text:
\section{Introduction}\label{section:introduction}
Phase retrieval refers to the problem of reconstructing an unknown vector $x_0 \in \mathbb{C}^n$ from $m$ measurements of the form
\begin{equation}\label{equ:measurements}
y_i = \vert \langle \xi^{\left(i\right)}, x_0 \rangle \vert^2 + w_i \quad \quad \left(\text{where } i \in \left[m\right] \right),
\end{equation}
where the $\xi^{\left(i\right)} \in \mathbb{C}^n$ are known measurement vectors and $w_i \in \mathbb{R}$ represents additive noise. Such problems are ubiquituous in many areas of science and engineering such as X-ray crystallography \cite{harrison1993phase,millane1990phase}, astronomical imaging \cite{fienup1987phase}, ptychography \cite{rodenburg2008ptychography}, and quantum tomography \cite{kueng2017low}.\\
The foundational papers \cite{candes2013phaselift,demanet2014stable,candes2014solving} proposed to reconstruct $x_0$ via the PhaseLift method, a convex relaxation of the original problem.
These papers have triggered many follow-up works since they were the first to establish rigorous recovery guarantees under the assumption that the measurement vectors $\xi^{\left(i\right)}$ are sampled uniformly at random from the sphere. Since then several papers have analyzed scenarios where the measurement vectors possess a significantly reduced amount of randomness, in particular spherical designs \cite{gross2015partial} and coded diffraction patterns \cite{candes2015phase,gross2017improved}. However, the theoretical results for coded diffraction patterns rely on the assumption that the modulus of the illumination patterns is varying. Indeed, it was shown in \cite{fannjiang1} that for certain illumination patterns with constant modulus ambiguities can arise, i.e., it is not possible to determine $x_0$ uniquely from the measurements $y_i$. In fact, such ambiguities can already arise in much simpler settings, where the measurement vectors $ \left( \xi^{\left(i\right)} \right)$ are i.i.d. subgaussian. For example, consider the case that $\xi^{\left(i\right)} = \left(  \varepsilon^{\left(i\right)}_{1} , \ldots, \varepsilon^{\left(i\right)}_{n} \right) $, where the $  \varepsilon^{\left(i\right)}_{j} $ are i.i.d. Rademacher random variables. That is, they only take the values $+1$ and $-1$ each with probability $\frac{1}{2}$. In this case the vector $x_0:=e_1=\left(1, 0, \ldots, 0\right)$ can never be distinguished from the vector $ \tilde{x}_0:=e_2=\left(0, 1, \ldots, 0\right)$. Note that in this scenario it holds that $\mathbb{E} \left[ \big\vert \xi^{\left(i\right)}_j \big\vert^4 \right]  = \mathbb{E} \left[ \big\vert \xi^{\left(i\right)}_j \big\vert^2 \right] $ and, hence, the vector $ \xi^{\left(i\right)} $ does not fulfill a small-ball probability assumption, which means that there is no constant $c>0$ such that for all $\varepsilon>0$ and for all vectors $x  $ it holds that
\begin{equation}
\mathbb{P} \left( \vert \langle x, \xi^{\left(i\right)}  \rangle \vert \le \varepsilon \Vert x \Vert  \right) \le c\varepsilon.
\end{equation}
When the signals are complex even additional classes of ambiguities can arise. For example, when the measurement vectors $ \xi^{\left(i\right)} $ are real, any signal $x$ and its complex-conjugate signal $ \overline{x} $ will result in identical observations.\\

For these reasons, previous works on phase-retrieval from subgaussian measurements (see, e.g., \cite{goldsmith1}) work with real signals and require that all entries of the vector $  \xi^{\left(i\right)} $ fulfill
\begin{equation}\label{ineq:nosmallball}
\mathbb{E} \left[ \big\vert \xi^{\left(i\right)}_j \big\vert^4 \right]  > \mathbb{E} \left[ \big\vert \xi^{\left(i\right)}_j \big\vert^2 \right]
\end{equation}
for all $ j \in \left[n\right] $ or make even stronger assumptions.\\
The only exception is \cite{krahmer2018phase} which shows for the real-valued case ($x_0 \in \mathbb{R}^n$ and $\xi^{\left(i\right)} \in \mathbb{R}^n $) PhaseLift recovers a large class of signals from subgaussian measurements even if estimates of the type \eqref{ineq:nosmallball} are not satisfied. More precisely, one obtains that all signals $x_0$ whose peak-to-average power ratio satisfies a mild bound of the form
\begin{equation}
\frac{\Vert x_0 \Vert_{\infty}}{\Vert x_0 \Vert} \le \mu < 1
\end{equation}
for some absolute constant $\mu>0$, can be recovered with high probability as long as $ m \gtrsim n $.  However, as the approach in \cite{krahmer2018phase} is intrinsically based on arguments in \cite{shahar_eldar} it cannot be generalized to the complex case in a straightforward manner. This paper provides an analysis both for real-valued and complex-valued signals. We believe that this understanding will be of importance for the subsequent study of structured scenarios such as coded diffraction patterns, which are also intrinsically complex in nature.\\

While the proofs in previous papers \cite{candes2015phase,gross2015partial,gross2017improved,krahmer2018phase} relied on the construction of a so-called dual certificate, our paper will employ a more geometric approach based on Mendelson's small ball method \cite{koltchinskii2015bounding,mendelson2014learning}. This is motivated by recent work \cite{kueng2017low,kabanava2016stable,krahmer2019convex}, which showed that a geometric analysis based on the descent cone of the trace norm can often yield additional insights compared to an approach based on dual certificates.

For the problem studied in this paper, however, the small-ball method cannot be applied directly to the entire descent cone or the entire  cone of directions in which positive semidefiniteness is preserved. Rather we divide the latter cone into two parts: One that contains all the problematic cases, but is small, and one that is larger, but easier to analyze. Then we control one of these cones using a restricted isometry property and one via the small-ball method.

We think that this novel viewpoint and also some of the techniques developed in this paper will be useful for the analysis of other interesting measurement scenarios, such as the case of heavy-tailed measurement vectors $\xi^{\left(i\right)} $ or the case that $\xi^{\left(i\right)}$ has only entries $0$ and $1$.\\

