# A geometric approach to spectrum interlacing

Ricardo S. Leite Departamento de Matemática, UFES<EMAIL_ADDRESS>Carlos Tomei Departamento de Matemática, PUC-Rio<EMAIL_ADDRESS>
###### Abstract

We provide a detailed description of the maps associated with spectral
interlacing in two scenarios, for rank one perturbations and bordering of
symmetric and Hermitian matrices. The arguments rely on standard techniques of
nonlinear analysis.

Keywords: Spectral interlacing, degree theory.

MSC-class: 15A29, 15A42, 15B57

##  1 Introduction

We recall two standard results, presented in more detail in [6]. Endow
${\mathbb{C}}^{n}$ with the Euclidean inner product
$\langle\cdot,\cdot\rangle$ which is anti-linear in the second coordinate. For
$v\in{\mathbb{C}}^{n}$, the linear rank one map $v\otimes v=vv^{\ast}$ is
$u\mapsto\langle u,v\rangle\ v$. Let $S$ be an $n\times n$ Hermitian matrix,
with ordered eigenvalues $\lambda_{1}\leq\ldots\leq\lambda_{n}$.

######  Theorem 1.1 (Cor. 4.3.9, Th. 4.3.21 [6])

For $v\in{\mathbb{C}}^{n}$, let the eigenvalues of $T=T(v)=S+v\otimes v$ be
$\mu_{1}\leq\ldots\leq\mu_{n}$. Then the eigenvalues of $S$ and $T$ interlace,

$\lambda_{1}\leq\mu_{1}\leq\lambda_{2}\leq\mu_{2}\leq\ldots\leq\lambda_{n}\leq\mu_{n}\
.$

Conversely, for a sequence $\\{\mu_{j}\\}$ interlacing $\\{\lambda_{k}\\}$ as
above, there is $v\in{\mathbb{C}}^{n}$ for which the eigenvalues of
$T=S+v\otimes v$ are $\\{\mu_{j}\\}$.

The second result is Cauchy’s interlacing theorem.

######  Theorem 1.2 (Th. 4.3.17 [6])

Let $v\in{\mathbb{C}}^{n}$, $c\in{\mathbb{R}}$,
$\mu_{1}\leq\mu_{2}\leq\ldots\leq\mu_{n+1}$ be the eigenvalues of the bordered
matrix

$T=T(v,c)=\begin{pmatrix}S&v^{\ast}\\\ v&c\end{pmatrix}.$

Then the eigenvalues of $S$ and $T$ interlace,

$\mu_{1}\leq\lambda_{1}\leq\mu_{2}\leq\lambda_{2}\leq\mu_{3}\leq\ldots\leq\lambda_{n}\leq\mu_{n+1}\
.$

For a sequence $\\{\mu_{j}\\}$ interlacing $\\{\lambda_{k}\\}$, there exist
$v\in{\mathbb{C}}^{n}$ and $c>0$ for which the eigenvalues of $T=T(v,c)$ are
$\\{\mu_{j}\\}$.

The results can be found in essentially any advanced book on linear algebra.
Extensions abound. Ionascu [8] indicates a number of references (from which we
emphasize [18]) proving related results for compact self-adjoint operators $S$
on a separable Hilbert space. Simon [16] considers finer aspects of the
spectrum of rank one perturbations of (mostly) Schrödinger operators with very
interesting applications, which are beyond the scope of this text.

In this text, we cast these results in geometric terms. Fix a normalized
eigenbasis ${\cal{Q}}=[q_{1},\ldots,q_{n}]$ of $S$, arranged as columns of a
unitary matrix $Q$. Let ${\cal{O}}_{Q}\in{\mathbb{C}}^{n}$, the positive
orthant associated with ${\cal{Q}}$, be the set of vectors of the form $Qp$,
where $p\in{\mathbb{R}}^{n}$ has nonnegative entries.

A vector $v\in{\mathbb{R}}^{n}$ is ordered if $v_{1}\leq\ldots\leq v_{n}$.
Write $\sigma_{o}(S)\in{\mathbb{R}}^{n}$ for the ordered vector with entries
given by the eigenvalues of $S$. For $r>0$, define the polytopes

${\cal{P}}_{F}=[\lambda_{1},\lambda_{2}]\times[\lambda_{2},\lambda_{3}]\times\ldots\times[\lambda_{n},\infty)\
,$
${\cal{P}}_{G}=(-\infty,\lambda_{1}]\times[\lambda_{1},\lambda_{2}]\times[\lambda_{2},\lambda_{3}]\times\ldots\times[\lambda_{n},\infty)\
,$

two half-open boxes, and the sphere
${\mathcal{S}}(r)=\\{v\in{\mathbb{C}}^{n},\|v\|=r\\}$.

###### Theorem 1.3

Let $S$ be an $n\times n$ Hermitian matrix, with spectrum
$\lambda_{1}<\ldots<\lambda_{n}$, an eigenbasis ${\cal{Q}}$ and positive
orthant ${\cal{O}}_{Q}$. The following maps are homeomorphisms, and
diffeomorphisms between the interior of their domain and image.

$\displaystyle F:{\mathcal{D}}_{F}={\cal{O}}_{Q}$ $\displaystyle\to\
{\cal{P}}_{F}\hskip 31.2982pt,\hskip 22.76228ptG:{\mathcal{D}}_{G}\ =\
{\cal{O}}_{Q}\times{\mathbb{R}}\to\ {\cal{P}}_{G}$ $\displaystyle v\ \mapsto\
$ $\displaystyle\sigma_{o}(S+v\otimes v)\hskip 51.21504pt(v,c)\
\mapsto\sigma_{o}(T(v,c))$

and their restrictions

$F^{r}:{\mathcal{D}}_{F}^{r}={\cal{O}}_{Q}\cap{\mathcal{S}}(r)\to{\cal{P}}_{F}^{r}={\cal{P}}_{F}\cap\\{\mu\in{\mathbb{R}}^{n},\
\ \sum_{j}\mu_{j}\ =\ r^{2}+\sum_{j}\lambda_{j}\\}\ ,$ $\quad
G^{c}:{\cal{O}}_{Q}\times\\{c\\}\to{\cal{P}}_{G}^{c}={\cal{P}}_{G}\cap\\{\mu\in{\mathbb{R}}^{n+1},\
\sum_{j}\mu_{j}\ =\ \sum_{k}\lambda_{k}+c\\}\ ,$ $\quad
G^{r,c}:({\cal{O}}_{Q}\cap({\mathcal{S}}(r))\times\\{c\\}\to{\cal{P}}_{G}^{c}\cap\\{\mu\in{\mathbb{R}}^{n+1},\
\sum_{j}\mu_{j}^{2}\ =\ \sum_{k}\lambda_{k}^{2}+2r^{2}+c^{2}\\}.$

If $S$ has distinct eigenvalues, Theorems 1.1 and 1.2 then follow. Full
generality is attained by taking limits.

