text:
\begin{document}

\title{Local Lidskii's theorems for unitarily invariant norms}
\author{ Pedro G. Massey $^{*}$, Noelia B. Rios $^{*}$ and Demetrio Stojanoff
\footnote{Partially supported by CONICET
(PIP 0150/14), FONCyT (PICT 1506/15) and  FCE-UNLP (11X681), Argentina. } \
 \footnote{ e-mail addresses: massey@mate.unlp.edu.ar, nbrios@mate.unlp.edu.ar, demetrio@mate.unlp.edu.ar}
\\ {\small Depto. de Matem\'atica, FCE-UNLP
and IAM-CONICET, Argentina  }}

\date{}
\maketitle
\begin{abstract}
Lidskii's additive inequalities (both for eigenvalues and singular values) can be interpreted
as an explicit description of global minimizers of functions that are built on unitarily invariant norms, with
domains consisting of certain orbits of matrices (under the action of the unitary group). In this paper, we show
that Lidskii's inequalities actually describe all global minimizers of such functions and that local minimizers
are also global minimizers. We use these results to obtain partial results related
to local minimizers of generalized frame operator distances in the context of finite frame theory.
\end{abstract}

\noindent  AMS subject classification: 42C15, 15A60.

\noindent Keywords: Lidskii's inequality, unitarily invariant norms, majorization,
frame operator distance.

\tableofcontents

\section{Introduction}

Lidskii's additive inequalities \cite{Lidskii}  are ubiquitous in matrix analysis. They are part of
the fundamental toolkit to deal with some of the most natural problems in this theory, such as matrix approximation problems
(matrix nearness problems) and singular values/eigenvalues inequalities (see \cite{Bhat,HJ1,HJ2} and the references therein).
Lidskii's inequalities are expressed in terms of an important pre-order between
real vectors called majorization. Since majorization is intimately related to tracial inequalities
involving convex functions, Lidskii's inequalities can be used to describe the structure of
matrices that are optimal with respect to families of entropic-like functionals (see \cite{dnp,mr2010,mrs2,mrs3}).
Lidskii's inequalities also provide some simple relations between the spectra of the sum of selfadjoint matrices and
its summands, related to the solution of Horn's conjecture \cite{A.Horn} on the spectra of the sum of selfadjoint matrices, based on the work
of A. Klyachko \cite{Klyach} and A. Knutson and T. Tao \cite{KT} (see \cite{Fulton} for a historical account and a comprehensive description of
the solution of Horn's conjecture).

\pausa
In the present paper, we consider local versions of Lidskii's inequalities with respect to unitarily invariant norms (u.i.n.).
To be more precise, consider a strictly convex u.i.n., denoted by N, on $\mat$ - the algebra of complex $d\times d$ matrices -
and fix a selfadjoint matrix $S\in \mat$. Fix $\mu \in \R^d$ and let
$\mathcal O_\mu=\{	U^*\, D_\mu\, U:\ U\in\matud\}$, where $\matud$ denotes the group of unitary matrices and $D_\mu\in\mat$ denotes the diagonal
matrix with main diagonal $\mu$. Then, we consider $$\Phi:\mathcal O_\mu\rightarrow \R_{\geq 0} \peso{given by} \Phi(G)=N(S-G)\,.$$
Using Lidskii's additive inequality for eigenvalues of selfadjoint matrices, we can construct
$G^{\rm op}\in \mathcal O_\mu$ such that $\Phi(G^{\rm op})\leq \Phi(G)$, for every $G\in\mathcal O_\mu$
(see Section \ref{sec prelis nuis} for details). That is, Lidskii's inequality allows us to construct (explicitly) global minimizers
of $\Phi$. It is natural to wonder about the structure of all possible minimizers of $\Phi$ in $\mathcal O_\mu$. Moreover, since
$\mathcal O_\mu$ has a natural metric (induced by the spectral norm) then we can ask about the structure of local minimizers of
$\Phi$ in $\mathcal O_\mu$. These local minimizers arise naturally when considering optimization of
$\Phi_2(G)=\|S-G\|_2$, i.e. when $N$ is the Frobenius norm. In this case, $\Phi_2^2$ is a smooth function defined on
a smooth manifold and thus we can apply (adapted) gradient descent algorithms to find minimizers of $\Phi_2$; notice that local minimizers of $\Phi_2$
are stability points of these algorithms and therefore their structure becomes part of the convergence analysis of these methods.
Thus, our first main problem is to study the structure of global and local minimizers of $\Phi$, for a general strictly convex u.i.n. $N$.
We carry out a similar analysis for Lidskii's inequality for singular values. In both cases we show that local minimizers are indeed global minimizers and
we compute their geometrical properties.

\pausa
Finite frame theory is a well established and rapidly growing area of research (see \cite{FinFram}).
It is well known by now that several fundamental results of finite frame theory are counter-parts of
well known results in matrix analysis. For example, the so-called frame design problem with prescribed frame operator and norms
- that has played a central role in finite frame theory - is equivalent to some formulations of
the Schur-Horn theorem (see the survey \cite{BowJas} and the reference therein).
So it is no surprise that our results have implications in this area of research. Indeed,
from the local version of Lidskii's theorem we
derive some partial results related to the structure of local minimizers of the generalized frame operator
distance (G-FOD) (see \cite{AC,dnp,strawn}).

\pausa
The paper is organized as follows. In Section \ref{sec prelis nuis} we recall several results from matrix analysis
that we use throughout the paper. In section \ref{lidskii nuis} we state and prove our main results
related to the local versions of Lidskii's theorem. Indeed, in Section \ref{sec 3.1}
we obtain complete results showing that local minimizers of functions that are built
on strictly convex u.i.n's (as above) are global minimizers.
In Section \ref{sec 3.2} we consider the
corresponding problem for Lidskii's singular value inequalities. In order to obtain these results, we consider
some (differential) geometrical properties of some auxiliary smooth maps.
In Section \ref{sec 4} we apply the results from the previous sections to the study of local
minimizers of the G-FOD induced by strictly convex u.i.n's.
We obtain some partial results regarding the general structure of these local minimizers and show that under
some further hypothesis, they are global minimizers of the G-FOD.

\section{Preliminaries}\label{sec prelis nuis}

In this section we introduce the notations, terminology and results from matrix analysis
that we will use throughout the
paper (see the texts \cite{Bhat,HJ1,HJ2}).

\pausa
{\bf Notation and terminology}. We let $\mathcal M_{k,d}(\C)$ be the space of complex $k\times d$ matrices and write $\mathcal M_{d,d}(\C)=\mat$ for the algebra of complex $d\times d$  matrices. We denote by $\matsad\subset \mat$ the real subspace of selfadjoint matrices and by $\matpos\subset \matsad$ the cone of positive semidefinite matrices. We let $\matud\subset \mat$ denote the group of unitary matrices.
For $d\in\N$, let $\I_d=\{1,\ldots,d\}$.
Given a vector $x\in\C^d$ we denote by $D_x$ the diagonal matrix in $\mat$ whose main diagonal is $x$.
Given $x=(x_i)_{i\in\I_d}\in\R^d$ we denote by $x\da=(x_i\da)_{i\in\I_d}$ the
vector obtained by rearranging the entries of $x$ in non-increasing order. We denote by
$(\R^d)\da=\{x\da:\ x\in\R^d\}$ and $(\R_{\geq 0}^d)\da=\{x\da:\ x\in\R_{\geq 0}^d\}$. Given a matrix $A\in\matsad$ we denote by
$\la(A)=\la(A)\da=(\la_i(A))_{i\in\I_d}\in (\R^d)\da$ the eigenvalues of $A$ counting multiplicities and arranged in
non-increasing order.
For $B\in\mat$ we let $s(B)=\la(|B|)$ denote the singular values of $B$, i.e. the eigenvalues of $|B|=(B^*B)^{1/2}\in\matpos$; we also let $\sigma(B)\subset \C$ denote the spectrum of $B$.
If $x,\,y\in\C^d$ we denote by $x\otimes y\in\mat$ the rank-one matrix given by $(x\otimes y) \, z= \langle z\coma y\rangle \ x$, for $z\in\C^d$.

\pausa Next we recall the notion of majorization between vectors, that will play a central role throughout our work.
\begin{fed}\rm
Let $x\in\R^k$ and $y\in\R^d$. We say that $x$ is
{\it submajorized} by $y$, and write $x\prec_w y$,  if
$$
\suml_{i=1}^j x^\downarrow _i\leq \suml_{i=1}^j y^\downarrow _i \peso{for every} 1\leq j\leq \min\{k\coma d\}\,.
$$
If $x\prec_w y$ and $\tr x = \sum_{i=1}^kx_i=\sum_{i=1}^d y_i = \tr y$,  then $x$ is
{\it majorized} by $y$, and write $x\prec y$.
\end{fed}

\begin{rem}\label{desimayo}
\pausa Given $x,\,y\in\R^d$ we write
$x \leqp y$ if $x_i \le y_i$ for every $i\in \mathbb I_d \,$.  It is a standard  exercise
to show that:
\begin{enumerate}
\item $x\leqp y \implies x^\downarrow\leqp y^\downarrow  \implies x\prec_w y $.
\item $x\prec y\implies |x|\prec_w|y|$, where $|x|=(|x_i|)_{i\in\I_d}\in\R_{\geq 0}^d$.
\item $x\prec y,\, |x|\da=|y|\da \implies x\da=y\da$.
\item If $\tr(x)=\sum_{i\in\I_d} x_i=t$ then $\frac{t}{d}\,\uno_d\prec x$.
\EOE
\end{enumerate}
\end{rem}

\pausa
Although majorization is not a total order in $\R^d$, there are several fundamental inequalities in
matrix theory that can be described in terms of this relation. As an example of this phenomenon we can consider
Lidskii's (additive) inequality for eigenvalues of sums of hermitians (see \cite{Bhat,HJ1,HJ2}). In the following result we also include the characterization of the case of equality obtained in \cite{mrs2}.

\begin{teo}[Lidskii's inequality]\label{mrs284}\rm
Let $A,\,B\in \matsad$ with eigenvalues $\la(A)$, $\la(B)\in (\R^d)^\downarrow$ respectively. Then
\ben
\item $\la(A)-\la(B) \prec \la(A-B)$.
\item $\paren{\la(A)-\la(B)}^{\downarrow}=\la(A-B)$ if and only if there exists
$\{v_i\}_{i\in\I_d}$ an ONB of $\C^d$ such that
\beq
A=\sum_{i\in\I_d}\la_i (A) \ v_i\otimes v_i \py B=\sum_{i\in\I_d}\la_i (B)\ v_i\otimes v_i\  .
\eeq
Notice that in this case, $A$ and $B$ commute.
\qed\een
\end{teo}
\pausa
Recall that a norm $\nui{\cdot}$ in $\mat$ is unitarily invariant if
$$ \nui{UAV}=\nui{A} \peso{for every} A\in\mat \py U,\,V\in\matud\,.$$
Examples of unitarily invariant norms (u.i.n.) are the spectral norm $\|\cdot\|$ and the $p$-norms $\|\cdot\|_p$, for $p\geq 1$.
It is well known that majorization relations between singular values of matrices are intimately related
with inequalities with respect to u.i.n's.
The following result summarizes these relations (see for example \cite{Bhat}):

\begin{teo}\label{teo intro prelims mayo}\rm
Let $A,\,B\in\mat$ be such that $s(A)\prec_w s(B)$. Then:
\ben
\item For every u.i.n. $N$ in $\mat$
we have that $N(A)\leq N(B)$.
\item If we assume that there exists a strictly convex
u.i.n. $N$ in $\mat$ such that $N(A)=N(B)$ then we have that $s(A)=s(B)$.
\een
\qed
\end{teo}

\section{Local Lidskii's theorems for unitarily invariant norms}\label{lidskii nuis}

Lidskii's additive inequalities (both for eigenvalues and singular values) can be interpreted
as an explicit description of global minimizers of functions that are built on unitarily invariant norms and whose
domains consist of certain orbits of matrices (under the action of the unitary group). In this section, we show
that Lidskii's inequalities actually describe all global minimizers of such functions, and that local minimizers
are also global minimizers. This last fact will play a central role in the next section, in which we state and study Strawn's
generalized conjecture.

\subsection{Selfadjoint matrices - eigenvalues}\label{sec 3.1}
\pausa
We begin with the following comments related to
 the classical Lidskii's inequality.
Fix $S\in\matsad$ and $\mu\in(\R^d)\da$, and consider $\cO_\mu$  given by
\beq\label{defi omu App nuis}
\cO_\mu=\{G\in\matsad:\la(G)=\mu\}=\{U^* D_{\mu}\,U:\ U\in\matud\}
\eeq
We consider the usual metric in $\cO_\mu$ induced by the operator norm; hence $\cO_\mu$ is a metric space.

\pausa
For $N$ a strictly convex u.i.n., let
\beq\label{defi nui App}
 \Phi=\phisn:\cO_\mu\rightarrow \R_{\geq 0} \peso{be given by} \Phi(G)=N(S-G).
\eeq
Using an ONB of eigenvectors of $S$ we can construct $G^{\rm op}\in \cO_\mu$ such that
$\lambda(S-G{\rm op})=(\la(S)-\mu)\da$. By Lidskii's inequality and Remark \ref{desimayo}, we see that for every
$G\in\cO_\mu$ we have that
\beq\label{eq submayo rel}
 \la(S-G^{\rm op})\prec \la(S-G) \implies  s(S-G^{\rm op})=|\la(S-G^{\rm op})|\prec_w |\la(S-G)|=s(S-G)\,.
\eeq
Hence, Theorem \ref{teo intro prelims mayo} implies that $\Phi(G^{\rm op})=N(S-G^{\rm op})\leq N(S-G)=\Phi(G)$, for $G\in\cO_\mu$.
Therefore, $G^{\rm op}$ is a global minimizer of $\Phi$ in $\cO_\mu$. Conversely, let $G\in \cO_\mu$ be a global minimizer of
$\Phi$ in $\cO_\mu$. The previous comments together with item 3. in Remark \ref{desimayo} show that
\beq\label{eq submayo rel2}
\la(S-G^{\rm op})\prec \la(S-G) \py  N(S-G^{\rm op})= N(S-G) \implies \la(S-G^{\rm op})= \la(S-G)\eeq
where we have used the fact that $N$ is strictly convex, the submajorization relation in Eq. \eqref{eq submayo rel}
and Theorem \ref{teo intro prelims mayo}. In turn, Eq. \eqref{eq submayo rel2} together with Theorem \ref{mrs284} imply that there exists
 an ONB $\{v_i\}_{i\in\I_d}$
of $\C^d$ such
$$
S=\sum_{i\in\I_d}\la_i \ v_i\otimes v_i \py G=\sum_{i\in\I_d}\mu_i\ v_i\otimes v_i\,,
$$where $(\la_i)_{i\in\I_d}=\la(S)\in (\R^d)\da$; that is, the global minimizer $G$ is obtained from $S$ as $G^{\rm op}$.