\section{Background and main results}

\subsection{Notation}

$\mathcal{S}^n$ denotes the vector space of all Hermitian matrices in $\mathbb{C}^{n \times n} $. By $\mathcal{S}^n_{+} \subset \mathcal{S}^n $ we will denote the set of all positive definite Hermitian matrices. For $A,B \in \mathcal{S}^n$ the Hilbert-Schmidt inner product is defined by $\innerproduct{A,B}:= \text{trace} \left(A^* B\right) $. The corresponding norm will be denote by $ \Vert \cdot \Vert_{HS}$. For a matrix $Z \in \mathcal{S}^n $ we will denote their eigenvalues by $\lambda_1 \left(Z\right), \lambda_2 \left(Z\right), \ldots, \lambda_n \left(Z\right) $, which are assumed to be arranged in decreasing order, i.e., $ \lambda_1 \left(Z\right) \ge \lambda_2 \left(Z\right) \ge \ldots \ge \lambda_n \left(Z\right) $. If no confusion can arise, we will suppress the dependence on $Z$ and write $\lambda_i  $ instead of $\lambda_i  \left(Z\right) $. By $\Vert Z \Vert_1$ we will denote the Schatten-$1$ norm of $Z$, i.e. $ \Vert Z \Vert_1 := \sum_{i=1}^{n} \vert \lambda_i \left(Z\right) \vert $. By $\text{diag} \left(Z\right) \in \mathcal{S}^n$ we denote the matrix, which we obtain by setting all off-diagonal entries of $Z$ equal to zero. We will write $a \lesssim b$ or $b \gtrsim a $ if there is a universal constant $C>0$ such that $ a \le Cb $.
\subsection{PhaseLift}
The PhaseLift method was first introduced in \cite{candes2013phaselift}. In this paper we focus on a variant \cite{candes2014solving,demanet2014stable} based on the observation that the measurements $y_i$ can be rewritten in the form
\begin{equation}
y_i = \text{trace} \left(\xi^{\left(i\right)} \left( \xi^{\left(i\right)} \right)^*  X_0  \right)+w_i,
\end{equation}
where $X_0 = x_0 x^*_0 $ is a rank-$1$ matrix encoding the signal to be recovered up to the true inherent phase ambiguity.
From this observation, PhaseLift relaxes the constraint that $X_0$ is of rank $1$ to obtain the optimization problem
\begin{equation}\label{opt:intern1}
\begin{split}
\text{minimize } \quad & \sum_{i=1}^{m} \big\vert \text{trace} \left( \xi^{\left(i\right)} (\xi^{\left(i\right)})^* X  \right) -y_i \big\vert\\
\text{ such that} \quad & X \in \mathcal{S}^n_{+}.
\end{split}
\end{equation}
In order to simplify notation we introduce the linear operator $\mathcal{A}: \mathcal{S}^n \rightarrow \mathbb{R}^m $ as
\begin{equation}\label{definition:Aop}
\mathcal{A} \left(Z\right) \left(i\right) := \innerproduct{ \xi^{\left(i\right)} (\xi^{\left(i\right)})^*, Z }.
\end{equation}
Hence, setting $y:= \left( y_1, \ldots, y_m \right) \in \mathbb{R}^m $, \eqref{opt:intern1} can be rewritten as
\begin{equation}\label{opt:SDP}
\begin{split}
\text{minimize } \quad & \Vert \mathcal{A} \left(X\right) -y \Vert_{\ell_1}\\
\text{such that} \quad & X \in \mathcal{S}^n_{+}.
\end{split}
\end{equation}
We note that while understanding the relaxation \eqref{opt:SDP} is an important benchmark approach and can be solved in polynomial time, it is typically not practical for applications, as lifting increases the number of optimization variables. For this reason, a very active line of research study recovery guarantees for algorithms that operate in the natural parameter domain such as alternating minimization (see, e.g., \cite{netrapalli2013phase,waldspurger2018alternating}), gradient-descent based formulations (see, e.g., \cite{candes_wirtinger,chen_wirtinger,soltanolkotabi2019structured,wright_phaseretrieval,chen2018gradient}), and \textit{anchored regression} \cite{bahmani2017phase,bahmani2017relax,bahmani2017anchored,goldstein2018phasemax}.
However, most of these guarantees have been shown under the assumption that the measurement vectors $\left\{ \xi^{\left(i\right)} \right\}_{i=1}^m $ are sampled i.i.d. from the unit sphere, so it will be a natural follow-up of this work to study to which extent our results generalize to the more practical nonconvex algorithms.
In particular, most reconstruction guarantees for these non-convex approaches require an appropriate initialization. For this reason, one needs to study which initialization schemes work for the measurements considered in this paper. A natural approach will be to try spectral initializations and recent generalizations that have been shown to be feasible for a basically minimal number of measurements \cite{lu2017phase,mondelli2020fundamental,dudeja2020analysis,luo2019optimal}. We expect that the analysis provided in this paper will prove useful for this endeavour as the spectral initialization is somewhat connected to trace-norm minimization.

\subsection{Subgaussian measurements}\label{section:subgaussian}
We consider random measurement vectors $ \left\{ \xi^{\left(i\right)} \right\}^m_{i=1} $ given as independent copies of a random vector $\xi $, whose entries $\xi_j$ are assumed to be i.i.d. subgaussian random variables with parameter $K$, expectation $\mathbb{E} \left[\xi_j\right] =0 $, and variance $ \mathbb{E} \left[ \vert \xi_j \vert^2 \right]  =1$. Recall that a random variable $X$ is subgaussian with parameter $K$, if and only if
\begin{equation}
\inf \left\{  t>0: \  \mathbb{E} \left[ \exp \left( X^2 / t^2 \right) \right]  \le 2  \right\} \le K < + \infty.
\end{equation}
It is well known (see, e.g., \cite{vershynin2016high})  that from this definition it follows for any (measurable) random variable $X$ that
\begin{equation}\label{ineq:subgaussianaux1}
\Vert X \Vert_{L_p} \lesssim \sqrt{p} K.
\end{equation}
Since $ \Vert \xi_1  \Vert_{L_2}^2   = \mathbb{E} \left[ \vert \xi_1 \vert^2 \right] =1 $ inequality \eqref{ineq:subgaussianaux1} immediately implies that
\begin{equation}\label{lemma:Kbound}
K \gtrsim 1.
\end{equation}
Moreover, it is well known (see, e.g., \cite{vershynin2016high}) that for all $ x\in \mathbb{C}^n$ the random variable $ \langle x,\xi \rangle $ is subgaussian with parameter $K \Vert x \Vert $.

\subsection{Previous work}
A number of previous works have studied phase retrieval with subgaussian measurements in the real-valued setting , i.e., $x_0 \in \mathbb{R}^n$ and $\xi \in \mathbb{R}^n$. For measurements fulfilling  $ \mathbb{E} \left[ \big\vert \xi_j \big\vert^4 \right]  > \mathbb{E} \left[ \big\vert \xi_j \big\vert^2 \right] $, \cite{goldsmith1} showed that PhaseLift admits order optimal uniform recovery guarantees.\footnote{In \cite{goldsmith1} instead of \eqref{opt:SDP} the original PhaseLift approach as in \cite{candes2013phaselift} is analysed.} Without the assumption $ \mathbb{E} \left[ \big\vert \xi_j \big\vert^4 \right]  > \mathbb{E} \left[ \big\vert \xi_j \big\vert^2 \right] $, in \cite{krahmer2018phase} the following result was proven, again for the real-valued case.
\begin{theorem}\label{thm:felixykai}\cite[Theorem V.1]{krahmer2018phase}
Let $\xi = \left( \xi_1, \ldots, \xi_n \right) \in \mathbb{R}^n $ be a random vector with i.i.d. subgaussian entries. Then there exist constants $C_1$, $C_2$, $C_3$, and $0 < \mu < 1 $, which depend only on the distribution of $\xi_1$, such that whenever
\begin{equation}
m \ge C_1 n,
\end{equation}
the following statement holds with probability at least $1-\exp \left(- C_2 n \right)$: For all signals $x_0 \in \mathbb{R}^n$ with $\Vert x_0 \Vert_{\infty} \le \mu \Vert x_0 \Vert $ and all noise vectors $ w \in \mathbb{R}^m $ any minimizer of \eqref{opt:SDP} fulfills
\begin{equation}
\Vert \hat{X} - x_0 x^*_0 \Vert_{HS} \le C_3 \frac{\Vert w \Vert_{\ell_1}}{m}.
\end{equation}
\end{theorem}