The general rank one Hermitian perturbation matrix is of the form $cv\otimes
v$ for a real unit vector $v$ and $c\in{\mathbb{R}}$. The sign of $c$
specifies if the perturbation of $S$ pushes the spectrum to the right (the
case $c>0$) or to the left ($c<0$). For the results above, $c\geq 0$: minor
alterations handle $c\leq 0$. Clearly, the interlacing property is associated
with the geometry of the polytopes ${\cal{P}}_{F}$ and ${\cal{P}}_{G}$.

It is rather intriguing that the interior of an orthant ${\cal{O}}_{Q}$ is
taken by $F$ to ${\cal{P}}_{F}$, a closed box with a face removed. As we shall
see in the proof, some faces of ${\cal{O}}_{Q}$ are creased by $F$, giving
rise to two faces of ${\cal{P}}_{Q}$. Something similar happens with $G$, but
now ${\cal{P}}_{G}$ is a box with two faces removed.

This is what happens for $n=2$. The horizontal axis is taken to the union of a
horizontal and a vertical segment. The vertical axis is sent to itself.

Figure 1: $F:{\mathcal{D}}\to{\mathbb{R}}^{2}$

The simple geometry of the maps $F$ and $G$ has implications to the
computation of their inverses, frequently described as an inverse problem
([6]). For $F$, given a symmetric matrix $S$ with ordered, simple, spectrum
$\lambda$ and an interlacing ordered $n$-tuple $\mu$, we look for a rank one
perturbation $cv\otimes v$ such that the spectrum of $S+cv\otimes v$ is $\mu$.
Theorem 1.3 shows that, in principle, the problem is solvable by numerical
continuation starting from any interior point of ${\mathcal{D}}_{F}$, as there
are no critical values there. The same argument proves that continuation from
an interior point of ${\mathcal{D}}_{G}$ obtains the inverse of $G$.

Given a function $f:X\to Y$, the preimages of $y\in Y$ are the points in the
set $f^{-1}(y)=\\{x\in X,\ f(x)=y\\}$. We now consider the preimages of the
maps in the previous theorem. We have to distinguish between matrices with
real or complex entries. Let ${\mathbb{K}}={\mathbb{R}}$ or ${\mathbb{C}}$ and
define

$\operatorname{abs}:{\mathbb{K}}^{n}\to{\cal{O}}_{Q},\quad
v=\sum_{j=1}^{n}c_{j}\ q_{j}\ \mapsto\ \sum_{j=1}^{n}|c_{j}|\ q_{j}\ .$

###### Theorem 1.4

Let $S$ as in the previous theorem.

  1. 1.

Say the entries of $S$ lie in ${\mathbb{K}}={\mathbb{R}}$ or ${\mathbb{C}}$,
$v\in{\mathbb{K}}^{n}$. Then $F$ and $G$ extend to

$\displaystyle\hat{F}:{\mathbb{K}}^{n}$ $\displaystyle\to\ {\cal{P}}_{F}\hskip
28.45274pt,\hskip 22.76228pt\hat{G}:{\mathbb{K}}^{n}\times{\mathbb{R}}\to\
{\cal{P}}_{G}\hskip 36.98866pt.$ $\displaystyle v\ \mapsto\ $
$\displaystyle\sigma_{o}(S+v\otimes v)\hskip 51.21504pt(v,c)\
\mapsto\sigma_{o}(T(v,c))$

Moreover,
$\hat{F}(v)=\hat{F}(w)\Leftrightarrow\hat{G}(v)=\hat{G}(w)\Leftrightarrow\operatorname{abs}(v)=\operatorname{abs}(w)$.
In particular, all preimages of a point belong to the same sphere
${\mathcal{S}}(r)$.

  2. 2.

If ${\mathbb{K}}={\mathbb{R}}$, a point $\mu\in{\cal{P}}_{F}$ belonging to
exactly $k$ faces has $2^{n-k}$ preimages under $\hat{F}$ or $\hat{G}$. If
${\mathbb{K}}={\mathbb{C}}$, for both functions the preimages of $\mu$ form a
product of $n-k-1$ circles: a torus.

Recently, Maciazek and Smilansky [14] considered analogous inverse problems
and pointed out the relevance of discrete information provided by strings of
signs. We believe our presentation sheds some light on the issue.

Theorems 1.3 and 1.4 are a strengthened version of a very special case of the
celebrated Horn’s conjecture [5], whose resolution, after work by several
authors ([10], [7], [12], [13]), is beautifully described in [11]. The
conjecture answers a question by Weyl [19]: what are the possible spectra of
the sum $A+B$ of two Hermitian matrices of given spectrum? Horn originally
provided a list of linear inequalities on the eigenvalues of the three
matrices which provide necessary and sufficient conditions relating their
spectra. For $A=S$, $B=v\otimes v$ with $\|v\|=r$, Horn’s conjecture states
that the image of the map $F^{r}$ is indeed ${\cal{P}}_{F}^{r}$.

Part of the statements in Theorems 1.3 and 1.4 may be deduced from the
sophisticated machinery of symplectic geometry. To give an idea of a more
familiar context, the Schur-Horn theorem for Hermitian matrices [4] is a
consequence of the powerful theorems about the convexity of the image of
moment maps of torus actions by Atiyah ([1]) and Guillemin-Sternberg ([3]).
The result for symmetric matrices then follows by an argument by Duistermaat
([2]). Similarly, the surjectivity of the maps $F,F^{r}$ and $G$ also follow
from convexity arguments, once the appropriate symplectic setting is
identified. Here, we take what Thompson ([17]) calls a low road in linear
algebra, but gain some information which does not follow directly from rote
application of these more general results.

The proof of Theorem 1.3 relies on a combination of well known facts of real
analysis, condensed in Lemma 2.1. The verification of the hypotheses of the
lemma is somewhat different for $F$ and $G$. In both cases, the theorem is
proved by induction on the dimension. In the inductive step, we see how faces
of the domain are ‘creased’ by either $F$ or $G$ so as to obtain the faces of
the image parallelotope. Theorem 1.4 is a simple consequence of Theorem 1.3.

The authors are supported by CNPq, CAPES, and FAPERJ. They are also grateful
to an anonymous reader of a previous version of this text, who indicated
errors and suggested a number of improvements.

##  2 A real analysis lemma

The outline of the proof of Theorem 1.3 is the same for the functions
$F:{\mathcal{D}}_{F}\to{\cal{P}}_{F}$ and
$G:{\mathcal{D}}_{G}\to{\cal{P}}_{g}$. In a nutshell, we must check the
hypotheses of the lemma below, which combines familiar arguments from real
analysis. We state it so as it applies directly to $F$. Let ${\mathcal{D}}$ be
${\cal{O}}_{I}$, the closed positive orthant of ${\mathbb{R}}^{n}$, and
${\cal{P}}$ be ${\cal{P}}_{F}$. Denote by $\operatorname{int}X$ the interior
of a set $X$.

###### Lemma 2.1

Let $\widetilde{H}:{\mathcal{D}}\to{\mathbb{R}}^{n}$ be a function satisfying
the following properties.

