations

$(x,y)R_{\phi}\left(\begin{smallmatrix}1&0\\\
0&r\end{smallmatrix}\right)R_{\phi}^{T}(x,y)^{T}=1=(x,y)R_{-\phi}\left(\begin{smallmatrix}1&0\\\
0&r\end{smallmatrix}\right)R_{-\phi}^{T}(x,y)^{T},$

and notes that being quadratic equations, they have no additional solutions
with $x=0$. If we want that the vertical intersection of $E_{1}$ and $E_{2}$
lies on a concentric circle $K_{\rho_{0}}$ with $\rho_{0}<1$ kept fixed, one
computes that the angle satisfies

(9)
$\phi=\frac{1}{2}\arccos\left(\frac{2-\rho_{0}^{2}(r+1)}{\rho_{0}^{2}(r-1)}\right).$

Step 8. Since we know that $\Phi$ maps points on $y$-axis again into points on
$y$-axis, and analogously for $x$-axis, we see that the above vertical point
is mapped to the vertical point of the intersection of $\Phi(E_{1})$ and
$\Phi(E_{2})$, that is to the point

$\left(0,\,\pm\frac{\sqrt{(a-1)+(b-1)\gamma^{2}\tan^{2}\phi}}{\sqrt{(a-1)b+(b-1)\gamma^{2}\tan^{2}\phi}}\right).$

Step 9. Likewise, the horizontal point of intersection $E_{3}\cap
E_{4}\ni\left(\pm\frac{1}{\sqrt{r\cos^{2}\phi+\sin^{2}\phi}},0\right)$ is
mapped to the horizontal point of intersection of $\Phi(E_{3})$ and
$\Phi(E_{4})$, that is, to the point

$\left(\pm\frac{\sqrt{(a-1)+(b-1)\gamma^{2}\cot^{2}\phi}}{\sqrt{(a-1)+a(b-1)\gamma^{2}\cot^{2}\phi}},\,0\right).$

We now interrupt the narrative with the following technical lemma which the
remaining steps rely upon.

######  Lemma 3.6.

Let $\gamma>0$ and let $\widehat{E_{0}}\colon a_{0}x^{2}+b_{0}y^{2}=1$ be an
ellipse inside ${\mathbf{K}}$. For every given angle
$\phi\in(0,\frac{\pi}{2})$ consider the uniquely given four ellipses
$\widehat{E_{1}},\widehat{E_{2}},\widehat{E_{3}},\widehat{E_{4}}$, contained
inside ${\mathbf{K}}$, touching ${\mathbf{K}}$, and with
$\widehat{E_{1}},\widehat{E_{2}}$ having longer axis symmetrically at angles
$\pm\arctan(\gamma\tan\phi)$ and $\widehat{E_{3}},\widehat{E_{4}}$ having
longer axis symmetrically at angles $\pm\arctan(\gamma\cot\phi)$, and such
that $\widehat{E_{1}}\cap\widehat{E_{2}}$ contains the intersection of
$\widehat{E_{0}}$ with $y$-axis and $\widehat{E_{3}}\cap\widehat{E_{4}}$
contains the intersection of $\widehat{E_{0}}$ with $x$-axis.

If for each $\phi\in(0,\frac{\pi}{2})$ there exists a standard ellipse,
contained inside
$\widehat{E_{1}},\widehat{E_{2}},\widehat{E_{3}},\widehat{E_{4}}$ and touching
all four of them, then $\frac{a_{0}-1}{b_{0}-1}=\gamma^{2}$.

Figure 2. Four ellipses with axes symmetrically at an angle $\phi$ relative to
$x$ and $y$, respectively, touching the same inner circle as well as the unit
circle, and intersecting vertically/horizontally at the same concentric circle
drawn in thick gray. Their images on the right picture must intersect at the
image of the gray thick circle, i.e., at the gray thick ellipse and must touch
the image of the inner circle, i.e., an inner ellipse.

###### Proof.

Let the touching ellipse $\widehat{E}$ (for an angle $\phi$) have an equation
$\widehat{E}\colon ax^{2}+by^{2}=1$. Note that $a=a(\phi)>1$ and
$b=b(\phi)>1$. Namely, if, for some $\phi$, $a(\phi)=1$, then the standard
ellipse $\widehat{E}$ would touch ${\mathbf{K}}$ and hence would either have
to intersect the slanted ellipses $\widehat{E}_{i}$ in four points or (if also
$b(\phi)=1$) would equal ${\mathbf{K}}$ and hence would not lie inside
$\widehat{E}_{i}$, a contradiction; similarly $b(\phi)=1$ is impossible. Set

$u=u(\phi)=a-1>0,\quad v=v(\phi)=b-1>0,\quad t=\tan{\phi}.$

As shown above, the four ellipses
$\widehat{E_{1}},\widehat{E_{2}},\widehat{E_{3}},\widehat{E_{4}}$, touching
$\widehat{E}$, and touching the unit circle at symmetrically placed points
with angles $\pm\arctan(\gamma\tan\phi)$ and $\pm\arctan(\gamma\cot\phi)$,
respectively, intersect at vertical points