\pausa
It is then natural to ask about the structure of local minimizers $G_0$ of the map $\Phi$ in $\cO_\mu$, which is our main problem in this section.
As we will see, these
local minimizers are actually global minimizers of $\Phi$ (see Theorem \ref{teo LLTApp} below).

\begin{fed}\label{muchas defis mil nuis}
Let $S,G_0\in\matsad$. We consider
\begin{enumerate}
\item The product manifold $\matud\times \matud$ endowed with the metric $$d((U_1,V_1),(U_2,V_2))=\max\{\|I-U_1^*U_2\|,\,\|I-V_1^*V_2\| \}\,.$$
\item
$\Gamma=\Gamma_{(S,G_0)}:\matud\times \matud\rightarrow \matsad_\tau\stackrel{\rm def}{=}\{M\in\matsad:\ \tr(M)=\tau\}$ for $\tau=\tr(S)-\tr(G_0)$, given by
$$ \Gamma(U,V)=U^* S\,U-V^*G_0\,V\peso{for} U,\, V\in\matud\,.$$
\item For a given u.i.n. $N$ on $\mat$, we consider
$\Delta_{(S,G_0)}^{N}=\Delta:\matud\times \matud\rightarrow \R_{\geq 0}$:
$$\Delta(U,V)=N(\Gamma(U,V)) \peso{for} U,\, V\in\matud\,.$$
\EOE\end{enumerate}
\end{fed}
\pausa
Our motivation for considering the previous notions comes from the following:
\begin{lem}
\label{lem equiv de los probs}
Let $S\in\matsad$, $\mu\in(\R^d)\da$, $G_0\in\cO_\mu$ and consider the notations from Definition \ref{muchas defis mil nuis}. Given a u.i.n. $N$ on $\mat$, the following conditions are equivalent:
\begin{enumerate}
\item $G_0$ is a local minimizer of $\Phi$ in $\cO_\mu$ (defined in Eq. \eqref{defi nui App});
\item $(I,I)$ is a local minimizer of $\Delta$ on $\matud\times \matud$.
\end{enumerate}
\end{lem}
\begin{proof}

1.$\implies$2. Consider $(U,W)\in \matud\times \matud$ such that
$$d((U,W),(I,I))=\max\{\,\|I-U^*\|\,,\,\|I-W^*\| \, \}:=\varepsilon\,.$$
Hence, if $Z=WU^*\in\matud\,$ then $ U^* S\,U-W^* G_0\,W=U^*( S- Z^* G_0\, Z)\,U$.
Notice that
$$\|Z-I\|=\|W \,(U^*-W^*)\|\leq \|U^*-I\|+\|I-W^*\|\leq 2\,\varepsilon\ \implies$$
$$\Delta(U,W)
=N(U^*( S- Z^* G_0\, Z)\,U )= \Phi(Z^*G_0Z)
\peso{with} \| Z^*G_0Z-G_0\|\leq 4\,\varepsilon\|G_0\|\,.$$

2.$\implies$1. This is a consequence of the fact that the map $\matud\ni Z\mapsto
Z^* G_0\,Z\in\cO_\mu$ is open
(see, for example,  \cite[Thm. 4.1]{AS}  or \cite{DF}).
\end{proof}

\pausa
In what follows, given $\mathcal S\subset\matsad$ we consider the commutant of $\mathcal S$, denoted $S '$,
that is the unital $^*$-subalgebra of $\mat$ given by
$$ \mathcal S'=\{\ C\in\mat:\ [C,D]=0\ \text{ for every }\ D\in\mathcal S\ \}\subset \mat\,,$$
where $[C,D]=CD-DC$ denotes the commutator of $C$ and $D$.

\pausa
Recall that $\matud$ has a natural smooth (differential) manifold structure. Hence, we can consider $\matud\times \matud$ as a smooth manifold, endowed with
the product structure.
\begin{lem}\label{lema gama sobre}
Consider the notations from Definition \ref{muchas defis mil nuis}. Then
$$
\Gamma
 \peso{is a submersion at} (I,I) \quad \iff \quad
\{S,\,G_0\}'=\C\cdot I \ .
$$
\end{lem}
\begin{proof}
The (exponential) map $\matsad\ni X\mapsto \exp(X)$
allows us to identify the tangent space $\cT_{I}\matud$ with
$i\cdot\matsad$. Since we consider the product structure on $\matud\times \matud$ we conclude that the differential of $\Gamma$ satisfies
$$D_{(I,I)}\Gamma (X,0)=[S,X] \py D_{(I,I)}\Gamma (0,X)=[X,G_0] \peso{for} X\in i\cdot \matsad\,. $$
Therefore $\Gamma$ is  not a submersion at $(I,I)$
if and only if there exists $0\neq Y\in\cT\matsad_\tau=\matsad_0$
(i.e. $Y\in\matsad$ such that $\tr\, Y=0$)
such that
\beq\label{orto}
\tr(Y\,[S,Z])=\tr(Y\,[Z,G_0])=0 \peso{for every}
Z\in i\cdot \matsad \ .
\eeq
Since $\tr(Y\,[S,Z])=\tr([Y,S]\,Z)$ and similarly
$\tr(Y\,[Z,G_0])=\tr(Z\,[G_0,Y])$,  we see that in this case
$$[Y,S]=0 =[G_0,Y]\in i\cdot \matsad\ .$$
Moreover, since $Y\neq 0$ and
$\tr \, Y = 0$, then $Y$ has some non-trivial spectral projection $P$  which also satisfies that $[P,S]=[P,G_0]=0$. Conversely, in case there exists a non-trivial projection $P$ such that
$[P,S]=[P,G_0]=0$, we can construct
$ Y = \frac{P}{\tr\, P}  - \frac{I-P}{\tr \,(I- P)} $
so that $\tr\, Y =0$. Then
$0\neq Y\in\cT\matsad_\tau$ and it satisfies Eq. \eqref{orto}, so that this matrix $Y$  is orthogonal to the range of the operator $D_{(I,I)}\Gamma$.
\end{proof}

\begin{pro}\label{pro al final commutan nomas}
Consider the notations from Definition \ref{muchas defis mil nuis} and assume that $N$ is a strictly convex u.i.n. If $(I,I)$ is a local minimizer of $\Delta$ in $\matud\times \matud$ then $[S,G_0]=0$.
\end{pro}
\begin{proof}
Assume that $[S,G_0]\neq 0$. Then there exists a minimal projection $P$ of the unital $^*$-subalgebra $\mathcal C=\{S,\,G_0\}'\subseteq\mat$ such that
$[P\,S,P\,G_0]\neq 0$.  Indeed, $I\in\mathcal C$ is a projection such that $[I\,S,I\,G_0]\neq 0$. If $I$ is not a minimal projection in $\mathcal C$ then
there exists $P_1,\,P_2\in\mathcal C$ non-zero projections such that $I=P_1+P_2$; hence $[P_iS,P_iG_0]\neq 0$ for $i=1$ or $i=2$. If the corresponding $P_i$ is not minimal in $\mathcal C$ we can repeat the previous argument (halving) applied to $P_i$. Since we deal with finite dimensional algebras, the previous procedure finds a minimal projection $P\in\mathcal C$ as above. By applying a convenient change of orthonormal basis we can assume that
$R(P)=\text{span}\{e_i: i\in\I_r\}$, where $r=\rk(P)>1$. Since $P$ reduces both $S$ and $G_0$
 we can consider $S_1=S|_{R(P)}\in\mathcal H(r)$ and
$G_1=G_0|_{R(P)}\in\mathcal H(r)$.
By minimality of $P$ we conclude that $\{S_1,\,G_1\}'=\C I_r \subset \cM_r(\C) $.
Using the case of equality of Lidskii's inequality (see Theorem \ref{mrs284}), we conclude that
$$ b:=(\la(S_1)-\la(G_1))\da\prec a:=\la(S_1-G_1) \py a\neq b\,.$$
If we let $\sigma=\tr(S_1-G_1)$ then, by Lemma \ref{lema gama sobre} the map
$$\cU(r)\times \cU(r)\ni(U,V)\mapsto U^*S_1\,U-V^*G_1\,V\in\cH(r)_\sigma $$
is a submersion at $(I_r,I_r)$. In particular, for every open neighborhood $\mathcal N $ of $(I_r,I_r)$ in $\cU(r)\times \cU(r)$ the set
$$ \cM:=\{ U^*S_1\,U-V^*G_1\,V: \ (U,V)\in\mathcal N \}$$ contains an open neighborhood of $S_1-G_1$ in $\cH(r)_\sigma$.
Consider $\rho:[0,1]\rightarrow (\R_{\geq 0}^{r})\da$ given by $\rho(t)=(1-t)\,a+t\,b$ for $t\in[0,1]$.
Notice that $\rho(t)\prec a$ and $\rho(t)\neq a$ for $t\in (0,1]$.
If we let $S_1-G_1=W^*D_{a}\,W$ for $W\in\cU(r)$ then the continuous curve $T(\cdot):[0,1]\rightarrow \cH(r)_\sigma$ given by
$T(t)=W^*D_{\rho(t)}\,W$ for $t\in[0,1]$ satisfies that $T(0)=S_1-G_1$, $\la(T(t))\prec a$ and $\la(T(t))\neq a$ for $t\in(0,1]$.
Therefore, there exists $t_0\in(0,1]$ such that $T(t)\in\cM$ for $t\in[0,t_0]$ so, in particular, there exists $(U,V)\in\mathcal N$ such that
$$ T(t_0)=U^*S_1\,U-V^*G_1\,V  \implies
\Delta(U\oplus P^\perp,V\oplus P^\perp)<\Delta(I_d,I_d)  \ ,
$$
because $N$ is a strictly convex u.i.n.,
where $U\oplus P^\perp,\, V\oplus P^\perp\in\matud$ act as the identity on $R(P)^\perp\subset \C^d$.
Since $\cN$ was an arbitrary neighborhood of $(I_r,I_r)$ we conclude that $(I_d,I_d)$ is not a local minimizer of $\Delta$ in $\matud\times \matud$,
which contradicts Lemma \ref{lem equiv de los probs}.
\end{proof}

\begin{teo}[Local Lidskii's theorem]\label{teo LLTApp}
Let $S\in\matsad$ and $\mu=(\mu_i)_{i\in\I_d}\in(\R^d)\da$. Assume that
$N$ is a strictly convex u.i.n. and that $G_0\in\cO_\mu$ is a local minimizer of
$\Phi=\phisn$ on $\cO_\mu\,$. Then, there exists an ONB
$\{v_i\}_{i\in\I_d}$ of $\C^d$ such that
\beq\label{eq base buenapp1}
S=\sum_{i\in\I_d}\la_i\ v_{i}\otimes v_{i} \py  G_0=\sum_{i\in\I_d}\mu_i\ v_{i}\otimes v_{i}\,,
\eeq
where $(\la_i)_{i\in\I_d}=\la(S)\in (\R^d)\da$.
 In particular, $\la(S-G_0)=(\la(S)-\la(G_0))\da$ so $G_0$ is also a global minimizer of $\Phi$ on $\cO_\mu\,$.
\end{teo}
\begin{proof}
By Lemma \ref{lem equiv de los probs} and Proposition \ref{pro al final commutan nomas} we conclude that $[S,G_0]=0$.
Notice that in this case there exists $\cB=\{v_i\}_{i\in\I_d}$ an ONB of $\C^d$ such that
$$
S=\sum_{i\in\I_d} \la_i \ v_i\otimes v_i
\ , \ G_0=\sum_{i\in\I_d} \nu_i \ v_i\otimes v_i \ \text{ with }\ \la=(\la_i)_{i\in\I_d}\in (\R_{\geq 0}^d)\da \ ,
$$
for some $\nu_1,\ldots,\nu_d\in \R$.
We now show that under a suitable permutation of the elements of $\I_d$ we can obtain a representation as in Eq. \eqref{eq base buenapp1} above.
Indeed, assume that $j\in\I_{d-1}$ is such that $\nu_j<\nu_{j+1}$. If we assume that $\la_j>\la_{j+1}$ then consider the continuous curve of unitary operators $U(t):[0,\pi/2)\rightarrow \matud$ given by
$$U(t)=\sum_{i\in\I_d\setminus\{j,\,j+1\}}v_i\otimes v_i + \cos(t)\ (v_j\otimes v_j
+v_{j+1}\otimes v_{j+1}) +\sin(t)\ (v_j\otimes v_{j+1}- v_{j+1}\otimes v_j)\ , \ \ t\in[0,\pi/2)\,.$$
Notice that $U(0)=I_d$. We now define the continuous curve $G(t)=U(t) \,G_0\,U(t)^* \in\cO_\mu$, for $t\in[0,\pi/2)$.
Then $G(0)=G_0$ and we have that
\beq\label{eq SFt}
 S-G(t)=\sum_{i\in\I_d\setminus\{j,\,j+1\}} (\la_i-\nu_i) \ v_i\otimes v_i + \sum_{r,s=1}^2 \gamma_{r,s}(t) \ v_{j+r}\otimes v_{j+s}\,,\eeq
where $M(t)=(\gamma_{r,s}(t))_{r,s=1}^2$ is determined by
$$ M(t)
= \begin{pmatrix} \la_j & 0 \\0 & \la_{j+1}\end{pmatrix}
- V(t)
\begin{pmatrix}\nu_{j} & 0 \\0 & \nu_{j+1}\end{pmatrix}
V(t)^* \py V(t)=
\begin{pmatrix}\cos(t) & \sin(t) \\-\sin(t) & \cos(t)\end{pmatrix}
\ , \ \ t\in[0,\pi/2)\,.
$$
Let us consider
\beq\label{defi R}R(t)=
V^*(t) \begin{pmatrix}
\la_j - \la_{j+1} & 0 \\
0 & 0
\end{pmatrix}V(t)
-
\begin{pmatrix}
\nu_j & 0 \\
0 & \nu_{j+1}
\end{pmatrix}
\implies M(t)=V(t) \, R(t)\, V^*(t) + \la_{j+1}\, I_2\,.
\eeq
We claim that $\la(R(t))\prec \la(R(0))$ and $\la(R(t))\neq \la (R(0))$ for $t\in(0,\pi/2)$ (i.e., the majorization relation is strict). Indeed, since
$R(t)$ is a curve in $\cH(2)$ such that $\tr(R(t))$ is constant, it is enough to show that the function $[0,\pi/2)\ni t\mapsto \tr(R(t)^2)$ is strictly decreasing in $[0,\pi/2)$.
Using that $\la_j-\la_{j+1}>0$ we have that
$$ V^*(t) \begin{pmatrix}\la_j - \la_{j+1} & 0 \\ 0 & 0 \end{pmatrix}V(t) = g(t)\otimes g(t) \peso{where} g(t)=(\la_j-\la_{j+1})^{1/2}(\cos(t),\sin(t)) \ , \ \ t\in[0,\pi/2)\,.
$$ If $D\in\cM_2(\C)$ is the diagonal matrix with main diagonal $(\nu_j,\,\nu_{j+1})$ then
$R(t)=g(t)\otimes g(t)-D$ so
$$\tr(R(t)^2)=\tr((g(t)\otimes g(t))^2)+\tr(D^2)-2\,\tr(g(t)\otimes g(t)\ D)=c-2\,\langle D\,g(t),\,g(t)\rangle$$
where $c=\|g(t)\|^4+\nu_j^2+\nu_{j+1}^2=(\la_j-\la_{j+1})^2+\nu_j^2+\nu_{j+1}^2\in\R$ is a constant and
$$\langle D\,g(t),\,g(t)\rangle =(\la_j-\la_{j+1})\ (\cos^2(t) \, \nu_j+\sin^2(t)\,\nu_{j+1})$$ is strictly increasing in $[0,\pi/2)$, since $\nu_j<\nu_{j+1}$. Thus, $\la(R(t))\prec \la(R(0))$ and $\la(R(t))\neq \la (R(0))$ for $t\in(0,\pi/2)$. Hence, by Eq. \eqref{defi R}, we see that
$$ \la(M(t))=\la(R(t))+\la_{j+1}\,\uno_2 \implies \la(M(t))\prec \la(M(0)) \ , \ \la(M(t))\neq \la(M(0)) \ ,\ \ t\in(0,\pi/2)\,.$$
Then, using Eq. \eqref{eq SFt} and Theorem \ref{teo intro prelims mayo}, for $t\in(0,\pi/2)$
$$
\la(S-G(t))\prec \la(S-G_0) \ \ , \ \ \la(S-G(t))\neq \la(S-G_0) \then N(S-G(t))<N(S-G_0).
$$
This last inequality, which is a consequence of the assumption $\la_j<\la_{j+1}$, contradicts
the local minimality of $G_0$ in $\cO_\mu$. Hence, since $\la_j\leq \la_{j+1}$ we see that $\la_j=\la_{j+1}$; in this
case, we can consider the basis $\cB'=\{v_i'\}_{i\in\I_d}$ obtained by transposing the vectors
$v_j$ and $v_{j+1}$ in the basis $\cB$. In this case $S\ v_i'=\la_i\ v_i'$ for $i\in\I_d$, $G_0\ v_i=\nu_i\ v_i'$ for
$i\in\I_d\setminus\{j,\,j+1\}$ and $G_0\ v_j'=\nu_{j+1}\, v_j'$, $G_0 \ v_{j+1}'=\nu_{j}\, v_{j+1}'$.
After performing this argument at most $d$ times we get the desired ONB.
\end{proof}

\subsection{Arbitrary matrices - singular values}\label{sec 3.2}

In this section we obtain results related to a local Lidskii's theorem for arbitrary matrices with respect to singular values.
As a consequence, we characterize the case of equality in the classical Lidskii's inequality for singular values.