  1. (H1)

$\widetilde{H}$ is a continuous, proper map, i.e.,
$\lim_{\|v\|\to\infty}\|\widetilde{H}(v)\|=\infty$.

  2. (H2)

The restriction of $\widetilde{H}$ to $\operatorname{int}{\mathcal{D}}$ is a
$C^{1}$ map with invertible Jacobians.

  3. (H3)

Some point of $\operatorname{int}{\mathcal{D}}$ is taken by $\widetilde{H}$ to
$\operatorname{int}{\cal{P}}$.

  4. (H4)

The restriction $\hat{H}:\partial{\mathcal{D}}\to\partial{\cal{P}}$ is a
homeomorphism.

  5. (H5)

No point of $\operatorname{int}{\mathcal{D}}$ is taken by $\widetilde{H}$ to
$\partial{\cal{P}}$.

Then the image of $\widetilde{H}$ is ${\cal{P}}$ and the function
$\widetilde{H}:{\mathcal{D}}\to{\cal{P}}$ obtained from $\widetilde{H}$ by
restricting its counterdomain is a homeomorphism which restricts to a
diffeomorphism between $\operatorname{int}{\mathcal{D}}$ and
$\operatorname{int}{\cal{P}}$.

In order to apply the lemma for the function $G$, consider
$\widetilde{H}:{\mathcal{D}}\to{\mathbb{R}}^{n+1}$, where
${\mathcal{D}}={\cal{O}}_{I}\times{\mathbb{R}}$ and set
${\cal{P}}={\cal{P}}_{G}$. The proof follows verbatim.

###### Proof.

We first show that points in the connected components of
${\mathbb{R}}^{n}\setminus\widetilde{H}(\partial{\mathcal{D}})={\mathbb{R}}^{n}\setminus\partial{\cal{P}}$
have the same number of preimages. Take
$\mu\in{\mathbb{R}}^{n}\setminus\partial{\cal{P}}$. By connectivity, it
suffices to show that, for a small open neighborhood $U$ of $\mu$, points in
$U$ have the same number of preimages. If $\mu$ has infinite preimages, by
properness (hypothesis (H1)) they have to accumulate at some preimage
$v_{\ast}$. Preimages in $\operatorname{int}{\mathcal{D}}$ are isolated, by
the inverse function theorem (use hypothesis (H2)), thus
$v_{\ast}\in\partial{\mathcal{D}}$, contradicting
$\widetilde{H}(v_{\ast})=\mu\in{\mathbb{R}}^{n}\setminus\partial{\cal{P}}$.

Thus $\mu$ has a finite number of isolated preimages, say
$v_{1},\ldots,v_{k}$. From the inverse function theorem, for every
sufficiently small open ball $B$ centered around $\mu$, there are open
disjoint sets $V_{i},i=1,\ldots,k$, each containing $v_{i}$, for which
$\widetilde{H}$ takes $V_{i}$ to $B$ diffeomorphically. Thus, points in $B$
have at least $k$ preimages.

If, for balls $B_{n}$ of radius $1/n$ there are points $\tilde{\mu}_{n}$ with
(at least) $k+1$ preimages, one preimage $w_{n}\in{\mathcal{D}}$ is outside
$\cup_{i=1}^{k}V_{i}$. By properness, they accumulate at
$w_{\ast}\notin\cup_{i=1}^{k}V_{i}$. But then $\widetilde{H}(w_{\ast})=\mu$,
contradicting the fact that $\mu$ has exactly $k$ preimages, all in
$\cup_{i=1}^{k}V_{i}$.

By (H3), there is $v\in\operatorname{int}{\mathcal{D}}$ such that
$\widetilde{H}(v)\in\operatorname{int}{\cal{P}}$. From the argument above and
$(H4)$, the set ${\cal{P}}$ lies in the image of $\widetilde{H}$. If, for some
$w\in\operatorname{int}{\mathcal{D}}$ we have
$\widetilde{H}(w)\notin{\cal{P}}$, then the segment joining $v$ in $w$, which
lies in $\operatorname{int}{\mathcal{D}}$, must contain a point whose image
lies in $\partial{\cal{P}}$, contradicting (H5). Thus, the image of
$\widetilde{H}$ is ${\cal{P}}$ and the associated function
$H:{\mathcal{D}}\to{\cal{P}}$ is well defined.

From (H1), $\widetilde{H}:{\mathcal{D}}\to{\mathbb{R}}^{n}$ is proper and has
a well defined topological degree $\deg(\widetilde{H},\mu)$ (an excellent
reference for degree theory is [15]) for any regular value
$\mu\in\operatorname{int}{\mathbb{R}}^{n}\setminus\partial{\cal{P}}$, i.e., a
point whose preimages are regular points (regular values are dense, by Sard’s
theorem). From (H4), for $\mu\in\operatorname{int}{\cal{P}}$,
$\deg(\widetilde{H},\mu)=\pm 1$. Moreover,

$\deg(\widetilde{H},\mu)\ =\
\sum_{v\in\widetilde{H}^{-1}(\mu)}\operatorname{sgn}\det D\widetilde{H}(v)$

and all preimages are counted with the same sign, by (H5). Indeed, the
determinant of the Jacobian $DH(v)$ is never zero for
$v\in\operatorname{int}{\mathcal{D}}$ and is continuous, by (H1). Thus every
point of ${\cal{P}}$ has a unique preimage.  $\blacksquare$

$\blacksquare$

We are left with proving the hypotheses of the lemma for the counterparts
$\widetilde{F}:{\mathcal{D}}_{F}\to{\mathbb{R}}^{n}$ and
$\widetilde{G}:{\mathcal{D}}_{G}\to{\mathbb{R}}^{n+1}$ of the functions $F$
and $G$.

##  3 Proof of Theorem 1.3 for $F$ and $F^{r}$

Without loss, suppose $S=D$, a diagonal matrix with eigenvalues

$D_{11}=\lambda_{1}<\ldots<D_{nn}=\lambda_{n}\ .$

We then take ${\cal{Q}}=[e_{1},\ldots,e_{n}]$ to consist of the canonical
vectors, so that $Q=I$ and ${\cal{O}}_{I}\subset{\mathbb{R}}^{n}$ is the usual
positive orthant. Consider

$\displaystyle\tilde{F}:{\mathcal{D}}$
$\displaystyle={\mathcal{D}}_{F}={\cal{O}}_{I}\to\ {\mathbb{R}}^{n}\hskip
8.5359pt,$ $\displaystyle v\ $ $\displaystyle\mapsto\sigma_{o}(D+v\otimes v)$

where now all numbers in sight are real. Complex numbers will return only in
the proof of Theorem 1.4 in Section 5.

