text:
\section{Introduction}

Assume that $H$ is a separable Hilbert space with inner product $\langle \, \cdot \,,\, \cdot \, \rangle$ and consider a self-adjoint operator~$A$ with simple discrete spectrum acting in~$H$. Our aim is to study spectral properties of the rank one perturbations of~$A$, i.e., of the operators $B$ of the form
\
    B=A + \langle \cdot, \varphi \rangle \psi,
\
where $\varphi$ and $\psi$ are nonzero elements of $H$.

Rank-one perturbations of operators and matrices have been actively studied in both mathematical and physical literature for the reason that, on the one hand, they are simple enough to allow description of the spectral properties of perturbed operators via closed-form formulae which then can be analysed using various techniques; on the other hand, such perturbations turn out to be general enough to produce various non-trivial effects.

One of the most general results in a finite-dimensional setting is given by Krupnik~\cite{Kru92} and states that a rank-one perturbation of an $n\times n$ matrix~$A$ can possess an arbitrary spectrum counting multiplicity. In other words, given any natural number $k$, any pairwise distinct complex numbers~$z_1, z_2, \dots, z_k$, and any natural numbers $m_1, m_2, \dots, m_k$ satisfying $m_1+ m_2 + \dots + m_k =n$, there is a rank-one perturbation~$B$ of~$A$ whose spectrum consists of the points $z_1, z_2, \dots, z_k$ of the corresponding algebraic multiplicities $m_1,m_2, \dots, m_n$. This statement is also specialized to cases when both~$A$ and the perturbed matrix~$B$ belong to the Hermitian, unitary, or normal classes. Savchenko~\cite{Sav03} studies the effect a generic rank-one perturbation has on the Jordan structure of a matrix~$A$; an interesting observation is that, typically, in each root subspace, only the Jordan chain of the largest length splits; in~\cite{Sav04}, this is further generalized to low-rank perturbations, cf.\ also~\cite{MorDop03}. Similar results in infinite-dimensional Banach spaces were earlier derived by H\"ormander and Melin in~\cite{HorMel94}. Bounds on the number of distinct eigenvalues of~$B$ in terms of some spectral characteristics of~$A$ are established in~\cite{Far16}.

Structured perturbations of matrices and matrix pencils have recently been thoroughly studied in a series of papers by Mehl a.o.~\cite{MehMehRanRod11, MehMehRanRod12, MehMehRanRod13, MehMehRanRod14, MehMehRanRod16, MehMehWoj17, SosMorMeh20}. Changes in the Jordan structures under perturbation within the classes of complex $J$-Hamiltonian and $H$-symmetric matrices and application in the control theory
is discussed in~\cite{MehMehRanRod11}; see~\cite{MehMehRanRod13, MehMehRanRod16} for further treatment in both the real and complex case. The class of $H$-Hermitian matrices, with (skew-)Hermitian $H$, is studied in~\cite{MehMehRanRod12, MehMehRanRod14} via the canonical form of the pair $(B, H)$. Rank-one perturbations of matrix pencils are discussed e.g.\ in~\cite{MehMehWoj17, GerTru17, BarRoc20}. A general perturbation theory for structured matrices is developed in the recent paper~\cite{SosMorMeh20}.

The above results typically exploit essentially matrix methods and thus are not directly applicable to the infinite-dimensional case (see, however, \cite{HorMel94}). Rank-one perturbations of bounded or unbounded operators in infinite-dimensional Hilbert spaces have been studied within the general operator theory. For instance, a comprehensive spectral theory for rank-one perturbations of unbounded operators in the self-adjoint case is developed in~\cite{Sim95}, where a detailed characterization of discrete, absolutely continuous, and singlularly continuous components of the spectrum of the perturbed operator is given. A thorough overview of the theory of singular point perturbations of Schr\"odinger operators (formally corresponding to additive Dirac delta-functions and their derivatives) is given in the monographs by Albeverio a.o.~\cite{AGHH, AlbKur00}. There has been much work devoted to the so-called singular and super-singular rank-one perturbations of self-adjoint operators, where the functions $\varphi$ and $\psi$ belong to the scales of Hilbert spaces $\operatorname{dom}(A^\alpha)$ with negative~$\alpha$, see e.g.~\cite{AlbKosKurNiz03, AlbKonKos05, AlbKos99, AlbKuzNiz08, Gol18, Kur04, KurLugNeu19, KuzNiz06, DudVdo16}; in this case, a typical approach is through the Krein extension theory of self-adjoint operators. Rank-one and finite-rank  perturbations of self-adjoint operators in Krein spaces have been recently discussed in e.g.~\cite{BehMoeTru14, BehLebPerMoeTru16}.

Despite the extensive research in the area, there seems to be no complete infinite-dimensional generalization of the results by Krupnik~\cite{Kru92}. The most pertinent work we are aware of include the papers by H\"ormader and Melin~\cite{HorMel94} and by Behrndt a.o.~\cite{BehLebPerMoeTru15}, which characterize possible changes in the Jordan structure of root subspaces of linear mappings in infinite-di\-men\-si\-o\-nal linear vector spaces under general finite-rank perturbations.

Our motivation in this work was to understand how the spectrum of an operator in an infinite-dimensional Hilbert space can change under a rank-one perturbation, both locally, i.e., on the level of root subspaces, and globally, i.e., on the level of eigenvalue asymptotics. This task is quite non-trivial even in the case when the unperturbed operator~$A$ is self-adjoint but has generic spectrum, cf.~\cite{Sim95}. Therefore, we decided to start with deriving  a complete spectral picture in the simplest case where the unperturbed operator~$A$ is self-adjoint and has simple discrete spectrum. Under this assumption, our main result (Theorem~\ref{thm:main}) shows that the rank-one perturbation~$B$ of~$A$ may get eigenvalues of arbitrary algebraic multiplicity in an arbitrary finite set of points; however, all sufficiently large eigenvalues remain simple and asymptotically close to the eigenvalues of~$A$. In the finite-dimensional case, our analysis leads to an extension of the result by Krupnik~\cite{Kru92}; Theorem~\ref{thm:finite-dim} states that one of the vectors~$\varphi$ or $\psi$ can be fixed arbitrarily in a ``generic'' set, and then one can find the other vector such that the perturbed matrix $B$  possesses the prescribed spectrum and, moreover, such choice is unique. We also specify this result in Theorem~\ref{thm:phi-arbitrary} to the case when $\varphi$ or $\psi$ is fixed arbitrarily. We also note that a complete characterization of the possible spectra of rank-one perturbations of self-adjoint operators in Hilbert space, including precise asymptotic distribution and the constructive algorithm of finding~$\varphi$ and $\psi$, is suggested in a subsequent paper~\cite{DobHry20}.

The structure of the paper is as follows. In the next section, we introduce the characteristic function of the perturbed operator~$B$ and discuss how it is related to its spectrum. In Section~\ref{sec:mult}, the algebraic multiplicities of eigenvalues are discussed, and in  Section~\ref{sec:asympt}, the asymptotic distribution of eigenvalues is established. In Section~\ref{sec:finite-dim}, we specialize the obtained results to the finite-dimensional case, and in the final section we discuss possible generalizations of the main results to wider classes of the operators~$A$.

\section{General spectral properties of $B$}\label{sec:general}

Throughout the paper, we make the following standing assumption on the operator~$A$:
\begin{itemize}
  \item[(A1)] the operator~$A$ is self-adjoint and has simple discrete spectrum.
\end{itemize}
The operator~$A$ is necessarily unbounded above or/and below; clearly, by considering $-A$ in place of~$A$, we reduce the case when $A$ is bounded above to the case when it is bounded below. Therefore, under assumption~(A1), the spectrum of~$A$ consists of real simple eigenvalues that can be listed in increasing order as $\lambda_n$, $n\in I$, with the index set~$I$ equal to~$\mathbb{N}$ in the case where $A$ is bounded below and to~$\mathbb{Z}$ otherwise.

The operator~$B$ is a rank one perturbation of the operator~$A$, i.e.,
\begin{equation}\label{eq:B}
    B=A + \langle \cdot, \varphi \rangle \psi
\end{equation}
with fixed nonzero vectors~$\varphi$ and $\psi$ in~$H$. Clearly, the operator~$B$
is well defined and closed on its natural domain $\dom (B)$ equal to $\dom (A)$.
Next, for $\lambda$ in the resolvent set $\rho(A)$ of~$A$, we introduce the \emph{characteristic function}
\begin{equation}\label{eq:F}
    F(\lambda) := \langle (A-\lambda)^{-1}\psi, \varphi \rangle + 1
\end{equation}
and denote by $\mathcal{N}_F$ the set of zeros of $F$.
Many spectral properties of the operator~$B$ of~\eqref{eq:B} will be derived from the explicit formula for its resolvent known as the Krein formula~(see, e.g., \cite[Sec. 1.1.1]{AlbKur00}); we include its proof for the sake of completeness and to derive some explicit relations to be used later on.

\begin{lemma}[The Krein formula]\label{lm:krein}
The set $\rho(A)\setminus \cN_F$ consists of resolvent points of the operator~$B$ and, for every~$\lambda\in\rho(A)\setminus \cN_F$,
\begin{equation}\label{eq:Krein}
    (B - \lambda)^{-1}
        = (A - \lambda)^{-1} - \frac{\langle \, \cdot \,, (A - \overline{\lambda})^{-1} \varphi \rangle}{F(\lambda)} \:
        (A-\lambda)^{-1} \psi.
\end{equation}
\end{lemma}

\begin{proof}
To prove that a fixed $\lambda\in\rho(A)\setminus \cN_F$ is a resolvent point of~$B$, we need to show that for every~$g\in H$ the equation
\begin{equation}\label{eq:krein-1}
    g = (B - \lambda) f
\end{equation}
can be uniquely solved for $f \in H$. Assuming such an~$f$ exists, writing the equality \eqref{eq:krein-1} as
\begin{equation}\label{eq:krein-2}
g = (A - \lambda) f + \langle f, \varphi \rangle \psi,
\end{equation}
and applying the resolvent of the operator $A$ to both sides, we obtain
\begin{equation}\label{eq:krein-3}
    (A - \lambda)^{-1} g
        =  f + \langle f, \varphi \rangle (A - \lambda)^{-1} \psi.
\end{equation}
Taking the inner product with $\varphi$ results in the equality
\begin{equation}\label{eq:krein-4}
    \langle (A - \lambda)^{-1} g, \varphi \rangle
    = \langle f, \varphi \rangle
        + \langle f, \varphi \rangle \langle (A - \lambda) ^ {- 1} \psi, \varphi \rangle
    = \langle f, \varphi \rangle F(\lambda),
\end{equation}
which on account of $F(\lambda)\ne0$ leads to
\begin{equation}\label{eq:krein-5}
    \langle f, \varphi \rangle =
    \frac {\langle (A - \lambda) ^ {-1} g, \varphi \rangle} {F (\lambda)}.
\end{equation}
Substituting now \eqref{eq:krein-5} in \eqref{eq:krein-3}, we derive the following formula for~$f$:
\begin{equation}\label{eq:krein-6}
    f = (A - \lambda)^{-1} g
        - \frac{\langle (A - \lambda)^{-1} g, \varphi \rangle}{F(\lambda)} (A-\lambda)^{-1}\psi.
\end{equation}
A direct verification shows that $f$ of~\eqref{eq:krein-6} belongs to~$\dom(B)=\dom(A)$ and is indeed a solution of equation~\eqref{eq:krein-1}.

Therefore the operator $B - \lambda$ is surjective. It is also injective since if an $f\in \dom(B)$ satisfies~\eqref{eq:krein-1}  with $g = 0$, then~\eqref{eq:krein-3} on account of~\eqref{eq:krein-5} gives~$f=0$. Thus the operator $B - \lambda$ is invertible and its inverse is equal to
$$
(B - \lambda)^{-1} = (A - \lambda)^{-1} - \frac{\langle \, \cdot \,, (A - \overline {\lambda})^{-1} \varphi \rangle} {F(\lambda)} \, (A - \lambda)^{-1} \psi
$$
as claimed. The proof is complete.
\end{proof}

The Krein formula shows that, for every $\lambda \in \rho(A) \setminus \cN_F$, the resolvent~$(B-\lambda)^{-1}$ is a rank one perturbation of the compact operator~$(A-\lambda)^{-1}$. Therefore, we get the following

\begin{corollary}\label{cr:krein}
The resolvent of the operator $B$ is compact, i.e., $B$ is an operator with discrete spectrum.
\end{corollary}

Next we denote by~$v_n$ a normalized eigenvector of~$A$ corresponding to its eigenvalue~$\lambda_n$; then the set~$\{v_n\}_{n\in I}$ is an orthonormal basis of~$H$. We also denote by $a_n$ and $b_n$ the Fourier coefficients of the vectors~$\varphi$ and $\psi$ with respect to this basis, so that%
\begin{footnote}
{In the case $I=\mathbb{Z}$, the summation will always be understood in the principal value sense}
\end{footnote}
\[
    \varphi = \sum_{n\in I} a_n v_n, \qquad \psi = \sum_{n\in I} b_n v_n.
\]

\begin{lemma}\label{lem:eig-B}
The following relations hold between the spectra of the operators $A$ and $B$:
\begin{itemize}
\item[a)]
for $\lambda\in\rho(A)$, $\lambda$ belongs to the spectrum of~$B$ if and only if~$\lambda \in \cN_F$;

\item[b)]
the eigenvalue $\lambda = \lambda_n$ of the operator $A$ belongs to spectrum of the operator $B$ if and only if $a_nb_n = 0$.
\end{itemize}
\end{lemma}

\begin{proof}
a) Let a point $\lambda \in \rho(A) $ belong to the spectrum of the operator~$B$. By Corollary \ref{cr:krein}, $\lambda$ is an eigenvalue of the operator~$B$, and we denote by~$y$ a corresponding eigenvector. Then \eqref{eq:krein-1} holds with~$g=0$ and with $y$ in place of~$f$, so that equations~\eqref{eq:krein-3} and~\eqref{eq:krein-4} can be recast as
\[
    y = - \langle y, \varphi \rangle (A - \lambda)^{-1} \psi
\]
and
\[
   \langle y ,\varphi \rangle F (\lambda) = 0,
\]
respectively. Since $y$ is a nonzero vector, we see from the former equality that $\langle y, \varphi \rangle\ne0$, and then the latter one yields $F(\lambda)=0$.