\section{Main results}
\subsection{Complex signals and complex measurement vectors}
In Theorem \ref{thm:felixykai} both the signal $x_0$ and the measurement vectors $\xi^{\left(i\right)}$ are assumed to be real. While for the measurement vectors this is often too restrictive, the signal $x_0$ is indeed typically real-valued in applications. This important special case will be discussed in Section \ref{section:realsignalcompmeas} below. Nevertheless, we find it still interesting
from a mathematical point of view under which assumptions recovery is possible for complex-valued signals. Our first result deals with this case.\\

As we have explained in Section \ref{section:introduction}, there are subgaussian distributions for which we cannot achieve uniform recovery of all signals $ x_0 \in \mathbb{C}^n$. For this reason, we define for all $0 < \mu \le 1 $ the set of all signals of mildly bounded peak-to-average power ratio
\begin{equation}\label{def:xmu}
\mathcal{X}_{\mu}:= \left\{ x_0 \in \mathbb{C}^n \setminus \left\{ 0\right\} :  \Vert x_0 \Vert_{\infty} \le \mu  \Vert x_0 \Vert    \right\}.
\end{equation}
Indeed, this restriction is very mild as $\mu$ will not depend on the dimension, whereas for a Gaussian random signal the ratio $\frac{\Vert x \Vert_{\infty}}{\Vert x \Vert} $ would scale like $ \sqrt{\frac{\log n}{n}}  $.
Now we are prepared to state the following theorem, which is our first main result.
\begin{theorem}\label{theorem:mainresult}
	Let the observation vector $y$ be given as in \eqref{equ:measurements}, where the random measurement vectors $ \left\{ \xi^{\left(i\right)} \right\}_{i=1}^m $ are defined as in Section \ref{section:subgaussian}. Assume that $ \vert \mathbb{E} \left[  \xi_1^2  \right] \vert^2 \le  1-\beta  $ for some $\beta \in \left(0,1\right) $ and that
	\begin{equation}\label{ineq:condition44}
	m \ge C_1 \frac {K^{20}}{\beta^{5/2}} n.
	\end{equation}
	Then for some probability parameter $p_{\beta}=1- \mathcal{O} \left( \exp \left( \frac{-m \beta^4}{C_2 K^{16}} \right) \right)$ the following two statements hold.
	\begin{enumerate}
		\item With probability at least $p_{\beta}$ one has that  for all vectors $x_0 \in \mathcal{X}_{1/81} $ and any noise vector $w\in \mathbb{R}^m$ any solution $\hat{X}$ of \eqref{opt:SDP} satisfies
		\begin{equation}\label{ineq:bound1}
		\Vert \hat{X} - x_0 x^*_0 \Vert_{1} \le C_3 \frac{K^8}{m \beta^{5/2}} \Vert w \Vert_{\ell_1}.
		\end{equation}
		\item If, in addition, $\mathbb{E} \left[ \vert \xi_1 \vert^4 \right] \ge 1+\beta $, then with probability at least $p_{\beta}$ inequality \eqref{ineq:bound1} holds for all $x_0 \in \mathbb{C}^n \setminus \left\{ 0 \right\} $.
	\end{enumerate}
	Here $C_1$, $C_2$, and $C_3$ are universal constants.
\end{theorem}
The first case of Theorem \ref{theorem:mainresult}, where one makes no assumption on the fourth moment of $\xi_1 $, can be applied also to certain scenarios, where  unique recovery is not possible without this assumption. One important example is that the entries $\xi_i$ are drawn from $ \left\{ z\in \mathbb{C}: \ \Vert z \Vert =1  \right\} $ uniformly at random. Note that these measurements will always yield the same observations $y$ for the two signals
\begin{align}
x_1&=\left(1, 0, \ldots, 0\right),\\
\tilde{x}_1&=\left(0, 1, \ldots, 0\right).
\end{align}
Such very sparse signals are exactly prevented by Condition 1, so there is no contradiction to the theorem's conclusion that unique recovery can be achieved via \eqref{opt:SDP} for all signals $x_0$ such that $\Vert x_0 \Vert_{\infty} \le \frac{1}{82} \Vert x_0 \Vert $.\\

Note that in the second scenario, where assumptions on the fourth moment of $\xi_1$ are available, we obtain a uniform recovery result over all $x_0 \in \mathbb{C}^n $. In the real-valued case a similar result has been shown in \cite{goldsmith1}.

\begin{remark}\label{remark1}
An assumption of the form $ \vert \mathbb{E} \left[  \xi_1^2  \right] \vert^2 \le  1-\beta $ cannot be avoided as the following argument shows. Indeed, if $ \vert \mathbb{E} \left[  \xi_1^2  \right] \vert^2 =1$ the assumption $ \mathbb{E} \left[ \vert   \xi_1^2 \vert  \right] =1  $ implies that $ \xi= \lambda \tilde{\xi} $ almost surely, where $ \lambda \in \left\{ z\in \mathbb{C}: \Vert z \Vert=1 \right\}$ is fixed and $\tilde{\xi} \in \mathbb{R} $ is a real random variable. We observe that
\begin{align}
\big\vert \langle \xi, x_0  \rangle \big\vert^2 &= \big\vert \langle \lambda \tilde{ \xi} , x_0 \rangle \big\vert^2\\
&= \big\vert \langle  \tilde{ \xi} , x_0 \rangle \big\vert^2\\
&= \big\vert \langle  \tilde{ \xi} , \overline{x_0} \rangle \big\vert^2\\
&=\big\vert \langle \xi, \overline{x_0}  \rangle \big\vert^2.
\end{align}
Consequently, $x_0$ and its complex-conjugate $\overline{x_0}$ will always lead to the same measurements.
\end{remark}

\subsection{Real signals and complex measurement vectors}\label{section:realsignalcompmeas}
We have seen in Remark \ref{remark1} that the assumption  $ \vert \mathbb{E} \left[  \xi_1^2  \right] \vert^2 \le  1-\beta $ is necessary to distinguish between a signal $x_0$ and $\overline{x_0}$. However, if, as in many practical applications, it is known a priori that the signal $x_0$ is real-valued then this ambiguity cannot arise and we can uniquely recover without additional assumptions via the following natural variant of the PhaseLift method, where we restrict the search space to real-valued matrices.
\begin{equation}\label{opt:SDP_2}
\begin{split}
\text{minimize } \quad & \Vert \mathcal{A} \left(X\right) -y \Vert_{\ell_1}\\
\text{such that} \quad & X \in \mathcal{S}^n_{+} \cap \mathbb{R}^{n \times n}.
\end{split}
\end{equation}
The following theorem shows that in this scenario the assumption $\vert \mathbb{E} \left[  \xi_1^2  \right] \vert^2 \le  1-\beta $ is indeed not necessary.

\begin{theorem}\label{theorem:mainresult2}
	Let the observation vector $y$ be given as in \eqref{equ:measurements}, where the random measurement vectors $ \left\{ \xi^{\left(i\right)} \right\}_{i=1}^m $ are as defined in Section \ref{section:subgaussian}.
	Then  the following two statements hold.