$\widehat{E_{1}}\cap\widehat{E_{2}}\cap\\{(x=0)\\}=\left(0,\pm\frac{\sqrt{(a-1)+(b-1)\gamma^{2}\tan^{2}\phi}}{\sqrt{(a-1)b+(b-1)\gamma^{2}\tan^{2}\phi}}\right)=\left(0,\pm\frac{\sqrt{u+v\gamma^{2}t^{2}}}{\sqrt{u(v+1)+v\gamma^{2}t^{2}}}\right)$

and at horizontal points

$\widehat{E_{3}}\cap\widehat{E_{4}}\cap\\{(y=0)\\}=\left(\pm\frac{\sqrt{(a-1)+(b-1)\gamma^{2}\cot^{2}\phi}}{\sqrt{{(a-1)+a(b-1)\gamma^{2}\cot^{2}\phi}}},0\right)=\left(\pm\frac{\sqrt{u+v\gamma^{2}/t^{2}}}{\sqrt{{u+(u+1)v\gamma^{2}/t^{2}}}},0\right).$

These points must belong to $\widehat{E_{0}}$, so that

$\frac{\sqrt{u+v\gamma^{2}/t^{2}}}{\sqrt{{u+(u+1)v\gamma^{2}/t^{2}}}}=\frac{1}{\sqrt{a_{0}}},\qquad\frac{\sqrt{u+v\gamma^{2}t^{2}}}{\sqrt{u(v+1)+v\gamma^{2}t^{2}}}=\frac{1}{\sqrt{b_{0}}}.$

Then, after squaring and inverting, these two equations simplify to

$a_{0}-1=\frac{uv\gamma^{2}}{ut^{2}+\gamma^{2}v},\qquad
b_{0}-1=\frac{uv}{\gamma^{2}vt^{2}+u}.$

The first equation implies

$u=u_{t}=\frac{(a_{0}-1)\gamma^{2}v}{\gamma^{2}v-(a_{0}-1)t^{2}},$

which we insert in the second equation to get

$v=v_{t}=\frac{\left(a_{0}-1\right)\left(1-t^{4}\right)}{\frac{a_{0}-1}{b_{0}-1}-\gamma^{2}t^{2}}.$

Since $v=v_{t}=b-1>0$ for each $t>0$ and the numerator of $v_{t}$ changes sign
at $t=1$, the denominator must do the same, so at $t=1$ the denominator
vanishes. This gives

$\frac{a_{0}-1}{b_{0}-1}=\gamma^{2}$

and the result follows. Let us remark for the future use that with this
$\gamma$ we have $v=(b_{0}-1)(1+t^{2})$ and $u=(a_{0}-1)(1+t^{2})$, which
implies that the parameters of a touching ellipse (with $t=\tan\phi$) are

(10) $a=a_{0}(1+t^{2})-t^{2},\qquad
b=b_{0}(1+t^{2})-t^{2}=\frac{1}{\gamma^{2}}(a_{0}-1)(1+t^{2})+1.\hskip
5.0pt\hbox{\hskip 5.0pt\vrule width=4.0pt,height=6.0pt,depth=1.5pt\hskip
1.0pt}$

We now continue with the steps.

Step 10. By Step 2, $\Phi$ maps the concentric circle $K_{1/2}$ of radius
$\rho_{0}=1/2$ into a standard ellipse

$\widehat{E}_{0}=\Phi(K_{1/2})\colon a_{0}x^{2}+b_{0}y^{2}=1.$

Let $\rho=1/\sqrt{r}\leq 1/2$ be the radius of a concentric circle $K_{\rho}$.
By Step 7, if we place ellipses $E_{1},E_{2},E_{3},E_{4}$, each having a long
axis of length $1$ and a short axis of length $\rho$, symmetrically (relative
to $x$-axis and $y$-axis) at an angle $\pm\phi$ (respectively, $\pi/2\pm\phi$
for $E_{3},E_{4}$), with $\phi$ given by (9), then they will intersect
vertically and horizontally, respectively, on a fixed circle $K_{\rho_{0}}$,
and moreover, all of them will touch ${\mathbf{K}}$ and will have $K_{\rho}$
as a common incircle. The point $(\cos\phi,\sin\phi)$ where $E_{1}$ touches
${\mathbf{K}}$ is mapped to a point
$\Phi((\cos\phi,\sin\phi))=(\mu\cos\phi,\lambda\sin\phi)$ where
$\widehat{E}_{1}=\Phi(E_{1})$ touches $\Phi({\mathbf{K}})={\mathbf{K}}$.

By Lemma 3.3, $\widehat{E}_{1}$ is contained in ${\mathbf{K}}$, and hence its
longer axis is of length $1$ at the angle
$\psi=\arctan(\tfrac{\lambda\sin\phi}{\mu\cos\phi})=\arctan(\gamma\tan\phi)$.
Likewise we see that $\widehat{E}_{i}=\Phi(E_{i})$, $i=1,2,3,4$ satisfy all
the assumptions on axis of Lemma 3.6. To see that
$\widehat{E}_{1}\cap\widehat{E}_{2}$ intersects
$\widehat{E}_{0}=\Phi(K_{1/2})$ on $y$-axis we use Step 3 (and likewise that
$\widehat{E}_{3}\cap\widehat{E}_{4}$ intersects $\widehat{E}_{0}$ on $x$-axis)
while $\widehat{E}=\Phi(K_{\rho})$ is their touching ellipse. Then, Lemma 3.6
implies that the parameters of $\widehat{E}_{0}=\Phi(K_{1/2})$ must satisfy
$(a_{0}-1)/(b_{0}-1)=\gamma^{2}$.