Conversely, if $F(\lambda)=0$ for some $\lambda \in \rho(A) $, then $y: = (A - \lambda)^{-1} \psi $ is an eigenvector of the operator $B$ for the eigenvalue $\lambda$, as is seen from the equalities
\[
    (A - \lambda) y + \langle y, \varphi \rangle \psi
        = [1 + \langle (A - \lambda)^{-1} \psi, \varphi \rangle]\psi
        = F(\lambda)\psi= 0.
\]
This completes the proof of part a).

b) Let $\lambda = \lambda_n $ belong to the spectrum of the operator $B$; then there is a vector $y \in \dom(B)$ such that $B y = \lambda_n y$, i.e.,
\begin{equation}\label{eq:1-2}
    (B-\lambda_n) y
    = (A - \lambda_n) y + \langle y, \varphi \rangle \psi = 0.
\end{equation}
Taking the inner product with $v_n$ results in
\begin{align*}
\langle (A - \lambda_n) y, v_n \rangle + \langle y, \varphi \rangle \langle \psi, v_n \rangle
         & = \langle y, (A - \lambda_n) v_n \rangle + \langle y, \varphi \rangle \langle \psi, v_n \rangle \\
         & = \langle y, \varphi \rangle \langle \psi, v_n \rangle = 0.
\end{align*}
Thus $\langle y, \varphi \rangle = 0$ or $\langle \psi, v_n \rangle = 0$. If $\langle y, \varphi \rangle = 0$, then $y = c v_n$ for some constant $c$ on account of~\eqref{eq:1-2}, so that $a_n = 0 $. If $\langle \psi, v_n \rangle = 0 $, then $b_n = 0$. Therefore the point $\lambda = \lambda_n $ belongs to the spectrum of $B$ only if $a_nb_n = 0$.

Conversely, let $a_nb_n = 0$; we need to prove that the point $\lambda = \lambda_n$ belongs to the spectrum of $B$. If $a_n = 0$, then
\[
    (B-\lambda_n) v_n = (A-\lambda_n)v_n + a_n \psi = 0
\]
so that $y = v_n$ is an eigenvector of~$B$ for the eigenvalue $\lambda_n$. If $b_n = 0$, then for all $y \in \dom(B)$
$$
    \langle (B - \lambda_n) y, v_n \rangle
    = \langle (A - \lambda_n) y, v_n \rangle
    = \langle  y, (A - \lambda_n)v_n \rangle =0,
$$
so that $B - \lambda_n$ is not surjective on $\dom(B)$ and the point $\lambda = \lambda_n$ belongs to the spectrum of the operator $B$. The proof is complete.
\end{proof}

We introduce the sets of indices
$$
I_0 \overset{\text{def}}{=} \{n \in I \, | \, a_nb_n = 0 \}, \quad
I_1 \overset{\text{def}}{=} \{n \in I \, | \, a_nb_n \neq 0 \}
$$
of cardinalities (possibly infinite) $N_0$ and $N_1$ respectively, and split the eigenvalues of~$A$ into the respective subsets
$$
    \sigma_0 (A) \overset {\text{def}}{=} \{\lambda_n \, | \, n \in I_0 \}
        \text {\qquad and \qquad}
    \sigma_1 (A) \overset {\text{def}}{=} \{\lambda_n \, | \, n \in I_1 \}.
$$
According to Lemma~\ref{lem:eig-B}, the spectrum of the operator $B$ consists of two parts: $\sigma_0(A) = \sigma(A) \cap \sigma(B)$, the common eigenvalues of $A$ and $B$, and the set $\cN_F$ of zeros of the function~$F$ in $\rho(A)$. Certainly, the latter part of $\sigma(B)$ is more interesting.

\section{Eigenvalue multiplicity}\label{sec:mult}

In this section we discuss multiplicity of eigenvalues of the operator~$B$.

First we recall that the \textit{geometric multiplicity} of an eigenvalue $\lambda$ of an operator~$T$ is the dimension of the corresponding eigenspace, i.e., the number $\dim \ker (T - \lambda)$ \cite[Ch.~5.1]{Kat95}, and its \textit{algebraic multiplicity} is the dimension of the corresponding root subspace, i.e., the rank of the corresponding spectral projector \cite[Ch.~5.4]{Kat95}. Note that for a selfadjoint operator geometric and algebraic multiplicities of every eigenvalue are equal.

Before proceeding, we recall that the function~$F$ was initially defined only on the resolvent set of the operator~$A$. However, using the spectral theorem for the operator~$A$, we can write the function $F$ as
\begin{equation}\label{eq:F-new}
    F(\lambda)
        = \sum_{n \in I_1} \frac {\overline{a_n} b_n}{\lambda_n - \lambda} + 1,
\end{equation}
and this formula gives an analytic continuation of~$F$ onto the set $\sigma_0(A)$. We shall denote this continuation by the same letter~$F$ but will write $\cN_F^0$ for the set of zeros of $F$ continued onto $\mathbb{C} \setminus \sigma_1(A)$.

\begin{lemma}\label{lem:geom-mult}
An eigenvalue $\lambda$ of~$B$ has geometric multiplicity larger than~$1$ if and only if there exists an integer $n$ such that $\lambda = \lambda_n$, $a_n = b_n = 0$, and $F(\lambda_n) = 0$. In that case, the geometric multiplicity of~$\lambda$ is equal to~$2$.
\end{lemma}

\begin{proof}
Assume that $\lambda\in\sigma(B)$ has geometric multiplicity larger than~$1$, and denote by $y$ any of the corresponding eigenvectors. Then
\[
    (B-\lambda) y = (A-\lambda) y + \langle y, \varphi \rangle \psi =0,
\]
and if $\lambda$ is a resolvent point of $A$, then $y$ must be collinear to the vector $(A-\lambda)^{-1}\psi$ and thus the geometric multiplicity of~$\lambda$ is one. Therefore $\lambda\in\sigma(A)$, so that $\lambda=\lambda_n$ for some $n\in I_0$.
Now, as in the proof of part b) of Lemma~\ref{lem:eig-B}, we find that
\[
      0=\langle (B-\lambda_n)y, v_n \rangle
        = \langle y, \varphi \rangle  \langle \psi, v_n \rangle
        = \langle y, \varphi \rangle   b_n,
\]
so that $\langle y, \varphi \rangle =0$ or $b_n=0$.

Assume that $b_n\ne0$; then $\langle y, \varphi \rangle =0$ and
$(B-\lambda_n)y =(A-\lambda_n) y =0$. Thus $y$ in that case must be collinear to $v_n$, and the geometric multiplicity of~$\lambda_n$ is then~$1$.
Therefore, $b_n=0$ and the vector $\psi$ belongs to the subspace $H_n:=H \ominus \langle v_n \rangle$. Since the nullspace of $B-\lambda_n$ is of dimension at least~$2$, there is an eigenvector $w$ in $H_n$. We denote by $A_n$ the restriction $A|_{H_n}$ of~$A$ onto its invariant subspace~$H_n$ and see that
\[
    (A_n - \lambda_n) w + \langle w, \varphi \rangle \psi =0.
\]
Note that $\lambda_n$ is a resolvent point of the operator~$A_n$, so that
the above equality implies that $w= c(A_n-\lambda_n)^{-1}\psi$ and that
\[
    \langle (A_n-\lambda_n)^{-1}\psi, \phi\rangle +1 = 0,
\]
i.e., that $F(\lambda_n)=0$. Therefore, there is at most one (up to a factor) eigenvector of~$B$ in the space~$H_n$, and thus its second eigenvector must be of the form $v_n + w_n$ with some $w_n \in H_n$. However, then
\[
    (B-\lambda_n)(v_n + w_n) = (A-\lambda_n)w_n
        + \langle v_n + w_n, \varphi \rangle \psi = 0
\]
so that $w_n$ is collinear to the eigenvector $(A_n-\lambda_n)^{-1}\psi$ found earlier, and thus $v_n$ must also be an eigenvector of~$B$. As $(B-\lambda_n)v_n = \langle v_n , \varphi \rangle \psi$, this requires that $a_n=0$.

Summing up, we see that the assumption that $\dim \ker (B-\lambda) > 1$ implies that $\lambda=\lambda_n$ for some~$n\in I$ and $b_n=0$; moreover, there is an eigenvector $w$ in the subspace~$H_n$ if and only if $F(\lambda_n)=0$, and then $w$ is collinear to~$(A_n-\lambda_n)^{-1}\psi$. The second eigenvector must be  $v_n$, which is possible if and only if $a_n=0$. Therefore all the conditions are necessary, and the geometric multiplicity is then equal to~$2$.

To prove that these conditions are also sufficient, we assume that $\lambda=\lambda_n$ is such that $a_n=b_n=0$ and $F(\lambda_n)=0$. Then, as shown above, $v_n$ and $w:= (A_n-\lambda_n)^{-1}\psi \in H_n$ are linearly independent eigenvectors of~$B$ for the eigenvalue~$\lambda_n$. The proof is complete.
\end{proof}

\begin{example}\rm
Let $\lambda$ and $\mu$ be distinct eigenvalues of an operator $A$ with corresponding normalized eigenvectors~$v$ and $w$; then for the operator
\(
	B := A + (\lambda - \mu)\langle \cdot, w\rangle w
\)
the number~$\lambda$ is an eigenvalue of geometric multiplicity two, $v$ and $w$ being the corresponding eigenvectors. As the above lemma shows, geometric multiplicity cannot be made larger by a rank-one perturbation of~$A$.
\end{example}

\begin{remark}\label{rem:multiplicity}
Assume that $a_n=b_n=0$, so that $\lambda_n$ is an eigenvalue of~$B$ with  eigenvector~$v_n$. Then $v_n$ is also an eigenvector of the adjoint operator~$B^*$, so that the subspaces $\langle v_n \rangle$ and $H_n$ are reducing for $B$. Moreover, the restrictions of $A$ and $B$ onto $\langle v_n\rangle$ coincide.

More generally, we denote by $H^{0}$ the closed linear space of all eigenvectors~$v_k$ of~$A$ for which $a_k=b_k=0$. Then the subspace $H^0$ is reducing for~$B$ and the restrictions of~$A$ and of $B$ onto $H^0$ coincide. Therefore, we can concentrate on the study of the restriction of the operator~$B$ onto its invariant subspace $H^1:=H \ominus H^0$. Without loss of generality, we shall assume that $H^0 = \{0\}$, so that $H=H^1$. Under this assumption, all eigenvalues of the operator~$B$ are geometrically simple.
\end{remark}

Next we discuss algebraic multiplicity of the eigenvalues of~$B$ in the resolvent set of~$A$. As every such an eigenvalue $\lambda$ is geometrically simple by Lemma~\ref{lem:geom-mult}, its algebraic multiplicity coincides with the largest length of chains of eigen- and associated vectors (also called Jordan chains). We recall that a sequence of vectors~$y_0, y_1, \dots, y_m$ forms a chain of eigen-  and associated vectors of~$B$ for an eigenvalue~$\lambda$ if every $y_k$ is in the domain of~$B$, $(B-\lambda)y_0=0$, and $(B-\lambda)y_k = y_{k-1}$ for $k=1,\dots,m$.
Chains of eigen- and associated vectors are not defined uniquely; however, for geometrically simple eigenvalues all such chains are closely related, as the next lemma demonstrates.

\begin{lemma}\label{lem:jordan-chains}
Assume that $\lambda$ is a (geometrically simple) eigenvalue of the operator~$B$ and $y_0,  y_1,\dots, y_m$ is a chain of eigen- and associated vectors corresponding to~$\lambda$.
\begin{itemize}
 \item[(i)] For every sequence of complex numbers $c_1, \dots, c_m$ introduce the vectors $\tilde y_0 = y_0$ and
 \begin{equation}\label{eq:CEAV}
    \tilde y_k = y_k + c_1 y_{k-1} + \cdots + c_k y_0
 \end{equation}
 for $k=1,\dots,m$. Then $\tilde y_0, \tilde y_1,\dots,\tilde y_m$ is a chain of eigen- and associated vectors of~$B$ corresponding to $\lambda$.
 \item[(ii)] Vice versa, assume that $\tilde y_0, \tilde y_1,\dots,\tilde y_m$ is another chain of eigen- and associated vectors of~$B$ corresponding to the eigenvalue $\lambda$ such that $\tilde y_0 = y_0$. Then there are constants $c_1, \dots, c_m$ such that for all $k=1,2,\dots,m$  relations \eqref{eq:CEAV} hold.
\end{itemize}
\end{lemma}

\begin{proof}
By definition of~$\tilde y_k$ and $y_k$, we find that
\[
    (B-\lambda)\tilde y_k =y_{k-1}+ c_1 y_{k-1} + \cdots + c_{k-1} y_0
        = \tilde y_{k-1}
\]
for $k\ge1$ thus establishing Part~(i).