The set $\partial{\mathcal{D}}$ consists of $n$ faces of
${\mathcal{D}}={\cal{O}}_{I}$,

$E_{i}\ =\ \\{v\in{\mathbb{R}}^{n}\ ,\ v_{i}=0\\},\quad i=1,\ldots,n\ .$

The parallelotope ${\cal{P}}\subset{\mathbb{R}}^{n}$ has $2n-1$ faces, which
we now describe. Set

$L_{i}\ =\
[\lambda_{1},\lambda_{2}]\times\ldots\times[\lambda_{i-1},\lambda_{i}]\ ,\quad
R_{i}\ =\ [\lambda_{i},\lambda_{i+1}]\times\ldots\times[\lambda_{n},\infty)\
,$

where sets using indices not in $\\{1,\ldots,n\\}$ are omitted. For $i>1$, as
we shall see, $\tilde{F}$ (and $F$) creases each face $E_{i}$, sending it to
two adjoining faces of $\partial{\cal{P}}$,

$F(E_{i})\ =\ \big{(}L_{i}\times\\{\lambda_{i}\\}\times R_{i+1}\big{)}\ \cup\
\big{(}L_{i-1}\times\\{\lambda_{i}\\}\times R_{i}\big{)}\ .$

Face $E_{1}$ is sent to a single face of $\partial{\cal{P}}$,
$\\{\lambda_{1}\\}\times R_{2}$. The reader is invited to check that the
formulas indeed describe five of the six faces of a parallelotope in
${\mathbb{R}}^{3}$.

Recall that a simple eigenvalue of a symmetric matrix varies smoothly with the
matrix [9]: in this case, if $Tw_{i}\ =\ \lambda_{i}w_{i}$ for a normalized
$w_{i}\in{\mathbb{R}}^{n}$, $\dot{\lambda}_{i}\ =\
\langle\dot{T}w_{i},w_{i}\rangle$.

We define three subsets of ${\mathcal{D}}$.

  1. .

${\mathcal{D}}_{d}$ is the set of points in which $\tilde{F}(v)$ has a double
eigenvalue.

  2. .

The critical set $\
{\mathcal{C}}\subset\operatorname{int}({\mathcal{D}}\setminus{\mathcal{D}}_{d})$
consists of points in which the Jacobian $D\tilde{F}$ is not invertible.

  3. .

The set of regular points is the complement
${\mathcal{D}}\setminus(\partial{\mathcal{D}}\cup{\mathcal{D}}_{d}\cup{\mathcal{C}})$.

###### Proposition 3.1

  * (i)

${\mathcal{D}}_{d}\subset\partial{\mathcal{D}}$. Thus, $\tilde{F}$ is
differentiable in $\operatorname{int}{\mathcal{D}}$.

  * (ii)

${\mathcal{C}}\ =\ \emptyset$.

  * (iii)

$\partial\tilde{F}({\mathcal{D}})\subset\tilde{F}(\partial{\mathcal{D}})$.

  * (iv)

The matrices $D$ and $D+v\otimes v$ share an eigenvalue $\lambda_{i}$ if and
only if $v\in E_{i}$. In particular,
$\widetilde{F}^{-1}(\partial{\cal{P}})\subset\partial{\mathcal{D}}$.

###### Proof.

We prove (i). A double eigenvalue $\lambda_{i}$ of $D+v\otimes v$ admits a
(nonzero) eigenvector $w$ in the subspace of eigenvectors associated with
$\lambda_{i}$ for which $w_{1}=0$. In the expression $(D-\lambda_{i})w\ =\
-(v\otimes v)w\ =\ -\langle v,w\rangle v$, equate first coordinates: either
$v_{1}=0$ or $\langle v,w\rangle=0$. In the first case, $v\in
E_{1}\subset\partial{\mathcal{D}}$ and we are done. Otherwise,
$(D-\lambda_{i})w=0$ and $w$ is a canonical vector, $w=e_{j}$. As $\langle
v,w\rangle=0$, we must have $v_{j}=0$ and then $v\in
E_{j}\subset\partial{\mathcal{D}}$.

To prove (ii), let $T\ =\ D+v\otimes v$, $Tw_{i}=\lambda_{i}w_{i},\
i=1,\ldots,n,\ \|w_{i}\|\ =\ 1$. The Jacobian of $\tilde{F}$ at a point $v$ is

$J(v)\dot{v}\ =\
\big{(}\langle\dot{T}w_{1},w_{1}\rangle,\ldots,\langle\dot{T}w_{n},w_{n}\rangle\big{)},$

where $\dot{v}\in{\mathbb{R}}^{n}$ and $\dot{T}\ =\ \dot{v}\otimes
v+v\otimes\dot{v}$. Let $\dot{V}$ be the vector space of such matrices. Write
the linear transformation $J(v)$ as a composition,

$J(v)\dot{v}\ =\ 2\big{(}\langle w_{1},v\rangle\langle
w_{1},\dot{v}\rangle,\ldots,\langle w_{n},v\rangle\langle
w_{n},\dot{v}\rangle\big{)}$ $\ =\ 2\operatorname{diag}(\langle
w_{1},v\rangle,\ldots,\langle w_{n},v\rangle)\ (\langle
w_{1},\dot{v}\rangle,\ldots,\langle w_{n},\dot{v}\rangle)^{T}\ .$

A point $v$ in the interior of ${\mathcal{D}}$ is critical if and only if
$J(v)$ not invertible. Clearly $\dot{v}\mapsto(\langle
w_{1},\dot{v}\rangle,\ldots,\langle w_{n},\dot{v}\rangle)$ is invertible, as
the vectors $\\{w_{i}\\}$ are linearly independent. Suppose by contradiction
that, for some $i$, we have $\langle w_{i},v\rangle\ =\ 0$. Equation
$(D+v\otimes v)w_{i}=\lambda_{i}w_{i}$ becomes $(D-\lambda_{i})w_{i}\ =\ 0$,
so that $w_{i}\ =\ te_{i},t\neq 0$. Now, $\langle v,w_{i}\rangle=0$ implies
$v_{i}=0$, and again $v\in E_{i}\subset\partial{\mathcal{D}}$.

By the inverse function theorem, $\tilde{F}$ is a local diffeomorphism at
regular points: this settles (iii).

To prove (iv), take a common eigenvalue $\lambda_{i}$ and eigenvectors
$e_{i},y\neq 0$, so that $De_{i}=\lambda_{i}e_{i}$ and $Dy+\langle v,y\rangle
v=\lambda_{i}y$ and $(D-\lambda_{i})(y-e_{i})\ =\ -\langle v,y\rangle v$. The
$i$-th entry of both sides of the last equation is zero. If $v_{i}\ =\ 0$, we
are done. Suppose $\langle v,y\rangle=0$: $y=te_{i}\in E_{i},t\neq 0$, and
then $v_{i}=0$. The converse is trivial.  $\blacksquare$

$\blacksquare$

For $v\in E_{i}$, $\lambda_{i}$ is in the spectrum of both $D$ and
$T=D+v\otimes v$, but we do not know yet where $\lambda_{i}$ sits among the
ordered eigenvalues of $T$. As we shall see, the study of
$\partial\tilde{F}(E_{i})$ requires the understanding of the map
$\tilde{F}_{\hat{D}}$ for the $(n-1)\times(n-1)$ matrix $\hat{D}$, obtained
from $D$ by removing the eigenvalue $\lambda_{i}$. Said differently, the proof
that $\tilde{F}$ takes $\partial{\mathcal{D}}$ to $\partial{\cal{P}}$
homeomorphically is by induction.