To see from here where a concentric circle of radius
$\rho=1/\sqrt{r}\leq\rho_{0}=1/2$ is mapped into, we insert into (10) back
again $t=\tan\phi$ (again, using (9) with $\rho_{0}=1/2$ for $\phi$) and
simplify using
$\tan\left(\frac{a}{2}\right)=\sqrt{\frac{1-\cos(a)}{1+\cos(a)}}$ to get that
the circle $K_{\rho}$ of radii $\rho=1/\sqrt{r}<1/2$ is mapped into an ellipse
$ax^{2}+by^{2}=1$ where

(11) $a=\frac{(a_{0}-1)(r-1)}{3}+1,\qquad
b=\frac{\left(a_{0}-1\right)(r-1)}{3\gamma^{2}}+1.$

Step 11. Let $z=x+iy\in\mathbb{C}=\mathbb{R}^{2}$ be a point inside $K_{1/2}$.
Then,

(12) $\Phi(x+iy)=\pm\frac{\sqrt{3}(x+i\gamma
y)}{\sqrt{\left(a_{0}-1\right)\left(1-x^{2}-y^{2}\right)+3\left(x^{2}+\gamma^{2}y^{2}\right)}}.$

Proof. Let us draw an ellipse $E_{z}$ with axis on a ray $[0,z]$ and passing
through $z$ at its extreme. Now let us draw an incircle $K_{\mathrm{inc}}$ to
this ellipse and an ellipse $E_{\mathrm{out}}$ with the same axis, the same
incircle, but also touching the unit circle. Moreover, let us draw a
concentric circle $K_{|z|}$ passing through $z$. The ellipse $E_{z}$ is mapped
into an ellipse touching $\Phi(E_{\mathrm{out}})$ and $\Phi(K_{\mathrm{inc}})$
at the same point.

Figure 3. Towards the proof of Step 11. The inner-most circle can be
arbitrarily small, so arbitrarily narrow ellipse touches a unit circle and has
axis on a ray passing through a given point. On the right are their
$\Phi$-images.

Now, we are free to take the ellipse $E_{z}$ inside ${\mathbf{K}}$ as long as
it touches the point $z$ at one of its axis. Its second axis can be
arbitrarily small, in fact there exists a sequence of ellipses $E_{z}$
converging to a line segment $[-z,z]$; these ellipses have arbitrarily small
inscribed circle so the corresponding sequence of outer ellipses
$E_{\mathrm{out}}$ touching the unit circle along the same axis and touching
the inscribed circles, are also converging to a line segment. The radius
$\rho_{\text{inc}}=1/\sqrt{r_{\text{inc}}}$ of inner inscribed circles
converge to zero. They are mapped into ellipses with parameters described in
(11) and clearly with $\rho_{\text{inc}}\to 0$, i.e., with
$r=r_{\text{inc}}\to\infty$, both parameters go to $\infty$ so that the
ellipse $\Phi(K_{\rho})$ shrinks to a point. This means that
$\Phi(E_{\text{out}})$ which touches the shrinking ellipses $\Phi(K_{\rho})$,
must also shrink to a line segment. Thus, $\Phi(E_{z})$, which always lies
inside $\Phi(E_{\text{out}})$, also shrinks to a line segment and it always
touches $\Phi(K_{|z|})$.

Notice that this line segment contains a point
$\Phi(E_{\text{out}}\cap{\mathbf{K}})=\Phi(\tfrac{x+iy}{|x+iy|})=\frac{\mu
x+i\lambda y}{|\mu x+i\lambda y|}$. This shows that $\Phi(z)$ is the
intersection of $\Phi(K_{|z|)})$ and a line segment through $\pm(\mu
x+i\lambda y)$.

From here, and using that $\Phi(K_{|z|})$ is an ellipse of the form
$aX^{2}+bY^{2}=1$ with parameters $a,b$ given in (11) (with
$r=\frac{1}{|z|^{2}}=\frac{1}{x^{2}+y^{2}}$ and $\gamma=\lambda/\mu$), one
computes the asserted form:

$\displaystyle\Phi(x+iy)$ $\displaystyle=$ $\displaystyle\pm\frac{\mu
x+i\lambda y}{\sqrt{a\mu^{2}x^{2}+b\lambda^{2}y^{2}}}=\pm\frac{x+i\gamma
y}{\sqrt{ax^{2}+b\gamma^{2}y^{2}}}$ $\displaystyle=$
$\displaystyle\pm\frac{\sqrt{3}(x+i\gamma
y)}{\sqrt{\left(4-a_{0}\right)x^{2}+y^{2}\left(1-a_{0}+3\gamma^{2}\right)+\left(a_{0}-1\right)}};\qquad|x+iy|\leq\tfrac{1}{2},$
$\displaystyle=$ $\displaystyle\pm\frac{\sqrt{3}(x+i\gamma
y)}{\sqrt{\left(a_{0}-1\right)\left(1-x^{2}-y^{2}\right)+3\left(x^{2}+\gamma^{2}y^{2}\right)}};\qquad|x+iy|\leq\tfrac{1}{2}.$

Step 12. The function on the right-hand side of (12), which we denote by
$\widehat{\Phi}$, is clearly defined for every $z=x+iy\in\mathbb{C}$ inside
the unit circle ${\mathbf{K}}$, and it coincides with $\Phi$ inside $K_{1/2}$.
We claim that this function maps quadrics into quadrics.