\
For part (ii), the proof is by induction. Since $(B-\lambda)(\tilde y_1 - y_1) = \tilde y_0 - y_0 =0$, it follows that there is $c_1 \in \mathbb{C}$ such that $\tilde y_1- y_1 = c_1 y_0$ thus establishing the base of induction. Assume that the claim has already been proved for $k =1, \dots, l-1<m$; then
\[
    (B-\lambda)(\tilde y_l - y_l) = \tilde y_{l-1} - y_{l-1}
        = c_1 y_{l-2} + \dots + c_{l-1} y_0
\]
and
\[
    (B-\lambda)(\tilde y_l - y_l - c_1 y_{l-1} - \dots - c_{l-1} y_1) = 0.
\]
Therefore, there exists a number $c_l \in \mathbb{C}$ such that $\tilde y_l - y_l - c_1 y_{l-1} - \dots - c_{l-1} y_1 = c_l y_0$, thus finishing the induction step and the proof of the lemma.
\end{proof}

\begin{lemma}\label{lem:alg-mult}
Let $ \lambda^* \in \rho (A)$ be an eigenvalue of the operator~$B$. Then its algebraic multiplicity coincides with the multiplicity of~$\lambda^*$ as a zero of the characteristic function $F$.
\end{lemma}

\begin{proof}
By Lemma~\ref{lem:eig-B}, $F(\lambda^*) = 0$, and the proof of that lemma shows that the vector $y_0:=(A-\lambda^*)^{-1}\psi$ is an eigenvector of~$B$ for the eigenvalue~$\lambda^*$.
Denote by $l+1$ the multiplicity of zero $\lambda=\lambda^*$ of~$F$ and set $y_k: = (A- \lambda ^ *) ^ {-(1+k)} \psi$ for $k=1,2,\dots,l$.
Recall~\cite[\S III.6]{Kat95} that the resolvent $(A-\lambda)^{-1}$ is differentiable on the set~$\rho(A)$ and that
\[
      \frac{d}{d\lambda} (A- \lambda)^{-1} = (A-\lambda)^{-2}.
\]
Observing now that
\begin{equation}\label{eq:alg-1}
    \frac1{k!}F^{(k)}(\lambda^*)
        = \langle (A - \lambda^*)^{-(1+k)} \psi, \varphi\rangle
        = \langle y_k, \varphi \rangle
\end{equation}
for $k\in\mathbb{N}$, we find that
\[
    (B- \lambda^*)y_k
    = (A - \lambda^*)y_k + \langle y_k, \varphi \rangle \psi
    = y_{k-1} + \frac1{k!} F^{(k)} (\lambda^*)\psi = y_{k-1}
\]
for $k=1,\dots, l$. Thus $y_0, y_1, \dots, y_l$ is a Jordan chain of the operator~$B$ for the eigenvalue~$\lambda^*$, so that the algebraic multiplicity of~$\lambda^*$ is at least~$l+1$.

Assume that $\tilde y_0, \tilde y_1, \dots, \tilde y_m$ is a Jordan chain of~$B$ for the eigenvalue~$\lambda^*$. Then $\tilde y_0$ is an eigenvector of~$B$, and without loss of generality we can assume that $\tilde y_0=y_0$.
Now we prove by induction that, with $c_k:= - \langle \tilde y_k, \varphi \rangle$ for $k=1,\dots, m$, we have
\begin{equation}\label{eq:tilde-y}
    \tilde y_k = y_k + c_1 y_{k-1} + \dots + c_k y_0
\end{equation}
and that
$F(\lambda^*) = F'(\lambda^*)=\dots = F^{(m)}(\lambda^*)=0$.

Indeed, $F(\lambda^*) =0$ by Lemma~\ref{lem:eig-B}, and the relation
\[
    (B-\lambda^*)\tilde y_1
        = (A-\lambda^*) \tilde y_1 + \langle \tilde y_1 , \varphi \rangle \psi = y_0
\]
shows that $\tilde y_1 = y_1 + c_1 y_0$. Taking now the scalar product with $\varphi$ and recalling that $\langle y_0,\varphi \rangle = -1$, we get the equality
$\langle y_1, \varphi \rangle = 0$ resulting in $F'(\lambda^*)=0$ in view of~\eqref{eq:alg-1} and establishing the base of induction.

Assuming that the relations
$F(\lambda^*)=F'(\lambda^*) = \dots = F^{(k)}(\lambda^*)=0$ and
$\tilde y_k = y_k + c_1 y_{k-1} + \dots + c_k y_0$ have already been proved for some $k<m$, we recast the equality $(B - \lambda^*)\tilde y_{k+1}=\tilde y_k$ as
\[
        (A - \lambda^*)\tilde y_{k+1}
            + \langle \tilde y_{k+1} , \varphi \rangle \psi
            = y_k + c_1 y_{k-1} + \dots + c_k y_0.
\]
Applying $(A-\lambda^*)^{-1}$ to both sides of the above equality leads to the relation
\[
    \tilde y_{k+1} = y_{k+1} + c_1 y_k + \dots + c_k y_1 +c_{k+1} y_0,
\]
which on account of~\eqref{eq:alg-1} and the induction assumption yields
\begin{align*}
    c_{k+1} &:= - \langle \tilde y_{k+1}, \varphi \rangle
            =  - \langle y_{k+1} + c_1 y_k + \dots + c_k y_1 +c_{k+1} y_0, \varphi \rangle\\
            &\hphantom{:}= -\frac1{(k+1)!} F^{(k+1)}(\lambda^*) + c_{k+1}
\end{align*}
and $F^{(k+1)}(\lambda^*)=0$, thus completing the induction step. Therefore,
$\lambda^*$ is a zero of~$F$ of multiplicity at least~$m+1$, and the proof is complete.
\end{proof}

\begin{example}
To demonstrate the above result, we show how to ``move'' any $m+1$ eigenvalues of $A$ into an arbitrary point $\lambda\not\in\sigma(A)$. For definiteness, we choose the eigenvalues $\lambda_0, \dots, \lambda_m$ and a point $\lambda = i$. We shall take both $\phi$ and $\psi$ in the linear span of the eigenvectors~$v_0, v_1,\dots, v_m$, i.e.,
\[
	\phi = \sum_{n=0}^m a_n v_n, \qquad \psi = \sum_{n=0}^m b_n v_n
\]
and set $c_n := \overline{a_n}b_n$ for $n=0, 1,\dots, m$. The corresponding characteristic function can be then written as
\[
	F(\lambda) = \sum_{n=0}^m \frac{c_n}{\lambda_n -\lambda} + 1 = \frac{p(\lambda)}{\prod_{n=0}^m (\lambda - \lambda_n)},
\]
with a monic polynomial~$p$ of degree $m+1$. According to the above lemma, $F$ must satisfy the equalities $F(i) = F'(i) = \dots = F^{(m)}(i) = 0$, which implies that
$p(\lambda) = (\lambda-i)^{m+1}$. Then we find that
\[
	c_n = - \operatornamewithlimits{res}_{\lambda=\lambda_n} F(\lambda)
			= - \frac{(\lambda_n - i)^{m+1}} 	{\prod_{k\ne n} (\lambda_n - \lambda_k)}
\]
and can choose, e.g., $a_n=\overline{c_n}$ and $b_n = 1$ for $n=0,1,\dots,m$.

In particular, $\phi = \sum_{n=0}^m \overline{c_n}v_n$, $\psi = \sum_{n=0}^m v_n$, and the vectors
\[
	y_k := (A-i)^{k+1}\psi = \sum_{n=0}^m \frac{v_n}{(\lambda_n - i)^{k+1}}, \qquad k=0,1,\dots,m,
\]
satisfy the relations
\begin{align*}
	(B-i)y_k &= (A-i)y_k + \langle y_k , \phi \rangle \psi \\
			 &= \sum_{n=0}^m \frac{v_n}{(\lambda_n - i)^{k}} + \sum_{n=0}^m \frac{c_n}{(\lambda_n - i)^{k+1}}\psi\\
			 &=\begin{cases}
			 F(i) \psi, \quad & k=0 \\
			 y_{k-1} + k!F^{(k)}(i), \quad & k=1, \dots, m.
			 \end{cases}
\end{align*}
In view of the relations $F(i) = F'(i) = \dots = F^{(m)}(i) = 0$, these vectors form a chain of eigen- and associated vectors of $B$ for the eigenvalue~$i$.
\end{example}

As we noted in Lemma~\ref{lem:eig-B}, every point $\lambda = \lambda_n$ of $\sigma_0(A)$ is also an eigenvalue of~$B$. We agreed earlier to exclude the non-interesting case where $a_n=b_n=0$, which by Lemma~\ref{lem:geom-mult} means that such a $\lambda_n$ is a geometrically simple eigenvalue of~$B$. However, its algebraic multiplicity may be greater than one, and we shall relate it to the multiplicity of $\lambda_n$ as a zero of the function~$F$; recall that $F$ was extended by continuity to the set~$\sigma_0(A)$ by  formula~\eqref{eq:F-new}.

\begin{lemma}\label{lem:alg-mult0}
  Assume that $\lambda_n\in \sigma_0(A)$ is an eigenvalue of $B$ of geometric multiplicity~$1$ and algebraic multiplicity~$m\ge1$. Denote by~$l$ multiplicity of $\lambda_n$ as a zero of the function~$F$; then $m=l+1$.
\end{lemma}

\begin{proof}
 Denote by $y_0$ an eigenfunction of~$B$ corresponding to the eigenvalue~$\lambda_n$. Since by assumption $\lambda_n$ is a geometrically simple eigenvalue, $y_0$ is determined uniquely up to a constant factor. As was shown in the proof of Lemma~\ref{lem:eig-B},
 \[
    0 = \langle (B-\lambda_n)y_0,v_n \rangle = \langle y_0, \varphi \rangle  \langle \psi, v_n \rangle
	=  \langle y_0, \varphi \rangle b_n
 \]
 so that either $ \langle y_0, \varphi \rangle =0$ or $b_n=0$. We shall analyse these two cases separately.

 \textbf{Case (a)}: $ \langle y_0, \varphi \rangle =0$. Then $(B-\lambda_n)y_0 = (A-\lambda_n)y_0 = 0$, so that $y_0$ can be taken equal to $v_n$. As a result, $\overline{a_n}=\langle v_n, \varphi \rangle =0$, i.e., $\varphi \in H_n := H \ominus v_n$.

 If a vector $y_1$ is associated to the eigenvector~$y_0$, then $y_1$ should satisfy the relation
 \begin{equation}\label{eq:jordan-y1}
    (B-\lambda_n) y_1 = (A- \lambda_n) y_1 + \langle y_1, \varphi \rangle \psi = y_0 = v_n
 \end{equation}
 and is determined up to the eigenvector~$y_0=v_n$. Therefore if such a vector~$y_1$ exists, it can be chosen orthogonal to~$v_n$, i.e., from $H_n$. Then after taking the scalar product of $(B-\lambda_n)y_1$ with $v_n$ and recalling that  $\ran (A-\lambda_n) = H_n$, we conclude from~\eqref{eq:jordan-y1} that
 \[
      \langle y_1, \varphi \rangle \langle \psi,v_n \rangle = \langle y_1, \varphi \rangle b_n = 1.
 \]
 Thus $b_n$ must be nonzero and $\langle y_1, \varphi \rangle =1/b_n$.

 Denote  by $P_n$ the orthogonal projector onto the subspace~$H_n$

 and by $A_n$ the restriction of the operator~$A$ onto the subspace~$H_n$; then $\lambda_n$ is a resolvent point of~$A_n$. Applying $P_n$ to \eqref{eq:jordan-y1}, we conclude that
 \[
    (A_n -\lambda_n) y_1 + \frac1{b_n} P_n \psi = 0,
 \]
 so that $y_1 = -\frac1{b_n}(A_n - \lambda_n)^{-1}P_n \psi$. The norming condition~$\langle y_1 , \varphi \rangle = 1/b_n$ can now be recast as
 \[
    \langle (A_n - \lambda_n)^{-1}P_n \psi, \varphi \rangle + 1
        =\langle \psi, (A_n - \lambda_n)^{-1}\varphi \rangle + 1
        = 0
 \]
 and amounts to the equality $F(\lambda_n)=0$. The conclusion is that an associated vector $y_1$ exists if and only if $b_n \ne 0$ and $\lambda_n$ is a zero of $F$ (i.e. $l>0$). In particular, $l=0$ is equivalent to $m=1$ (recall that the case $a_n = b_n = F(\lambda_n)=0$ was excluded), and the equality $m=l+1$ is then satisfied.