Statements (i) and (ii) above imply hypothesis (H2) of Lemma 2.1.

We verify hypotheses (H1) and (H3).

###### Proposition 3.2

Hypothesis (H1) of Lemma 2.1 holds: the map $\tilde{F}$ is proper.

###### Proof.

With the Frobenius norm, for a matrix $T=D+v\otimes v$ with eigenvalues
$\\{\mu_{i}\\}$, we have $\|D+v\otimes v\|^{2}\ =\ \sum_{i}\mu_{i}^{2}$.
$\blacksquare$

$\blacksquare$

###### Proposition 3.3

Hypothesis (H3) also holds: for some $v\in\operatorname{int}{\mathcal{D}}$,
$\widetilde{F}(v)\in\operatorname{int}{\cal{P}}$.

###### Proof.

Let ${\bf 1}=(1,1,\ldots,1)$, $t>0$. For eigenvalues $\lambda_{i}$ of
$D+tv\otimes v$, we have $\dot{\lambda}_{i}(t=0)=\langle{\bf 1}\otimes{\bf 1}\
e_{i},e_{i}\rangle=1>0$: $\sigma_{o}(D+t\ {\bf 1}\otimes{\bf 1})$ enters
${\cal{P}}$ for $t>0$ small.  $\blacksquare$

$\blacksquare$

Hypotheses (H4) and (H5) require an inductive argument, presented below.

Proof of Theorem 1.3 for $F$ and $F^{r}$: We first prove by induction the
claim about $F$, and then we handle $F^{r}$. The case $n\ =\ 2$ contains the
gist of the proof. For $v\in E_{1}$, $v=(0,c)$, so that

$T\ =\ D+v\otimes v\ =\ \begin{pmatrix}\lambda_{1}&0\\\
0&\lambda_{2}+c^{2}\end{pmatrix}\ .$

As $\lambda_{1}<\lambda_{2}$, we also have $\lambda_{1}<\lambda_{2}+c^{2}$, so
that

$\tilde{F}(T)=(\lambda_{1},\lambda_{2}+c^{2})\in\\{\lambda_{1}\\}\times[\lambda_{2},\infty)\in\partial{\cal{P}}\
.$

If instead $v\in E_{2}$, $v=(c,0)$ and

$T\ =\ D+v\otimes v\ =\ \begin{pmatrix}\lambda_{1}+c^{2}&0\\\
0&\lambda_{2}\end{pmatrix}\ .$

There are two possibilities. If $\lambda_{1}+c^{2}\leq\lambda_{2}$, in
accordance with Figure 1,

$\tilde{F}(T)\ =\
(\lambda_{1}+c^{2},\lambda_{2})\in[\lambda_{1},\lambda_{2}]\times\\{\lambda_{2}\\}\in\partial{\cal{P}}\
.$

Otherwise

$\tilde{F}(T)\ =\
(\lambda_{2},\lambda_{1}+c^{2})\in\\{\lambda_{2}\\}\times[\lambda_{2},\infty)\in\partial{\cal{P}}\
.$

As $\tilde{F}:{\mathcal{D}}\to{\mathbb{R}}^{n}$ is proper, the restriction
$F:\partial{\mathcal{D}}\to\partial{\cal{P}}$ is a homeomorphism: $F$
satisfies hypothesis (H4) of Lemma 2.1. From Lemma 3.1(iv), (H5) also holds.
The first step of the induction argument is complete.

We assume the claim for $\widetilde{F}$ acting on $(n-1)\times(n-1)$ matrices.
For a diagonal $n\times n$ matrix $D$, we consider
$\tilde{F}(\partial{\mathcal{D}})$. For $v\in E_{i}$ in a face of the orthant
${\mathcal{D}}$, the $i$-th column and row of the matrix $D+v\otimes v$ equal
$\lambda_{i}e_{i}^{T}$ and $\lambda_{i}e_{i}$, so that $\lambda_{i}$ is a
common eigenvalue of $D$ and $D+v\otimes v$. The remaining eigenvalues of
$D+v\otimes v$ belong to the spectrum of $\hat{D}+\hat{v}\otimes\hat{v}$,
where $\hat{D}$ is obtained by removing the $i$-th row and column of $D$ and
$\hat{v}$ is obtained from removing the $i$-th entry of $v$.

In order to apply the inductive hypothesis, at the risk of being pedantic,
identify $E_{i}$ with the positive orthant
$\hat{\mathcal{D}}\subset{\mathbb{R}}^{n-1}$, so that
$\tilde{F}:E_{i}\to{\mathbb{R}}^{n-1}$ is identified with
$\tilde{F}_{\hat{D}}:\hat{\mathcal{D}}\to{\mathbb{R}}^{n-1}$ which, by
induction, induces a homeomorphism
$F_{\hat{D}}:\hat{\mathcal{D}}\to\hat{\cal{P}}$, where, for $i>1$,

$\hat{\cal{P}}\ =\
[\lambda_{1},\lambda_{2}]\times\ldots\times[\lambda_{i-2},\lambda_{i-1}]\times[\lambda_{i-1},\lambda_{i+1}]\times[\lambda_{i+1},\lambda_{i+2}]\times\ldots\times[\lambda_{n},\infty).$

Notice that the two intervals containing $\lambda_{i}$ in the definition of
${\cal{P}}$ were replaced by a single interval
$[\lambda_{i-1},\lambda_{i+1}]$. Split
$[\lambda_{i-1},\lambda_{i+1}]=[\lambda_{i-1},\lambda_{i}]\cup[\lambda_{i},\lambda_{i+1}]$,
and then

$\hat{\cal{P}}\ =\
\big{(}[\lambda_{1},\lambda_{2}]\times\ldots\times[\lambda_{i-2},\lambda_{i-1}]\times[\lambda_{i-1},\lambda_{i}]\times[\lambda_{i+1},\lambda_{i+2}]\times\ldots\times[\lambda_{n},\infty)\big{)}$
$\cup\
\big{(}[\lambda_{1},\lambda_{2}]\times\ldots\times[\lambda_{i-2},\lambda_{i-1}]\times[\lambda_{i},\lambda_{i+1}]\times[\lambda_{i+1},\lambda_{i+2}]\times\ldots\times[\lambda_{n},\infty)\big{)}.$

In order to compute $\tilde{F}(E_{i})$, we insert $\lambda_{i}$ among the
ordered eigenvalues in $\hat{\cal{P}}$.