Proof. To see this note that a general quadric takes the form
$ax^{2}+bxy+cy^{2}=1$ for some parameters $a,b,c$. One can solve it for $y$
and insert the $\widehat{\Phi}$-image of parametrization $(x,y(x))$ into a
polynomial $AX^{2}+BXY+CY^{2}$. By letting

(14) $\displaystyle A$
$\displaystyle=\frac{1}{3}\left(a\left(a_{0}-1\right)-a_{0}+4\right),$
$\displaystyle B$ $\displaystyle=\frac{\left(a_{0}-1\right)b}{3\gamma},$
$\displaystyle C$
$\displaystyle=\frac{\left(a_{0}-1\right)(c-1)}{3\gamma^{2}}+1$

the polynomial simplifies into a constant $1$, as claimed.

Step 13. The function $\hat{\Phi}$ can identify two points only if they belong
to the same line through the origin. Moreover,

(15) $a_{0}\leq 4\qquad\hbox{and}\qquad a_{0}\leq 1+3\gamma^{2}.$

Proof. The first claim is a consequence of the fact that

$\frac{\mathrm{Re}(\hat{\Phi}(x+iy))}{\mathrm{Im}(\hat{\Phi}(x+iy))}=\frac{x}{\gamma
y}.$

In particular, if a quadric $Q$ intersects the interior or $K_{1/2}$, then
$\hat{\Phi}(Q\cap\mathrm{int}K_{1/2})$ contains infinitely many points, so by
Step 12 belongs to a unique quadric.

To prove the second statement, let us consider first a point
$T_{\alpha}=(\alpha,0)$ with $\alpha>1$. Each ellipse of the family
$E_{c}\colon x^{2}/\alpha^{2}+cy^{2}=1$ ($c>4$) contains $T_{\alpha}$ and
passes through the interior of $K_{1/2}$. Hence, the ellipse $\Phi(E_{c})$ is
uniquely determined by its subset
$\Phi(E_{c}\cap\mathrm{int}K_{1/2})=\hat{\Phi}(E_{c}\cap\mathrm{int}K_{1/2})$.
In fact, it equals the quadric determined by coefficients (14), that is, the
quadric

$\frac{\left(\frac{1}{\alpha^{2}}\left(a_{0}-1\right)-a_{0}+4\right)}{3}x^{2}+\frac{\left(a_{0}-1\right)(c-1)+3\gamma^{2}}{3\gamma^{2}}y^{2}=1,$

which contains a point $\Phi(\alpha,0)$ for each $c>1$. Letting $c\to\infty$
we see that $\Phi(\alpha,0)=(x_{\alpha},0)$ where
$x_{\alpha}=\pm\frac{\sqrt{3}}{\sqrt{\frac{(a_{0}-1)}{\alpha^{2}}+4-a_{0}}}$.
This is clearly defined for each $\alpha>1$ and letting $\alpha\to\infty$ we
see that $4-a_{0}\geq 0$ (in order that the denominator in $x_{\alpha}$ is
never imaginary).

Similarly, the point $T^{\prime}_{\beta}=(0,\beta)$ (with $\beta>1$) belongs
to each ellipse of the family $E^{\prime}_{\beta}\colon
ax^{2}+y^{2}/\beta^{2}=1$, here with $a>4$. Arguing as before,
$\Phi(E^{\prime}_{\beta})$ is an ellipse with the equation

$\frac{\left(a\left(a_{0}-1\right)-a_{0}+4\right)}{3}x^{2}+\frac{\left(a_{0}-1\right)(\frac{1}{\beta^{2}}-1)+3\gamma^{2}}{3\gamma^{2}}y^{2}=1$

and it always contains $\Phi(0,\beta)$. For $a\to\infty$ we get
$\Phi(0,\beta)=(0,y_{\beta})$ with
$y_{\beta}=\frac{\sqrt{3\gamma^{2}}}{\sqrt{\left(a_{0}-1\right)(\frac{1}{\beta^{2}}-1)+3\gamma^{2}}}$.
This is defined for each $\beta>1$ and with $\beta\to\infty$ we must have
$-(a_{0}-1)+3\gamma^{2}\geq 0$ in order the the denominator is never
imaginary.

Step 14. It follows from Step 13 and (3) that $\widehat{\Phi}(z)$ is defined
for every $z\in\mathbb{C}$. We claim that $\Phi(z)=\widehat{\Phi}(z)$ for
every $|z|\geq 0$ and not just for $|z|\leq 1/2$.