 Assume therefore that $l>0$ and introduce the vectors
 \[
      y_k:= - \frac1{b_n}(A_n - \lambda_n)^{-k}P_n \psi, \qquad k\ge1.
 \]
 Then one sees that
 \[
    (B-\lambda_n) y_k = (A-\lambda_n) y_k + \langle y_k ,\varphi \rangle \psi
	= y_{k-1} + \langle y_k ,\varphi \rangle \psi
 \]
 and
 \[
    \langle y_k ,\varphi \rangle = - \frac1{b_n} \langle (A_n - \lambda_n)^{-k}P_n \psi ,\varphi \rangle
	= -\frac1{b_n(k-1)!}F^{(k-1)}(\lambda_n).
 \]
 It follows that the vectors $y_1, y_2,\dots,y_l$ form a chain of vectors associated to the eigenvector~$y_0$, so that the algebraic multiplicity $m$ of the eigenvalue~$\lambda_n$ is at least $l+1$.

 Conversely, as in the proof of Lemma~\ref{lem:alg-mult} one can show that in any chain $\tilde y_0, \tilde y_1, \dots, \tilde y_{m-1}$ of eigen- and associated vectors for $B$ the vectors $\tilde y_1, \dots, \tilde y_{m-1}$ are related to the above-constructed vectors $y_1, \dots, y_{m-1}$ via~\eqref{eq:tilde-y} and that $F(\lambda_n) = F'(\lambda_n) = \dots = F^{(m-2)}(\lambda_n)=0$. This shows that $l \ge m-1$ and completes the proof in the case (a).

 \textbf{Case (b)}: $b_n = 0$. Then $\psi$ belongs to $H_n = H \ominus v_n$ and thus the range
 $\ran(B - \lambda_n)$ of $B-\lambda_n$ is contained in $H_n$.
 We look for an eigenvector $y_0$ of~$B$ of the form $\alpha_0 v_n + z_0$ with $z_0 \in H_n$. Then $(A-\lambda_n)y_0 = (A_n - \lambda_n)z_0$, and $(B-\lambda_n)y_0=0$ can be written as
 \[
    (A_n - \lambda_n) z_0 + \langle y_0 , \varphi \rangle \psi =0,
 \]
 so that $z_0 = c(A_n - \lambda_n)^{-1}\psi$ with an appropriate constant~$c$. Substituting this $z_0$ into the above equation results in the relation
 \[
    c \psi + \bigl[\alpha_0 \overline{a_n} + c \langle (A_n - \lambda_n)^{-1}\psi , \varphi \rangle \bigr] \psi =0,
 \]
 yielding the equality
 \begin{equation}\label{eq:alpha0}
    c F(\lambda_n) + \alpha_0 \overline{a_n} =0.
 \end{equation}

 In order that for the eigenvector~$y_0$ there could exist an associated vector~$y_1$, it is necessary that $y_0 = (B-\lambda_n)y_1$ belong to $H_n$ and thus that $\alpha_0 = 0$ and $y_0=z_0$. Equation~\eqref{eq:alpha0} then yields $cF(\lambda_n)=0$, and since $c=0$ would lead to the contradiction that $y_0=z_0=0$, we conclude that necessarily $F(\lambda_n)=0$. In particular, $l=0$ gives $m=1$ as stated.

 Assume therefore that $l>0$, so that $F(\lambda_n)=0$. As the case $a_n=b_n =0$ was excluded earlier, we have $a_n\ne0$ and thus $\alpha_0=0$ by~\eqref{eq:alpha0} and $y_0= z_0 := (A_n - \lambda_n)^{-1}\psi$.

 We first show that $m\ge l+1$ by constructing a chain $y_1, \dots, y_l$ of vectors associated to this~$y_0$. Namely, take
  $y_k := (A_n - \lambda_n)^{-(1+k)}\psi$
 for $k=1,\dots, l-1$ and
  $y_l := \alpha_l v_n + (A_n - \lambda_n)^{-(1+l)}\psi$
 with an~$\alpha_l$ to be determined later. As in the proof of Case~(a) we find that
 \[
        (B-\lambda_n) y_k
            = (A_n - \lambda_n) y_k + \langle y_k,\varphi\rangle \psi
            = y_{k-1} + \frac{1}{k!} F^{(k)}(\lambda_n) \psi
            = y_{k-1}
 \]
 for $k =1, 2, \dots, l-1$. For $k=l$ we get
 \[
        (B-\lambda_n) y_l
            = (A_n - \lambda_n) y_l + \langle y_l,\varphi\rangle \psi
            = y_{l-1}
                + \bigl[\alpha_l \overline{a_n}
                        +\frac{1}{l!} F^{(l)}(\lambda_n)\bigr] \psi,
 \]
 and the equality $(B-\lambda_n) y_l = y_{l-1}$ is guaranteed by taking (recall that $a_n \ne0$)
 \[
    \alpha_l := - \frac{1}{\overline{a_n}l!} F^{(l)}(\lambda_n).
 \]

 It remains to show that $l \ge m-1$. We take a chain of eigen- and associated vectors $\tilde y_0, \dots, \tilde y_{m-1}$ of the maximal possible length~$m>1$. The equalities $(B-\lambda_n) \tilde y_k = \tilde y_{k-1}$ for $k=1, \dots, m-1$ show that the vectors $\tilde y_0, \dots, \tilde y_{m-2}$ belong to $H_n$. Without loss of generality we may assume that $\tilde y_0 = y_0$ and then prove by induction that with $c_k:= - \langle \tilde y_k, \varphi \rangle$ for $k=0,1,\dots, m-2$ we have
 \
    \tilde y_k = y_k + c_1 y_{k-1} + \dots + c_k y_0
 \
 with $y_k$ defined above and that $F(\lambda_n) = F'(\lambda_n)=\dots = F^{(k)}(\lambda_n)=0$.

 The base of induction was already set up: $\tilde y_0 = y_0$ and $F(\lambda_n) = 0$. Assume therefore that the claim holds for all indices $k$ less than $j$ with $0<j<m-2$ and rewrite the equality $(B-\lambda_n) \tilde y_j = \tilde y_{j-1}$ as
 \[
    (A_n-\lambda_n) \tilde y_j + \langle \tilde y_j, \varphi \rangle \psi = \tilde y_{j-1}.
 \]
 It follows that $\tilde y_j = (A_n - \lambda_n)^{-1} \tilde y_{j-1} - \langle y_j, \varphi \rangle y_0$, which by the induction assumption can be recast as
 \[
     \tilde y_j = y_j + c_1 y_{j-1} + \dots + c_j y_0.
 \]
 Since
 \[
    \frac1{k!}F^{(k)}(\lambda_n) = \langle y_k, \varphi \rangle
 \]
 for $k\in \mathbb{N}$ and the equalities $F(\lambda_n) = F'(\lambda_n) = \cdots = F^{(j-1)}(\lambda_n) = 0$ hold by assumption, we find that $\langle y_0, \varphi \rangle = -1$ and, by taking the scalar product with $\varphi$ in the above formula for $\tilde y_j$ that
 \[
    -c_j = \langle \tilde y_j, \varphi \rangle
        = \langle y_j, \varphi \rangle - c_j.
 \]
 Therefore $\langle y_j, \varphi \rangle = 0$ yielding the relation $F^{(j)}(\lambda_n) = 0$.

 This completes the induction step and shows that $\lambda_n$ is a zero of~$F$ of multiplicity at least~$m-1$. The proof is complete.
\end{proof}

\begin{example}\label{ex:multiple-EV}\rm
In the Hilbert space~$L_2(0,2\pi)$, we consider a self-adjoint operator
\[
		A = \frac1{i}\frac{d}{dx}
\]
subject to the periodic boundary condition $y(0)=y(2\pi)$. The spectrum of $A$ coincides with the set $\mathbb{Z}$, and an eigenfunction $v_n$ corresponding to the eigenvalue~$\lambda_n:=n$ is equal to $e^{inx}/\sqrt{2\pi}$.

For every $m\in\mathbb{N}$, we shall construct a rank one perturbation~$\langle \,\cdot\,,\phi\rangle \psi$ so that the perturbed operator~$B$ has an eigenvalue~$\lambda_0=0$ of algebraic multiplicity $2m+1$ and simple eigenvalues $\mu_n = \lambda_n$ if $|n|>m$. More precisely, we take
\[
	\phi(x) = \sum_{k=-m}^m e^{ikx} = \frac{\sin(m+\tfrac12)x}{\sin(\tfrac12x)}
\]
and
\[
	\psi(x) = \sum_{k=1}^m d_k \sin(kx)
\]
with coefficients $d_k$ to be determined.  Since $\langle \psi, v_0 \rangle = 0$, the corresponding chain of eigen- and associated vectors can be formed as in Case (b) of the above theorem. Namely, with $A_0$ standing for the restriction of~$A$ onto the space $H_0:= H \ominus v_0$, we take
\[
	y_k := A_0^{-(k+1)}\psi, \qquad k=0,\dots, 2m-1,
\]
and
\[
	y_{2m}:= d_0v_0 + A_0^{-(2m+1)}\psi
\]
for a suitable $d_0$. We next show that there is a unique set of $d_0,\dots,d_m$ for which the above $y_0,\dots, y_{2m}$ form a chain of eigen- and associated vectors of~$B$ and that there is no longer chains of eigen- and associated vectors corresponding to~$\lambda_0$.

Notice that
\[
	A_0^{-2l}\psi(x) = \sum_{k=1}^m \frac{d_k}{k^{2l}}\sin(kx)
\]
and
\[
	A_0^{-2l+1} = -i \sum_{k=1}^m \frac{d_k}{k^{2l-1}}\cos(kx).
\]
It then follows that $y_{2l+1}$ are odd functions for all $l=0,\dots,m-1$, and as~$\phi$ is an even function, we find that $By_{2l+1} = A y_{2l+1} = y_{2l}$. On the other hand, the equalities
$By_{2l} = y_{2l-1}$ for $l=0,\dots,m$ amount to a non-singular system of $m+1$ linear equations in $m+1$ variables $d_0,d_1,\dots, d_m$,
\begin{equation}\label{eq:example-system}
	\sum_{k=1}^m \frac{d_k}{k^{2l+1}} = f_l, \quad l=0,1,\dots,m,
\end{equation}
with $f_0 = -i/(2\pi)$, $f_1 = \dots = f_{m-1} = 0$, and $f_m = -id_0/\sqrt{2\pi}$.

Note that $d_0 \ne0$ as otherwise the system would be inconsistent, so that $y_{2m}$ does not belong to $H_0$ and thus the chain cannot be extended further. In view of Lemma~\ref{lem:jordan-chains}, this is true of any other chain of EAV's for the eigenvalue~$\lambda_0$. As $a_0\ne0$, geometric multiplicity of $\lambda_0=0$ is equal to one by Lemma~\ref{lem:geom-mult}; therefore, $\lambda_0$ is a geometrically simple eigenvalue of~$B$ of algebraic multiplicity~$2m+1$.

The explicit form of $\phi$ and $\psi$ yields their Fourier coefficients: $a_n = b_n = 0$ if $|n|>m$, $a_n = \sqrt{2\pi}$ for $|n| \le m$, and, finally, $b_n = \sqrt{2\pi}d_n/2i$ for $n=1,\dots,m$, $b_n = -b_{-n}$ for $n=-m,\dots,-1$, and $b_0=0$. Then the characteristic function,
\[
	F(z) =  \sum_{n=-m}^m\frac{\overline{a_n}b_n}{n-z} + 1
		 =  \sqrt{2\pi}\sum_{n=1}^m\frac{2nb_n}{n^2-z^2} + 1
		 = \frac{2\pi}i\sum_{n=1}^m\frac{nd_n}{n^2-z^2} + 1
\]
is a rational function of the form $P(z)/Q(z)$ with $P$ and $Q$ polynomials of degree at most $2m$. Therefore, $F$ has at most $2m$ zeros counting with multiplicity. On the other hand, it is straightforward to verify that equations~\eqref{eq:example-system} amount to the relations
\[
	F(0) = F'(0) = \dots = F^{(2m-1)}(0) = 0,
\]
so that $z = 0$ is a zero of~$F$ of multiplicity~$2m$. This implies that $F$ has no other zeros. In particular, $F(n)\ne0$ if $n\ne0$, and thus $\lambda_n = n$ is an algebraically simple eigenvalue of the operator~$B$ whenever $|n|>m$.