	\begin{enumerate}
	\item Assume that
	\begin{equation}\label{ineq:condition44_2}
	m \ge C_1 {K^{20}} n.
	\end{equation}
	Then, with probability at least $1- \mathcal{O} \left( \exp \left( \frac{-m }{C_2 K^{16}} \right) \right)$ one has that for all vectors $x_0 \in \mathcal{X}_{1/81}\cap \mathbb{R}^n$ and any noise vector $w\in \mathbb{R}^m$ any solution $\hat{X}$ of \eqref{opt:SDP_2} satisfies
	\begin{equation}\label{ineq:bound1_2}
	\Vert \hat{X} - x_0 x^*_0 \Vert_{1} \le C_3 {K^8}m  \Vert w \Vert_{\ell_1}.
	\end{equation}

	\item If, in addition, it holds that $\mathbb{E} \left[ \vert \xi_1 \vert^4 \right] \ge 1+\beta $ for some  $\beta \in (0,1] $, then, under the refined assumption
	\begin{equation}\label{ineq:condition44_2}
		m \ge C_1 \frac {K^{20}}{\beta^{5/2}} n,
	\end{equation}
	 one has a more general bound. Namely it holds that with probability at least $1- \mathcal{O} \left( \exp \left( \frac{-m \beta^4}{C_2 K^{16}} \right) \right)$  and for all vectors $x_0 \in \mathbb{R}^n \backslash \left\{0\right\}$, again for arbitrary noise vectors $w\in \mathbb{R}^m$, any solution $\hat{X}$ of \eqref{opt:SDP_2} satisfies
		\begin{equation}\label{ineq:bound1_2}
			\Vert \hat{X} - x_0 x^*_0 \Vert_{1} \le C_3 \frac{K^8}{m \beta^{5/2}} \Vert w \Vert_{\ell_1}.
		\end{equation}
	\end{enumerate}
	Here $C_1$, $C_2$, and $C_3$ are universal constants.
\end{theorem}

\begin{remark}
In comparison to Theorem \ref{thm:felixykai} the probability bound in Theorem \ref{theorem:mainresult} and Theorem \ref{theorem:mainresult2} is slightly better, as it improves from $1-\exp \left(- \Omega \left( n\right) \right) $ to $ 1-\exp \left(- \Omega \left( m\right) \right)  $.
Moreover, note that in contrast to Theorem \ref{thm:felixykai} the dependence on the subgaussian distribution of $\xi$ is not hidden in the constants. Also note that in our result the dependence on $\beta$ is stated explicitly. However, we do not know whether these bounds are optimal with respect to $K$ and $\beta$.
\end{remark}

\section{Proof of main results}

\subsection{Proof of Theorem \ref{theorem:mainresult}}\label{subsection:proof1}

Our goal is to show that with high probability the matrix $x_0 x_0^*$ is close to the minimizer $\hat{X}$ of the expression $\Vert \mathcal{A}(W)-y\Vert_{\ell_1}$ over all $W\in \mathcal{S}^n_{+}$. A common proof strategy that we will also follow is to establish that all $X \in \mathcal{S}^n_{+}$ with
\begin{equation}\label{equation:suffcondition1}
\Vert \mathcal{A}(X)-y\Vert_{\ell_1} \le \Vert \mathcal{A}(x_0 x_0^*)-y\Vert_{\ell_1} = \Vert w\Vert_{\ell_1}
\end{equation}
are sufficiently close to the true solution in $\Vert \cdot \Vert_{1} $-norm. More precisely, a sufficient condition for inequality \eqref{ineq:bound1} is that every $X$ fulfilling condition \eqref{equation:suffcondition1} satisfies
\begin{equation}\label{ineq:bound2}
\Vert X - x_0 x^*_0 \Vert_{1} \le C_3 \frac{K^8}{m \beta^{5/2}} \Vert w \Vert_{\ell_1}.
\end{equation}
Setting $Z= X- x_0 x_0^*$,  equation \eqref{equation:suffcondition1} reads
\begin{equation}\label{equation:suffcondition2}
 \Vert \mathcal{A}(Z)-w\Vert_{\ell_1} \le  \Vert w\Vert_{\ell_1}.
\end{equation}
By the triangle inequality this implies that
\begin{equation}
\Vert \mathcal{A}(Z)\Vert_{\ell_1} < 2 \Vert w \Vert_{\ell_1}.
\end{equation}
Hence, the upper bound \eqref{ineq:bound2} that we aim to establish directly follows from an appropriate lower bound for $\Vert \mathcal{A} \left(Z\right) \Vert_{\ell_1}/ \Vert Z \Vert_1 $.
Here $Z \in \mathcal{S}^n$ ranges over those matrices for which $x_0 x^*_0 + Z  $ is positive semidefinite. This set is convex, so it is locally well-approximated by a convex cone. To establish a uniform recovery result over all $x_0 \in \mathcal{X}_{\mu} $, we need to study the union of the corresponding cones as given by
\begin{equation}
\mathcal{M}_{\mu} := \text{cone} \left\{ Z \in \mathcal{S}^n  : \exists x_0\in \mathcal{X}_{\mu} \text{ such that } x_0x^*_0 +  Z\in \mathcal{S}^n_{+}   \right\}.
\end{equation}
We will refer to this set as the \emph{cone of admissible directions}.

With this notation, our proof strategy can be summarized as establishing a lower bound for
\begin{equation}
\lambda_{\min} \left( \mathcal{A}, \mathcal{M}_{\mu} \right) := \underset{Z \in  \mathcal{M}_{\mu} \setminus \left\{ 0\right\}  }{\inf} \frac{\Vert \mathcal{A} \left(Z\right) \Vert_{\ell_1} }{\Vert Z \Vert_1},
\end{equation}
which in the literature is commonly referred to as the minimum conic singular value (see, e.g., \cite{tropp2015convex,krahmer2019convex}).
Except for the precise nature of the cone under consideration, this strategy is exactly analogous to a number of works in the recent literature on linear inverse problems \cite{chandrasekaran2012convex,kueng2017low}. In particular, the following lemma, which summarizes our motivating considerations above, can be seen as a variant of
 \cite[Proposition 2.2]{chandrasekaran2012convex}.
\begin{lemma}\label{lemma:chandrasekaranvariant}
Let $\mathcal{A}$ be the operator defined in (\ref{definition:Aop}). Assume that $y= \mathcal{A} \left( x_0 x^*_0 \right) +w $. Then the minimizer $\hat{X}$ of \eqref{opt:SDP} satisfies
\begin{equation}
\Vert \hat{X}-x_0 x^*_0 \Vert_{1} \le \frac{2 \Vert w \Vert_{\ell_1}}{\lambda_{\min} \left( \mathcal{A}, \mathcal{M}_{\mu} \right) }.
\end{equation}
\end{lemma}
In the following, our goal will be to derive an appropriate lower bound for $\lambda_{\min} \left( \mathcal{A}, \mathcal{M}_{\mu} \right)$. One difficulty in the analysis is that not all matrices belonging to $\mathcal{M}_{\mu} $ are positive semidefinite. Indeed, in this scenario one could use that for positive semidefinite matrices an approximate $\ell_1$ isometry holds (see, e.g. \cite[Section 3]{candes2013phaselift}). While not all matrices in $\mathcal{M}_{\mu}$ are positive semidefinite the following lemma states that each matrix belonging to $\mathcal{M}_{\mu} $ possesses at most one negative eigenvector.
\begin{lemma}\label{lemma:onenegativeeigen}
Suppose that $Z\in \mathcal{M}_{\mu}$. Then $Z$ has at most one strictly negative eigenvalue.
\end{lemma}
\begin{proof}
Let $Z\in \mathcal{M}_{\mu}$. By definition of $ \mathcal{M}_{\mu} $ we can find $x_0 \in \mathcal{X}_{\mu}$ and $t>0 $ such that
\begin{equation}\label{ineq:intern1}
x_0x^*_0 + tZ \in \mathcal{S}^n_{+}.
\end{equation}
Suppose now by contradiction that $Z$ has two (strictly) negative eigenvalues with corresponding eigenvectors $ z_1, z_2 \in \mathbb{C}^n$.  Then we can find a vector $u \in \text{span} \left\{ z_1, z_2 \right\} \backslash \left\{ 0 \right\}  $ such that $ \langle  u,x_0 \rangle =0 $. This implies that for any $ t >0 $ we have that
\begin{equation}
u \left(  x_0 x^*_0 + t Z \right) u^* = t u^*Zu < 0,
\end{equation}
which is a contradiction to (\ref{ineq:intern1}).
\end{proof}