\pausa
Recall that if $A,B\in\mat$ then Lidskii's singular value inequality states that
\begin{equation}\label{lidskii vs sing}
\abs{s(A)-s(B)}\prec_w s(A-B).
\end{equation}
In what follows we fix $A\in\mat$, $s\in(\R^d_{\geq 0})\da$, $s\neq 0$, and $N$ a strictly convex u.i.n. We consider the set of matrices whose vector of singular values is $s$, i.e.
$$
\cV_s:=\llav{C\in\mat:\, s(C)=s},
$$ endowed with the usual metric, induced by the spectral norm. We further consider the function
$$\Psi_{(N,\,A,\,s)}=\Psi:\cV_s\rightarrow \R_{\geq 0} \peso{given by}
\Psi(C)=N(A-C).
$$
With an argument similar to that in the beginning of Section \ref{sec 3.1}, now
based on the singular value decomposition (SVD) and Lidskii's inequality in Eq. \eqref{lidskii vs sing}, we can explicitly construct global minimizers
 of $\Psi$ on $\cV_s$. As before, we are interested in the structure of local minimizers of $\Psi$ in
$\cV_s$.

\pausa
We will describe the structure of local minimizers of
$\Psi$ in $\cV_s$ and show that local minimizers are
 actually global minimizers. In order to do this we consider the following well known matrix construction:
for $C\in\mat$, let $\widehat{C}\in\cH(2d)$ be given by
$$
\widehat{C}=
\begin{pmatrix}
0 & C\\
C^* & 0
\end{pmatrix}.
$$
Let $U,V\in \cU(d)$ be such that $C=V^* D_{s(C)} U$,
 and define $W\in \cU(2d)$ given by
$$
W=\frac{1}{\sqrt{2}}
\begin{pmatrix}
V & U\\
-V & U
\end{pmatrix}.
$$
Then $\widehat{C}= W^*\, (D_{s(C)} \oplus-D_{s(C)}) \,W$, which implies that
\beq\label{eq orden}
\la(\widehat{C})_i= \begin{cases}  \quad \quad
 s_i(C) & \peso{if} \ 1\leq i\leq d \\
-s_{d-i+1}(C)
& \peso{if}\ d+1\leq i\leq 2d\ .
\end{cases}
\eeq

\medskip

\begin{fed}
Let $A\coma B \in \mat$, let $N$ be a u.i.n. on $\mat$ and $s\in (\R^d_{\geq 0})\da$, $s\neq 0$. We consider:
\begin{enumerate}
\item The real space $\mathcal S=\{\widehat C:\ C\in\mat\}\subset H(2d)$.
\item The map $\Pi_{(A\coma B)}=\Pi:\matud^4\rightarrow \mathcal S$ given by
\begin{eqnarray}\label{eq defi Pi}\Pi(U_1\coma U_2\coma V_1\coma V_2)&=&(U_1\oplus V_1)^* \, \widehat A  \ (U_1\oplus V_1) -
(U_2\oplus V_2)^* \,\widehat B  \ (U_2\oplus V_2)\\ \nonumber &=& \widehat{U_1^*\,A\,V_1-U_2^*\, B\, V_2}\,.
\end{eqnarray}
\item The map $\Xi_{(A\coma B)}:\matud^4\rightarrow \R_{\geq 0}$ given by
\beq \label{eq defi Xi}
\Xi(U_1\coma U_2\coma V_1\coma V_2)=N(U_1^*\,A\,V_1-U_2^*\, B\, V_2)\,.
\eeq \EOE
\end{enumerate}
\end{fed}

\begin{lem}\label{lem equiv de los probs para nuis}
Let $A\in\mat$, $s\in(\R_{\geq 0}^d)\da$, $s\neq 0$, and $B\in\cV_s$.
Given a u.i.n. $N$ on $\mat$, the following conditions are equivalent:
\begin{enumerate}
\item $B$ is a local minimizer of $\Psi_{(N\coma A\coma s)}$ in $\cV_s$;
\item $(I,I,I,I)$ is a local minimizer of $\Xi_{(A\coma B)}$ on $\matud^4$.
\end{enumerate}
\end{lem}
\begin{proof} An argument similar to that in the proof of Lemma \ref{lem equiv de los probs} shows the equivalence of the items above.
\end{proof}

\pausa Next we develop some geometric properties of $\Pi$. As before, we consider $\matud^4$ as a smooth manifold, endowed with the product structure.
\begin{lem}\label{lema Pi sobre} Let $A,\, B\in\mat$ and let $\Pi$ be as Eq. \eqref{eq defi Pi}.
Then the following conditions are equivalent:
\begin{enumerate}
\item $\Pi_{(A\coma B)}$ is a submersion at $(I,I,I,I)$;
\item Whenever $Z\in\mat$ is such that $A^*Z\coma AZ^*\coma B^*Z\coma BZ^*\in\matsad$, then $Z=0$.
\end{enumerate}
\end{lem}
\begin{proof}
Notice that since $\Pi$ is a smooth function, item 1. holds if and only if the differential map
$$D=D\Pi_{(I\coma I\coma I\coma I)}:(i\cdot \matsad)^4 \rightarrow \mathcal S\subset \mathcal H(2d)\peso{is surjective .}$$
We now check that $D$ is {\it not} surjective if and only if there exists $Z\in\mat$, $Z\neq 0$, such that
$A^*Z\coma AZ^*\coma B^*Z\coma BZ^*\in\matsad$. Indeed, it is straightforward to compute
$$D(X_1\coma X_2\coma Y_1\coma Y_2)= \widehat{-X_1\,A+A\,Y_1  + X_2\,B-B\,Y_2 }\peso{for} X_1\coma X_2\coma Y_1\coma Y_2\in i\cdot\matsad\,.$$
Hence, $D$ is not surjective if and only if there exists $Z\in\mat$, $Z\neq 0$, such that
\begin{equation}\label{eq cond perp}
\widehat Z\ \ \perp  \ \ \widehat{-X_1\,A+A\,Y_1  +X_2\,B-B\,Y_2 }\peso{for} X_1\coma X_2\coma Y_1\coma Y_2\in i\cdot\matsad\,.
\end{equation}
In this case (setting $X_2=Y_2=0$) we have that
\begin{equation}\label{eq cond perp2}
0=\tr(\widehat Z \ (\widehat{-X_1\,A+A\,Y_1}))=2\,\Preal[\tr(Z^*(-X_1\,A+A\,Y_1))]
\peso{for} X_1\coma Y_1\in i\cdot\matsad\,.
\end{equation}
Using that $\Preal[\tr(C)]=\tr(\Preal [C])$ and the tracial property, we see that Eq. \eqref{eq cond perp2} is equivalent to
\begin{equation}\label{eq cond perp3}
0=\tr(X_1\,(AZ^*-ZA^*)) + \tr(Y_1\,(A^*Z-Z^*A)) \peso{for} X_1\coma Y_1\in i\cdot\matsad\,.
\end{equation}
Since $(AZ^*-ZA^*)\coma (A^*Z-Z^*A)\in i\cdot \matsad$, Eq. \eqref{eq cond perp3} holds if and only if
$$AZ^*-ZA^*=0\py A^*Z-Z^*A=0 \implies AZ^*\coma A^*Z\in\matsad\,.$$
Similarly, by setting $X_1=Y_1=0$ in Eq. \eqref{eq cond perp} and arguing as before, we conclude that
$BZ^*\coma B^*Z\in\matsad$.

\pausa
Conversely, assume that there exists $Z\in\mat$, $Z\neq 0$, such that $A^*Z\coma AZ^*\coma B^*Z\coma BZ^*\in\matsad$.
Then, arguing as before, it follows that $Z$ verifies the perpendicularity condition in Eq. \eqref{eq cond perp}; thus, $D$ is not surjective in this case.
\end{proof}

\begin{pro}\label{pro asumiendo que pi es sumersion}
Fix $A\in \mat$, a strictly convex u.i.n. $N$ on $\mat$ and $s\in (\R^d_{\geq 0})\da$, $s\neq 0$. If $B\in\cV_s$ be a local minimizer of
$\Psi=\Psi_{(N\coma A\coma s)}$ then, $\Pi_{(A\coma B)}$ is not a submersion at $(I\coma I\coma I\coma I)$.
\end{pro}
\begin{proof}
Assume that $\Pi_{(A\coma B)}$ is a submersion at $(I\coma I\coma I\coma I)$.
Assume further that any of the conditions $A^*B\coma  A\,B^*\in\matsad$ does not hold.
In this case it is straightforward to check that $\widehat A$ and $\widehat B$ do not commute.
In particular, by Theorem \ref{mrs284} $$b:=(\la(\widehat A)-\la(\widehat B))\da\prec
a:=\la(\widehat A - \widehat B) \py a \neq b\,.$$
Now, by Eq. \eqref{eq orden} we see that if we let
$$\tilde b:=(s(A)-s(B)\coma -[s(A)- s(B)]) \ , \ \tilde a:=(s(A-B)\coma -s(A-B))\implies b=(\tilde b) \da \ , \ a=(\tilde a)\da\,. $$
Hence, if we let $\rho:[0,1]\rightarrow \R^{2d}$ be given by $\rho(t)= (1-t)\, \tilde a + t\, \tilde b$ , for $t\in [0,1]$ then:
\begin{enumerate}
\item $\rho(t)\prec a$ and $\rho(t)\da \neq a$, for every $t\in (0,1]$ ;
\item $\rho(0)=\tilde a$ ;
\item For every $t\in[0,1]$ there exists $c_t\in \R^d$ such that $\rho(t)=(c_t\coma -c_t)$.
\end{enumerate}
In order to see item 1. above, recall that $$ \rho(t)=(1-t)\, \tilde a + t\, \tilde b\prec (1-t)\, (\tilde a)\da
 + t\, (\tilde b)\da= (1-t)\,  a + t\, b \prec a$$ and, $((1-t)\,  a + t\, b)\da=(1-t)\,  a + t\, b\neq a$ (since $a\neq b$),
for $t\in (0,1]$. Consider a SVD for $A-B= V^*\, D_{s(A-B)} \, U$, for some $U\coma V\in\matud$, and define
$$
W=\frac{1}{\sqrt{2}}
\begin{pmatrix}
V & U\\
-V & U
\end{pmatrix}\in \mathcal U(2d)\,.
$$
Then $\widehat{A-B}= W^*\, (D_{s(A-B)} \oplus-D_{s(A-B)}) \,W=W^*\ D_{\tilde a}\ W$; Let us consider
$T(t)= W^* \, D_{\rho(t)}\, W$ for $t\in [0,1]$.
Then, using item 3. above, we see that $T(t)\in \mathcal S$ for $t\in [0,1]$. By the hypothesis on $\Pi_{(A\coma B)}=\Pi$,
for every open neighborhood of $I\in \mathcal N\subset \matud$, the set
$$ \mathcal M=\{ \Pi(U_1\coma U_2\coma V_1\coma V_2):\ U_i\coma V_i\in\mathcal N\, , \ i=1,2\}$$
contains an open neighborhood of $\widehat{A-B}$ in $\mathcal S$. Since $T:[0,1]\rightarrow \mathcal S$ is
a continuous curve such that $T(0)=\widehat{A-B}$, then there exists $t_0\in (0,1)$ such that $T(t)\in \mathcal M$, for
$t\in [0,t_0]$. In particular, there exist $U_i\coma V_i\in\mathcal N$, for $i=1,2$ such that
$$T(t_0)=\widehat{U_1^*\,A\, V_1 - U_2^*\, B\, V_2} \ \ , \ \ \la(T(t_0))=\rho(t)\da \prec a \ \ , \  \ \la(T(t_0))\neq  a\,.$$
Hence, $s( U_1^*\,A\, V_1 - U_2^*\, B\, V_2)\prec_w s(A-B)$ and $s( U_1^*\,A\, V_1 - U_2^*\, B\, V_2)\neq  s(A-B)$. Using that
$N$ is a strictly convex u.i.n. we conclude that $$ \Xi_{(A\coma B)}(U_1\coma U_2\coma V_1\coma V_2)
=N(U_1^*\,A\, V_1 - U_2^*\, B\, V_2)< N(A-B)=\Xi_{(A\coma B)}(I\coma I\coma I\coma I)\,.$$
Since $\mathcal N$ is an arbitrary neighborhood of $I$ in $\matud$ we see that $(I\coma I\coma I\coma I)$ is not a local
minimizer of $\Xi_{(A\coma B)}$, which contradicts Lemma \ref{lem equiv de los probs para nuis}.

\pausa
The previous argument shows that $A^*B\coma AB^*\in\matsad$. If we set $Z=B\in\mat$, we see that
$$Z\neq 0\py A^*Z\coma AZ^*\coma B^*Z\coma BZ^*\in\matsad\,.$$
Now, Lemma \ref{lema Pi sobre} implies that $\Pi$ is not a submersion at $(I\coma I\coma I\coma I)$, which contradicts
our assumption on $\Pi$; this last fact proves the result.
\end{proof}

\begin{rem}\label{rem sobre la condicion EY}
Let $A\coma B\in\mat$ be such that $A^*B,\, AB^*\in\matsad$. Then, Eckart and Young \cite{EY} claimed that there exist
 matrices $U,\, V\in \matud$ such that
$$U^* A\,V=A \py U^*B\,V=D_\beta \peso{with} \beta\in\R^d\,.$$
Indeed, notice that the hypothesis also holds for $X^*AY$ and $X^*BY$, for any $X\coma Y\in\matud$. Thus, by considering a SVD of $A$ and the previous
comment, we can assume that
$A=\oplus_{i=1}^k\alpha_i\,I_i$ with $I_i\in\mathcal M_{d_i}(\C)$ the identity matrix, $d_1+\ldots+d_k=d$ and $\alpha_1>\ldots>\alpha_k\geq 0$.
Let  $\C^d=\oplus_{i=1}^k \C^{d_i}$, and consider the block representation of $B$ with respect to this decomposition,
$B=(B_{ij})_{i,j=1}^k$.
Under the previous assumption on $A$, we have that $A\,B, AB^*\in\matsad$; then,
$A\,B=(A\,B)^*=B^*\,A$ and $AB^*=(AB^*)^*=BA$.
These equations imply that
$$ B_{ji}^*\,\alpha_j=\alpha_i\, B_{ij} \py  \alpha_i\,B_{ji}^*=B_{ij}\,\alpha_j \peso{for}1\leq i,j\leq k\,.$$
In particular, if $i\neq j$ and $\alpha_i\neq 0$ we get
$$\alpha_i \, B_{ij} = \alpha_j\, B_{ji}^*=\frac{\alpha_j^2}{\alpha_i}  B_{ij}\implies B_{ij}=0\,.$$
In case that $i\neq j$ and $\alpha_i=0$ then
$\alpha_j\,B_{ij}=0 \implies B_{ij}=0$ because $\alpha_j\neq \alpha_i=0$.
And if $\alpha_i\neq 0$ then
$$\alpha_i\, B_{ii}^*=\alpha_i \,B_{ii}\implies B_{ii}=B_{ii}^*\,.$$
Thus $B=\oplus_{i=1}^k B_{ii}$. Let note that if $\alpha_k=0$, the block $B_{kk}\in\cM_{d_k}(\C)$ is arbitrary. Consider now the unitary matrices
$U_i\in\mathcal U(d_i)$ such that $U^*B_{ii}U=D_{\gamma_i}$, with $\gamma_i\in\R^{d_i}$ for $\alpha_i\neq 0$ (that includes $1\leq i\leq k-1$),
and eventually (when $\alpha_k=0$), a SVD $U_k^*B_{kk}V=D_{\gamma_k}$ for
$U_k,\, V_k\in\mathcal U(d_k)$, with $\gamma_k\in \R_{\geq 0}^{d_k}$. Then, taking
$$ U=\oplus_{i=1}^k U_i \py V=\oplus_{i=1}^{k-1}U_i\oplus V_k,$$
and setting $\beta=(\gamma_1\coma \ldots\coma \gamma_k)\in\R^d$, we get
$$ U^*AV=A \py  U^*BV=\oplus_{i=1}^k D_{\gamma_i}=D_{\beta}\,.$$
\EOE
\end{rem}

\begin{pro}\label{estruct de min loc de Lidskki para val sing}
Fix $A\in \mat$, a strictly convex u.i.n. $N$ on $\mat$ and $s\in (\R^d_{\geq 0})\da$, $s\neq 0$.
Let $B\in\cV_s$ be a local minimizer of
$\Psi=\Psi_{(N\coma A\coma s)}$. Then,
$A^*B\coma  A\,B^*\in\matsad$.
\end{pro}
\begin{proof} We argue by induction on the dimension $d\geq 1$. Indeed, in case $d=1$ then
the result follows from the fact that, given $a\in\C$, any local minimizer $b$ of the function $f(c)=|a-c|$ for $c\in\{z\in\C:\ |z|=s>0\}$
satisfies that $\bar a\cdot b\in\R$, and then also $a\cdot \bar b\in\R$.