To sum up, the operator~$B$ has an eigenvalue~$\lambda_0 = 0$ of algebraic multiplicity~$2m+1$ and simple eigenvalues $\lambda_n$ for $|n|>m$. Loosely speaking, the rank one perturbation shifts the eigenvalues $\lambda_{-m}, \dots, \lambda_{-1}$, $\lambda_1, \dots, \lambda_m$ towards $\lambda_0$ respectively enlarging the multiplicity of the latter.
\end{example}

\section{Spectral localization of the operator~$B$}\label{sec:asympt}

We next turn to the question, what spectra the rank-one perturbations~$B$ of a given self-adjoint operator~$A$ can have. Keeping in mind the most important and interesting applications to the differential operators, in addition to~$(A1)$ we assume that
\begin{itemize}
	\item[(A2)] the eigenvalues of~$A$ are separated, i.e.,
	\begin{equation}\label{eq:dist}
	\inf_{n \in I} |\lambda_{n+1} - \lambda_n| =: d > 0.
	\end{equation}
\end{itemize}

We next localize the spectrum of~$B$ by studying its characteristic function
\begin{equation*}\label{eq:F1}
	F(z) = \sum_{k \in I_1} \frac {\overline{a_k} b_k}{\lambda_k - z} + 1.
\end{equation*}
As the Fourier coefficients $a_k$ and $b_k$ of the functions~$\phi$ and $\psi$ are in $\ell_2(I)$, the sequence $\overline{a_k}b_k$ is summable and, due to the Cauchy--Bunyakowsky--Schwarz inequality, its $\ell_1$-norm is bounded by $\|\varphi\|\|\psi\|$.

\begin{lemma}
	The spectrum of $B$ lies in the strip
	\[
		\Pi := \{z \in \bC \mid |\myIm z| \le \|\varphi\|\|\psi\|\}.
	\]
\end{lemma}

\begin{proof}
	If $z\not\in \Pi$, then $|\lambda_k - z| \ge |\myIm z| > \|\varphi\|\|\psi\|$, so that
	\[
		\sum_{k \in I_1} \biggl|\frac {\overline{a_k} b_k}{\lambda_k - z}\biggr|
			< \sum_{k \in I_1} |\overline{a_k} b_k|/(\|\varphi\|\|\psi\|)
			< 1
	\]
	so that $F(z) \ne 0$.
\end{proof}

Next, for an $\varepsilon>0$ we denote by $C_n(\varepsilon)$ the open circle
\[
	C_n(\varepsilon) := \{ z \in \bC \mid |z - \lambda_n| < \varepsilon\}
\]
and set
\[
	R_{N, \varepsilon}:= \Bigl\{z \in \bC \mid |\myRe z| \ge  N\}
		\setminus \Bigl(\bigcup\nolimits_{n\in I} C_n(\varepsilon) \Bigr)\Bigr\}
\]

\begin{lemma}\label{lem:RN}
For every $\varepsilon>0$ there is $N>0$ such that $R_{N,\varepsilon}$ belongs to the resolvent set of the operator~$B$.
\end{lemma}

\begin{proof}
For an $\varepsilon >0$, we choose $N'\in\mathbb{N}$ so that%
\begin{footnote}
	{Throughout this section, the symbol $\sum{\hspace*{-2pt}\vphantom{\sum}}^{(1)}$  denotes summation over the index set $I_1$}
\end{footnote}
	\[
		\sumI_{|k|\ge N'} |\overline{a_k} b_k| \le \frac{\varepsilon}4;
	\]
	then, for $z$ outside every circle $C_n(\varepsilon)$,
	\[
		\Bigl|\sumI_{|k|\ge N'} \frac{\overline{a_k} b_k}{\lambda_k - z}\Bigr|
			\le \frac1\varepsilon \sum_{k\in I_1, |k|\ge N'} |a_k b_k| \le \frac14.
	\]
	We now take $N''\in\mathbb{N}$ such that  $N''\ge N' + 4 \|\varphi\|\|\psi\|/d$ and choose $N\in \mathbb{N}$ such that $N\ge |\lambda_{N''}|$ and $N \ge |\lambda_{-N''}|$ if $-N'' \in I$. Due to Assumption~$(A2)$ it holds that $|\lambda_k - \lambda_m| \ge d|k-m|$; therefore,
	$|\lambda_k - z| \ge d (N'' - N') \ge 4\|\varphi\|\|\psi\|$ whenever $z \in R_{N,\varepsilon}$ and $|k| \le N'$, so that
	\[
	  	\Bigl|\sumI_{|k| < N'} \frac{\overline{a_k} b_k}{\lambda_k - z}\Bigr|
	  		\le \frac14
	\]
	for such $z$.
	As a result, for all $z \in R_{N,\varepsilon}$ it holds
	\[
			|F(z)| \ge 1 - 	\Bigl|\sum_{k\in I_1} \frac{\overline{a_k} b_k}{\lambda_k - z}\Bigr|
				\ge \frac12;
	\]
	by Lemma~\ref{lem:eig-B} the set $R_{N,\varepsilon}$ is in the resolvent set of~$B$, and the proof is complete.
\end{proof}

Combining the above two lemmata, we conclude that the spectrum of $B$ is localized in the circles $C_n(\varepsilon)$ and in the rectangular domain
\[
	\{z \in \bC \mid |\myRe|\le N, \ |\myIm z| \le \|\varphi\|\|\psi\|\},
\]
with $N=N(\varepsilon)$ from Lemma~\ref{lem:RN}.

\begin{lemma}\label{lem:EVinCn}
	For every $\varepsilon>0$ there is $K=K(\varepsilon)$ such that for each $n\in I$ with $|n| > K(\varepsilon)$ the circle $C_n(\varepsilon)$ contains precisely one eigenvalue of~$B$.
\end{lemma}

\begin{proof}
	By Lemma~\ref{lem:RN}, for all $n$ with large enough $|n|$, the boundary $\partial C_n(\varepsilon)$ of~$C_n(\varepsilon)$ is in the resolvent set of~$B$. We next show that the Riesz spectral projections for $A$ and $B$ corresponding to $C_n(\varepsilon)$ are of the same rank (and thus of rank~$1$) for large enough~$|n|$.

	 For every $n$ with $\partial C_n(\varepsilon) \subset \rho(B)$, we denote by $P_n$ and $P'_n$ the Riesz spectral projectors for $A$ and $B$ respectively on the root subspaces corresponding to the eigenvalues inside $C_n(\varepsilon)$,
	\[
		P_n  = \frac1{2\pi i} \int_{C_n(\varepsilon)} (A - z)^{-1}\,dz, \qquad
		P'_n = \frac1{2\pi i} \int_{C_n(\varepsilon)} (B - z)^{-1}\,dz.
	\]
	By the Krein resolvent formula~\eqref{eq:Krein}, we get
	\[
		P_n - P'_n = \frac1{2\pi i} \int_{C_n(\varepsilon)}\frac{dz}{F(z)} \langle \, \cdot \,, (A - \overline{z})^{-1} \varphi \rangle
		(A-z)^{-1} \psi.
	\]
	As the norm of a rank-one operator $\langle \, \cdot \, u \rangle v$ is equal to $\|u\|\|v\|$ and, as proved in Lemma~\ref{lem:RN}, $|F(z)|\ge 1/2$ on $C_n(\varepsilon)$ for large enough $|n|$, we conclude that
	\[
		\|P_n - P'_n\| \le d \max_{z\in C_n(\varepsilon)} \|(A - \overline{z})^{-1} \varphi\| \|(A - {z})^{-1} \psi\|
	\]
	for such $n$. Observe now that for every vector $u = \sum c_k v_k$ we have
	\[
		\|(A - z)^{-1} u \|^2 = \sum_{k\in I} \frac{|c_k|^2}{|\lambda_k - z|^2};
	\]
	applying the Lebesgue dominated convergence theorem, we conclude that
	\[
		\max_{z\in C_n(\varepsilon)} \|(A - z)^{-1} u \|^2 \to 0
	\]
	as $|n| \to \infty$. Therefore, $\|P_n - P'_n\| \to 0$ as $|n|\to\infty$; as a result~\cite[\S IV.2]{Kat95}, the ranks of the Riesz projectors $P_n$ and $P'_n$ coincide for all $n$ with large enough $|n|$, and the proof is complete.
\end{proof}

Therefore, the operator~$B$ has at most finitely many nonsimple eigenvalues; we next prove that there are no other restrictions on them.

\begin{lemma}\label{lem:nonrealEV}
	Fix an arbitrary $n\in\mathbb{N}$, an arbitrary sequence $z_1, z_2, \dots, z_n$ of pairwise distinct complex numbers, and an arbitrary sequence $m_1$, $m_2$, $\dots$, $m_n$ of natural numbers. Then there is a rank-one perturbation~$B$ of the operator~$A$ such that, for every $j=1,2,\dots, n$, the number $z_j$ is an eigenvalue of~$B$ of algebraic multiplicity~$m_j$.
\end{lemma}

\begin{proof}
	For simplicity, we assume that none of $z_j$ is in the spectrum of~$A$; the changes to be made otherwise are not very significant, cf.~Lemma~\ref{lem:alg-mult0} and Example~\ref{ex:multiple-EV}.

	Set $N:= m_1 + m_2 + \dots + m_n$; we will construct a rank-one perturbation~$B$ of $A$ with
	\[
		\varphi = \sum_{k=1}^N a_k v_k, \qquad
		\psi = \sum_{k=1}^N b_k v_k.
	\]
	According to Lemma~\ref{lem:eig-B}, it suffices to choose $a_k$ and $b_k$ in such a way that the characteristic function~$F$ of~\eqref{eq:F-new} has zeros $z_1, z_2, \dots, z_n$ of multiplicity $m_1, m_2, \dots, m_n$ respectively. Set $c_k := \overline{a_k}b_k$, $k=1,2,\dots,n$; then
	\[
		F(z) = \sum_{k=1}^n \frac{c_k}{\lambda_k - z} + 1,
	\]
	and the equalities $F(z_k) = F'(z_k) = \dots = F^{(m_k-1)}(z_k) = 0$ lead to an inhomogeneous system of $N$ equations in the variables $c_1, c_2, \dots, c_N$:
	\begin{equation}\label{eq:system}
		\sum_{k=1}^N \frac{c_k}{(\lambda_k - z_j)^m} + \delta_{m1}= 0, \quad j = 1,2, \dots, n, \quad m = 1, 2, \dots, m_j,
	\end{equation}
	with $\delta_{m1}$ being the Kronecker delta. By Lemma~\ref{lem:Cauchy} below, the coefficient matrix of the above system is non-singular; therefore, the system possesses a unique solution~$c_1,c_2, \dots, c_N$. It remains to take $a_k =1$ and $b_k = c_k$ for $k=1,2,\dots, N$, and the proof is complete.
\end{proof}

\begin{lemma}\label{lem:Cauchy}
	The coefficient matrix of system~\eqref{eq:system} is non-singular.
\end{lemma}

\begin{proof}
	For pairwise distinct numbers $\omega_1, \omega_2, \dots, \omega_N$ from the resolvent set of~$A$, we introduce the Cauchy matrix $M$ with entries
	\[
	(M)_{jk} = \frac{1}{\lambda_k - \omega_j}.
	\]
	It is non-singular and has determinant equal to
	\begin{equation}\label{eq:Cauchy}
	D(\omega_1,\omega_2,\dots,\omega_{N}) = \frac{\prod\prod_{j>k}(\lambda_j - \lambda_k)(\omega_j-\omega_k)} 		{\prod_j\prod_k(\lambda_j-\omega_k)}.
	\end{equation}
	We set $C:= \prod\prod_{j>k}(\lambda_j - \lambda_k)$ for brevity.

	Taking the derivative of that determinant in $\omega_2$ and setting $\omega_2 = \omega_1 = z_1$, we get the determinant of the matrix~$M_2$, whose first and second rows have entries
	\[
			\frac1{\lambda_k - z_1 } \quad \text{and} \quad \frac1{(\lambda_k - z_1)^2}, \qquad k = 1, 2, \dots, N,
	\]
	respectively, and the other rows are as in the matrix~$M$. By~\eqref{eq:Cauchy}, we have
	\[
		D(\omega_1,\omega_2,\dots,\omega_{N}) = (\omega_2 - \omega_1) D_2(\omega_1,\omega_2,\dots,\omega_{N}),
	\]
	so that
	\[
		\frac{\partial}{\partial \omega_2}D(z_1,\omega_2,\dots,\omega_{N})\Bigr|_{\omega_2 = z_1}
			= D_2(z_1, z_1, \omega_3, \dots, \omega_N).
	\]
	Explicit calculations give
	\begin{multline*}
		\det M_2 = D_2(z_1, z_1, \omega_3, \dots, \omega_N) \\
			= C \prod_{j>2}(\omega_j-z_1)^2 \frac{\prod\prod_{j>k>2}(\omega_j-\omega_k)}
			{\prod_j (\lambda_j-z_1)^2\prod_{k>2}(\lambda_j-\omega_k)}
			\ne 0.
	\end{multline*}

	Next, we take the second derivative of $D_2(z_1, z_1, \omega_3, \dots, \omega_N)$ in $\omega_3$ and set $\omega_3 = z_1$; this becomes the determinant $D_3(z_1, z_1, z_1, \omega_4, \dots, \omega_N)$ of the matrix $M_3$ that is $M_2$ with its third row replaced by
	\[
		\frac2{(\lambda_k - z_1)^3}, \qquad k = 1, 2, \dots, N.
	\]
	On the other hand,
	\begin{multline*}
		\det M_3 = D_3(z_1, z_1, z_1, \omega_4, \dots, \omega_N)
			= \frac{\partial^2}{\partial \omega^2_3}D(z_1, z_1, \omega_3, \dots,\omega_{N})\Bigr|_{\omega_3 = z_1} \\
			= 2C \prod_{j>3}(\omega_j-z_1)^3 \frac{\prod\prod_{j>k>3}(\omega_j-\omega_k)}
				{\prod_j (\lambda_j-z_1)^3\prod_{k>3}(\lambda_j-\omega_k)}
			\ne 0.
	\end{multline*}
	On each next step, we repeat a similar procedure with the next row and variable until we reach row number $m_1$.