Recall that for a matrix $Z \in \mathcal{S}^n$ we denoted its eigenvalues by $\left\{ \lambda_i \left(Z\right) \right\}^n_{i=1} $ in decreasing order. By the previous lemma it holds that $ \lambda_i \left(Z\right)  \ge 0$ for all $ i \in \left[n-1\right] $ and all $Z \in \mathcal{M}_{\mu} $. For the proof we will partition $\mathcal{M}_{\mu}$ into two sets. Namely, for $ \alpha > 0 $ we define
\begin{align}
\mathcal{M}_{1,\mu, \alpha} &:= \left\{Z\in \mathcal{M}_{\mu}:  -\lambda_n \left(Z\right) \le \alpha  \sum_{i=1}^{n-1} \lambda_i \left( Z \right)  \right\},\\
\mathcal{M}_{2,\mu, \alpha}& := \left\{Z\in  \mathcal{M}_{\mu}:   -\lambda_n \left(Z\right) > \alpha  \sum_{i=1}^{n-1} \lambda_i \left( Z \right)   \right\}.
\end{align}
The two sets can be interpreted in the following way. If we would suppose that $\alpha=1 $ it would follow that $\text{trace} \left(Z\right) < 0 $ for all matrices $Z\in \mathcal{M}_{2,\mu, \alpha} $. In particular, this implies that there is $x_0 \in \mathcal{X}_{\mu}$ such that $Z$ is in the descent cone of the function $\text{trace} \left( \cdot \right)$ at the point $x_0 x^*_0 $. Hence, for $ \alpha < 1 $ we can interpret $\mathcal{M}_{2,\mu, \alpha}  $ as a slightly enlarged union of descent cones. In order to bound $ \underset{Z \in \mathcal{M}_{2,\mu, \alpha} }{\inf} \Vert \mathcal{A} \left(Z\right) \Vert_{\ell_1}  / \Vert Z \Vert_1 $ from below we will rely on the following lemma, which is proven in Section \ref{section:smallballapplied}.
\begin{lemma}\label{lemma:smallballapplied}
	Assume that one of following two conditions is satisfied for $ \beta \in (0,1] $:
	\begin{enumerate}
		\item It holds that $ \vert \mathbb{E} \left[  \xi_1^2  \right] \vert^2 \le  1-\beta  $. In this case we set $\mu=1/81$.
		\item In addition to $ \big\vert \mathbb{E} \left[  \xi_1^2  \right] \big\vert^2 \le  1-\beta $, the inequality  $\mathbb{E} \left[ \vert \xi_1 \vert^4 \right] \ge 1+\beta $, is fulfilled. In this case we set $ \mu =1 $.
	\end{enumerate}
	Moreover, assume that
	\begin{equation}\label{ineq:condition2}
	m \ge C_1 \frac {K^{20}}{\beta^{5}} n.
	\end{equation}
	Then with probability at least $1- 2\exp \left( \frac{-m\beta^4}{C_2 K^{16}}   \right) $ it holds that
	\begin{equation}\label{ineq:key22}
	\underset{Z\in \mathcal{M}_{2,\mu, \alpha}  \setminus \left\{ 0 \right\}}{\inf} \frac{\Vert \mathcal{A} \left(Z\right) \Vert_{\ell_1}}{ \Vert Z\Vert_{1}} \ge C_3 \frac{\beta^{5/2}}{K^8} m ,
	\end{equation}
	where $ \alpha = 4/5 $. Here $C_1$, $C_2$, and $C_3$ are universal constants.
\end{lemma}
The proof of Lemma \ref{lemma:smallballapplied} makes use of the fact that the set $ \mathcal{M}_{2,\mu, \alpha}  $ has low complexity in the sense that the matrices in $ \mathcal{M}_{2,\mu, \alpha}  $ are approximately low-rank.\\

In contrast, the set $ \mathcal{M}_{1,\mu, \alpha} $ has rather high complexity. For example, note that $\mathcal{S}^n_{+} \subset \mathcal{M}_{1,\mu, \alpha}  $. Nevertheless, the quantity $	\underset{Z\in \mathcal{M}_{1,\mu, \alpha}  \setminus \left\{ 0 \right\}}{\inf} \Vert \mathcal{A} \left(Z\right) \Vert_{\ell_1} / \Vert Z\Vert_{1} $ can be bounded from below, because the measurement matrices $\xi^{\left(i\right)} (\xi^{\left(i\right)})^* $  are positive semidefinite and the matrices in $\mathcal{M}_{1,\mu, \alpha} $ also have a dominant positive semidefinite component. This is achieved by the following lemma, whose proof can be found in Section \ref{section:oneestimate}.
\begin{lemma}\label{lemma:oneestimate}
	Let $0 < \mu \le 1$, $ \alpha >0$, and $ \delta>0$. Assume that
	\begin{equation}
	m \ge  \frac{C_1K^4n}{\delta^2}.
	\end{equation}
	Then with probability at least $1- \mathcal{O} \left(  \exp \left( -\frac{m}{C_2K^4}  \right) \right) $ for all $Z\in \mathcal{M}_{1,\mu,\alpha} $ it holds that
	\begin{equation}\label{ineq:intern4}
	\frac{1}{m} \Vert \mathcal{A} \left(Z\right) \Vert_{\ell_1} \ge \frac{1-\delta-\alpha-\alpha\delta}{1+ \alpha} \Vert Z \Vert_1.
	\end{equation}
	Here $C_1$ and $C_2$ are absolute constants.
\end{lemma}
We remark that Lemma \ref{lemma:oneestimate} would no longer hold if the measurement matrices $\xi^{\left(i\right)} (\xi^{\left(i\right)})^* $ would be replaced by  symmetric matrices with i.i.d. Gaussian entries (see \cite[Proposition 1]{slawski2015regularization}).\\