\pausa
We assume that the result holds for all dimension $\tilde d$ such that $1\leq \tilde d\leq d-1$. Let $A\coma B\in\mat$ be such that $B$ is a local minimizer
of $\Psi$ in $\cV_s$. Notice that by Proposition \ref{pro asumiendo que pi es sumersion},
$\Pi_{(A\coma B)}$ is not a submersion at $(I\coma I\coma I\coma I)$.
By Lemma \ref{lema Pi sobre}, we conclude that there exists $Z\in\mat$, $Z\neq 0$, such that
$A^*\,Z\coma A\,Z^*\coma B^*\,Z\coma B\,Z^*\in\matsad$. Consider a SVD, $D_{s(Z)}=U^*\, Z\, V$, for $U,\, V\in \matud$.
By replacing $A$ and $B$ by $U^*\, A\, V$ and $U^*\, B\, V$ we can further assume that $Z=D_{s(Z)}$,
where $s(Z)=(s_i(Z))_{i\in\I_d}\in (\R^d_{\geq 0})\da$. We let:
\begin{enumerate}
\item $\sigma(Z)=\{\sigma_1>\ldots>\sigma_k\}$ be the distinct eigenvalues of $Z=D_{s(Z)}\in \matpos$.
\item $I_j=\{i\in\I_d:\ s_i(Z)=\sigma_j\}$ and $m_j=\# (I_j)$, for $j\in\I_k$.
\end{enumerate}
Notice that since $Z\neq 0$ then $\sigma_1>0$. Using that $A^*Z\coma AZ^*\coma B^*Z\coma BZ^*\in\matsad$ with
 $Z=\oplus_{j\in\I_k} \sigma_j\, I_j$ and Remark \ref{rem sobre la condicion EY}, we conclude that:
\beq \label{eq rep a,b,a-b}
A=\oplus_{j\in\I_k} A_j \py B=\oplus_{j\in\I_k} B_j\ \ \implies \ \ A-B=\oplus_{j\in\I_k} A_j -B_j \, ,
\eeq
where $I_j\coma A_j\coma B_j \in\mathcal M_{m_j}(\C)$, for $j\in\I_k$; moreover, $A_j\coma B_j \in\mathcal H(m_j)$, whenever $\sigma_k\neq 0$,
for $j\in\I_k$. Using the fact that $B$ is a local minimizer of $\Psi$ on $\cV_s$ we see that
$$B_j \ \text{is a local minimizer of} \ \Psi_{(N_j\coma A_j\coma s(B_j))} \  , \ \text{for}  \ j\in\I_k\, ,$$
where $N_j$ is the strictly convex u.i.n. on $M_{m_j}(\C)$ given by $N_j(C)=N(C\oplus 0_{d-m_j})$.
In turn, this last fact shows that for each $j\in\I_k$ for which $\sigma_j\neq 0$ - which includes all $1\leq j\leq \max\{k - 1\coma 1\}$ -
$B_j$ is a local minimizer of $\Phi_{(N_j\coma A_j\coma \la(B_j))}$; Theorem
\ref{teo LLTApp} shows that $A_j$ and $B_j$ commute, so
$A_j^*B_j\coma A_j\, B_j^*\in \mathcal H(m_j)$, for $j\in\I_k$ such that $\sigma_j\neq 0$.
Therefore, we consider two possible cases: on the one hand, if $\sigma_k\neq 0$ then
the previous remarks show that $$A^*\,B=\oplus_{j\in\I_k} A_j^*\,B_j\in \mathcal H(d)\,,$$ and similarly, $A\,B^*\in\matsad$.

\pausa
On the other hand, if $\sigma_k=0$, notice that $m_k=\dim \ \ker Z<d$ (since $Z\neq 0$), and
$B_k\in M_{m_k}(\C)$ is a local minimizer of  $\Psi_{(N_k\coma A_k\coma s(B_k))}$.
In this case we can apply the inductive hypothesis and conclude that $A_k^*\,B_k\coma A_k\, B_k^*\in\mathcal H(m_k)$.
Since we have already showed that $A_j^*\,B_j\coma A_j\, B_j^*\in\mathcal H(m_j)$, for $1\leq j\leq k-1$, we now see that
$A^*\,B\coma A\, B^*\in\mathcal H(d)$.
\end{proof}
\begin{teo}\label{desc en vs simultanea}
Let $A\in\mat$, $s\in(\R^d_{\geq 0})\da$ and $N$ be a strictly convex u.i.n.
If $B$ is a local minimizer of $\Psi$ in $\cV_s$ then
$A$ and $B$ have a joint SVD i.e.,
there exist $U,V\in\cU(d)$ such that
$$
A=U^*D_{s(A)}V \py B=U^*D_{s(B)}V.
$$ In particular, $s(A-B)=|s(A)-s(B)|\da$ and $B$ is a global minimizer of $\Psi$ in $\cV_s$.
\end{teo}
\begin{proof}
 Notice that if $B$ is a local minimizer of $\Psi$ in $\cV_s$ and $X^*AY=D_{s(A)}$ is a SVD
 of $A$ for some $X,\, Y\in\matud$, we can replace  $A$ by $D_{s(A)}$ and $B$ by $X^*BY$ to get
$$
N(A-B)=N(D_{s(A)}-X^*BY).
$$
Since $\cV_s\ni C\mapsto X^*CY\in \cV_s$ is a homeomorphism of $\cV_s$ then we can assume,
 without loss of generality, that $A=D_{s(A)}$. By Proposition \ref{estruct de min loc de Lidskki para val sing} we get that $A\,B,\, A\,B^*\in\matsad$;
then by \cite{EY} (see Remark \ref{rem sobre la condicion EY}) there exist matrices $U,\, V\in \matud$ such that
$$ U^* A\,V=D_{s(A)}\,(=A) \py U^*B\,V=D_\beta\peso{with} \beta \in\R^d\,. $$
Suppose now that $\beta\notin \R_{\geq 0}^{d}$, so there exists $1\leq \ell \leq d$ such that $\beta_\ell<0$.
Notice that the function $f(t):[0,\pi]\rightarrow \R_{\geq 0}$ given by
$$f(t)=|s_\ell(A)- e^{it}\beta_\ell| \peso{for} t\in[0,\pi]$$ is strictly decreasing. Let
$W(t)=(w_{jk})_{j,\,k\,\in\I_d}\in\matud$ be the diagonal matrix whose main diagonal is  given
 by $w_{jj}=1$ for all $j\neq \ell$, and $w_{\ell\ell}=e^{it}$ for $t\in[0,\pi]$;
hence $W(0)=I$. Define
$$
B(t)=U\,W(t)D_\beta V^* \peso{for} t\in [0,\pi]\,.
$$
Then $B(t)$ is a continuous curve such that $B(0)=B$, $B(t)\in \cV_{s}$ for $t\in [0,\pi]$, and
$$
\Psi(B(t))= N(U\,(D_{s(A)} -D_{\beta(t)})\, V^*)=N(D_{|\alpha-\beta(t)|}) \ ,
$$ where $\beta(t)=s(B(t))$ for $t\in [0,\pi]$. Hence, $\beta_j(t)=\beta_j$ for $j\neq \ell$ and $\beta_\ell(t)=e^{it}\,\beta_\ell$. Therefore,
$|\alpha_j-\beta_j(t)|=\alpha_j-\beta_j$ is constant for $j\neq \ell$ and $|\alpha_\ell-\beta_\ell(t)|=f(t)$ for $t\in[0,\pi]$. Since $f$ is strictly decreasing, we conclude that $$|\alpha-\beta(t)|\prec_w  |\alpha-\beta| \ \implies \ \Psi(B(t)) \ \text{is strictly decreasing for} \ t\in [0,\pi]\,.$$ This last fact contradicts the assumption of $B$.
Therefore $\beta \in\R_{\geq 0}^{d}$.

\pausa
Suppose now that $\beta\neq s=s(B)$ i.e. $\beta\neq \beta\da$; since $A=D_{s(A)}$ with $s(A)=s(A)\da$ then,
by Theorem \ref{teo LLTApp}, $D_\beta$ is not a local minimizer
of $\Phi=\Phi_{(N,\,D_{s(A)},\,s)}$ on $\cO_{s}\,$. Then, there exists a continuous curve
$\delta(t):[0,1]\rightarrow \matud$ such that $\delta(0)=I$ and
$h(t)=N(D_{s(A)}-\delta(t)^*\, D_\beta\, \delta(t)\,)$,
is strictly decreasing in $[0,1]$. Therefore, if we let $\tilde B(t)=U\,\delta(t)^*\, D_\beta\,\delta(t)\,V^*$ for
$t\in [0,1]$ then $\tilde B(0)=B$, $\tilde B(t)\in \cV_s$ for $t\in [0,1]$, and the function $\Psi(\tilde B(t))=h(t)$
is strictly decreasing in $[0,1]$. These facts contradict our assumption that $B$ is a local minimizer of $\Psi$ in $\cV_s$.
Hence, $U^*BV=D_{s}$ and then, $s(A-B)=\abs{s(A)-s}\da$ which implies that $B$ is a global minimizer of $\Psi$ in $\cV_s$.
\end{proof}

\begin{cor}[Equality in Lidskii's inequality for singular values]\label{coro caract eq in desi Lidskii val sin} Let $A,\, B\in\mat$.
Then $|s(A)-s(B)|\da=s(A-B)$ if and only if $A$ and $B$ have a joint SVD.
\end{cor}
\begin{proof} In case $A\coma B\in \mat$ have a joint SVD, then it is straightforward to show that $|s(A)-s(B)|\da=s(A-B)$.
Conversely, assume that $|s(A)-s(B)|\da=s(A-B)$ and choose your favorite strictly convex u.i.n. $N$ on $\mat$. By the comments
at the beginning of this section, we see that $B$ is a global minimizer of $\Psi_{(N\coma A\coma s(B))}=\Psi$ on $\cV_{s(B)}$. In particular, $B$ is a
local minimizer of $\Psi$ in  $\cV_{s(B)}$; the result now follows from Theorem \ref{desc en vs simultanea}.
\end{proof}

\section{Application: Generalized Strawn's conjecture}\label{sec 4}

In this section we consider some problems within the theory of finite frames (see the texts \cite{FinFram, Chr}.
for general references on this topic). It is worth pointing out that our results can be also be described as
the solution to certain matrix nearness problems, following the scheme of \cite{High} (see Remark \ref{rem equiv probs nearness}).

\pausa
In what follows we adopt the following:

\pausa
{\bf Notation and terminology}: let $\cF=\{f_i\}_{i\in\I_k}$ be a finite sequence in $\C^d$. Then,
\ben
\item $T_\cF\in \cM_{d,k}(\C)$ denotes the synthesis operator of $\cF$ given by $T_\cF\cdot(\alpha_i)_{i\in\I_k}=\sum_{i\in\I_k}\alpha_i\, f_i$.
\item $T_\cF^*\in \cM_{k,d}(\C)$ denotes the analysis operator of $\cF$ and it is given by $T_\cF^*\cdot f=(\langle f,f_i\rangle)_{i\in\I_k}$.
\item  $S_\cF\in \matpos$ denotes the frame operator of $\cF$ and it is given by $S_\cF=T_\cF\,T_\cF^*$. Hence, $Sf=\sum_{i\in\I_k} \langle f,f_i\rangle f_i=\sum_{i\in\I_k} f_i\otimes f_i (f)$ for $f\in\C^d$.
\item We say that $\cF$ is a frame for $\C^d$ if it spans $\C^d$; equivalently, $\cF$ is a frame for $\C^d$ if $S_\cF$ is a positive invertible operator acting on $\C^d$.
\EOE\een

\subsection{Generalized frame operator distances}
Let $S\in\matpos$ and $\asubi$. In this case we consider
$$
\tcal := \llav{ \cG=\{g_i\}_{i\in\I_k} \in  (\C^d)^{k} \  :
\norm{g_{i}}^2=a_i  \,\mbox{, for every }   i \in \IN{k}}\,.
$$ By definition, $\tcal$ is the (Cartesian) product of spheres in $\C^d$;
hence, we consider the product metric of the Euclidean metrics in each of these spheres, namely $$d(\cG\coma \cG')=\max\{\|g_i-g_i'\|:\ i\in\I_k\} \peso{for}
\cG=\{g_i\}_{i\in\I_k},\, \cG'=\{g_i'\}_{i\in\I_k}\in\tcal\,.$$  Notice that
$\tcal$ is a compact metric space with the product metric.
Given a strictly convex u.i.n $N$ on $\mat$, we can consider
the generalized frame operator distance (G-FOD) in $\tcal$ (see \cite{dnp}) given by
$$\Theta_{(N,\, S,\, \ca)}=\Theta: \tcal \rightarrow \R_{\geq0} \peso{given by} \Theta(\cG)=N(S-S_{\cG})\,$$
where $S_\cG=\sum_{i\in\I_k} g_i\otimes g_i$ denotes the frame operator of a family $\cG\in\tcal$.
This notion is based on the frame operator distance (FOD) $\Theta_{(\|\cdot\|_2,\, S,\, \ca )}$ introduced by Strawn in \cite{strawn}, where
$\|A\|_2^2=\tr(A^*A)$ denotes the Frobenius norm, $A\in\mat$.

\pausa
Based on his work and on numerical evidence, Strawn conjectured in \cite{strawn} that
local minimizers of $\Theta_{(\|\cdot\|_2,\, S,\, \ca )}:\tcal\rightarrow \R_{\geq 0}$ are global minimizers. In
\cite{dnp} we settled Strawn's conjecture in the affirmative; indeed, we obtained the following results
related to the more general G-FOD induced by a strictly convex u.i.n.:
\begin{teo}[See\cite{dnp}]\label{teo dnp} Let $S\in\matpos$ and $\asubi$. Then, there exists $\nu^{\rm op}=
\nu^{\rm op}(S\coma \ca)\in (\R_{\geq 0}^d)\da$ (that can be computed explicitly) such that:
\begin{enumerate}
\item There exists $\cG^{\rm op}\in \tcal$ such that $\lambda(S_{\cG^{\rm op}})=\nu$. In this case, if $N$ is a u.i.n. in $\mat$ then
\begin{equation}\label{eq nuopop}
\Theta_{(N,\, S,\, \ca)}(\cG^{\rm op})\leq \Theta_{(N,\, S,\, \ca)}(\cG) \quad \text{for} \ \cG\in\tcal\,.
\end{equation}
\item If $N$ is a strictly convex u.i.n. and $\cG_0$ is a global minimizer of $\Theta_{(N,\,S,\, \ca)}$ on $\tcal$ then $\lambda(S_{\cG_0})=\nu^{\rm op}$.
\item If $\cG_0$ is a local minimizer of $\Theta_{(\|\cdot\|_2,\, S, \,  \ca)}$ on $\tcal$ then $\lambda(S_{\cG_0})=\nu^{\rm op}$; hence $\cG_0$
is a global minimizer of $\Theta_{(\|\cdot \|_2,\, S,\, \ca )}$.
\qed
\end{enumerate}
\end{teo}
\pausa
We point out that Theorem \ref{teo dnp} is obtained in terms of a translation of G-FOD problems into frame completion
problems with prescribed norms. Roughly speaking, given $S\in\matpos$ and $\asubi$ as above, there exists a family $\cF_0=\{f_i\}_{i\in\I_d}\in\C^d$
such that for each $\cG\in\tcal$ the operator difference
\beq\label{eq dnp trans}S-S_{\cG}=\|S\|\,I-(S_{\cF_0}+S_\cG)\,.\eeq
Notice that if we let $\cF=(\cF_0\coma \cG)\in (\C^d)^{d+k}$ be the finite sequence obtained by juxtaposition of $\cF_0$ and $\cG$ then
$S_{\cF_0}+S_\cG=S_{\cF}$. In \cite{dnp} any such $\cF$ is called a completion of $\cF_0$ by a family $\cG$, with norms prescribed by the sequence $\ca$.
Eq. \eqref{eq dnp trans} can be used to show items 1. and 2. in Theorem \ref{teo dnp} for a u.i.n. $N$. In order to get information about
local minimizers of $\Theta_{(N,\, S, \, \ca)}$ from Eq. \eqref{eq dnp trans} we should assume further that $N$ is the Frobenius norm.
This obstruction to the general case of item 3. (for a strictly convex u.i.n. $N$) seems to be a limitation of the
reduction methods from \cite{dnp}. Hence, we state the following:
\begin{conj}\label{conjetura gfod}Given $S\in\matpos$ and $\asubi$ and a strictly convex u.i.n. $N$ in $\mat$, let
$\Theta_{(N,\, S, \, \ca)}:\tcal\rightarrow \R_{\geq 0}$ be given by $\Theta_{(N,\, S, \, \ca)}(\cG)=N(S-S_\cG)$.
If $G_0$ is a local minimizer $\Theta_{(N,\, S, \, \ca)}$ in $\tcal$ then:
\begin{enumerate}
\item $\la(S-S_{\cG_0}) \prec \la(S-S_{\cG})$, for every $\cG\in\tcal$;
\item In particular, $\cG_0$ is a global minimizer of $\Theta_{(\tilde N,\, S, \, \ca)}$, for every u.i.n. $\tilde N$ on $\mat$.
\EOE
\end{enumerate}
\end{conj}
\pausa
We point out that item 2. in Conjecture \ref{conjetura gfod} is a consequence of item 1. Nevertheless, item 2. is directly related
with the possible applications of the solution of Conjecture \ref{conjetura gfod} for the G-FOD problems.