$F(E_{i})\ =\
\big{(}[\lambda_{1},\lambda_{2}]\times\ldots\times[\lambda_{i-2},\lambda_{i-1}]\times[\lambda_{i-1},\lambda_{i}]\times\\{\lambda_{i}\\}\times[\lambda_{i+1},\lambda_{i+2}]\times\ldots\times[\lambda_{n},\infty)\big{)}$
$\cup\
\big{(}[\lambda_{1},\lambda_{2}]\times\ldots\times[\lambda_{i-2},\lambda_{i-1}]\times\\{\lambda_{i}\\}\times[\lambda_{i},\lambda_{i+1}]\times[\lambda_{i+1},\lambda_{i+2}]\times\ldots\times[\lambda_{n},\infty)\big{)}$
$\ =\ \big{(}L_{i}\times\\{\lambda_{i}\\}\times R_{i+1}\big{)}\ \cup\
\big{(}L_{i-1}\times\\{\lambda_{i}\\}\times R_{i}\big{)}$

in the notation introduced in the beginning of the section. Thus $F$ indeed
creases faces $E_{i}$, $i>1$, giving rise to two faces of $\partial{\cal{P}}$.
Moreover, $F$ is a homeomorphism between the remaining faces
$E_{1}\subset\partial{\mathcal{D}}$ and $\big{(}\\{\lambda_{1}\\}\times
R_{2}\big{)}\subset\partial{\cal{P}}$: the details are left to the reader
(simply omit intervals containing the index $i-1$).

Thus, $F:\partial{\mathcal{D}}\to\partial{\cal{P}}$ is surjective, and
injective on the restriction to each face $E_{i}\times{\mathbb{R}}$. We are
left with showing injectivity on the union of the faces. Let $v_{i}\in E_{i}$
and $v\in{\mathcal{D}}$ such that $\tilde{F}(v_{i})=\tilde{F}(v)$. As
$\lambda_{i}$ is an eigenvalue of $\tilde{F}(v_{i})$, by Proposition 3.1(iv)
we must have $v\in E_{i}$. As the restriction of $\tilde{F}$ to $E_{i}$ is
injective, global injectivity in $\partial{\mathcal{D}}$ follows. A simple
argument then shows that $F:\partial{\mathcal{D}}\to\partial{\cal{P}}$ is a
homeomorphism, so that hypothesis (H4) of Lemma 2.1 holds. From Lemma 3.1
(iv), (H5) also holds. Item (1) now follows from Lemma 2.1.

We now consider $F^{r}$. Again, $S=D$. Since

$\operatorname{tr}(S+v\otimes v)\ =\ \operatorname{tr}D+\langle
v,v\rangle^{2}$

and $F:{\mathcal{D}}\to{\cal{P}}$ is a homeomorphism, we have that $F^{r}$ is
also a homeomorphism. When restricting to the interior of ${\mathcal{D}}$,
$F^{r}$ takes one hypersurface to another and the Jacobian at each point is
easily seen to be invertible, showing that $F^{r}$ is indeed a diffeomorphism
between interiors.  $\blacksquare$

##  4 Proof of Theorem 1.3 for $G$, $G^{c}$ and $G^{r,c}$

Again, without loss, $S\ =\ D$, a diagonal matrix with eigenvalues
$\lambda_{1}<\ldots<\lambda_{n}$. Now ${\mathcal{D}}\ =\
{\cal{O}}_{I}\times{\mathbb{R}}$ has faces of the form
$E_{i}\times{\mathbb{R}}$ and the box

${\cal{P}}={\cal{P}}_{G}=(-\infty,\lambda_{1}]\times[\lambda_{1},\lambda_{2}]\times[\lambda_{2},\lambda_{3}]\times\ldots\times[\lambda_{n},\infty)\subset{\mathbb{R}}^{n+1}$

has $2n$ faces. Recall that all numbers in sight are real. Define

$T=T(v,c)\ =\ \begin{pmatrix}D&v\\\ v^{\ast}&c\end{pmatrix}\ .$

We must show that the map $\tilde{G}:{\mathcal{D}}\to{\mathbb{R}}^{n+1},\
(v,c)\mapsto\sigma_{o}(T(v,c))=\sigma_{o}(T)$ defines a homeomorphism
$G:{\mathcal{D}}\to\partial{\cal{P}}$. This time, as we shall see, $G$ takes
every face of $\partial{\mathcal{D}}$ to two adjoining faces of
$\partial{\cal{P}}$.

As before ${\mathcal{D}}_{d}$ consists of the points $(v,c)\in{\mathcal{D}}$
for which $T(v,c)$ has a double eigenvalue, the critical set ${\mathcal{C}}$
is the set of points in the interior of
${\mathcal{D}}\setminus{\mathcal{D}}_{d}$ in which $G$ is differentiable with
not invertible Jacobian, and its complement in ${\mathcal{D}}$ is the set of
regular points. The counterpart of Proposition 3.1 still holds.

###### Proposition 4.1

(i) ${\mathcal{D}}_{d}\subset\partial{\mathcal{D}}$, so that $\widetilde{G}$
is differentiable in $\operatorname{int}{\mathcal{D}}$, (ii) ${\mathcal{C}}\
=\ \emptyset$, (iii)
$\partial\tilde{G}({\mathcal{D}})\subset\tilde{G}(\partial{\mathcal{D}})$,
(iv) The matrices $D$ and $T(v,c)$ share an eigenvalue $\lambda_{i}$ if and
only if $v\in E_{i}$ and $(v,c)\in\partial{\mathcal{D}}$. Thus
$\tilde{G}^{-1}(\partial{\mathcal{D}})\subset\partial{\mathcal{D}}$.

###### Proof.

For (i), take a double eigenvalue $\rho$ and an associated eigenvector
$w\in{\mathbb{R}}^{n+1}$ with $w_{n+1}\ =\ 0$. Expanding $(T(v,c)-\rho)w=0$ we
have that $\rho=\lambda_{i}$ for some $i=1,\ldots,n$, is an eigenvalue of $D$
and $w=te_{i},t\neq 0$. But then $(T(v,c)-\lambda_{i})e_{i}\ =\ 0$ implies
that $v_{i}=0$, so that $v\in E_{i}$. For (ii), imitate the argument in the
previous section:

$DG(v,c)(\dot{v},\dot{c})\ =\ \big{(}\langle
w_{1},\dot{T}w_{1}\rangle,\ldots,\langle
w_{n+1},\dot{T}w_{n+1}\rangle\big{)},$

where $T(v,c)\ w_{j}=\lambda_{j}\ w_{j},\ j=1,\ldots,n+1,\ \|w_{j}\|\ =\ 1$
and

$\dot{T}\ =\ {\dot{T}}(v,c)(\dot{v},\dot{c})\ =\ \begin{pmatrix}0&\dot{v}\\\
\dot{v}^{\ast}&\dot{c}\end{pmatrix}\ .$

Let $\dot{V}\simeq{\mathbb{R}}^{n+1}$ be the vector space spanned by the
matrices $\dot{T}$.