Having gathered all the necessary ingredients we can prove the main result  of this manuscript.
\begin{proof}[Proof of Theorem \ref{theorem:mainresult}]
	Set $\alpha =4/5$. The proof of the two statements is analogous, except that for the first statement we set $ \mu =1/81 $ whereas for the second statement we set $ \mu =1 $. By Lemma \ref{lemma:oneestimate} and assumption (\ref{ineq:condition44}) it follows that with probability at least $1-\mathcal{O} \left( \exp \left( -\frac{m}{CK^4} \right) \right) $
	\begin{equation}\label{ineq:intern48}
	\underset{Z\in \mathcal{M}_{1,\mu, \alpha}  \setminus \left\{ 0 \right\}	}{\inf} \frac{\Vert \mathcal{A} \left(Z\right) \Vert_{\ell_1}}{ \Vert Z \Vert_1} \gtrsim m.
	\end{equation}
	Furthermore, by Lemma \ref{lemma:smallballapplied} we have with probability at least $1- 2\exp \left( \frac{-m\beta^4}{C_2 K^{16}}   \right) $ that
	\begin{equation}\label{ineq:intern49}
	\underset{Z \in \mathcal{M}_{2,\mu, \alpha}  \setminus \left\{ 0 \right\}}{\inf} \frac{\Vert \mathcal{A} \left(Z\right) \Vert_{\ell_1}}{ \Vert Z \Vert_{1}} \gtrsim \frac{\beta^{5/2}}{K^8}m
	\end{equation}
	holds.\\

	Set $Z:= \hat{X} - x_0 x^*_0$. Note that by definition we have that $Z$ is an admissible direction, i.e., $Z\in \mathcal{M}_{\mu}$. It follows by (\ref{ineq:intern48}), (\ref{ineq:intern49}), and $\mathcal{M}_{\mu}= \mathcal{M}_{1,\mu, \alpha}  \cup \mathcal{M}_{2,\mu, \alpha}  $ that
	\begin{equation}
	\lambda_{\min} \left(  \mathcal{A}, \mathcal{M}_{\mu}  \right) \gtrsim \min \left\{ 1; \frac{\beta^{5/2}}{K^8}   \right\} m \ge  \frac{\beta^{5/2}}{K^8}  m,
	\end{equation}
	where in the last inequality we used (\ref{lemma:Kbound}) and $0<\beta  \le 1 $. It follows by Lemma \ref{lemma:chandrasekaranvariant} that
	\begin{equation}
	\Vert \hat{X} - x_0 x^*_0 \Vert_1 \le \frac{2 \Vert w \Vert_{\ell_1}}{ 	\lambda_{\min} \left( \mathcal{M}_{\mu}  \right)  } \lesssim  \frac{K^8}{m \beta^{5/2}}   \Vert w \Vert_{\ell_1},
	\end{equation}
	which finishes the proof.
\end{proof}

\subsection{Proof of Theorem \ref{theorem:mainresult2}}
The proof of Theorem \ref{theorem:mainresult2} is in large parts analogous to the proof of Theorem \ref{theorem:mainresult}. For this reason,  we will only highlight the main differences. Replacing $\mathcal{X}_{\mu} $ by $ \mathcal{X}_{\mu} \cap \mathbb{R}^n $ and $ \mathcal{M}_{\mu} $ by $\mathcal{M}_{\mu} \cap  \mathbb{R}^{n \times n} $ we can argue analogously to Section \ref{subsection:proof1} with the only difference that Lemma \ref{lemma:smallballapplied} has to be replaced by the following variant.
\begin{lemma}\label{lemma:smallballapplied_variant}
	Assume that one of following two conditions is satisfied for $ \mu, \beta \in (0,1] $:
	\begin{enumerate}
		\item It holds that $\mu  = \frac{1}{81} $ and $ \beta =1 $.
		\item It holds that $\mathbb{E} \left[ \vert \xi_1 \vert^4 \right] \ge 1+\beta $ and $\mu=1 $.
	\end{enumerate}
	Moreover, assume that
	\begin{equation}
	m \ge C_1 \frac {K^{20}}{\beta^{5}} n.
	\end{equation}
	Then with probability at least $1- 2\exp \left( \frac{-m\beta^4}{C_2 K^{16}}   \right) $ it holds that
	\begin{equation}
	\underset{Z\in (\mathcal{M}_{2,\mu, \alpha}  \cap \mathbb{R}^{n \times n} ) \setminus \left\{ 0 \right\}}{\inf} \frac{\Vert \mathcal{A} \left(Z\right) \Vert_{\ell_1}}{ \Vert Z\Vert_{1}} \ge C_3 \frac{\beta^{5/2}}{K^8} m ,
	\end{equation}
	where $ \alpha = 4/5 $. Here $C_1$, $C_2$, and $C_3$ are universal constants.
\end{lemma}
In order to prove Lemma \ref{lemma:smallballapplied_variant} we can proceed similarly as in the proof of Lemma \ref{lemma:smallballapplied}, see Section \ref{section:smallballapplied}, where in the proof of Lemma \ref{lemma:smallballapplied} we have highlighted the necessary modifications.

\section{Proof of Lemma \ref{lemma:oneestimate}}\label{section:oneestimate}

\begin{proof}
Note that for any $z\in \mathbb{C}^n $ we have that $\Vert \mathcal{A} \left(zz^*\right) \Vert_{\ell_1} = \sum_{i=1}^{m} \vert \langle \xi_i , z \rangle \vert^2 $. Let $A\in \mathbb{C}^{m\times n}$ be the matrix whose rows are given by $ \left\{ \xi_i \right\}_{i=1}^m $. It follows that $ \Vert \mathcal{A} \left(zz^*\right) \Vert_{\ell_1}  = \Vert Az \Vert^2 $. It follows from  \cite[Theorem 4.6.1]{vershynin2016high} that due to our assumption on $m$ with probability at least $ 1- \mathcal{O} \left(  \exp \left( -\frac{m}{CK^4}  \right) \right) $ for all $z\in \mathbb{C}^m $ it holds that
\begin{equation}
\left(1-\delta\right) \Vert z \Vert^2 \le \frac{1}{m} \Vert Az \Vert^2 \le \left(1+\delta\right) \Vert z \Vert^2.
\end{equation}
Due to the observation above this is equivalent to
\begin{equation}\label{equ:rank1rip}
\left(1-\delta\right) \Vert z \Vert^2 \le \frac{1}{m} \Vert \mathcal{A} \left(zz^*\right) \Vert_1 \le \left(1+\delta\right) \Vert z \Vert^2
\end{equation}
for all $z \in \mathbb{C}^n $. We will assume in the following that (\ref{equ:rank1rip}) holds for all $z \in \mathbb{C}^m$.\\