\pausa
In what follows, we will describe the first features of local minimizers of $\Theta_{(N,\, S, \, \ca)}:\tcal\rightarrow \R_{\geq 0}$,
for an arbitrary strictly convex u.i.n. $N$ in $\mat$.
We will also show that Conjecture \ref{conjetura gfod} holds under some further hypothesis on the spectral structure of local minimizers.

\pausa
We end this section with the following remark, in which we show the connection between G-FOD problems and matrix nearness problems.
\begin{rem}\label{rem equiv probs nearness}
Let $S\in\matpos$ and consider a strictly convex u.i.n. $N$ in $\mat$. Let $\mu_j\in(\R^d)\da$, for $j\in\I_k$, and consider the orbits
$$\mathcal O_{\mu_j}=\{G\in\matsad:\ \lambda(G)=\mu_k\}\ , \ j\in\I_k\,.$$
We can then consider the matrix nearness problem (as described in \cite{High}, see also \cite{LiPoon})
\begin{equation}\label{eq equiv prob aprox}
\text{argmin}\ \{N(S-H)):\ H\in\mathcal O_{\mu_1}+ \ldots + \mathcal O_{\mu_k} \}\,.
\end{equation}
Let $\asubi$ and consider the particular case: $\mu_j=a_j\, e_1$, for $j\in\I_k$, where $\{e_i\}_{i=1}^d$ denotes the canonical basis of $\C^d$.
Then $G\in \mathcal O_{\mu_j}$ if and only if $G=g\otimes g$ for some $g\in\C^d$ with $\|g\|^2=a_j$, $j\in\I_k$. Hence, the matrix nearness problem
in Eq. \eqref{eq equiv prob aprox} coincides with the problem of computing global minimizers on $\Theta_{(N\coma S\coma \ca)}$ in $\mathbb T_d(\ca)$. Similarly,
the study of local minimizers of the matrix nearness problem corresponds to the study of local minimizers of $\Theta_{(N\coma S\coma \ca)}$.
It is worth pointing out that for the Frobenius norm, local minimizers of the matrix nearness problem arise naturally as stability points of
(effective) gradient descent algorithms, as those considered in \cite{LiPoon}. Hence, settling Conjecture \ref{conjetura gfod} in the affirmative would be a
relevant result from an applied point of view. \EOE

\end{rem}

\subsection{Properties of local minimizers of the G-FOD on $\tcal$}

In this section we consider the following

\begin{nota}\label{notaciones strawn}
Fix $S\in\matpos$,  $\asubi$ and a strictly convex u.i.n. $N$ on $\mat$. We consider
\begin{enumerate}
\item $\Theta_{(N,\, S, \, \ca)}=\Theta: \tcal \rightarrow \R_{\geq0}$ given by $\Theta(\cG)=N(S-S_{\cG})$.
\item A local minimizer $\cG_0=\llav{g_i}_{i\in\I_k}\in\tcal$ of $\Theta_{(N,\, S, \, \ca)}$, with frame operator
$S_0=S_{\cG_0}$.
\item For $\mu\in(\R^d)\da$, the unitary orbit  $\cO_\mu$  given by
$$\cO_\mu=\{G\in\matsad:\la(G)=\mu\}=\{U^* D_{\mu}\,U:\ U\in\matud\}\,, $$
with the usual metric, induced by the operator norm;
\item The function $\Phi=\phisn:\cO_\mu\rightarrow \R_{\geq 0}$ given by
$ \Phi(G)=N(S-G).$
\end{enumerate}
\end{nota}

\begin{teo}\label{teo applic1}
Consider Notation \ref{notaciones strawn}. Then,
\begin{enumerate}
\item $S-S_0$ and $g_j \otimes g_j$ commute, for $j\in\I_k$. Hence, $g_j$ is an eigenvector of $S-S_0$, for $j\in\I_k$.

\item There exists $\{v_i\}_{i\in\I_d}$ an ONB of $\C^d$ such that
$$
S=\sum_{i\in\I_d} \la_i(S) \ v_i\otimes v_i
\py  S_0=\sum_{i\in\I_d} \la_i(S_0) \ v_i\otimes v_i
\,.
$$
In particular, we have that $\la(S-S_0)=\big[\la(S)-\la(S_0)\,\big]\da$.
\end{enumerate}
\end{teo}
\begin{proof}
For $j\in\I_k$ define  $$S_{[j]}=S-\sum_{i\neq j} g_i \otimes g_i\,\in\cH(d) \py \mu_{[j]}=a_j e_1\,\in\R^d_{\geq 0}\,.$$
Then, $\cO_{\mu_{[j]}}=\llav{g\otimes g: \norm{g}^2=a_j}$
 and it is straightforward to check that $g_j \otimes g_j$ is a local minimizer of $\Theta_{(N,\, S_{[j]},\, \mu_{[j]})}$ in $\cO_{\mu_{[j]}}$. Thus,
by Theorem \ref{teo LLTApp},
 $g_j \otimes g_j$ commutes with $S_{[j]}$, for $j\in\I_k$.
This last fact implies that $S-S_0$ and $g_j \otimes g_j$ commute, for $j\in\I_k$, which proves item 1.

\pausa
Since $\cG_0$ is a local minimizer of $\Theta$ in $\tcal$, there exists $\varepsilon>0$ such that
\beq\label{eq loc para Phi}
U \in B_{(I\coma \eps)}\igdef
\{U\in\cU(d): \|I-U\|<\epsilon\}\  \implies  \ \Phi_{(N,\, S,\, \mu)}(U\,S_{0}\,U^*)\geq  \Phi_{(N,\, S,\, \mu)}(S_{0})\, ,
\eeq where we are using Notation \ref{notaciones strawn}, with $\mu=\lambda(S_0)$. Indeed, let $\varepsilon>0$ be such that for
$\cG'\in \tcal$ with $d(\cG_0,\cG')<\varepsilon$ we have that $\Theta(\cG')\geq \Theta(\cG_0)$. Notice that if $U\in
B_{(I\coma \eps)}$ then $U\cdot\cG_0=\{U\,g_i\}_{i\in\I_k}\in\tcal$ is such that $d(\cG_0\coma U\cdot\cG_0)<\epsilon$.
Therefore, $$
\Phi_{(N,\, S,\, \mu)}(U\,S_{0}\,U^*)=\Theta(U\cdot\cG_0)
\geq \Theta(\cG_0)=\Phi_{(N,\, S,\, \mu)}(S_{0})\,.
$$
Now, the map 	$\pi : \matud \to \cO_\mu$ given by $\pi(U) = U\,(S_{0})\,U^*$ is open
(see \cite[Thm 4.1]{AS}), so that $\pi (B_{(I\coma \eps)})$ is an open
neighborhood of $S_{0}$ in $\cO_\mu\,$, and
$S_{0}$ is a local minimum for the map $\phisn$ on $\cO_{\mu}\,$.
Item 2 now follows from
Theorem \ref{teo LLTApp} and the fact that $\mu=\la(S_0)\in(\R^d)\da$.
\end{proof}

\begin{cor}\label{cor consecuencias1}
Consider Notation \ref{notaciones strawn}. Let $W=R(S_0)\subset \C^d$; then,
\begin{enumerate}
\item $W$ reduces $S-S_0\in\cH(d)$; hence, $D:=(S-S_0)|_W\in L(W)$ is a selfadjoint operator;
\item Let $\sigma(D)=\{c_1\coma \ldots\coma c_p\}$ be such that $c_1<c_2<\ldots<c_p$ and let
$$J_j=\{\ell \in \I_k: \ D\,g_\ell=c_j\, g_\ell\} \peso{for} j\in\I_p\,.$$Then $\I_k$ is the disjoint union of $\{J_j\}_{j\in\I_p}$;
\item If we let $W_j=\text{span}\{g_\ell:\ \ell\in J_j\}$ then $W_j$ reduces both
$S$ and $S_0$, for $j\in\I_p$. Moreover, $W=\oplus_{j\in\I_p} W_j$.
\end{enumerate}
\end{cor}
\begin{proof}
Notice that $W=\text{span}\{g_i:\ i\in\I_k\}$; on the other hand, by Theorem \ref{teo applic1}, $g_i$ is an eigenvector of $S-S_0$, for each $i\in\I_k$.
These two facts show that $W$ is an invariant subspace of $S-S_0$; since $S-S_0$ is selfadjoint, $W$ reduces $S-S_0$. Thus, the restriction
$D=(S-S_0)|_W\in L(W)$ is a well defined selfadjoint operator acting on $W$. The previous remarks also show that
$\I_k$ is the disjoint union of $\{J_i\}_{i\in\I_p}$.

\pausa
Let $j,\,\ell \in \I_p$ with $j\neq \ell$ and let $r\in J_j$ and $s\in J_\ell$. Then, $g_r\perp g_s$, since these vectors are eigenvectors
of a selfadjoint operator, corresponding to different eigenvalues. Hence, $W_j\perp W_\ell$ and
$$ S_0 \,g_r=\sum_{u\in J_j} \langle g_r\coma g_u\rangle\ g_u\in W_j \,.$$ Thus, in particular, $W_j$ reduces $S_0$; using that $W_j$ also reduces $S-S_0$ we conclude that
$W_j$ reduces $S=(S-S_0)+S_0$, for $j\in\I_p$. On the other hand, since $W=\sum_{j\in\I_p}W_j$ then
$W=\oplus_{j\in\I_p} W_j$.
\end{proof}

\begin{teo}\label{gen teo gen 4.6}
Consider Notation \ref{notaciones strawn}. Let $W=R(S_0)$ and let $\sigma((S-S_0)|_W)=\{c_1\coma\ldots\coma c_p\}$ as in Corollary
\ref{cor consecuencias1}. Let $j\in \I_p$ and assume that there exists $c\in \sigma(S-S_0)$ such that $c_j<c$. Then, the family $\{g_j\}_{j\in J_j}$ is linearly independent.
\end{teo}
\begin{proof}
Suppose that for some $j\in\I_p$ the family $\{g_i\}_{i\in J_j}$ is linearly dependent.
Hence there exist coefficients $z_l\in \C$, $l\in J_j$ (not all
zero) such that every $|z_l|\leq 1/2$ and
\begin{equation}\label{eq1}
\sum_{l\in J_j}\overline{z_l} \ a_l\rai \ g_l=0\ .
\end{equation}
Let $I_j \inc J_j$ be given by $I_j =\{l\in J_j:\ z_l\neq 0\}$. Assume that there exists
$c\in\sigma(S-S_0)$ such that $c_j<c$
and let $h\in \C^d$ be such that $\|h\|=1$ and  $(S-S_0)\, h=c \,h$.
For $t\in (-1/2,1/2)$ let
$\cG(t)=\{g_i(t)\}_{i\in\IN{k}}$ be given by
$$
g_l(t) = \begin{cases}  \ (1-t^2\,|z_l|^2)^{1/2} g_l+t\,z_l\,a_l\rai \, h
& \mbox{if} \ \ l\in I_j\,;  \\%&\\
\quad \quad\quad g_l & \mbox{if} \ \ l\in \I_k\setminus I_j  \  .
\end{cases}
$$ Notice that $\cG(t)\in \tcal $ for $t\in (-1/2,1/2)$.
 Let $\Preal(A)= \frac {A+A^*}{2}$ denote the real part of
$A \in \mat$. For $l\in I_j$ then
$$
g_l(t)\otimes g_l(t)
=(1-t^2\,|z_l|^2)\ g_l\otimes g_l+ t^2\,|z_l|^2\,a_l \ h\otimes h
+ 2\,(1-t^2\,|z_l|^2)^{1/2}\,t \ \Preal(h\otimes \overline{z_l}\, a_l\rai\, g_l)
$$
Notice that $\cG(t)$ is a continuous curve in $\tcal$ such that $\cG(0)=\cG_0$.
Let $S(t)$ denote the frame operator of $\cG(t)\in \tcal $, so that $S(0)=S_0$, and let
$T(t)=S-S(t)$ for $t\in (-1/2\coma 1/2)$.
Note that
$$
T(t)=S-S_0+ t^2 \sum_{l\in I_j }  |z_l|^2 \left(  g_l\otimes g_l - a_l \ h\otimes h \right) + R(t)
$$
where $R(t)=- 2 \suml_{l\in I_j }(1-t^2\,|z_l|^2)^{1/2}\,t \ \Preal(h\otimes a_l\rai\,\overline{z_l}\, g_l)$.
Then $R(t)$ is a smooth function such that
$$
R(0) = 0 \ \ , \ \
R'(0)=- \sum_{l\in I_j } \Preal(h\otimes \overline{z_l}\, a_l\rai\,g_l)
=- \Preal(h\otimes \sum_{l\in I_j } \overline{z_l}\,a_l\rai\, g_l) \stackrel{\eqref{eq1}}{=} 0 \ ,
$$
and such that $R''(0)=0$. Therefore
$\lim\limits_{t\rightarrow 0} \ t^{-2}\ R(t)=0 $.
We now consider
$$
V=\gen\,\big(\,\{g_l:\ l\in I_j \}\cup \{h\}\,\big)
=\gen\,\big\{\,g_l:\ l\in I_j \,\big\}\stackrel{\perp}{\oplus} \C\cdot h\ .
$$
Then $\dim V=s+1$,  for $s=\dim\gen\{g_l:\ l\in I_j \}\geq 1$.
By construction, the subspace $V$ reduces $S-S_0$ and $T(t)$ in such a way that $(S-S_0)|_{V^\perp}=T(t)|_{V^\perp}$,
for $t\in(-1/2\coma 1/2)$. On the other hand
\beq\label{adet}
T(t)|_{V}=(S-S_0)|_V+t^2 \sum_{l\in I_j }  |z_l|^2 \left(  g_l\otimes g_l
- a_l \ h\otimes h \right) + R(t) = A(t)+R(t)\in L(V)\ ,
\eeq
where we use the fact that the ranges of the selfadjoint operators
in the second and third term in the formula above clearly lie in $V$.
Then $\la\big((\,S-S_0)|_V\,\big)=\big(\,c\coma   c_j \, \uno_s\, \big)
\in (\R^{s+1}_{>0})\da $ and
$$\barr{rl}
\la \Big(\, \sum_{l\in I_j }  |z_l|^2  g_l\otimes g_l\,\Big) &
=(\gamma_1 \coma \ldots \coma \gamma_s \coma 0)
\in (\R^{s+1}_{\geq 0})\da \peso{with} \gamma_s>0 \ ,\earr
$$
where we have used the definition of $s$ and the fact that $|z_l|>0$ for $l\in I_j \,$ (and the known fact that
if $	S\coma T\in \matpos \implies R(S+T) = R(S)+R(T)\,$).
Hence, for sufficiently small $t$,
the spectrum of the operator  $A(t)\in L(V)$
defined in Eq. \eqref{adet} is
$$\barr{rl}
\la\big(\, A(t)\,\big)
&=\big(\, c
-t^2 \, \sum_{l\in I_j }a_l\,|z_l|^2 \coma c_j+t^2\,\gamma_1 \coma \ldots \coma c_j+t^2 \,\gamma_s \,\big) \in (\R^{s+1}_{\geq 0})\da \ , \earr
$$
where we have used the fact that $\langle g_l \coma h\rangle=0$ for every $l\in I_j \,$.
Let us now consider
$$
\la\big(\, R(t)\,\big)
=\big(\,\delta_1(t) \coma \ldots \coma \delta_{s+1}(t)\, \big)
\in (\R^{s+1}_{\geq 0})\da\peso{for} t\in \R \ .
$$
Recall that in this case $\lim\limits_{t\rightarrow 0}t^{-2} \delta_j(t)=0$
for $1\leq j\leq s+1$. Using Weyl's inequality
on Eq. \eqref{adet},  we now see that
\beq\label{eq pro nueva 0}
\lambda \big(\,T(t)|_V\,\big)\prec \la\big(\, A(t)\,\big)
+\la\big(\, R(t)\,\big)\igdef \rho(t)\in (\R^{s+1}_{\geq 0})\da\,.\eeq We know that
$$
\barr{rl}
\rho(t)&= \big(\, c-t^2 \, \sum_{l\in I_j }a_l\,|z_l|^2  +\delta_{1}(t)\coma c_j+t^2\,\gamma_1+\delta_2(t) \coma
\ldots \coma c_j+t^2 \,\gamma_s+\delta_{s+1}(t) \, \big)
\\ &\\
&=
\Big(\, c-t^2 \, (\sum_{l\in I_j }a_l\,|z_l|^2  +\frac{\delta_{1}(t)}{t^2})\coma
c_j+t^2\,(\gamma_1+\frac{\delta_2(t)}{t^2}) \coma \ldots \coma
c_j+t^2 \,(\gamma_s+\frac{\delta_{s+1}(t)}{t^2}) \,\Big)  \ .
\earr
$$
Since by hypothesis $c_j<c$ then, the previous remarks show that there exists $\varepsilon>0$ such that
if $t\in(0,\varepsilon)$ then, for every $i\in \I_s$
$$
c>c-t^2 \, (\sum_{l\in I_j }a_l\,|z_l|^2+\frac{\delta_{1}(t)}{t^2})>
c_j+t^2(\gamma_{i}+\frac{\delta_{i+1}(t)}{t^2}) >c_j\,.$$
The previous facts show that for $t\in(0,\varepsilon)$ then
 $\rho(t) \prec
\lambda((S-S_0)|_V)=\big(\,c\coma   c_j \, \uno_s\, \big)$ strictly.
Therefore,
\begin{eqnarray*}
\lambda(T(t))&=&\big(\lambda((S-S_0)|_{V^\perp})\coma T(t)|_V\,\big)\da \stackrel{\eqref{eq pro nueva 0}}{\prec}
\big(\lambda((S-S_0)|_{V^\perp})\coma \rho(t)\,\big)\\ &\prec& \big(\lambda((S-S_0)|_{V^\perp})\coma \lambda(S-S_0)|_V\,\big)\da=\lambda(S-S_0)\,,
\end{eqnarray*}
where the second majorization relation is strict (i.e.
$(\lambda((S-S_0)|_{V^\perp})\coma \rho(t))\da\neq (\lambda((S-S_0)|_{V^\perp})\coma \lambda(S-S_0)|_V)\da$).
Since $N$ is strictly convex,
for every $t\in(0,\varepsilon)$ we have that
$$
\Theta(\cG(t))=N(T(t))<N(S-S_0)=\Theta(\cG)\,.
$$
This last fact contradicts
the assumption that $\cG_0$ is a local minimizer of $\Theta$ in $\tcal $.
\end{proof}