To see this, let us consider two distinct ellipses $E_{z}^{1}$ and $E_{z}^{2}$
both of which contain a point $z$ and points from $\mathrm{int}K_{1/2}$. Then,

(16) $\Phi(z)\in\Phi(E_{z}^{1})\cap\Phi(E_{z}^{2}).$

By the assumptions, $\Phi(E_{z}^{i})$ is an ellipse while
$\widehat{\Phi}(E_{z}^{i})$ is a quadric, which coincides with the ellipse
$\Phi(E_{z}^{i})$ on infinitely many points (the $\hat{\Phi}$-images of
$E_{z}^{i}\cap\mathrm{int}K_{1/2})$. It follows that
$\widehat{\Phi}(E_{z}^{i})=\Phi(E_{z}^{i})$, and the claim follows from (16).

Step 15. Since $\Phi=\widehat{\Phi}$ is surjective, for every $u+iv$ there
exists $x+iy$ with $\hat{\Phi}(x+iy)=u+iv$ (recall that the action of
$\hat{\Phi}$ is described in (3)). Thus, we need to solve a system of two
equations with real variables $x,y$. By squaring both equations we get a
system

$\displaystyle\frac{3x^{2}}{\left(4-a_{0}\right)x^{2}+y^{2}\left(1-a_{0}+3\gamma^{2}\right)+a_{0}-1}$
$\displaystyle=u^{2},$
$\displaystyle\frac{3\gamma^{2}y^{2}}{\left(4-a_{0}\right)x^{2}+y^{2}\left(1-a_{0}+3\gamma^{2}\right)+a_{0}-1}$
$\displaystyle=v^{2}$

with solution

$\displaystyle x^{2}$
$\displaystyle=\frac{\left(a_{0}-1\right)\gamma^{2}u^{2}}{\left(a_{0}-4\right)\gamma^{2}u^{2}+v^{2}\left(a_{0}-1-3\gamma^{2}\right)+3\gamma^{2}},$
$\displaystyle y^{2}$
$\displaystyle=\frac{\left(a_{0}-1\right)v^{2}}{\left(a_{0}-4\right)\gamma^{2}u^{2}+v^{2}\left(a_{0}-1-3\gamma^{2}\right)+3\gamma^{2}}.$

Since $a_{0}>1$, the numerator is in both cases nonnegative, so denominator
must be nonnegative as well. Taking $v=0$ and $u\to\infty$ we see that
$a_{0}-4\geq 0$, that is, $a_{0}\geq 4$. Taking $u=0$ and $v\to\infty$ we get
$a_{0}-1-3\gamma^{2}\geq 0$. So $\Phi$ is surjective if and only if

$a_{0}\geq 4\qquad\hbox{and}\qquad a_{0}\geq 1+3\gamma^{2}.$

Combining this with (15) we deduce

$a_{0}=4\quad\hbox{ and }\quad\gamma=1.$

Thus, $T={\mathrm{d}iag\,}(\mu,\lambda)={\mathrm{d}iag\,}(\mu,\gamma\mu)$ is a
scalar matrix. Moreover, from the equation for $\Phi$, we see that
$\Phi(x+iy)=x+iy$ is the identity.

##  4\. Preservers of the minus partial order

Let $A,B\in M_{n}(\mathbb{F)}$. It is known (see, e.g. [9, page 149]) that

(17) $A\leq^{-}B\quad\text{if and only
if}\quad\mathop{\mathrm{I}m}B=\mathop{\mathrm{I}m}A\oplus\mathop{\mathrm{I}m}(B-A)\quad\text{if
and only if}\quad RAL\leq^{-}RBL$

for any invertible $R,L\in M_{n}(\mathbb{F)}$. Let $A,B\in M_{n}(\mathbb{C)}$.
If there exists an invertible matrix $S\in M_{n}(\mathbb{C)}$ such that

  * a)

$B=$ $SAS^{T}$, then we say that $A$ and $B$ are congruent;

  * b)

$B=$ $SAS^{\ast}$, then we say that $A$ and $B$ are *congruent.

By Sylvester’s law of inertia (see [8, page 282]) two (Hermitian) matrices
$A,B\in H_{n}(\mathbb{C)}$ are *congruent if and only if they have the same
inertia, i.e., they have the same number of positive eigenvalues and the same
number of negative eigenvalues. Two (real symmetric) matrices $A,B\in
H_{n}(\mathbb{R)}$ are *congruent via a complex matrix if and only if they are
congruent via a real matrix [8, page 283]. So, Sylvester’s law for the real
case states that $A,B\in H_{n}(\mathbb{R)}$ are congruent via an invertible
$S\in M_{n}(\mathbb{R)}$ (i.e., $B=$ $SAS^{T}$) if and only if $A$ and $B$
have the same number of positive eigenvalues and the same number of negative
eigenvalues. Note that congruent (respectively, *congruent) matrices have the
same rank [8, page 281].

Denote by $E_{ij}$ the $n\times n$ matrix with all entries equal to zero
except the $(i,j)$-entry which is equal to one. Let
$E_{k}=E_{11}+E_{22}+\ldots+E_{kk}$. For $A,B\in M_{n}(\mathbb{R)}$ we will
write $A<^{-}B$ when $A\leq^{-}B$ and $A\neq B$. We now state and prove our
main result.