	After that, we set $\omega_{m_1+1} = z_2$, take the derivative in $\omega_{m_1+2}$ at $\omega_{m_1 + 2} = z_2$, and repeat with the subsequent rows until we reach row number $m_1 + m_2$. Clearly, the operations described above can be performed on separate groups of variables $\omega_l$ with $l=m_1 + \dots + m_j + 1, m_1 + \dots + m_j + 2, \dots, m_1 + m_2 + \dots + m_{j+1}$ independently. At the end, the determinant of the coefficient matrix of the system~\eqref{eq:system} is found explicitly to be
	\[
		\frac{\prod_{j=k+1}^N\prod_{k=1}^N(\lambda_j - \lambda_k)	\prod_{j=k+1}^n\prod_{k=1}^n(z_j-z_k)^{m_j + m_k}}
		{\prod_{j=1}^N\prod_{k=1}^n(\lambda_j-z_k)^{m_j}} \ne 0,
	\]
	and the proof is complete.
\end{proof}

\begin{remark}
	In the paper~\cite{DobHry20}, it is proved that the operators $A$ and $B$ have the same number of eigenvalues in special increasing  rectangles exhausting the whole complex plane~$\bC$. Combined with the results of Lemmata~\ref{lem:RN} and \ref{lem:EVinCn}, this allows an enumeration of the eigenvalues of~$B$ as $\mu_n$, $n\in I$, such that each value $\mu_n$ is repeated according to its multiplicity and $\mu_n- \lambda_n \to0$ as $|n|\to\infty$.
\end{remark}

We summarize the above results in the following theorem.

\begin{theorem}\label{thm:main}
	Assume that $A$ is an operator in a Hilbert space~$H$ satisfying assumptions~$(A1)$ and $(A2)$ and $B$ is its rank-one perturbation~\eqref{eq:B}. Then
	\begin{itemize}
		\item[(i)] all eigenvalues of $B$ of sufficiently large absolute value are localized within $\varepsilon$-neighbourhood of the eigenvalues of~$A$ and thus are simple;
		\item[(ii)] the eigenvalues of~$B$ can be enumerated as $\mu_n$, $n\in I$, so that $\mu_n-\lambda_n \to 0$ as $|n|\to\infty$;
		\item[(iii)] geometric multiplicity of every eigenvalue of~$B$ is at most~$2$, and multiplicity~$2$ is only possible when the corresponding eigenspace of~$A$ is reducing for $B$.
	\end{itemize}
	Moreover, for every prescribed finite set $z_1, z_2, \dots, z_n$ of pairwise distinct complex numbers, and an arbitrary sequence $m_1$, $m_2$, $\dots$, $m_n$ of natural numbers there exists a~$B$ such that each $z_j$, $j=1,2,\dots, n$, is an eigenvalue of~$B$ of algebraic multiplicity~$m_j$.
\end{theorem}

\section{Finite-dimensional case}\label{sec:finite-dim}

The analysis of Section~\ref{sec:mult} allows to essentially complement the results in the fi\-ni\-te-di\-men\-si\-o\-nal case. Namely, assume that $A$ is a Hermitian matrix in $\bC^n$ with pairwise distinct eigenvalues $\lambda_1, \lambda_2, \dots, \lambda_n$ and normalized (column) eigenvectors~$\mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_n$ and define the \emph{generic set} $\mathcal{G}(A)$ of $A$ as
\[
	\mathcal{G}(A) = \{ \mathbf{x} \in \mathbb{C}^n  \mid \langle \mathbf{x}, \mathbf{v}_k\rangle_{\bC^n} \ne 0, \quad k = 1, 2, \dots,n  \}.
\]
Then we have the following generalization of the result of~\cite{Kru92}.

\begin{theorem}\label{thm:finite-dim}
	Under the above assumptions, let $\bm{\varphi}$ be a vector from the generic set~$\mathcal{G}(A)$. Then for any natural number $k$, any pairwise distinct complex numbers~$z_1, z_2, \dots, z_k$, and any natural numbers $m_1, m_2, \dots, m_k$ satisfying $m_1+ m_2 + \dots + m_k =n$, there is a unique vector $\bm{\psi} \in \bC^n$ such that the rank-one perturbation  $B = A + \bm{\psi}\bm{\varphi}^\top$ of the matrix~$A$ has eigenvalues $z_1, z_2, \dots, z_k$ of corresponding multiplicities $m_1, m_2, \dots, m_k$.

	Similarly, for every fixed $\bm{\psi} \in \mathcal{G}(A)$ there is a unique $\bm{\varphi}\in \bC^n$ such that $B$ has the eigenvalues $z_j$ of prescribed multiplicities $m_j$, $j = 1,2, \dots, k$.
\end{theorem}

\begin{proof}
Denote by $\sigma_0(A)$ the common part of the spectrum $\sigma(A)$ of $A$ and the set $\{z_1, z_2, \dots, z_k\}$, by $\sigma_1(A):=\sigma(A) \setminus\sigma_0(A)$ the remaining part of $\sigma(A)$, and let $I_\ell := \{ j \mid \lambda_j \in \sigma_p(A)\}$, $\ell=0,1$, be the corresponding index sets. We update the multiplicities $m_j$ to
\begin{equation}\label{eq:reduce-mult}
	m'_j := \begin{cases}
		m_j - 1, & \qquad z_j \in \sigma(A); \\
		m_j, & \qquad z_j \not\in \sigma(A);
	\end{cases}
\end{equation}
and set
\begin{equation}\label{eq:F-prod}
	F(z):= \frac{\prod_{j=1}^k (z - z_j)^{m'_j}}{\prod_{j \in I_1} (z - \lambda_j)}.
\end{equation}
Denoting by $-c_j$ the residue of the function~$F$ at the point $z=\lambda_j$, $j\in I_1$, we conclude that $F$ can be written in the form
\begin{equation}\label{eq:F-sum}
	F(z) = \sum_{j \in I_1} \frac{c_j}{\lambda_j-z} + 1.
\end{equation}
Denote by $a_j = \langle \bm{\varphi}, \mathbf{v}_k\rangle_{\bC^n}$, $j= 1,2, \dots, n$, the coefficients of the vector $\bm{\varphi}$ in the basis $\mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_n$. By assumption, no $a_j$ vanishes, and we set $b_j:= c_j / \overline{a_j}$ for $j \in I_1$ and $b_j = 0$ for $j \in I_0$, and define the vector $\psi$ via
\[
	\bm{\psi}  = \sum_{j=1}^n b_j \mathbf{v}_j = \sum_{j \in I_1} b_j \mathbf{v}_j.
\]
It follows from the results of Section~\ref{sec:mult} that the characteristic function of the matrix $B = A + \bm{\psi}\bm{\varphi}^\top$ coincides with the above function~$F$; therefore, the matrix $B$ has eigenvalues $z_1, z_2, \dots, z_k$ and the multiplicity of the eigenvalue $z_j$ is $m_j'$ if $z_j \not \in \sigma(A)$ or $m_j'+1$ otherwise.

The second part is proved in a similar manner, by interchanging the roles of $a_n$ and $b_n$.
\end{proof}

If the vector $\bm{\varphi}$ is not in the generic set~$\mathcal{G}(A)$ of $A$, the above theorem has the following analogue.

\begin{theorem}\label{thm:phi-arbitrary}
	Under the above assumptions on the matrix~$A$, take a nonzero vector $\bm{\varphi} = \sum _{j=1}^n a_j \mathbf{v}_j \in \bC^n$ and set $I_0 := \{j \mid a_j = 0\}$ and $\sigma_0(A):=\{\lambda_j \mid j \in I_0\}$. Then for every natural $k$, every set $S=\{z_1, z_2,\dots, z_k\}$ of $k$ pairwise distinct complex numbers obeying $S \cap \sigma(A) = \sigma_0(A)$, and every sequence $m_1, m_2, \dots, m_k$ of natural numbers with $m_1 + m_2 + \dots + m_k = n$ there is a vector $\bm{\psi}\in\bC^n$ such that the matrix $B = A + \bm{\psi}\bm{\varphi}^\top$ has eigenvalues $z_1, z_2, \dots, z_k$ of multiplicities $m_1, m_2 , \dots, m_k$ respectively.

	A similar statement holds with the r\^oles of $\bm{\varphi}$ and $\bm{\psi}$ interchanged.
\end{theorem}

\begin{proof}
	The fact that the set $S$ is in the spectrum of~$B$ is proved in Lemma~\ref{lem:eig-B}. We denote by~$\sigma_1(A)$ the spectrum of~$A$ not in $\sigma_0(A)$ and set $I_1$ to be the corresponding set of indices. Reducing by $1$ the multiplicity of each $z_j$ from $S$ and denoting the resulting multiplicities by $m'_j$ as in~\eqref{eq:reduce-mult}, we construct the function~$F$ of~\eqref{eq:F-prod} and observe that it assumes the form~\eqref{eq:F-sum}, with uniquely determined residues~$-c_j$, $j\in I_1$. Then we define $b_j$ for such $j$ from the relation~$\overline{a}_jb_j = c_j$, and fix arbitrarily $b_j$ for $j \in I_0$.

	By Lemmata~\ref{lem:alg-mult} and \ref{lem:alg-mult0}, the numbers $z_j$ not in $\sigma_0(A)$ are eigenvalues of the matrix~$B$ of multiplicity~$m_j'$, while those in~$\sigma_0(A)$ have multiplicity $m_j'+1$. The proof is complete.
\end{proof}

\begin{remark}
	We can conclude from the above proof that the coordinates of the vector $\bm{\psi}$ in the basis $\mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_n$ for $j\in I_0$ are not fixed; therefore, there is an $|I_0|$-dimensional affine set of such vectors producing the required spectrum.
\end{remark}

\section{Concluding remarks}

It should be noted that some restrictions imposed on~$A$ can be relaxed. For instance, self-adjointness of~$A$ is not essential; the proof with minor amendments will work for rank-one perturbations of every normal operator with simple discrete spectrum, or even in the case when the eigenvectors of~$A$ can be chosen to form a Riesz basis of~$H$. Simplicity of the eigenvalues of~$A$ can also be dropped; however, this will result in a more complicated Jordan structure of the root subspaces of~$B$, cf.~\cite{BehMoeTru14}. Also, the operator $A$ may possess, in addition to an infinite discrete spectrum, a non-trivial essential component; the results we proved have natural generalization to this case as well.

Finally, this study has found its continuation in~\cite{DobHry20}, in which a complete characterization of all possible spectra of rank-one perturbations~\eqref{eq:B} of self-adjoint operators~$A$ with simple discrete spectrum is given.

redpajama_set_name: RedPajamaArXiv


text:
Alana Chenevert (Los Angeles, 26 de maio de 1987), conhecida artisticamente como The Last Artful, Dodgr, é uma rapper norte-americana.

Rappers dos Estados Unidos

redpajama_set_name: RedPajamaWikipedia


text:
Lea Schüller, née le  à Tönisvorst, est une footballeuse allemande évoluant au poste de milieu de terrain. Elle évolue dans l'équipe du Bayern Munich au sein du championnat d'Allemagne féminin, et avec l'équipe nationale d'Allemagne.

Biographie

Elle participe aux éliminatoires de la Coupe du monde 2019 et marque cinq buts lors de la phase de groupe.

Le , elle s'engage pour trois saisons à partir de l'été 2020 en faveur de Bayern Munich.

Le , elle prolonge son contrat jusqu'en 2026.

Statistiques
Le tableau ci-dessous retrace la carrière de Lea Schüller depuis ses débuts :

En club

Palmarès

En club
  Bayern Munich
 Championnat d'Allemagne (1)
 Champion :2021

Sélection
  Équipe du Allemagne
 Championnat d'Europe
 Finaliste : 2022

Notes et références

Liens externes

Naissance en novembre 1997
Naissance à Tönisvorst
Footballeuse internationale allemande
Joueuse du SG Essen-Schönebeck
Joueuse du Bayern Munich


text:
Q: Travis CI with Clang 3.4 and C++11 Is it possible to get Travis CI working with Clang that is capable of C++11? (I want Clang, not GCC, I already have GCC 4.8 working in Travis CI.) It appears that the version that is there pre-installed is not C++11 capable. All my attempts at installing any newer version end up failing because of this:
In file included from /usr/lib/gcc/x86_64-linux-gnu/4.8/../../../../include/c++/4.8/bits/move.h:57:
/usr/lib/gcc/x86_64-linux-gnu/4.8/../../../../include/c++/4.8/type_traits:269:39: error:
use of undeclared identifier '__float128'
struct __is_floating_point_helper<__float128>

I have seen the -D__STRICT_ANSI__ trick but that clashes with other things for me.
Is it possible to get it working? See also my .travis.yml.

A: There is now a better way to do this.
sudo: false
dist: trusty
language: cpp
os:
  - linux
compiler:
  - gcc
  - clang
install:
# /usr/bin/gcc is 4.6 always, but gcc-X.Y is available.
- if [[ $CXX = g++ ]]; then export CXX="g++-4.9" CC="gcc-4.9"; fi
# /usr/bin/clang has a conflict with gcc, so use clang-X.Y.
- if [[ $CXX = clang++ ]]; then export CXX="clang++-3.5" CC="clang-3.5"; fi
addons:
  apt:
    sources:
    - ubuntu-toolchain-r-test
    - llvm-toolchain-precise-3.5 # not sure why we needed this
  packages:
    - gcc-4.9
    - g++-4.9
    - clang-3.5

(The explicit sudo: false will let it build in Docker (for speed) even if you have a pre-docker repo, according to Travis support.)
Thanks to solarce at Travis support for noticing my error and fixing the docs.