\subsection{Some special cases of Conjecture \ref{conjetura gfod}}

Consider Notation \ref{notaciones strawn} and assume that $k\geq d$; if we let $W=R(S_0)\subset \C^d$ then,
as shown in Corollary \ref{cor consecuencias1}, $W$ reduces the self-adjoint
operator $S-S_0\in\matsad$. In this section we show that in case $W$ is an eigenspace of $S-S_0$ then Conjecture \ref{conjetura gfod} holds for $\cG_0$ i.e.,
$\cG_0$ is a global minimizer of $\Theta_{(N\coma S\coma \ca)}$
in $\tcal$. In order to tackle this particular case, we introduce the following

\begin{rem}[A naive model]\label{un modelo naivo}
Fix $S\in\matpos$,  $\asubi$ and a strictly convex u.i.n. $N$. We let $t=\tr(\ca)=\sum_{i\in\I_k}a_i>0$.
 If $\cG\in\tcal$ then it is clear that
$$ S_\cG\in \matpos \py \tr(S_\cG)=\sum_{i\in \I_k} \|g_i\|^2=\tr(\ca)=t\,.$$
Hence, we consider
\beq\label{defi N}
\matpos_t=\{ A:\ A\in\matpos\coma \tr(A)=t\} \supset \{S_\cG:\ \cG \in\tcal\}\,,
\eeq endowed with the metric induced by the operator norm.
Moreover, we consider the map
\begin{equation}\label{eq defi D}
\cD_{(N,\,S,\, t)}=\cD:\matpos_t\rightarrow \R_{\geq 0} \peso{given by} \cD(A)=N(S-A)\,.
\end{equation}
By Eq. \eqref{defi N} we see that
\beq\label{eq just N}
\min\{\cD(A):\ A\in\matpos_t\} \leq
\min\{\Theta(\cG):\ \cG\in\tcal\}\,.
\eeq The inequality in Eq. \eqref{eq just N} can be strict. Yet, we will show that under some additional hypothesis equality holds
in Eq. \eqref{eq just N}. Moreover, since $\matpos_t$ is a (larger but) simpler set, we are able to compute those $A\in\matpos_t$
that attain the minimum in the left hand side of Eq. \eqref{eq just N} (see Theorem \ref{teo caso co-fea} below); these facts together will allow us to prove Conjecture \ref{conjetura gfod} in some special cases.
\qed
\end{rem}

\begin{teo}\label{teo caso co-fea}
Let $S\in \matpos$, $\la(S)=(\la_i)_{i\in \I_d}\in(\R_{\geq 0}^d)\da$, $t>0$ and let $N$ be a u.i.n.
Consider $\llav{v_i}_{i\in\I_d}$ an ONB of $\C^d$ such that $S\, v_i=\la_i\ v_i$, for $i\in\I_d$.
Let $c\leq \la_1$ be uniquely determined by
$\sum_{i\in\I_d} (\la_i-c)^+=t$ and set
$$ A^{\rm op}= \sum_{i\in\I_d} (\la_i-c)^+ \,v_i\otimes v_i \in\matpos_t\peso{so that} \lambda(S-A^{\rm op})=(\min\{c\coma \lambda_i\})_{i\in\I_d}\in (\R^d)\da\,.$$
Then, $A^{\rm op}$ is a global minimizer of $\cD$, defined as in Eq. \eqref{eq defi D}.
\end{teo}
\begin{proof}
By construction we see that $\la(S-A^{\rm op})=(\min\{c\coma \la_i\})_{i\in\I_d}$. Let $A\in\matpos_t$ be arbitrary; we consider the following cases:

\pausa
In case $c\leq \la_d$ then we see that $\la(S-A^{\rm op})=c\,\uno_d$. Since $\tr(A)=t$, then
$\tr(\la(S-A))=\tr(S-A)=\tr(S)-t=\tr(\la(S-A^{\rm op}))$. Thus, in this case we have (see item 4. in Remark \ref{desimayo}) that $\la(S-A^{\rm op})=c\, \uno_d\prec \la(S-A)$.
Hence, we conclude that $\cD(A^{\rm op})=N(S-A^{\rm op})\leq N(S-A)=\cD(A)$.

\pausa
In case $c>\la_d$, there exists $r\in\I_{d-1}$ such that $\la_{r}\geq c>\la_r+1$. Then,
$$(\gamma_i)_{i\in\I_d}:=\la(S-A^{\rm op})=(c\,\uno_{r}\coma \la_{r+1} \coma \ldots\coma \la_d )\in(\R^d)\da\,.$$
 If we let $\la(A)=(\alpha_i)_{i\in\I_d}\in (\R^d_{\geq 0})\da$ then,
by Lidskii's additive inequality, we get that
\beq\label{ap defi delta}
(\delta_i)_{i\in\I_d}:= ((\la_i-\alpha_i)_{i\in\I_d})\da=(\la(S)-\la(A))\da\prec \la(S-A)\,.
\eeq
We now show that $(\gamma_i)_{i\in\I_d}\prec (\delta_i)_{i\in\I_d}$;
by construction $\tr((\gamma_i)_{i\in\I_d})=\tr((\delta_i)_{i\in\I_d})$ that is
\begin{equation}\label{trazagam}
\tr(\gamma)=\sum_{j=1}^d \gamma_j= r\, c + \sum_{j=r+1}^d \la_j =
\tr(\delta)=\sum_{j=1}^d \delta_j= \sum_{j=1}^d (\la_j -\al_j)\,.
\end{equation}
 Thus, in order to show that
$(\gamma_i)_{i\in\I_d}\prec (\delta_i)_{i\in\I_d}$ we need to prove that
$\sum_{j=k}^d \gamma_j\geq \sum_{j=k}^d \delta_j$, for every $k\in\I_d$, since the vectors are arranged in non-increasing order.
Notice that $\la_i\geq \la_i-\alpha_i$, for every $i\in\I_d$; then, by Remark \ref{desimayo}, we conclude that
$\la_i\geq \delta_i$, for $i\in\I_d$. This guarantees that, for $r+1\leq k\leq d$,
\begin{equation}\label{ultimos}
\sum_{j=k}^d \gamma_j= \sum_{j=k}^d \la_j\geq  \sum_{j=1}^k \delta_j.
\end{equation}
We now define $\beta=\sum_{j=r+1}^d (\gamma_j - \delta_j)$ and notice that Eq. \eqref{ultimos} shows that $\beta\geq 0$.
By Eq. \eqref{trazagam},
$$
\sum_{j=1}^r \delta_j = r\,(c+\beta/r),
$$ which implies that $(c+\beta/r)\uno_{r} \prec (\delta_{1},\cdots,\delta_r)$. Hence, if $1\leq k\leq r$ then
\beq\label{eq desi para lem ap1}
\sum_{j=k}^r \delta_j \leq (r-k+1)\,(c+\beta/r)\leq (r-k+1)\, c + \beta\,.
\eeq
Therefore, for $1\leq k\leq r$,
\beq\label{eq desi para lem ap2}
\sum_{j=k}^{d} \gamma_j -\sum_{j=k}^{d} \delta_j
= (r-k+1)\,c + \beta -\sum_{j=k}^r \delta_j
\stackrel{\eqref{eq desi para lem ap1}}{\geq} 0\,.
\eeq
Then, Eqs. \eqref{trazagam}, \eqref{eq desi para lem ap1} and \eqref{eq desi para lem ap2} show that $\gamma\prec \delta$.
Finally, if $N$ is a (strictly convex) u.i.n. then
$$N(S-A^{\rm op})=N(D_\gamma)\leq N(D_\delta)\stackrel{\eqref{ap defi delta}}{\leq} N(S-A)$$ so $A^{\rm op}$ is a global minimizer of $\cD$ in
$\matpos_t$.
\end{proof}

\pausa
The next result verifies Conjecture \ref{conjetura gfod} under some additional assumptions on the spectral structure of
local minimizers.

\begin{teo}\label{teo conj caso esp}
Let $S\in\matpos$,  $\asubi$, with $k\geq d$, and let $N$ be a strictly convex u.i.n. in $\mat$. Let $\cG_0=\{g_i\}_{i\in\I_d}$ be a local minimizer of
$\Theta$ in $\tcal$ such that there exists $c_1\in \R$ that satisfies $(S-S_{\cG_0}) g_i=c_1\,g_i$, for $i\in\I_k$.
Then there exists an ONB $\{v_i\}_{i\in\I_d}$ of $\C^d$ such that
\beq\label{eq teo rep s y s0}
S=\sum_{i\in\I_d} \la_i\ v_i\otimes v_i \py S_{\cG_0}=\sum_{i\in\I_d} (\la_i-c_1)^+\ v_i\otimes v_i\,,
\eeq
where $(\la_i)_{i\in\I_d}=\la(S)\in (\R_{\geq 0}^d)\da$. Moreover, $\la(S-S_{\cG_0})\prec \la(S-S_\cG)$ for $\cG\in\tcal$. In particular, $\cG_0$ is a global minimizer of $\Theta$ in $\tcal$.
\end{teo}
\begin{proof} Let $S_0=S_{\cG_0}$. By Theorem \ref{teo applic1}
there exists an ONB $\{v_i\}_{i\in\I_d}$ of $\C^d$ such that
\beq\label{eq rep s y s0}
S=\sum_{i\in\I_d} \la_i \ v_i\otimes v_i
\py  S_0=\sum_{i\in\I_d} \la_i(S_0) \ v_i\otimes v_i
\,.
\eeq In particular, $\la(S-S_0)=(\la(S)-\la(S_0))\da$. Let $W=R(S_0)=\text{span}\{g_i:\ i\in\I_k\}$,
 which reduces $S-S_0$ by Corollary \ref{cor consecuencias1}.  Then, by hypothesis we have that $\sigma((S-S_0)|_W)=\{c_1\}$. We consider the following two cases:

\pausa
Assume that $W=\C^d$. In this case $\sigma(S-S_0)=\{c_1\}$ and therefore $\la(S-S_0)=c_1\,\uno_d$. Thus, $\la_i-\la_i(S_0)=c_1$ which implies that
$\la_i(S_0)=(\la_i-c_1)^+$, for $i\in\I_d$.  Notice that for every $\cG\in\tcal$ we have that $\tr(S-S_\cG)=\tr(S)-\tr(\ca)$; then we see that
$c_1\,\uno_d=\tr(\la(S-S_0))=\tr(S-S_\cG)$ which shows (see item 4. in Remark \ref{desimayo}) that $\la(S-S_0)\prec \la(S-S_{\cG})$ for every $\cG\in\tcal$. This last fact implies that
$\Theta(S_{\cG_0})=N(S-S_0)\leq N(S-S_{\cG})=\Theta(\cG)$, for every $\cG\in\tcal$. Thus, $\cG_0$ is a global minimizer of $\Theta$ in $\tcal$.

\pausa
Assume now that $W\neq \C^d$. Hence, $d>\dim W=\text{span}\{g_i:\ i\in\I_k\}$ which shows that $\cG_0$ is a linearly dependent family, since $k\geq d$.
Then, Theorem \ref{gen teo gen 4.6} implies that $c\leq c_1$ for every $c\in\sigma(S-S_0)$. Let $1\leq r\leq d-1$ be such that $\dim W=r$.
Hence, $\la(S_0)=(\la_{1}(S_0)\coma \ldots\coma \la_{r}(S_0)\coma 0_{d-r})$ and $W=\text{span}\{v_i: 1\leq i\leq r\}$.
Therefore, using Eq. \eqref{eq rep s y s0} and the previous facts we conclude that
$$S-S_0=\sum_{i=1}^r(\la_i - \la_i(S_0))
\ v_i\otimes v_i + \sum_{i=r+1}^d \la_i\ v_i\otimes v_i
= c_1\ \sum_{i=1}^r  \ v_i\otimes v_i +  \sum_{i=r+1}^d  \la_i\ v_i\otimes v_i $$
Thus, $\sigma(S-S_0)\ni \la_i\leq c_1$ and $\la_i(S_0)=0$, for $r+1\leq i\leq d$; hence, $\la_i(S_0)=(\la_i(S)-c_1)^+$, for $r+1\leq i\leq d$.
Then, $\la_i(S_0)=(\la_i-c_1)^+$ for $i\in\I_d$ and therefore we obtain the representation of $S_0$ as in Eq. \eqref{eq teo rep s y s0}.