######  Theorem 2.

Let $n\geq 3$ be an integer. Then $\Phi:H_{n}^{+}(\mathbb{R})\rightarrow
H_{n}^{+}(\mathbb{R})$ is a surjective bi-monotone map with respect to the
minus partial order $\leq^{-}$ if and only if there exists an invertible
matrix $S\in M_{n}(\mathbb{R)}$ such that

$\Phi(A)=SAS^{T}$

for every $A\in H_{n}^{+}(\mathbb{R})$.

###### Proof.

Let $\Phi:H_{n}^{+}(\mathbb{R})\rightarrow H_{n}^{+}(\mathbb{R})$ be of the
form $\Phi(A)=SAS^{T}$, $A\in H_{n}^{+}(\mathbb{R})$, where $S\in
M_{n}(\mathbb{R)}$ is an invertible matrix. Then $\Phi$ preserves by (17) the
order $\leq^{-}$ in both directions and is clearly surjective.

Conversely, let $\Phi:H_{n}^{+}(\mathbb{R})\rightarrow H_{n}^{+}(\mathbb{R})$
be a surjective map that preserves the order $\leq^{-}$ in both directions. We
will split the proof into several steps.

Step 1. $\Phi$ is bijective. Let $\Phi(A)=\Phi(B)$ for $A,B\in
H_{n}^{+}(\mathbb{R})$. The order $\leq^{-}$ is reflexive so
$\Phi(A)\leq^{-}\Phi(B)$ and $\Phi(B)\leq^{-}\Phi(A)$. Since $\Phi$ preserves
the order $\leq^{-}$ in both directions, we have $A\leq^{-}B$ and
$B\leq^{-}A$. It follows that $A=B$, since $\leq^{-}$ is antisymmetric. Thus,
$\Phi$ is injective and therefore bijective.

Step 2. $\Phi(0)=0$. Note that $0\leq^{-}A$ for every $A\in
H_{n}^{+}(\mathbb{R})$. So, on the one hand $0\leq^{-}\Phi(0)$ and on the
other hand, since $\Phi^{-1}$ has the same properties as $\Phi$,
$0\leq^{-}\Phi^{-1}(0)$ and thus $\Phi(0)\leq^{-}0$.

Step 3. $\Phi$ preserves the rank, i.e.,
$\mathop{\mathrm{r}ank}(A)=\mathop{\mathrm{r}ank}(\Phi(A))$ for every $A\in
H_{n}^{+}(\mathbb{R})$. Let $A\in H_{n}^{+}(\mathbb{R})$ with
$\mathop{\mathrm{r}ank}(A)=k$. By Sylvester’s law of inertia there exists an
invertible matrix $R\in M_{n}(\mathbb{R)}$ such that $E_{k}=RAR^{T}$. Clearly
(see (1)),

$0<^{-}E_{1}<^{-}E_{2}<^{-}\ldots<^{-}E_{n}=I.$

Since congruence preserves rank, we have by (17)

$0<^{-}R^{-1}E_{1}(R^{-1})^{T}<^{-}R^{-1}E_{2}(R^{-1})^{T}<^{-}\ldots<^{-}R^{-1}E_{k}(R^{-1})^{T}<^{-}\ldots<^{-}R^{-1}E_{n}(R^{-1})^{T}.$

From $(R^{-1})^{T}=(R^{T})^{-1}$ and since $\Phi$ preserves the order
$\leq^{-}$ and it is injective, we obtain

(18)
$0<^{-}\Phi(R^{-1}E_{1}(R^{T})^{-1})<^{-}\Phi(R^{-1}E_{2}(R^{T})^{-1})<^{-}\ldots<\Phi(A)<^{-}\ldots<^{-}\Phi(R^{-1}(R^{T})^{-1}).$

Let $C,D\in M_{n}(\mathbb{R)}$ with $C<^{-}D$ and
$\mathop{\mathrm{r}ank}(C)=\mathop{\mathrm{r}ank}(D)$. Then by (1),
$\mathop{\mathrm{r}ank}(D-C)=0$ and therefore $D=C$, a contradiction. So, if
$C<^{-}D$, then $\mathop{\mathrm{r}ank}(C)<\mathop{\mathrm{r}ank}(D)$.

Every succeeding matrix in (18) has the rank that is strictly greater then its
predecessor. Since $\mathop{\mathrm{r}ank}\Phi(R^{-1}(R^{T})^{-1})\leq n$, it
follows that $\mathop{\mathrm{r}ank}\Phi(R^{-1}(R^{T})^{-1})=n$ and therefore
$\mathop{\mathrm{r}ank}(\Phi(A))=k$.

Step 4. We may without loss of generality assume that $\Phi(I)=I$. By the
previous step, $\Phi(I)=B$ where $B\in H_{n}^{+}(\mathbb{R})$ is an invertible
(positive definite) matrix. It follows that there exists a positive definite
matrix $\sqrt{B}\in H_{n}^{+}(\mathbb{R})$ such that
$\Phi(I)=\sqrt{B}\sqrt{B}$. Let $\Psi:$ $H_{n}^{+}(\mathbb{R})\rightarrow
H_{n}^{+}(\mathbb{R})$ be defined by

$\Psi(A)=\left(\sqrt{B}\right)^{-1}\Phi(A)\left(\sqrt{B}\right)^{-1}.$

Then $\Psi$ is a bijective map that preserves the order $\leq^{-}$ in both
directions. Also, $\Psi(I)=I$. We will thus from now on assume that