Let $Z\in \mathcal{M}_{1,\mu, \alpha}$ with corresponding eigenvalue decomposition $ Z = \sum_{i=1}^{n} \lambda_i v_i v^*_i  $. We observe that
\begin{align}
 \frac{1}{m}  \Vert \mathcal{A} \left(Z\right) \Vert_{\ell_1} &=  \frac{1}{m}  \sum_{j=1}^{m} \vert  \left(\xi^{\left(j\right)}\right)^*  Z \xi^{\left(i\right)} \vert \\
& \ge  \frac{1}{m}  \sum_{j=1}^{m} \left(\xi^{\left(j\right)}\right)^* Z \xi^{\left(j\right)}\\
& =  \frac{1}{m}   \sum_{j=1}^{m} \left(\xi^{\left(j\right)}\right)^*\left( \sum_{i=1}^{n} \lambda_i v_i v^*_i   \right) \xi^{\left(j\right)}\\
&= \frac{1}{m} \sum_{i=1}^{n} \lambda_i \sum_{j=1}^{m} \left(\xi^{\left(j\right)}\right)^* v_i v^*_i \xi^{\left(j\right)}\\
&= \frac{1}{m} \sum_{i=1}^{n} \lambda_i  \Vert  \mathcal{A} \left(v_i v^*_i\right) \Vert_{\ell_1}.
\end{align}
By Lemma \ref{lemma:onenegativeeigen} we know that $Z$ has at most one negative eigenvalue. If all eigenvalues $\lambda_i \left(Z\right) $ are positive, this inequality chain and inequality (\ref{equ:rank1rip}) imply that
\begin{equation}
\frac{1}{m} \Vert \mathcal{A} \left(Z\right) \Vert_{\ell_1} \ge \left(1-\delta\right) \sum_{i=1}^{n} \lambda_i = \left(1-\delta\right) \Vert Z \Vert_1,
\end{equation}
which shows (\ref{ineq:intern4}). Now suppose that $\lambda_n \left(Z\right) <0$. By (\ref{equ:rank1rip}) and $ -\lambda_n \left(Z\right) \le \alpha \sum_{i=1}^{n-1} \lambda_i \left(Z\right) $, which is due to $Z\in \mathcal{M}_{1,\mu,\alpha} $, we obtain that
\begin{equation}\label{ineq:intern2}
\begin{split}
 \frac{1}{m}  \Vert \mathcal{A} \left(Z\right) \Vert_1 &\ge  \left( 1-\delta \right) \sum_{i=1}^{n} \lambda_j  + \left(1+\delta \right) \lambda_n\\
 &\ge  \left(1-\delta - \alpha \left(1+\delta\right)\right)  \sum_{i=1}^{n-1} \lambda_i.
\end{split}
\end{equation}
Again using the relation $ -\lambda_n \left(Z\right) \le \alpha \sum_{i=1}^{n-1} \lambda_i \left(Z\right) $ we can also observe that
\begin{equation}\label{ineq:intern3}
\Vert Z \Vert_1 =  \sum_{j=1}^{n-1} \lambda_j  -\lambda_n \le  \left(1+\alpha \right) \sum_{i=1}^{n-1} \lambda_j.
\end{equation}
Combining (\ref{ineq:intern2}) and (\ref{ineq:intern3}) shows (\ref{ineq:intern4}), which finishes the proof.
\end{proof}

\section{Proof of Lemma \ref{lemma:smallballapplied} and Lemma \ref{lemma:smallballapplied_variant}}\label{section:smallballapplied}

In order to prove Lemma \ref{lemma:smallballapplied} and Lemma \ref{lemma:smallballapplied_variant} we will use the following version of Mendelson's small ball method \cite{koltchinskii2015bounding,mendelson2014learning}, a tool for deriving a lower bound for nonnegative empirical process.

\begin{lemma}\cite[Lemma 1]{dirksen2018gap}\label{thm:smallballmethod}
	Let $\mathcal{Z} \subset \mathcal{S}^n $ and let $\xi^{(1)}, \xi^{(2)}, \ldots, \xi^{(m)} $ be i.i.d. random vectors. Let $u>0$ and $t>0$ and define
	\begin{equation}\label{definition:Q}
	Q_{\mathcal{Z}} \left( u \right) := \underset{Z \in \mathcal{Z}}{\inf} \mathbb{P} \left( \vert \innerproduct{  \xi^{\left(1\right)} \left(\xi^{\left(1\right)}\right)^*, Z } \vert   \ge u \right).
	\end{equation}
	Then, with probability at least $1-2\exp \left( -2t^2 \right) $, it holds that
	\begin{align}
	&\underset{Z \in \mathcal{Z}}{\inf}   \left( \frac{1}{m} \sum_{i=1}^{m} \big\vert \innerproduct{ \xi^{\left(i\right)} \left(\xi^{\left(i\right)}\right)^*, Z} \big\vert \right)\\
	 \ge & u \left(  Q_{\mathcal{Z}} \left( 2u \right) - \frac{4}{u} \mathbb{E} \left[ \underset{Z \in \mathcal{Z}}{\sup} \Big\vert  \frac{1}{m} \sum_{i=1}^{m} \varepsilon_i \innerproduct{ \xi^{\left(i\right)} \left(\xi^{\left(i\right)}\right)^* , Z }   \Big\vert \right]       -\frac{t}{\sqrt{m}}   \right),
	\end{align}
	where $ \left( \varepsilon_{i} \right)^m_{i=1} $ are independent, symmetric, $\left\{ -1,1 \right\} $-valued random variables that are independent of $ \left(X_i \right)^m_{i=1} $.
\end{lemma}
Our goal is to apply Lemma \ref{thm:smallballmethod} to $\mathcal{Z}= \mathcal{M}_{2,\mu, \alpha}  \cap \left\{Z \in \mathcal{S}^n: \ \Vert Z \Vert_F =1 \right\} $. The following key lemma  shows that matrices in $ \mathcal{M}_{2,\mu, \alpha}  $ have two favorable properties: They are approximately low-rank and their mass with respect to the Frobenius norm is not concentrated on the diagonal for $\mu$ is small. The first property follows directly from the fact that the negative eigenvalue is rather small, the second property requires the spectral flatness of $x_0$, i.e., that $\mu$ is bounded.
\begin{lemma}\label{lemma:key1}
Let $ \alpha >0 $ and $ 0 < \mu \le 1 $. Assume that $Z \in \mathcal{M}_{2,\mu, \alpha}$. Then it holds that
\begin{enumerate}
\item
\begin{equation}\label{ineq:lowrank}
\Vert Z \Vert_1 \le \left(1+  \frac{1}{\alpha} \right) \Vert Z \Vert_{\text{HS}},
\end{equation}
\item
\begin{align}
\Vert \diagonal{Z} \Vert_{HS}  \le \left(  \sqrt{1 - \frac{1}{ \left( 1 + \alpha^{-1} \right)^2 }}  + 3\mu   \right) \Vert Z \Vert_{HS}.
\end{align}
\end{enumerate}
\end{lemma}

\begin{proof}
Let $Z \in  \mathcal{M}_{2,\mu, \alpha} $. By definition of $\mathcal{M}_{2,\mu, \alpha}  $ we have that $ \alpha \sum_{i=1}^{n-1} \lambda_i \left(Z\right) < - \lambda_n \left(Z\right)  $, which implies that
\begin{equation}\label{ineq:interna2}
\Vert Z \Vert_1 = \sum_{i=1}^{n-1} \lambda_i \left(Z\right)   -\lambda_n \left(Z\right) \le - \left( 1 + \frac{1}{\alpha}   \right) \lambda_n \left(Z\right) \le    \left( 1 + \frac{1}{\alpha}    \right)  \Vert Z \Vert_{\text{HS}}.
\end{equation}
This proves inequality (\ref{ineq:lowrank}).\\