Using Frobenius inner products, the Jacobian $DG:\dot{V}\to{\mathbb{R}}^{n+1}$
becomes

$DG(v,c)(\dot{v},\dot{c})\ =\ \big{(}\operatorname{tr}(w_{1}\otimes
w_{1})\dot{T},\ldots,\operatorname{tr}(w_{n+1}\otimes w_{n+1})\dot{T}\big{)}$
$\ =\ \big{(}\langle\dot{T},w_{1}\otimes
w_{1}\rangle,\ldots,\langle\dot{T},w_{n+1}\otimes w_{n+1}\rangle\big{)}.$

Thus, a point $(v,c)\in\operatorname{int}{\mathcal{D}}$ is critical if and
only if every linear combination of the eigenprojections $w_{k}\otimes w_{k}$
is orthogonal to some nonzero matrix $\dot{T}\in\dot{V}$. As $T$ has simple
spectrum (by (i)), such linear combination is a polynomial in $T$,

$\sum_{j=1}^{n+1}c_{j}\ w_{j}\otimes w_{j}\ =\ \sum_{k=0}^{n}d_{k}\ T^{k}\ .$

Thus there is $\dot{T}$ orthogonal to all polynomial functions $p(T)$. The
inner product of $\dot{T}\in\dot{V}$ with an arbitrary real symmetric matrix
$M$ is simply

$\langle\dot{T},\begin{pmatrix}*&y\\\ y^{\ast}&x\end{pmatrix}\rangle\ =\
\langle\begin{pmatrix}0&\dot{v}\\\
\dot{v}^{\ast}&\dot{c}\end{pmatrix},\begin{pmatrix}*&y\\\
y^{\ast}&x\end{pmatrix}\rangle\ =\ \langle(2\dot{v},\dot{c}),(y,x)\rangle\ =\
\langle(2\dot{v},\dot{c}),Me_{n+1}\rangle\ ,$

where $e_{n+1}\ =\ (0,\ldots,1)\in{\mathbb{R}}^{n+1}$ is canonical. Thus, a
point $(v,c)$ corresponding to a matrix $T\ =\ T(v,c)$ is critical if and only
if there is a matrix $\dot{T}$ associated with a nonzero $(\dot{v},\dot{c})$
such that $e_{n+1},Te_{n+1},\ldots,T^{n}e_{n+1}$ are orthogonal to
$(\dot{v},\dot{c})$.

Thus $T$ is critical if and only if the vectors
$e_{n+1},Te_{n+1},\ldots,T^{n}e_{n+1}$ are linearly dependent, i.e., $e_{n+1}$
is not a cyclic vector of $T$. Diagonalize $T\ =\ Q_{T}^{T}D_{T}Q_{T}$, where
the rows of the orthogonal matrix $Q_{T}$ are the eigenvectors of $T$ and
$D_{T}$ has simple spectrum, from (i). The vectors
$e_{n+1},Te_{n+1},\ldots,T^{n}e_{n+1}$ are linearly dependent if and only if
the vectors $Q_{T}e_{n+1},D_{T}Q_{T}e_{n+1},\ldots,D_{T}^{n}Q_{T}e_{n+1}$ are.
Let $M$ be the matrix having such vectors as columns and define

$q=Q_{T}e_{n+1}\ ,\quad D_{T}=\operatorname{diag}(d_{1},d_{2},\ldots,d_{n+1})\
.$

Then

$M=\begin{pmatrix}q_{1}&0&0&\ldots&0\\\ 0&q_{2}&0&\ldots&0\\\ &&\ldots&&\\\
0&0&0&\ldots&q_{n+1}\\\
\end{pmatrix}\begin{pmatrix}1&d_{1}&d_{1}^{2}&\ldots&d_{1}^{n}\\\
1&d_{2}&d_{2}^{2}&\ldots&d_{2}^{n}\\\ &&\ldots&&\\\
1&d_{n+1}&d_{n+1}^{2}&\ldots&d_{n+1}^{n+1}\\\ \end{pmatrix}$

Since the $d_{i}$’s are distinct, the (Vandermonde) determinant of the matrix
on the right is nonzero and $\det M$ is zero if and only if some coordinate
$q_{i}$ of $Q^{T}e_{n+1}$ is. Said differently, the last coordinate of some
eigenvector of $T$ is zero. Say $w\ =\ (\tilde{w},0)$ satisfies
$(T-\lambda_{i})w=0$. Then $(D-\lambda_{i})\tilde{w}=0$, so that $\lambda_{i}$
is also an eigenvalue of $D$. Since $D$ has simple spectrum, we must have
$\tilde{w}=\alpha e_{k}$ for some $\alpha\neq 0$ and
$e_{k}\in{\mathbb{R}}^{n}$ a canonical vector, Thus, without loss, $w\ =\
(e_{k},0)$. Equating the $(n+1)$-th entry $(n+1)$ of $(T-\lambda_{i})w=0$, we
obtain $T_{k,n+1}=T(v,c)_{k,n+1}=v_{k}\ =\ 0$. Thus $v_{k}\in E_{k}$ and
$(v,c)\in\partial{\mathcal{D}}$. The proof of (ii) is complete.

Item (iii) again follows from (ii) and the inverse function theorem.

To prove (iv), simply expand $\det(T(v,c)-\lambda_{i}I)$ along row $i$.
$\blacksquare$

$\blacksquare$

Again, statements (i) and (ii) above imply hypothesis (H2) of Lemma 2.1.
Hypothesis (H1) is proved mimicking Proposition 3.2, but (H3) is more
delicate.

###### Proposition 4.2

(H3) holds for $G$: for some $v\in\operatorname{int}{\mathcal{D}}$,
$G(v)\in\operatorname{int}{\cal{P}}$.

###### Proof.

Fix $\lambda_{n+1}=c>\lambda_{n}$ and consider ${\bf
1}=(1,1,\ldots,1)\in{\mathbb{R}}^{n}$, $t>0$,

$T=T(t)\ =\ \begin{pmatrix}D&0\\\
0&\lambda_{n+1}\end{pmatrix}+t\begin{pmatrix}0&{\bf 1}\\\ {\bf
1}^{\ast}&0\end{pmatrix}\ .$

For eigenvalues $\lambda_{j}(t)$, $j=1,\ldots,n+1$, of $T(t)$,

$\dot{\lambda}_{j}(t=0)=\langle\begin{pmatrix}0&{\bf 1}\\\ {\bf
1}^{\ast}&0\end{pmatrix}\ e_{j},e_{j}\rangle=0$

and we must compute second derivatives. Define normalized eigenvectors
$w_{j}(t)$ such that $T(t)w_{j}(t)=\lambda_{j}(t)w_{j}(t)$, $w_{j}(0)=e_{j}$.
Then ([9])

$\dot{\lambda}_{j}(t)=\langle\begin{pmatrix}0&{\bf 1}\\\ {\bf
1}^{\ast}&0\end{pmatrix}\ w_{j}(t),w_{j}(t)\rangle\ ,\quad\dot{w}_{j}(t)\
=-(T(t)-\lambda_{j}(t))^{-1}(\dot{T}(t)-\dot{\lambda}_{j}(t))w_{j}(t)\ ,$

where the map being inverted is the restriction
$T(t)-\lambda_{j}(t):\\{w_{i}\\}^{\perp}\to\\{w_{j}\\}^{\perp}$. For $t=0$, as
$\dot{\lambda}_{j}(0)=0$ for $j=1,\ldots,n+1$,