$\Phi(I)=I.$

Step 5. There exists a bijective, linear map
$T:\mathbb{R}^{n}\rightarrow\mathbb{R}^{n}$ such that for every $P\in
P_{n}(\mathbb{R})$ the matrix $\Phi(P)$ is the projector onto
$T(\mathop{\mathrm{I}m}P)$, i.e.,

$\mathit{\ }\Phi(P)=P_{T(\mathop{\mathrm{I}m}P)}.$

Let $P\in M_{n}(\mathbb{R})$ be an idempotent matrix, i.e., $P^{2}=P$. Then
$\mathbb{R}^{n}=\mathop{\mathrm{I}m}P\oplus\mathop{\mathrm{K}er}P=\mathop{\mathrm{I}m}P\oplus\mathrm{\mathop{\mathrm{I}m}}(I-P)$
and therefore by (17), $P\leq^{-}I$. Moreover, if $Q\in M_{n}(\mathbb{R)}$ is
an idempotent matrix and if $A\leq^{-}Q$ for $A\in M_{n}(\mathbb{R)}$, then
by, e.g. [11, Lemma 2.9], $A^{2}=A$. Thus for $P\in M_{n}(\mathbb{R})$ we have

$P\leq^{-}I\quad\text{if and only if}\quad P^{2}=P.$

Let now $P\in$ $P_{n}(\mathbb{R})$, i.e., $P$ is a symmetric and idempotent
matrix. It follows that $P\leq^{-}I$ and therefore $\Phi(P)\leq^{-}\Phi(I)=I$.
So, $\Phi(P)$ is an idempotent matrix and by the definition of the map $\Phi$
also symmetric, i.e., $\Phi(P)\in$ $P_{n}(\mathbb{R})$. Since $\Phi^{-1}$ has
the same properties as $\Phi$, we may conclude that

$P\in P_{n}(\mathbb{R})\quad\text{if and only if}\quad\Phi(P)\in
P_{n}(\mathbb{R}),$

i.e., $\Phi$ preserves the set of all projectors. Recall that we may identify
subspaces of $\mathbb{R}^{n}$ with elements of $P_{n}(\mathbb{R})$. Let
$\mathcal{C}(\mathbb{R}^{n})$ be the lattice of all subspaces of
$\mathbb{R}^{n}$. It follows that the map $\Phi$ induces a lattice
automorphisms, i.e., a bijective map
$\tau:\mathcal{C}(\mathbb{R}^{n})\rightarrow\mathcal{C}(\mathbb{R}^{n})$ such
that

$M\subseteq N\quad\text{if and only if}\quad\tau(M)\subseteq\tau(N)$

for all $M,N\in\mathcal{C}(\mathbb{R}^{n})$. In [10, page 246] (see also [4,
pages 820 and 823] or [15, page 82]) Mackey proved that for $n\geq 3$ every
such a map is induced by an invertible linear operator, i.e., there exists an
invertible linear operator $T:$ $\mathbb{R}^{n}\rightarrow\mathbb{R}^{n}$ such
that $\tau(M)=T(M)$ for every $M\in$ $\mathcal{C}(\mathbb{R}^{n})$. For the
map $\Phi$ it follows that

(19) $\Phi(P)=P_{T(\mathop{\mathrm{I}m}P)}$

for every $P=P_{\mathop{\mathrm{I}m}P}\in P_{n}(\mathbb{R})$.

Step 6. There exist orthogonal matrices $U,V\in M_{n}(\mathbb{R)}$, such that
$U^{T}\Phi(V^{T}AV)U=A$ for every rank-one $A\in H_{n}^{+}(\mathbb{R})$. Note
that for $T$ from Step 5 there exists a singular value decomposition $T=UDV$,
with $U,V$ orthogonal and
$D=\mathrm{diag\,}(\lambda_{1},\lambda_{2},\ldots,\lambda_{n})$ being a
diagonal matrix with $\lambda_{1}=\max\lambda_{i}$ and
$\lambda_{2}=\min\lambda_{i}$. By replacing the map $\Phi$ with the map
$X\mapsto U^{T}\Phi(V^{T}XV)U$, $X\in M_{n}(\mathbb{R)}$, we can assume that
$\Phi(I)=I$ and $\Phi(x\otimes
x^{*}/\|x\|^{2})=(Dx)\otimes(Dx)^{*}/\|Dx\|^{2}$ for every nonzero
$x\in\mathbb{R}^{n}$. In particular, $\Phi(E_{ii})=E_{ii}$. By (19) and since
$T$ is now diagonal we have $\Phi(I_{2}\oplus 0_{n-2})=I_{2}\oplus 0_{n-2}$.