\pausa
Notice that $c_1=\la_1-(\la_1-c_1)^+$, since $W\neq \{0\}$. This shows that $c_1\leq \la_1$. Moreover, if we let $\tr(\ca)=t>0$ then
$$ \sum_{i\in\I_d}(\la_i-c_1)^+=\tr(S_{\cG_0})=\tr(\ca)=t\,.$$
Using Remark \ref{un modelo naivo} and Theorem \ref{teo caso co-fea} we now see that for every $\cG\in\tcal$ we have that
$\la(S-S_{\cG_0})\prec \la(S-S_\cG)$; in particular,
$$\Theta(\cG)=N(S-S_\cG)=\cD(S_\cG) \geq \cD(S_0)=N(S-S_0) \peso{since} S_\cG\in\matpos_t\,.$$
Thus, $\cG_0$ is a global minimizer of $\Theta$ in $\tcal$.
\end{proof}

\pausa
{\bf Acknowledgment}. We would like to thank Professor Eduardo Chiumiento for fruitful conversations related to the content of this work.


text:
\section{Achievable Rate Analysis}

In this section, the overall full-duplex system performance using the proposed digital cancellation technique is investigated in terms of the achievable rate gain compared to conventional half-duplex systems. The performance is investigated in the same three operating scenarios described in Section III-E. Both full-duplex and half-duplex system performances are investigated in the presence of all transmitter and receiver impairments. The performance is investigated at two transmit power values: 20dBm, and 5dBm.

Generally, the achievable rate for both full-duplex and half-duplex systems can calculated as

\begin{equation}\label{eq:41}
R^{FD}=E\left[log_2\left(1+SINR^{FD}\right)\right]\text{,}
\end{equation}

\begin{equation}\label{eq:42}
R^{HD}=\frac{1}{2} E\left[log_2\left(1+SNR^{HD}\right)\right]\text{,}
\end{equation}

where $E[\ ]$ denotes expectation process, $R^{FD}$, $R^{HD}$ is the full-duplex and half-duplex system's achievable rate, $SINR^{FD}$ is the signal-of-interest to interference plus noise ratio in full-duplex systems, and $SNR^{HD}$ is the signal to noise ratio in half-duplex systems. The factor of $1/2$ in the half-duplex rate equation is due to the fact that the resources are divided between the two communicating nodes.

Figure~\ref{Fig8Label}(a)-(c) shows the achievable rate for the three scenarios at different SNR values. The conclusions from these results are multifold: first, the results show that the proposed technique significantly outperforms the conventional digital cancellation technique in all operating scenarios. The only exception is the case of 5dBm transmit power in the third scenario, where both the proposed technique and the conventional digital cancellation technique achieves the same performance. The reason is that, at 5dBm transmit power with the 60dB passive suppression assumed in the third scenario, only 35dB of digital cancellation are required to suppress the self-interference signal to the noise floor. Those 35dB could be easily achieved using conventional digital cancellation techniques. However, as the transmit power increases, the proposed technique will be able to achieve more self-interference cancellation, while the cancellation achieved using the conventional techniques will saturate. Second, figure~\ref{Fig8Label}(a) shows that, in the first scenario where the passive suppression amount is relatively low, the proposed technique can only be used in low transmit power applications.

Third, in the second and third scenarios, the proposed technique achieves significant rate improvement compared to conventional half-duplex systems, especially in high SNR regimes. In order to quantify the achievable rate gain of the proposed full-duplex system compared to the conventional half-duplex system, we calculate the average rate gain over the whole SNR range from 0 to 40dB for all scenarios, and the results are shown in table~\ref{Table1Label}. The results in table~\ref{Table1Label} are average results; however, exact rate gain for each SNR value could be obtained from figure~\ref{Fig8Label}(a)-(c). Negative rate gain means that operating in full-duplex mode will degrade the overall system performance.

config: arxiv


TITLE: $\lim_{n \to +\infty}\frac {3^n}{n^32^n}$
QUESTION [0 upvotes]: I'm trying to evaluate $$\lim_{n \to +\infty}\frac {3^n}{n^3 \cdot 2^n}$$
Can someone help me?
It should be 0

REPLY [2 votes]: Prove that $$\left(\frac{3}{2}\right)^n>n^4$$ for all $n\geq40.$


TITLE: What would happen in a single slit diffraction experiment if the slit ends are made of photo sensitive material?
QUESTION [0 upvotes]: In a single slit diffraction experiment if the slit ends are made of a photosensitive material,
And if a light wave falls on the ends of the slit, will it get diffracted or will it lead to ejection of photoelectron?
Diffraction shows the wave nature of light and photoelectric effect shows particle nature of light.
In the situation stated above will it show wave nature or particle nature?

REPLY [2 votes]: If done with regular light sources it will show both, provided that the frequency is above the threshold. However if you send in photons one at a time, you’ll either see the photoelectric effect, or a detection on the screen. But not both simultaneously. Keep in mind that the detection on the screen is still a blip localised at some point. But over time, from the buildup of detections on the screen the diffraction pattern will emerge.
This is equivalent to saying that the photon will either be detected at the obstruction (slit) or will pass through the slit (detection at the screen). You might be interested in this answer.


TITLE: Definition of Property: Function vs Subset
QUESTION [1 upvotes]: Wikipedia says:

In mathematics, a property is any characteristic that applies to a given set. Rigourously, a property $p$ defined for all elements of a set $X$ is usually defined as a function $p:X\to \{\text{true},\text{false}\}$, that is true whenever the property holds; or equivalently, as the subset of $X$ for which $p$ holds; i.e. the set $\{x:p(x)=\text{true}\}$; $p$ is its indicator function.

Technically speaking, aren't the function and the subset definitions different? I think a function from $X$ to $\{\text{true},\text{false}\}$ is a subset of the Cartesian product of $X$ and $\{\text{true},\text{false}\}$, not just a subset of $X$. If I'm wrong, what am I missing?

REPLY [1 votes]: Yes, the definitions of a function and a subset are different. As you say, a function $f: A \to B$ is often defined as a subset $S$ of $A \times B$, that satisfies the condition that for $x_1 \in A$, there is only one element in $S$ of the form $(x_1, y)$.
However, the definition you've got from Wikipedia doesn't seem to be defining a property as a function generally, but rather as a function $f$ whose preimage of $\{\text{True}\}$ is precisely the subset on which the property holds. Since $f$ is a function into a two element set, the preimage of $\{\text{True}\}$, which is a subset of $X$, completely characterizes the function. It is in this sense that the two definitions are equivalent.


TITLE: type theory that does not treat the terms of $\mathrm{Prop}$ as types
QUESTION [2 upvotes]: In type theory there is a type $\mathrm{Prop}$ that contains every proposition, so $p\colon\mathrm{Prop}$ (in words, "$p$ is of type $\mathrm{Prop}$") where $p$ is a proposition. In all type theories I have seen, any proposition $p$ is treated as a type whose terms are its proofs. I find that unnatural because I don't think it is satisfying to identify a propositions (such as $\forall m\colon \mathbb{N}. \forall n\colon \mathbb{N}. m+n = n+m$) with the collection of its proofs.
So here is my question:
Are there (HOL) type theories that do not treat propositions as types? Do you can give some references?

REPLY [6 votes]: Before I actually answer the question asked, let me try to explain one way of thinking about proofs as elements of propositions. It is not the only way, but it should appeal to a mathematician with a taste for abstraction.
A long time ago mathematicians thought of functions as symbolic expressions, and many high-school students and beginning undergraduates still think of them that way. This sort of thinking had some unfortunate consequences. For example, it was sometimes assumed that every function can be differentiated because every expression one wrote down could obviously be differentiated symbolically.
Modern mathematics has gone far beyond such a naive view of functions. Today you take it for granted to say things like "let $V$ and $W$ be any Banach spaces and consider the space $L(V,W)$ of continuous linear functions from $V$ to $W$". A mathematician from the past would have a hard time understanding what this means because he would be used to thinking only about a single function at a time (but $L(V,W)$ is an entire space of functions), and the only concrete expression denoting a map $V \to W$ that he could actually write down would be $x \mapsto 0$.
Now let us consider the situation with proofs. Traditional logic tells us that proofs are syntactic entities: sequences (or trees) of statements each of which is an axiom or a consequence of the previous ones. Mathematicians do not study proofs at all, at least not in the same way that they study numbers, curves, and other mathematical objects. They mostly think about one proof at a time. Our understanding of proofs is where our understanding of functions was a couple of centuries ago.
When type theory says that "the elements of a proposition are its proofs" you should read that as "let us consider spaces of abstract proofs", just like we consider spaces of abstract functions. A function need not be an expression, and  a proof need not be a sequence (or a tree) of statements. Once this is accepted we are able to conceive of an entire space of proofs as a single mathematical entity (a type), and then we can think of constructions of such entities, and so on.
Alfred Tarski's cylindrical algebras are another example of how proofs in first-order logic can be made abstract.
I can express my point with an equation:
$$\frac{\text{mathematical logic}}{\text{type theory}} =
  \frac{\text{Newton's calculus}}{\text{functional analysis}}.$$
This equation is wrong in many respects, but I hope it communicates the point I am trying to make.
Finally, to answer your question: Church's theory of types is classical and has the type bool of truth values which are not proofs. It is used by the HOL family of proof assistants.


TITLE: Derive logitboost using the logistic loss function
QUESTION [1 upvotes]: An additive model constructed using the exponential loss function
$$L(y, f (x)) = \exp(−yf (x))$$
gives   Adaboost. How can we derive the corresponding additive model (known as logitboost) using the logistic loss function
$$L(y, f (x)) = \log(1 + \exp(−yf (x)))$$
What steps I should take to do the above proof?

REPLY [2 votes]: To get you answer,you may wanna look at this paper  http://dept.stat.lsa.umich.edu/~gmichail/ada_final.pdf.
Algorithm 2 summarizes the step you have to take to derive a boosting algorithm from any given convex loss including logitboost.


TITLE: Irreducible polynomial method.
QUESTION [0 upvotes]: I have found a question where I have to determine the irreducibility of a polynomial $f$ in $\textbf{Q}[X]$. Namely,
$$ f=2X^3-X-6.$$
The method used which I do not understand is;
Suppose there is a rational zero , $r=\frac{m}{n}$ where $n$ is bigger than or equal to one,  subbing in for $r$ and multiplying up by $n$ we get ,
$$2m^{3}-mn^2-6n^3,$$ if $p|n$ then from the equation $p|2m^3$. Since $p$ does not divide $m$, we have $p=2$, But then $4|2m^3$. So that $m$ and $n$ are both even; which is a contradiction. Thus $n=1$ and $m^{3}-m-6=0$ Hence $m|6$ and so we need to check the divisors of 6. Which shows it is irreducible over $\textbf{Q}$.
I understand the last bit once we have deduced $2m^{3}-m-6=0$ but it is the part before, please may someone explain this method.

REPLY [1 votes]: Assume that a root is $\frac{m}{n}$ where $\gcd(m,n)=1$. Then as you show, we get
$$2m^3 - mn^2 - 6n^3 = 0,\text{ so that }2m^3 = mn^2 + 6n^3 = n(mn+6n^2).$$
If $n\ne 1$, choose some prime $p\mid n$. Then $p$ divides the right-hand side above, so $p\mid 2m^3$. Since $\gcd(m,n) = 1$, we must have $p\mid 2$, so that $p = 2$. But then the right-hand side, $n(mn+6n^2)$, consists of two even factors, so is divisible by $4$. Thus $4\mid 2m^3$ and therefore $2\mid m$, a contradiction.
This implies $n=1$, so that (as you point out) we are reduced to looking for integral roots of $2m^3-m-6$.


TITLE: Lowest index giving half of the sum
QUESTION [6 upvotes]: Numbers $x_1,x_2,\ldots,x_n$ are drawn independently and uniformly from the interval $[0,1]$. Order them as $y_1\ge y_2\ge\dots\ge y_n$, and let $S$ be their sum. Let $k$ be the smallest index such that $y_1+\dots+y_k\geq S/2$.
What is $E[k]$, and what are other known properties about $k$ (e.g., variance)? Is there any reference to this? I don't know what terms to use to search.

REPLY [1 votes]: Here's an approximate answer: let $U_1,U_2,\ldots,U_n$ be the i.i.d. uniform random variables. I will consider them ordered from smallest to largest because it makes the arithmetic cleaner. We're then asking adding from smallest to largest, how many do we need to sum in order that $U_{i_1}+\ldots +U_{i_k}>\frac S2$, where $S=U_1+\ldots+U_n$.
Define additional random variables:
$$
\begin{align*}
X&=\#\{i\colon U_i<1/\sqrt 2\};\\
Y&=\sum_{U_i<1/\sqrt 2} U_i-n/4;\\
Z&=\sum_i U_i-n/2.
\end{align*}
$$
It's not hard to check that $\mathbb EX=n/\sqrt 2$, $\mathbb EY=\mathbb EZ=0$.
The ideal situation is when $Z=2Y$; that is the terms below $1/\sqrt 2$ have exactly the same sum as those above. To compensate for the difference, we need to move terms totalling $\frac Z2-Y$ into the "small number subset" (if $Z>2Y$)
or move terms totalling $Y-\frac Z2$ into the large number subset (if $Z<2Y$).
Since the terms on the boundary are of size approximately $\frac 1{\sqrt 2}$, one expects that the number such that $U_{i_1}+\ldots+U_{i_k}=S/2$ is well-approximated by $M=X+\sqrt 2(\frac Z2-Y)$.
From the definition, one sees
$$
M=\sum_{i=1}^n \Big( \mathbf 1_{U_i<1/\sqrt 2} + (1/\sqrt 2)U_i - \sqrt 2U_i\mathbf 1_{U_i<1/\sqrt 2}\Big).
$$
A calculation shows that $\mathbb EM=n/\sqrt 2$ and $\text{Var}(M)=n(9\sqrt 2-2)/12$. I anticipate that the expectation is correct to $O(1)$ and the variance is correct to $O(\sqrt n)$.
BTW: If anyone can tell me how make this rigorous, I'd very much like to know.


TITLE: Question involving angles and the unit circle
QUESTION [0 upvotes]: Given $\tan A = \sqrt3$ and $\csc B = -\sqrt2$ where $A,B \in[\pi,\frac{3}{2}\pi]$,
evaluate
$\cos^2A + (\cot B)(\sin A)$.
I haven't been able to figure out the angle for $B$.

REPLY [2 votes]: $$\csc B = -\sqrt 2 \implies \frac 1 {\sin B}=-\sqrt 2\implies\frac {-1} {\sqrt 2}=\sin B$$
Certainly now you can figure out what $B$ is.

REPLY [1 votes]: Using the identity $1+\tan^2A=\sec^2A$, you have that $\sec^2A=4$. i.e $\cos^2A=\frac{1}{4}$. So $\cos A=-\frac{1}{2}$, since $A$ is in the third quadrant. From this we get $\sin A=-\frac{\sqrt3}{2}$, again since $A$ is in the third quadrant.
Now, using the identity $1+\cot^2B=\csc^2B$, you get $\cot B=1$. This is positive since $\tan$ is positive in the third quadrant (which means $\cot$ is also positive), and so the result follows.

REPLY [1 votes]: Hints: We can find an explicit expression without knowing the angles (the angles are, however, not hard to find).
For example, recall the identity $1+\tan^2 A=\sec^2 A$.  If you know $\tan A$, then you know $\sec^2 A$, and therefore you know $\cos^2 A$.
If you know $\cos^2 A$, you know $\sin^2 A$. Therefore you almost know $\sin A$, apart from sign.  But since $A$ is in the third quadrant, you know the sign of $\sin A$.
You know $\csc B$, and want to know $\cot B$. There are various approaches. For example, from $\sin^2x+\cos^2 x=1$, dividing both sides by $\sin^2 x$, you get $1+\cot^2 x=\csc^2 x$. If you know $\csc B$, you can find $\cot^2 B$, and then, $\cot B$. Again, you have to worry about sign.  But that's not hard, $\tan$ is positive in the third quadrant.
Or else you know $\csc B$, so you know $\sin^2 B$, so you know $\cos^2 B$, so you know $\cot^2 B$.
Comment: We know that $\csc B=-\sqrt{2}$. Thus $\sin B=-1/\sqrt{2}$, a familiar quantity. We know that the sine of $\pi/4$ is $1/\sqrt{2}$. So halfway between $\pi$ and $3\pi/2$, the sine is $-1/\sqrt{2}$.