$\dot{w}_{j}(0)\ =-\begin{pmatrix}D-\lambda_{j}&0\\\
0&\lambda_{n+1}-\lambda_{j}\end{pmatrix}^{-1}\begin{pmatrix}0&{\bf 1}\\\ {\bf
1}^{\ast}&0\end{pmatrix}e_{j}$

so that, for $j=1,\ldots,n$ and $j=n+1$ we have, respectively,

$\dot{w}_{j}(0)=-\frac{1}{\lambda_{n+1}-\lambda_{j}}\ e_{n+1}\ \ \hbox{or}\ \
\dot{w}_{n+1}(0)=-\begin{pmatrix}(D-\lambda_{n+1})^{-1}{\bf 1}\\\
0\end{pmatrix}\ .$

We are ready to compute the second derivative of $\lambda_{j}$ at $t=0$,

$\ddot{\lambda}_{i}(0)\ =\ 2\langle\begin{pmatrix}0&{\bf 1}\\\ {\bf
1}^{\ast}&0\end{pmatrix}\ \dot{w}_{i}(0),w_{j}(0)\rangle\ .$

For $j=1,\ldots,n$, as $\lambda_{n+1}>\lambda_{i}$,

$\ddot{\lambda}_{j}(0)\ =\ -2\langle\begin{pmatrix}0&{\bf 1}\\\ {\bf
1}^{\ast}&0\end{pmatrix}\ \frac{1}{\lambda_{n+1}-\lambda_{j}}\
e_{n+1},e_{j}\rangle\ =\ \frac{-2}{\lambda_{n+1}-\lambda_{i}}<0\ .$

For $j=n+1$,

$\ddot{\lambda}_{n+1}(0)=-2\ \langle\begin{pmatrix}0&{\bf 1}\\\ {\bf
1}^{\ast}&0\end{pmatrix}\ \begin{pmatrix}(D-\lambda_{n+1})^{-1}{\bf 1}\\\
0\end{pmatrix},e_{n+1}\rangle=(-2)\langle{\bf 1},(D-\lambda_{n+1})^{-1}{\bf
1}\rangle>0$

and $\sigma_{o}(T(t))$ indeed belongs to $\operatorname{int}{\cal{P}}$ for
small $t>0$.  $\blacksquare$

$\blacksquare$

Proof of Theorem 1.3 for $G$, $G^{c}$ and $G^{r,c}$: We first use induction to
prove the claim for $G$. For $n\ =\ 1$, ${\mathcal{D}}\ =\
[0,\infty)\times{\mathbb{R}}$, so that
$\partial{\mathcal{D}}=\\{(0,c),c\in{\mathbb{R}}\\}$. The eigenvalues of
$T(0,c)$ are $\\{\lambda_{1},c\\}$ and must be ordered. If $c<\lambda_{1}$
then

$G(0,c)=(c,\lambda_{1})\in(-\infty,\lambda_{1}]\times\\{\lambda_{1}\\}\ .$

If $c>\lambda_{1}$ then
$G(0,c)=(\lambda_{1},c)\in\\{\lambda_{1}\\}\times[\lambda_{1},\infty)$. If $c\
=\ \lambda_{1}$, $G(0,c)$ lies in the common subface, a single point of double
spectrum associated with the diagonal matrix $\lambda_{1}I$. Again, it is the
ordering which creases $\partial{\mathcal{D}}$, a straight line, so as to
cover both faces of ${\cal{P}}$. The first inductive step is complete.

Take a diagonal $n\times n$ matrix $D$: we consider
$G(\partial{\mathcal{D}})$. The $i$-th face of $\partial{\mathcal{D}}$ is
$E_{i}\times{\mathbb{R}}$. Let

$L_{i}\ =\ (-\infty,\lambda_{1}]\times\ldots\times[\lambda_{i-1},\lambda_{i}]\
,\ R_{i}\ =\
[\lambda_{i},\lambda_{i+1}]\times\ldots\times[\lambda_{n},\infty)\ ,$

where again sets with indices not in $\\{1,\ldots,n\\}$ are omitted in
formulas. Following the argument in the previous section, for each face
$E_{i}\times{\mathbb{R}},\ i=1,\ldots,n$,

$G(E_{i})\ =\ \big{(}L_{i}\times\\{\lambda_{i}\\}\times R_{i+1}\big{)}\ \cup\
\big{(}L_{i-1}\times\\{\lambda_{i}\\}\times R_{i}\big{)}\ .$

Thus, $G:\partial{\mathcal{D}}\to\partial{\cal{P}}$ is surjective, and
restricts injectively to each face $E_{i}\times{\mathbb{R}}$. We are left with
showing injectivity on the union of the faces.

For $(v,c)\in E_{i}\times{\mathbb{R}}$ set $M=T(v,c)$ and consider another
bordered matrix $N=T(w,d)$ with the same ordered spectrum. As $\lambda_{i}$
belongs to $\lambda(M)$, we must have $\lambda_{i}\in\lambda(N)$ and, from the
previous lemma, $w_{i}=0$: $(w,v)\in E_{i}\times{\mathbb{R}}$. Global
injectivity now follows from injectivity of $G$ restricted to
$E_{i}\times{\mathbb{R}}$.

The claim about $G^{c}$ follows the proof of the claim about $F^{r}$: compare
traces. For $G^{r,c}$, compare traces of squares,

$\sum_{j}\mu_{j}^{2}=\operatorname{tr}T(v,c)^{2}=\operatorname{tr}D^{2}+2\langle
v,v\rangle+c^{2}=\sum_{k}\lambda_{k}^{2}+2r^{2}+c^{2}$

and imitate the rest of the argument relating two surfaces of codimension 2.
$\blacksquare$

##  5 Proof of Theorem 1.4

We consider $\hat{F}:{\mathbb{K}}^{n}\to{\cal{P}}_{F}$, the other cases being
analogous.

From Theorem 1.3, as $F$ is injective,
$F(\operatorname{abs}(v))=F(\operatorname{abs}(w))$ if and only if $v=w$. From
the surjectivity of $F$, given $\mu\in{\cal{P}}_{F}$, there is a (unique)
$v\in{\cal{O}}_{Q}$ for which $F(v)=\mu$. Hence,
$\hat{F}^{-1}(\mu)=\operatorname{abs}^{-1}\circ
F^{-1}(\mu)=\operatorname{abs}^{-1}(v)$.

Each nonzero coordinate $v_{k}$ of $v$ gives rise to a circle
$e^{i\theta_{k}}v_{k}$ of possible values for the $k$-th coordinate of
$\operatorname{abs}^{-1}(v)$. Clearly $z_{k}=0$ if and only if $z\in E_{k}$.
$\blacksquare$

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