Consider a rank-two matrix $A\in H_{2}^{+}(\mathbb{R})\oplus 0_{n-2}$. Note
that the set $\Xi_{A}=\\{x\otimes x^{*}\leq^{-}A\\}$ is mapped by $\Phi$ onto
$\Xi_{\Phi(A)}$ since $\Phi$ is bi-monotone. By Lemma 2.2, the set $\Xi_{A}$
is in a bijective correspondence with the set of unordered pairs of antipodal
points on a concentric planar ellipse
$\mathfrak{E}_{A}=\left\\{x\in\mathop{\mathrm{I}m}A:x^{\ast}A^{\dagger}x=1\right\\}\subseteq\mathbb{R}^{2}\oplus
0_{n-2}$, see (3). By Step 3, $\Phi(A)$ is also a rank-two matrix. Hence
$\Phi$ induces a map, which we still denote by $\Phi$, that maps concentric
ellipses from $\mathbb{R}^{2}$ into concentric planar ellipses. By Lemma 3.1,
they all lie in the same plane, which is equal to
$\mathop{\mathrm{I}m}\Phi(I_{2}\oplus 0_{n-2})=\mathop{\mathrm{I}m}I_{2}\oplus
0_{n-2}=\mathbb{R}^{2}\oplus 0_{n-2}$. Therefore the induced $\Phi$
bijectively maps the points from $\mathbb{R}^{2}$ onto $\mathbb{R}^{2}$,
modulo identifying pairs of points symmetric to the origin, and maps
concentric ellipses onto concentric ellipses. Notice also that
$\mathfrak{E}_{I_{2}\oplus 0_{n-2}}$ is a unit circle by Lemma 2.2. Since
rank-one projectors which are below $I_{2}\oplus 0_{n-2}$ are identified with
(antipodal) points on the unit circle, we see that
$\Phi(\mathfrak{E}_{I_{2}\oplus 0_{n-2}})=\mathfrak{E}_{I_{2}\oplus 0_{n-2}}$.
Also, a point $(c,s)$ on the unit circle corresponds to a projector
$(c,s,0,\ldots,0)\otimes(c,s,0,\ldots,0)^{\ast}$ which is mapped to a
projector

$\displaystyle\Phi((c,s,0,\ldots,0)\otimes(c,s,0,\ldots,0)^{\ast})$
$\displaystyle=\frac{D(ce_{1}+se_{2})\otimes(D(ce_{1}+se_{2}))^{\ast}}{\|D(ce_{1}+se_{2})\|^{2}}$
$\displaystyle=\frac{(\lambda_{1}ce_{1}+\lambda_{2}se_{2})\otimes(\lambda_{1}ce_{1}+\lambda_{2}se_{2})^{\ast}}{\|\lambda_{1}ce_{1}+\lambda_{2}se_{2}\|^{2}}$

that corresponds to a point
$\frac{(\lambda_{1}c,\lambda_{2}s)}{\|(\lambda_{1}c,\lambda_{2}s)\|}$ also on
the unit circle. Hence induced $\Phi$ satisfies all the assumptions of Section
3 and therefore $\lambda_{1}=\lambda_{2}$ and thus $D$ is a scalar matrix. So,
$\Phi$ fixes all projectors. Moreover, by Section 3 the induced map $\Phi$ is
an identity on $\mathbb{R}^{2}$ so the map $\Phi$ fixes all rank-one matrices
which are below $I_{2}\oplus 0_{n-2}$. Now we consider an arbitrary rank-one
matrix $R\in H_{n}^{+}(\mathbb{R})$ and choose any rank-two matrix $B\in
H_{n}^{+}(\mathbb{R})$ that majorizes $R$ with respect to the minus partial
order. There exists an orthogonal matrix $W\in M_{n}(\mathbb{R})$ such that
$\mathop{\mathrm{I}m}(WBW^{T})=\mathbb{R}^{2}\oplus 0_{n-2}$. Note that
$\mathop{\mathrm{I}m}B=\mathop{\mathrm{I}m}P_{\mathop{\mathrm{I}m}{B}}$ where
$P_{\mathop{\mathrm{I}m}{B}}$ is the projector onto the image of $B$, so again
by Lemma 3.1,
$\mathop{\mathrm{I}m}\Phi(B)=\mathop{\mathrm{I}m}\Phi(P_{\mathop{\mathrm{I}m}{B}})$.
Since we know that $\Phi$ fixes all the projectors, we see that
$\mathop{\mathrm{I}m}\Phi(B)=\mathop{\mathrm{I}m}B$. By repeating the above
arguments on the map $X\mapsto W\Phi(W^{T}XW)W^{T}$ which also fixes all
projectors and maps $\mathbb{R}^{2}\oplus 0_{n-2}$ to $\mathbb{R}^{2}\oplus
0_{n-2}$, we see that $\Phi$ also fixes the rank-one matrix $R$.

Step 7. Conclusion of the proof.  Let $A\in H_{n}^{+}(\mathbb{R})$ be of rank-
one. By Step 6, $\Phi(A)=A$. It follows immediately by Remark 2.3 that then
$\Phi(A)=A$ for every $A\in H_{n}^{+}(\mathbb{R})$. Taking into account all
the assumptions we may conclude that $\Phi:H_{n}^{+}(\mathbb{R})\rightarrow
H_{n}^{+}(\mathbb{R})$ is a surjective bi-monotone map with respect to the
minus partial order $\leq^{-}$ if and only if there exists an invertible
matrix $S\in M_{n}(\mathbb{R)}$ such that

$\Phi(A)=SAS^{T}$

for every $A\in H_{n}^{+}(\mathbb{R})$.

## References

  * [1] R. Bhatia, Positive definite matrices, Princeton Series in Applied Mathematics, Princeton University Press, Princeton, New Jersey, 2007.
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