TITLE: Show that $1/T$ has the same distribution as $T$
QUESTION [0 upvotes]: Let T be the ratio X/Y of two i.i.d. N (0, 1) r.v.s. X, Y . This is the Cauchy distribution and, as shown  it has PDF $f_T(t)= \frac{1}{\pi (1+t^2)}$
(a) Show that $1/T$ has the same distribution as $T$ using calculus, after first finding the CDF of $1/T$.
(b) Show that $1/T$ has the same distribution as $T$ without using calculus, in 140 characters or fewer.

a) $Y= \frac 1T$, $P(Y \le y)=P(\frac1T \le y)=P(T \le 1/y)= \int_{-\infty}^{1/y} \frac{1}{\pi} \frac {1}{(1+t^2)}dt=\frac {tan^{-1}(1/y)}{\pi}$; $y \in (-\infty;\infty)$
$f_Y(y)= \frac {-1}{\pi x^2+\pi}$
b) Can you help me with $b$ and check $a$?

REPLY [0 votes]: If $y>0$ then $1/T\le y$ if $T<0$ or $T\ge1/y,$ so $$F_Y(y) = 1/2 + P(T>1/y)= 1- \frac{1}{\pi}\tan^{-1}(1/y).$$ If $y<0$ then $1/T \le y$ if $1/y \le T < 0.$ So $$F_Y(y) = P(1/y <T<0) = -\frac{1}{\pi}\tan^{-1}(1/y).$$ This can be condensed into $$ F_Y(y) = 1/2+\frac{1}{\pi}\cot^{-1}(1/y) $$ for all $y\ne 0$ and $1/2$ for $y=0.$


TITLE: part of proof about stochastic processess
QUESTION [2 upvotes]: I need help proving a part of a proof. The exercise in the book( Exercise 3.14 in Bernt Øksendal: Stochastic Differential equations) is given as to prove something, but I will only talk about the first part of the proof in this post:
The proof I am talking about:

Assume that you have a stocastic process $B_t$ in $\mathbb{R}^n$ given
on the probability space $(\Omega, \mathcal{A}$, P). The filtration
$\mathcal{F}_t$ is defined as the sigma algebra generated by the sets
$\{B_s^{-1}{(A}): A \in \mathcal{B}(\mathbb{R}^n), s \le t\}$.(This is
supposed to be the natural filtration, or the filtration generated by
the process.
Then:
$h$ is $\mathcal{F}_t$ -measurable iff it can be written as the
pointwise limit, of sums off functions of the form
$g_1(B_{t_1})g_2(B_{t_2})\cdots g_k(B_{t_k})$, where each g is
continous and all the $t_j\le t$.

Now the first part of the proof, which I am stuck on, is to explain why it is enough for h to be bounded?
Here is what I tried:
My first attempt is to assume that h is not bounded, and assume that the theorem holds for bounded functions and then try to approximate.
I tried to divide the range in partitions according to the absolute value of h. For instance $h_1=h\mathcal{X}_{(\{\omega: |h(\omega)|<1\}} $, $h_2=h\mathcal{X}_{(\{\omega: |h(\omega)|<2\}} $. These functions converges pointwise to h, and I can use the theorem to get a sequence of the form I like on each of them. The only way to continue here is with some sort of sequence to works for all the functions, but the problem with using a classical diagonal argument is that I can not choose one sequence to be a subsequence of the earlier ones. So it doesn't really work.
My last idea was to work with $h_1=h\mathcal{X}_{(\{\omega: |h(\omega)|<1\}} $, $h_2=h\mathcal{X}_{(\{\omega: 1\le|h(\omega)|<2\}} $ etc. Now the $h_n$ won't converge but the sums of $h_n$ will converge to $h$. If I use the theorem separately on each of the $h_n$, I am close to having my result, but the problem is this: If for instance $f^1_k\rightarrow h_1, f^2_k\rightarrow h_2$ etc., where $f_k=f_{k-1}+\prod_{i=1}^k g_i$, where the g's are given by the theorem. Then it is difficult showing that for instance we can pick up a function that converges and have our properties, because we only have that they converge pointwise to $0$ where they are cut off, and when I introduce more terms, I don't see how to get the convergence correct. Also if I introduce $\mathcal{X}_{\{\omega|j-1\le|h(\omega)|<j\}}$ to the sequence $f_k^j$ then I lose the continuity part.
Any tips or hints here?

REPLY [0 votes]: As you say, it does not seem to be true (in general) that, rephrasing it, if $h_{n}^j\to h_n$ a.e. (or pointwise) as $j\to \infty$ and $h_n\to h$ a.e. (or pointwise) as $n\to \infty$, then there is a subsequence that $h_{n_k}^{j_k}\to h$ a.e. (or pointwise) as $k\to \infty$. The problem is that the subsequence can vary for each point.
Nevertheless I would say that Exercise 3.14 can be solved using $L^1$ convergence (and then taking an almost everywhere convergent subsequence). In particular, taking $h_n^\pm:=\max\{h^\pm,n\}$, $h_n=h_n^+-h_n^-$ are bounded and converge in $L^1$ to $h$ by the Monotone Convergence Theorem. Then, if for every $\varepsilon>0$ there is a $g=g_{n,\varepsilon}$ (as described in the exercise) such that $\|h_n-g\|_{L^1(P)}<\varepsilon/2$, then
$$
\|h-g\|_{L^1(P)}<\varepsilon
$$
if $n$ is large enough. Thus, it is enough to find that $g$ for bounded Borel functions.


TITLE: usage of this condition
QUESTION [5 upvotes]: A very widely stated result:
A sequence $x_n \to x$ iff every subsequence $x_{n^{\prime}}$ of $x_n$ contains a further subsequence $x_{n^{\prime\prime}}$ such that $x_{n^{\prime\prime}}\to x$.
My question is that there is no need to pass on to sub-sub sequence. The statement is true for sub sequence replacing sub-sub sequence. Then why is this statement useful?? Is it the case that in several places its difficult to find a convergent sub-sequence but easy to get a convergent sub-sub sequence??

REPLY [8 votes]: It is a useful criterion. One may not use it every day, but it comes up not too rarely.
The typical use-case that I come across, is that one has a sequence of which it is known that every subsequence has a convergent subsequence - a sequence in  a sequentially compact space; for example a normal family, or the closed unit ball of a reflexive Banach space in the weak topology  - and one can show that all convergent subsequences must converge to the same limit. Then the criterion gives the convergence of the full sequence.
Example:
Suppose you have a connected open $\Omega \subset \mathbb{C}$, and a holomorphic $f \colon \Omega \to \Omega$ that has an attractive fixed point $z_0 \in \Omega$, i.e. $f(z_0) = z_0$ and $\lvert f'(z_0)\rvert < 1$.
It is then clear that the sequence $f^n$ (exponent means composition here) converges uniformly to the constant function $c_{z_0} \colon z \mapsto z_0$ in a neighbourhood $U$ of $z_0$. But we don't know a priori how the sequence $\bigl(f^n\bigr)_{n\in\mathbb{N}}$ behaves on all of $\Omega$.
By the identity theorem, if a subsequence of $\bigl(f^n\bigr)_{n\in\mathbb{N}}$ converges locally uniformly, its limit must be the constant $c_{z_0}$.
Now, if $\Omega$ is nice (bounded, a half plane; $\mathbb{C}\setminus\Omega$ contains at least two points, but this one is deep), the family $\mathscr{F} = \lbrace f^n : n \in \mathbb{N}\rbrace$ is a normal family, that is, every sequence in $\mathscr{F}$ has a locally uniformly convergent subsequence.
In that case, pick any subsequence $\bigl(f^{n_k}\bigr)_{k\in\mathbb{N}}$ of $\bigl(f^n\bigr)_{n\in\mathbb{N}}$. Since that is a sequence in the normal family $\mathscr{F}$, it has a locally uniformly convergent subsequence $\bigl(f^{n_{k_m}}\bigr)_{m\in\mathbb{N}}$. By the identity theorem, as mentioned above, $f^{n_{k_m}}$ can only converge to $c_{z_0}$. So

every subsequence $\bigl(f^{n_k}\bigr)_{k\in\mathbb{N}}$ of $\bigl(f^n\bigr)_{n\in\mathbb{N}}$
has a subsequence $\bigl(f^{n_{k_m}}\bigr)_{m\in\mathbb{N}}$ that converges (locally uniformly) to $c_{z_0}$.

By the criterion, $\bigl(f^n\bigr)_{n\in\mathbb{N}}$ itself converges locally uniformly to $c_{z_0}$.
Note that we need to pass to sub-subsequences. If we directly apply normality to $\bigl(f^{n_k}\bigr)_{k\in\mathbb{N}}$, all we get is that $\bigl(f^{n_k}\bigr)_{k\in\mathbb{N}}$ has (at least) one subsequence converging locally uniformly to $c_{z_0}$. That wouldn't rule out the possibility that $\bigl(f^{n_k}\bigr)_{k\in\mathbb{N}}$ has subsequences that don't converge locally uniformly.
Another use of the criterion is here.


TITLE: Octal System Conversion
QUESTION [1 upvotes]: So I am trying to change base ten 1,398 to an octal equivalent. I had assumed that I would find it by taking 1,398 and changing it like base-four. So 1,398 is $1*8^3 + 3*8^2 +9*8^1 + 8*8^0*$ or $512+192+72+8=784$ but I guess the answer is 2566. Where am I wrong?

REPLY [0 votes]: To methodically perform the conversion:
$1398 \mod 8 = 6$  ($8^0$ place octal digit)
$\lfloor 1398 / 8 \rfloor = 174$
$174 \mod 8 = 6$ ($8^1$ place octal digit)
$\lfloor 174 / 8 \rfloor = 21$
$21 \mod 8 = 5$ ($8^2$ place octal digit)
$\lfloor 21 / 8 \rfloor =2 $
$2 \mod 8 =2$ ($8^3$ place octal digit)
$\lfloor 2/8\rfloor =0$
So $1398_{10} = 2566_8$


TITLE: Intersections of circles drawn on vertices of regular polygons
QUESTION [10 upvotes]: Using only a compass, draw all possible circles on the vertices of a regular $n$-sided polygon.
(That is, in every ordered pair of vertices one is the center, and their distance is the radius.)
How many intersections are there?

Let $a(n)$ be the intersection count for given $n\in\mathbb N$.
First three terms $a(1),a(2),a(3)=0,2,6$ are simple. The next three terms are:

Notice that the circle set (given by a $n$-sided polygon) can be split into $n$ symmetric regions.
Let $A_n$ count intersections inside one of the $n$ regions. Let $\delta_n\in\{0,1\}$ compensate for when there is one extra central intersection. This implies we can write every term as:
$$a(n)=nA_n+\delta_n$$
The first $20$ terms should be (constructed in GeoGebra):
$$\begin{array}{}
a(1) &= \space\space0  &= \space\space1\cdot0       \\
a(2) &= \space\space2  &= \space\space2\cdot1       \\
a(3) &= \space\space6  &= \space\space3\cdot2       \\
a(4) &= 40 &= \space\space4\cdot10      \\
a(5) &= 55 &= \space\space5\cdot11       \\
a(6) &= 145&= \space\space6\cdot24  +  1 \\
a(7) &= 238&= \space\space7\cdot34       \\
a(8) &= 584&= \space\space8\cdot73      \\
a(9) &= 612&= \space\space9\cdot68      \\
a(10) &= 1350&= 10\cdot135       \\
a(11) &= 1804&= 11\cdot164       \\
a(12) &= 2401&= 12\cdot200+1       \\
a(13) &= 3523&= 13\cdot271      \\
a(14) &= 5180&= 14\cdot370      \\
a(15) &= 6150&= 15\cdot410       \\
a(16) &= 9312&= 16\cdot582      \\
a(17) &= 11101&= 17\cdot653       \\
a(18) &= 13645&= 18\cdot758+1      \\
a(19) &= 17746&= 19\cdot934       \\
a(20) &= 22300&= 20\cdot1115 \\
\dots
\end{array}$$
We can notice that it seems $\delta_n=1$ if and only if $n$ is a multiple of $6$.

Are there any other patterns? Is it possible to find a closed form for $a(n)$?

My attempt:
WLOG, Let $V_1,V_2,\dots,V_n$ be vertices of a regular $n$-sided polygon with circumradius $1$.
We can take $V_i=(x_i,y_i)=(\cos(\frac{2i\pi}{n}),\sin(\frac{2i\pi}{n})),i=1,2,\dots,n$.
The $k$-th diagonal from some vertex $V$ of the polygon will have length $2\sin(\frac{k\pi}{n})$.
At each vertex $V$, we will have $c=1,2,\dots,\left\lfloor\frac{n}{2}\right\rfloor$ circles$^{[1]}$ with radii $r_c=2\sin(\frac{c\pi}{n})$.
$$
(i,c)\text{-Circle}\dots\space\space \left(x-\cos\frac{2i\pi}{n}\right)^2+\left(y-\sin\frac{2i\pi}{n}\right)^2=\left(2\sin\frac{c\pi}{n}\right)^2
$$
Is it possible to derive a closed form for the number of intersections from this?
I believe I managed to solve a simpler problem:

"At each vertex $V$, consider only one circle of radius $r$."
Then the number of such intersections $I(n,r)$ should be:
$$
I(n,r)=\begin{cases}
(n-1)n, & r \gt 1\\
(n-1)n - n\left\lfloor\frac{n}{2}\right\rfloor+1, & r=1\\
(n-2)n, & \sin(\frac{\left(\frac{n}{2}-1\right)\pi}{n})\lt r\lt\sin(\frac{\pi}{2})=1\\
(n-3)n, & r = \sin(\frac{\left(\frac{n}{2}-1\right)\pi}{n})\\
\dots & \dots \\
(2k)n, & \sin(\frac{k\pi}{n}) \lt r \lt \sin(\frac{(k+1)\pi}{n})\\
(2k-1)n, & r = \sin(\frac{k\pi}{n})\\
\dots & \dots \\
4n, & \sin(\frac{2\pi}{n}) \lt r \lt \sin(\frac{3\pi}{n})\\
3n, & r = \sin(\frac{2\pi}{n})\\
2n, & \sin(\frac{\pi}{n}) \lt r \lt \sin(\frac{2\pi}{n})\\
1n, & r = \sin(\frac{\pi}{n})\\
0, & r \lt \sin(\frac{\pi}{n})
\end{cases}
$$

This solves the problem of intersections for any $r\in\mathbb R_{+}$ but for only one layer of circles.
In the original problem, we have $\left\lfloor\frac{n}{2}\right\rfloor$ layers of circles with different radii on each layer. The radii of circles between layers have specific ratios (determined by $n$): radii are diagonals of the regular $n$-sided polygon.
My idea was to use $I(n,r_c),c=1,2,\dots,\left\lfloor\frac{n}{2}\right\rfloor$ to get to $a(n)$. But, I get lost when trying to add and subtract all of the unique and duplicate intersections.
How can we solve the original problem and find $a(n)$?

REPLY [1 votes]: We can find an upper bound by considering the case $n \to \infty :$ There are $\left \lfloor \frac{n}2 \right \rfloor \leq \frac{n}2$ radii. Let the distance between the points be $1$, every circle with radius $k$ has two unique intersections with $1+2(k-1)$ circles with the same radius so in total
$$2n \sum_{k=1}^{n/2} 1+2(k-1) = \frac{n^3}2$$

Every circle with radius $k$ has four unique intersections with $1+2(k-1)$ circles for every bigger radius:
$$4n \sum_{k=1}^{n/2} \bigl( 1+2(k-1) \bigr) \left( \frac{n}2 - k \right) = \frac{n^2}6 (n-1)(n-2)$$

In order to not overcount the starting points we consider them seperately.
The upper bound is shown below as well as the fit $a(n) \approx 0.089 \, n^{4.14}$
$$a(n) \leq n + \frac{n^3}2 + \frac{n^2}6 (n-1)(n-2) = n + \frac{n^2}3 + \frac{n^4}6$$
$\hspace{1cm}$