*

*http://docs.travis-ci.com/user/apt/

*http://genbattle.bitbucket.org/blog/2016/01/17/c++-travis-ci/

*https://github.com/open-source-parsers/jsoncpp/blob/master/.travis.yml (for up-to-date example)

A: Here is a part of my .travis.yml files (mostly taken from this file).
language: cpp

compiler:
  - clang
  - gcc

before_install:
  # g++4.8.1
  - if [ "$CXX" == "g++" ]; then sudo add-apt-repository -y ppa:ubuntu-toolchain-r/test; fi

  # clang 3.4
  - if [ "$CXX" == "clang++" ]; then sudo add-apt-repository -y ppa:h-rayflood/llvm; fi

  - sudo apt-get update -qq

install:
  # g++4.8.1
  - if [ "$CXX" = "g++" ]; then sudo apt-get install -qq g++-4.8; fi
  - if [ "$CXX" = "g++" ]; then export CXX="g++-4.8"; fi

  # clang 3.4
  - if [ "$CXX" == "clang++" ]; then sudo apt-get install --allow-unauthenticated -qq clang-3.4; fi
  - if [ "$CXX" == "clang++" ]; then export CXX="clang++-3.4"; fi

script:
  - $CXX --version

EDIT because it can be very useful to add libc++ for travis. Up to my knowledge, there is no Linux package for libc++, so one has to compile it "by hand". Do not forget -stdlib=libc++ in CXXFLAGS while compiling with clang.
install:
  # clang 3.4
  - if [ "$CXX" == "clang++" ]; then sudo apt-get install --allow-unauthenticated -qq clang-3.4; fi
  - if [ "$CXX" == "clang++" ]; then export CXXFLAGS="-std=c++0x -stdlib=libc++"; fi
  - if [ "$CXX" == "clang++" ]; then svn co --quiet http://llvm.org/svn/llvm-project/libcxx/trunk libcxx; fi

  - if [ "$CXX" == "clang++" ]; then cd libcxx/lib && bash buildit; fi
  - if [ "$CXX" == "clang++" ]; then sudo cp ./libc++.so.1.0 /usr/lib/; fi
  - if [ "$CXX" == "clang++" ]; then sudo mkdir /usr/include/c++/v1; fi
  - if [ "$CXX" == "clang++" ]; then cd .. && sudo cp -r include/* /usr/include/c++/v1/; fi
  - if [ "$CXX" == "clang++" ]; then cd /usr/lib && sudo ln -sf libc++.so.1.0 libc++.so; fi
  - if [ "$CXX" == "clang++" ]; then sudo ln -sf libc++.so.1.0 libc++.so.1 && cd $cwd; fi

  - if [ "$CXX" == "clang++" ]; then export CXX="clang++-3.4"; fi

A: It appears that the clang developers fixed this when you build in -std=gnu++11 mode.
Are you able to build with that flag instead of __STRICT_ANSI__?

*

*http://llvm.org/bugs/show_bug.cgi?id=13530

*http://llvm.org/viewvc/llvm-project?view=revision&revision=184476

redpajama_set_name: RedPajamaStackExchange


text:
Documents reveal Canadians were spied upon through free airport Wi-Fi
By Jeff Belanger - February 2, 2014 - 4 comments Email article | Print article
It appears that Canada's spy agency has monitored Canadians through their access to free Wi-Fi at an undisclosed airport.
This information comes to us through another document released by Edward Snowden. It seems that the Communications Security Establishment Canada (CSEC) was testing out new software that it had developed in coordination with the National Security Agency (NSA). This software allowed them to track people, through the free Wi-Fi that is offered at most airports, and follow the person's movements for several days after.
CSEC, in a statement, said that they are "mandated to collect foreign signals intelligence to protect Canada and Canadians. And in order to fulfill that key foreign intelligence role for the country, CSEC is legally authorized to collect and analyze metadata." They also claim that "no Canadian communications were (or are) targeted, collected or used."
CSEC's chief, John Forster, has also recently said that "I can tell you that we do not target Canadians at home or abroad in our foreign intelligence activities, nor do we target anyone in Canada. In fact, it's prohibited by law. Protecting the privacy of Canadians is our most important principle."
Despite these claims, the document itself points to this being untrue, and it is more than likely that metadata was collected on Canadians as well.
"I can't see any circumstance in which this would not be unlawful, under current Canadian law, under our Charter, under CSEC's mandates." said Ronald Deibert,who is renowned cyber-security expert in Canada. He sees the actions made by CSEC as an "indiscriminate collection and analysis of Canadians' communications data."
In response to this news, Ann Cavoukia, who is the privacy commissioner for Ontario, is "blown away." She also states that "it is really unbelievable that CSEC would engage in that kind of surveillance of Canadians. Of us. I mean that could have been me at the airport walking around… This resembles the activities of a totalitarian state, not a free and open society."
[via CBC News]
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About Jeff Belanger
I'm a Canadian, I like to write, I like to play video games, watch movies and read books and all that. Also interested in current affairs and comic books as well.
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PLEASE NOTE	Because this item is made to order, its estimated delivery date includes a longer lead time.
Established in 1997, Robin Bruce curates upholstered collections for the decor enthusiast using green materials and building methods. Each unique piece of Robin Bruce furniture is built with pride by local craftspeople in Elliston, Virginia.
This club chair is crafted of kiln-dried hardwood with a classic barrel-style silhouette and Crypton velvet upholstery--an eco-friendly, high-performance fabric that repels stains, spills, bacteria, and mold. A tufted seat adds a thoughtful finishing touch, while ample padding ensures long-lasting comfort and support. Made in the USA.
Dressed in a versatile neutral hue, this midcentury-style club chair features plush velvet upholstery and boxy track arms. A hardwood frame and tapered legs provide sturdy support, kiln-dried for added strength and durability. Made in the USA.


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Talented students from 24 NRF member schools were nominated to represent the next generation of retail leaders. The nominees below received a travel scholarship to attend Retail's BIG Show, Student Program and the inaugural NRF Foundation Gala in New York City.
Retail executives from nearly 20 companies selected five finalists to receive scholarships. The top recipient received a $25,000 scholarship and four finalists each received $10,000 scholarships. Meet the finalists and see more scenes from the NRF Foundation's inaugural Gala.
In her own words, "No great success is ever achieved without some form of failure along the way." After struggling to find an internship her sophomore year, Sarah Kendzior landed two consecutive internships and eventually a full-time job offer from Kohl's. She now knows that retail "is always changing. It is revolutionary. It is innovative. It is filled with opportunity." But most importantly, it is precisely where she wants to be in the future.
Biancha Felix
Independent and self-reliant at 16, Biancha Felix discovered her love for retail when she started working part-time at Charlotte Russe to pay her college tuition. "It fascinated me. It distracted me. It became a place that felt like home … It consumed me." Her drive is evident as she is a full-time student and intern with a part-time job. In retail, she has "found not only her true career aspiration, but also her life's calling."
Courtney Hunter
Courtney Hunter has learned that in order to be a strong leader you must first be an effective communicator. She believes that success is impossible without passion, persistent determination, confidence in your instincts and the courage to pursue innovation. Her undeniable drive and ability to grasp concepts quickly has led to recognition by retailers: She'll be interning at QVC during the Spring 2015 semester, and Burlington Stores has offered her a second summer internship.
Garrett Ledbetter
For two years, Garrett Ledbetter has balanced a full course load while working full-time at Macy's — no easy task. Through these experiences he has realized that teamwork is not only necessary, but also the best way to achieve goals. "Each individual brings a unique set of skills to the table and combining everyone's strengths helps the group to succeed." IU Senior Lecturer Janis Shaffer notes that Garrett is 'the most curious and well-read student I've ever had in my 22 years of teaching. He is passionate about the industry and always wants to learn more."
Ashley Nudge
Through her work at the Banana Republic Factory Store for the past two years, Ashley Nudge has learned the best way to lead is by example. UNT Senior Lecturer Marissa Zorola writes, "Her enthusiasm to expand her experiences is refreshing and her desire to succeed makes her a role model for her fellow classmates." Ashley aspires to be a creative director and believes having great communication skills to inspire and lead those around her is key to bringing any vision to life.
Deema Aridi
Evelin Castillo Chang
Haelee Catchpole
Kayla Cline
Christen Finkbeiner
Kasia Gibbs
Jenna Gorman
Samantha Hale
Michelle Herlihy
Johnson & Wales University - Providence
Samantha Lockhart
Lasell College
Dylan Mack
Karla Maldonado
FIDM/Fashion Institute of Design & Merchandising
Lucille Reynolds
Kaila Richardson
Cody Villanueva
Anna Watson
Courtney Whitaker


From June 10th to June 14th we are at TNC18 which is hosted at Scandic Lerkendal Hotel in Trondheim, Norway. Nuno & Sven will be on site for you appointment.
Agenda for all talks and Sign-Up (Standard ticket: 1.000,- € incl. VAT) can be found at tnc18.geant.org.
TNC is the largest and most prestigious European research networking conference, with more than 650 participants attending this annual event. It brings together decision makers, managers, networking and collaboration specialists, plus identity and access management experts from all major European networking and research organizations, universities, worldwide sister institutions, as well as many industry representatives.


RPS, a leading multi-sector global professional services firm, today announces its third quarter trading update for the three months ended 30 September 2018 ('Q3-2018' or the 'period'). RPS also provides updated guidance on the outlook for the Group for the financial years ending 31 December 2018 ('FY-2018') and 31 December 2019 ('FY-2019').
Group fee income for Q3-2018 was £139 million (Q3-2017: £141 million, at constant currency £138 million) and was marginally below Board expectations. For the nine months ended 30 September 2018 fee income was £428 million (nine months ended 30 September 2017: £422 million, at constant currency £412 million) and in line with Board expectations.
Profit before tax and amortisation ('PBTA') for the Group in Q3-2018 was £12.8 million (Q3-2017; £14.3 million, at constant currency £13.9 million) and below Board expectations. For the nine months ended 30 September 2018 PBTA was £40.1 million (nine months ended 30 September 2017: £41.5 million, at constant currency £40.2 million) and below Board expectations.
The Group's net bank debt at 30 September 2018 was £86.2 million (30 September 2017: £90.6 million; 31 December 2017: £80.6 million). RPS' conversion of profit into cash from operations improved during Q3-2018 and for the year to date is 58%, (year to date September 2017: 58%). For the twelve months ended 30 September 2018 cash conversion was 93% (last twelve months ended 30 September 2017: 82%).
The Board's expectations for FY-2018 are that fee income will be slightly above FY-2017 at constant currency, and PBTA will be slightly below. The Board anticipates that fee income will be marginally below market expectations*; whilst PBTA will be below market expectations. The Board reiterates the previously stated dividend policy and accordingly expects the dividend for the full year to be in line with the previous year.
The current view of the Board is that Group fee income will be slightly higher in FY-2019 than in FY-2018 and total segment profit will be higher. However, necessary investment in the business will impact Group PBTA in 2019. Group level costs in FY-2019 are likely to be higher compared to FY-2018 in the order of £2.5 million. The majority of these will be ongoing. Additionally, we will incur one-off global brand relaunch expense totaling approximately £2.0 million. Taking into account the additional costs, the current view is that PBTA will be broadly similar to FY-2018.
The Board continues to invest in RPS in support of its strategic priorities that are essential to improve the longer-term performance of the Group.
Our people: The loss of key staff from recently acquired businesses after the settlement of deferred consideration is affecting business performance this year. Investing in our people is vital for the future long-term health of the business. Following the appointment of RPS' first Group People Director the business has made considerable headway in developing an HR structure that supports growth, attracts high calibre talent, engages our people and builds on our high-performance culture which enables staff to build meaningful careers.
Telling our story better and exploiting revenue synergies where they exist: The Group is making good progress in respect of its priority to improve the coherence and connectivity of our brand across sectors where RPS has deep expertise and capability. We are transforming the way we penetrate the market and differentiate ourselves from our competitors. By so doing we will improve Group performance and increase profit.
Energy: The recovery in the oil price and in the wider industry has helped us grow and reinvigorate the business and further improvement is possible in 2019.
Selective acquisitions: The Group is focused on achieving organic growth, but the Board continue to consider selective acquisitions to add depth to the services that RPS already provides. Our strategy is to fund potential acquisitions through existing bank facilities.
As reported at the interim results, investment in the Group's IT systems is required to underpin these strategic priorities. We have appointed a new Chief Information Officer ('CIO') who will commence imminently. A major component of this investment will be in a global ERP system that will be developed by the end of 2019 and with deployment completed in 2021. Our new CIO will review the project before significant funds are committed to it, but the Board anticipates a total capital investment of approximately £14 million which will be mainly incurred in 2019 and 2020.
In Energy trading in Q3-2018 was encouraging, with fee income and segment profit both growing well and maintaining the momentum exhibited in H1-2018. These conditions are expected to continue for the rest of this year and in 2019. This may result in FY-2019 showing good fee and profit growth before bad debt recovery. Significant bad debt recoveries are unlikely in FY-2019.
Consulting UK and Ireland's market conditions in Q3-2018 were comparable to those in H1-2018, although its performance continued to be held back by recruitment challenges, especially in the UK. We anticipate fee and segment profit growth in FY-2019.
Services UK and Netherlands performed ahead of management's expectations due to good performance in its Water Services business in the UK. The current Asset Management Plan ('AMP') applicable to the England and Wales water industry ends in March 2020 and it is anticipated that demand for the Water Services business will reduce in 2019 before rising again. This will temper the performance of this segment in FY-2019, but we expect recovery in 2020 when the new AMP cycle commences.
The co-location of the businesses in Norway will complete in early 2019. One-off costs of approximately £0.8 million will be incurred in FY-2018 suppressing segment profit this year. Management anticipates fee growth next year and a much-improved profit performance following the completion of the integration of our businesses.
The Group's North America segment had a difficult Q3-2018 following a disappointing first half suffering project overruns in our Design & Development business. Management anticipates this business will recover during FY-2019 and for the growth of our Oceans and Coastal business to continue. However, due to recent senior staff departures there is some risk that growth in our Environmental business may be tempered in FY-2019.
Australia Asia Pacific ('AAP') performance was below Board's expectations in Q3-2018. In 2019 we anticipate conditions to remain strong in private and public infrastructure markets and defence spending to remain high. Overall, fee and profit growth in this segment is anticipated in FY-2019.
"RPS is going through a fundamental shift from being a conglomerate of small consulting and service businesses providing local expertise, to a truly global firm that uses its expertise to deliver professional services around the world. We are developing a clear sense of purpose and identity in our chosen markets and have a greater focus on organic growth and embracing technology. Inevitably this shift has consequences for shorter term financial performance as it requires investment, but it will drive longer term, sustainable growth, for the future. We are confident that these changes will drive increased shareholder returns in the medium term. We will provide an update on our progress at the time of the Group's final results in February 2019."
* The Board considers market expectations to be the consensus fee income and PBTA published in the notes of those analysts who regularly follow and interact with the Group. For FY-2018 consensus fee income is £582 million and PBTA is £55.4 million. For FY-2019 consensus fee income is £603 million and PBTA is £59.6 million.
The financial information given in respect of Q3-2018 is derived from management accounts that are unaudited.
A conference call for analysts will be held at 9.30am this morning, which will be accessible via the details below.
A recording of the conference call and a transcript will be available in due course at ir.rpsgroup.com.
Founded in 1970, RPS is a leading global professional services firm of 5,600 consultants and service providers. Located in 125 countries across all six continents RPS defines, designs and manages projects that create shared value for a complex, urbanising and resource scarce world.
The Group delivers a broad range of services in six sectors: property, energy, transport, water, defence & government services and resources. Services provided across RPS' six sectors cover twelve service clusters: project & programme management, design & development, water services, environment, advisory & management consulting, exploration & development, planning & approvals, heath, safety & risk, oceans & coastal, laboratories, training and communication & creative services.
RPS's London Stock Exchange ticker is RPS.L. For further information, please visit www.rpsgroup.com.


I started noticing it a few years ago: Seemingly every young person on the internet was selling something. Not in the way that we're all selling ourselves now, cultivating our personal brands on social media. They were selling actual stuff. Shipping hoverboards from China and unloading them in their American high schools. Running parody Twitter accounts that fed into viral ad networks. Posting sponsored content on Instagram for pocket money and freebies.
Occasionally these youthful businesses evidenced a casual relationship with grift. The hoverboards literally blew up. The Twitter accounts stole jokes from struggling comics. Teenage influencers started concocting fake brand deals so they could appear worthy of corporate sponsorship.
The sprites behind the schemes could come across as craven; an online crew of viral marketers referred to non-entrepreneurial peers as "peasants." But most of them seemed sincere. They wanted to change the world, and they believed that the way to do that was to start their own businesses. Immediately. If corners were cut along the way, they were cut honestly. They were just kids.
I've been thinking of them as last year's "summer of scam" has extended into a never-ending, all-weather "scam season." We are held rapt by tales of big-time millennial grifters, entrepreneurs whose dizzy rises are rivaled only by their disastrous ends. Among them is Billy McFarland (born in 1991), the impresario who dreamed up the doomed Bahamas-based Fyre music festival; Elizabeth Holmes (b. 1984), creator of the at-home blood-testing service Theranos, which was once valued at more than $9 billion but never really worked; and Anna Delvey (b. 1991), the so-called SoHo Scammer, who is said to have passed herself off as a German heiress in a bid to build a plush Manhattan arts club named after herself.
Now their tales are fueling a genre of grifter-related entertainment. Earlier this month, Hulu and Netflix dropped rival streaming documentaries on Fyre. Soon HBO will air a documentary about Holmes directed by Alex Gibney, while ABC News has its own documentary and a podcast, too. An Anna Delvey show written by Shonda Rhimes is on its way to Netflix.
Stories of scammers are reliably captivating. Mark Twain's fiction held a soft spot for the flimflam man. The terms grifter and scam have their beginnings in the carnival circuit of the early 20th century, and scam is the root of the term scamp — as in, lovable rogue. There's a reason we call them con artists.
These people have always exploited a certain proximity to America's most storied figures. "The Inventor," HBO's Theranos documentary, draws comparisons between Holmes and one of her idols, Thomas Edison, who may have been better at cultivating his public image as an inventor than he was at actually inventing. The math behind the American dream — a nobody, combined with pure grit, transforms into a somebody — is already a reach. The grifter just stretches the concept a little further.
But perhaps no moment has been so primed for grift as our current one. As Jia Tolentino wrote in The New Yorker last year: "At some point between the Great Recession, which began in 2008, and the terrible election of 2016, scamming seems to have become the dominant logic of American life." The mantras of Silicon Valley — Think big! Think different! Move fast and break things! — encourage harebrained scheming. Venture capital requires business ideas to reap more than profits; they must promise explosive growth. And moving fast is presented not as a shortcut to getting rich but a necessary aspect of any successful venture. The grandiose expectations placed on actual children have grown wildly out of proportion with the economic reality into which they've been born. Only a scam could bridge the gap.
Meanwhile, the tools of grift have rarely been more attainable. Early chat rooms introduced a generation of kids to the pastime of pretending to be other people online. Now, everyone with a smartphone can create and manipulate images. And in this hyper-visual culture, constructing an image of something can feel like the most important step in conjuring the thing itself. Just ask the Chinese manufacturers that hawk clothes online with images of glamorous magazine spreads, then ship out disfigured imitations to unsuspecting customers.
Holmes was obsessed with the image of a finger-prick drop of blood unlocking a wealth of medical information, and many powerful people liked the picture, too. ("The Inventor" makes much of a photo shoot in which Holmes holds a tiny red vial between her fingertips.) The only thing the Fyre Festival did well was create a viral ad featuring famous models cavorting on yachts and swimming with pigs. At some level, we are all growing familiar with this kind of sleight of hand. On Instagram, we are training to create beautiful images of ourselves with no possibility — or really, expectation — that they be replicated in person.
McFarland may be the one in prison for fraud, but the most fascinating thing about Fyre was how deep the scam reached. The festival was marketed by Jerry Media, an online advertising agency that began as a popular Instagram account that cribbed other peoples' memes without credit. It was aggressively promoted by Instagram influencers — those model/advertiser/guru hybrids whose entire project rests on smoke and mirrors. Even the documentaries on the festival seemed compromised. Hulu paid McFarland for an interview, while Elliot Tebele, the founder of Jerry Media, is an executive producer of the Netflix documentary. It was scams all the way down.
And at the top? The closest-held secret of modern scam artists is how woefully unimaginative they really are. Our oversaturated scam culture seems to have produced something new: utterly unappealing con men and women who could not credibly be referred to as "artists" at all.
Though the Fyre documentaries are stocked with associates who swear to McFarland's charm, he appears to have all the appeal of a sentient airport lounge. Holmes, built up as a striking female entrepreneur with a heartfelt personal story, is revealed as stilted and vampiric onscreen; "The Inventor" includes a haunting scene in which she celebrates a company win by dancing robotically to MC Hammer. And since her unmasking, Delvey has been dragged on the internet for her tellingly bad hair.
There's not a lovable scamp or a master chameleon in the bunch. They're just young people who wanted to make something incredible, failed and couldn't accept it. In a word, unexceptional. These days, there's a scammer born every minute.


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Photo courtesy of @MOFAICUAE via Twitter
Israel, UAE And Bahrain Sign Historic Peace Agreements At White House
UNITED STATES—On Tuesday, September 15, President Donald Trump presided over a ceremony at The White House with leaders of Israel, the United Arab Emirates and Bahrain for the signing of historical deals between the nations.
Israeli Prime Minister Benjamin Netanyahu, United Arab Emirates Foreign Affairs Minister Abdullah bin Zayed bin Sultan Al Nahyan and Bahrain's Foreign Minister Dr. Abdullatif bin Rashid Al-Zayan signed the Abraham Accords. The Emiratis and the Israelis signed their own bilateral document, as did the Bahrainis and the Israelis. Representatives from the nations also signed a document. The United States signed the deals as a witness.
These declarations intend to normalize diplomatic relations between Israel and the two Gulf nations. Under the deal, the nations will establish embassies, exchange ambassadors and cooperate on issues like trade, healthcare and security. It also includes "efforts to achieve a just, comprehensive, and enduring resolution to the Israeli-Palestinian conflict."
Foreign Minister Dr. Abdullatif bin Rashid Al-Zayan, Prime Minister Benjamin Netanyahu, President Donald Trump and Foreign Affairs Minister Abdullah bin Zayed bin Sultan Al Nahyan. Photo courtesy of @arabnews via Twitter.
"We're here this afternoon to change the course of history. Together these agreements will serve as the foundation for a comprehensive peace across the entire region…The nations of the regions had to decide what kind of a future they wanted for their children and for their families, and for their nation itself. No one could make that choice for them. They had to do that themselves. Today the world sees that they are choosing cooperation over conflict, friendship over enmity, prosperity over poverty and hope over despair. They are choosing a future in which Arabs and Israelis, Muslims, Jews and Christians can live together, pray together and dream together, side by side in harmony, community and peace," said Trump.
Israel's Netanyahu called the agreements "a pivot of history" that "heralds a new dawn of peace."
Bahrain's Al Zayani stated: "For too long, the Middle East has been set back by conflict and mistrust, causing untold destruction and thwarting the potential of generations of our best and brightest."
When asked if other potential nations might join the agreement, Trump said the U.S. is "very far down the road with five additional countries."
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Paper Tales Review
Sheet Fighter
If you ever found yourself taking a long, deep breath preparing to explain to new players how science cards work in 7 Wonders, you will find Paper Tales immediately appealing. It is 7 Wonders-lite, subtracting a good chunk of playtime, complicated point calculations and focusing on just one of its core ideas – fighting.
Paper Tales still follows the familiar staples of card-drafting games; players begin with a hand of five unit cards, choose one and pass the rest along. The same strategic concerns are also present: a push and pull between taking the best card for yourself or denying your opponents a useful card, although it might not fit as well with your own layout.
However, while in most games of this type and length the game swiftly proceeds to the next round after the cards are revealed, in Paper Tales there is still a lot to do. Drafting cards is just the tip of the iceberg, followed by fighting, collecting coins, building and aging units, each action filled with its own set of decisions equally as important as the initial pick.
Fighting in Paper Tales, while absolutely the core of the game, is surprisingly non-confrontational. Players gain points depending on the numbers of wars they win that round, but there are no penalties for losing. Therefore, it never feels aggressive or targeted, and there is no pressure in choosing units with predominantly attack-focused skills every round just to keep up. You can concentrate on drafting units that produce resources or victory points one round, knowing there will be no repercussions to the card engine itself and you can always catch up on the point track next turn.
As with any game where cards have unique abilities, knowing roughly what is available in the deck is an advantage. There is a certain level of luck with each draw, as all the unit cards are shuffled together and drawn randomly, which makes creating an engine a little bit of a gamble. In general, there is a good overlap of powers, so it is quite easy to make one unit's ability chain off another for added bonuses. Paper Tales has an interesting mechanic of aging cards, making them viable generally for a maximum of two rounds before they have to be replaced. This encourages players to switch and adapt to new strategies and also prevents one player staying overpowered.
Money is a very precious resource in Paper Tales. Some units cost nothing to deploy, but better, more powerful cards are expensive. Players can also play building cards that provide permanent resources and bonuses; while the first does not require payment in coins, future ones get progressively more expensive. As Paper Tales lasts only four rounds, building later in the game becomes almost counter-productive, as the permanent effects do not provide enough of a return and the coin could be better spent on more powerful units.
The land of kings, knights, manticores and krakens takes on a beautiful, modernised look on the Paper Tales box and is equally stunning on the cards, retaining the fantastical feel, while not looking dated. The theme is more window dressing than an integral part of the game, but still fits comfortably within the palette of character abilities and the mechanics of the game.
For a game with a relatively small playtime, Paper Tales manages to subtly fit a good depth of strategy behind streamlined, simple mechanics. It brings in just enough complexity to keep matches tense and interesting, but its rules and scoring mechanics never feel convoluted or overtaxing.
ALEX SONECHKINA
Elegantly simple, Paper Tales is still strategic and competitive, and, for a game about fighting, you won't be aggrieved by anyone around the table.
Buy your copy here.
Designer: Masato Uesugi
Artist: Christine Alcouffe
This review originally appeared in the April 2018 issue of Tabletop Gaming. Pick up the latest issue of the UK's fastest-growing gaming magazine in print or digital here – or subscribe to make sure you never miss another issue.
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