\noindent In order to prove the second inequality note that by definiton of $\mathcal{M}_{2,\mu,\alpha} \subset \mathcal{M}_{\mu} $ we can choose $x_0 \in \mathcal{X}_{\mu}  \cap S^{n-1} $ such that there exists $t>0$ with $x_0x^*_0 + tZ$ positive semidefinite. For this choice of $x_0$ we can decompose $Z$ uniquely into
\begin{equation}
Z= \underset{=: Z_1}{\underbrace{ -\lambda x_0x^*_0 + ux^*_0+ x_0u^*} }+ Z_2,
\end{equation}
where $ \lambda \in \mathbb{R} $, $ \langle u,x_0 \rangle =0$, and $Z_2x_0=0$. We observe that
\begin{align}\label{ineq:interna5}
\Vert \diagonal{Z} \Vert_{HS} &\le  \Vert \diagonal{Z_1 } \Vert_{HS}  + \Vert \diagonal{Z_2}\Vert_{HS}.
\end{align}
We will bound the two summands separately. We begin with $ \Vert \diagonal{Z_1} \Vert_{HS} $ and observe that
\begin{equation}\label{ineq:interna6}
\begin{split}
\Vert \diagonal{Z_1} \Vert_{\text{HS}} &\le \vert \lambda \vert \Vert \text{diag} \left(x_0 x^*_0\right) \Vert_{HS} + 2 \Vert \text{diag} \left(ux^*_0\right) \Vert_{\text{HS}}\\
&= \vert \lambda \vert \sqrt{\sum_{i=1}^{n} \big\vert \left(x_0\right)_i \big\vert^4  } + 2 \sqrt{\sum_{i=1}^{n} \big\vert \left(x_{0}\right)_i \big\vert^2 \big\vert u_i \big\vert^2  }\\
&\le \mu \vert \lambda \vert + 2\mu \Vert u \Vert \\
&\le 3\mu \Vert Z_1 \Vert_{\text{HS}}\\
&\le 3\mu \Vert Z \Vert_{\text{HS}}.
\end{split}
\end{equation}
In the first inequality we used the triangle inequality and in the third line we used that $\Vert x_0 \Vert_{\infty} \le \mu \Vert x_0 \Vert= \mu  $ due to $x_0 \in \mathcal{X}_{\mu}\cap S^{n-1}$. In the fourth line we used that $\vert \lambda \vert \le \Vert Z_1 \Vert_{HS} $ and $ \Vert u \Vert \le \Vert Z_1 \Vert_{HS} $, which follows from the fact that the summands of $Z_1= -\lambda x_0 x^*_0 + ux^*_0 + x_0 u^* $ are orthogonal to each other. In the last line we again used that $\Vert Z_1 \Vert_{HS} \le \Vert Z \Vert_{HS} $ as $Z$ is decomposed orthogonally into $Z=Z_1+Z_2$. \\

In order to bound $ \Vert \diagonal{Z_2}  \Vert_{HS}$ we note first  that $Z_2$ is positive semidefinite. Indeed, suppose by contradiction that $Z_2$ is not positive semidefinite. Then there would exist a vector $v\in \mathbb{C}^n$ such that $ \langle v, x_0 \rangle =0$ and $ v^* Z_2 v <0$. In particular, this would imply that $v^* \left( x_0 x^*_0 + t Z \right)v <0 $ for all $ t >0$, which is a contradiction to our choice of $x_0$.\\
Now let $w\in \mathbb{C}^n$ be the normalized (i.e., $ \Vert w \Vert=1 $) eigenvector corresponding to the eigenvalue $ \lambda_n \left(Z\right)$. Then we obtain that
\begin{align}\label{ineq:interna3}
\lambda_n \left( Z \right) = w^* Z w = w^* Z_1 w + w^*Z_2w \ge w^* Z_1 w \ge  - \Vert Z_1 \Vert  \ge - \Vert Z_1 \Vert_{HS} ,
\end{align}
where the first inequality follows from the fact that $Z_2$ is positive semidefinite. Using this observation we obtain that
\begin{equation}
\begin{split}
\Vert \diagonal{Z_2}\Vert_{\text{HS}} &\le \Vert Z_2 \Vert_{\text{HS}}\\
&= \sqrt{ \Vert Z \Vert^2_{\text{HS}} - \Vert Z_1 \Vert_{\text{HS}}^2 }\\
& \overset{(\ref{ineq:interna3})}{\le}  \sqrt{ \Vert Z \Vert^2_{\text{HS}} - \lambda^2_n \left(Z\right) } \\
&\le \sqrt{ \Vert Z \Vert^2_{\text{HS}} -   \frac{1}{ \left( 1 + \alpha^{-1}  \right)^2} \Vert Z \Vert_{1}  }\\
& \le \Vert Z \Vert_{\text{HS}} \sqrt{1 - \frac{1}{ \left( 1 + \alpha^{-1}  \right)^2 }},
\end{split}
\end{equation}
where in the fourth line we used that $ - \lambda_n \left(Z\right) \ge      \frac{1}{1+\alpha^{-1}}  \Vert Z \Vert_{1} $, which is a consequence of the first inequality of (\ref{ineq:interna2}). Combining this estimate with (\ref{ineq:interna5}) and (\ref{ineq:interna6}) shows part (2), which finishes the proof.

\end{proof}
In analogy to \cite{koltchinskii2015bounding} we bound $Q_{\mathcal{Z}} \left(2u\right) $ using the following lemma, whose proof is based on the Paley-Zygmund inequality. A key difference is that we use the Hanson-Wright inequality to control the fourth moment $\mathbb{E}  \vert \xi^* A \xi \vert^4  $ appropriately.
\begin{lemma}\label{lemma:PRsmallballprobabilities}
	Let $A \in  \mathcal{S}^n$ and let $ \xi = \left( \xi_1, \ldots, \xi_n \right) $ be a random vector with independent and identically distributed entries $ \xi_i  $ taking values in $ \mathbb{C} $ such that $ \mathbb{E} \xi_i = 0 $, $ \mathbb{E} \vert \xi_i \vert^2 =1 $, and $\Vert \xi_i  \Vert_{\psi_2} \le K $. Then we have that
	\begin{equation}\label{equ:smallballphaseretrieval}
	\mathbb{P} \left(  \vert  \xi^* A \xi \vert^2 \ge \frac{\mathbb{E} \vert \xi^* A \xi \vert^2}{2}  \right) \ge  \frac{ \left( \mathbb{E} \vert \xi^* A \xi \vert^2 \right)^2 }{ C \left( K^8 \Vert A \Vert^4_{HS} +  \left( \text{trace}\left(A\right) \right)^4    \right) }.
	\end{equation}
	Here $C>0 $ is an absolute constant.
\end{lemma}

\begin{proof}
	Note that by the Paley-Zygmund inequality (see, e.g., \cite{decoupling}) we have that for all $ 0 < t \le  \mathbb{E} \vert \xi^* A \xi \vert^2$
	\begin{equation}
	\mathbb{P} \left(  \vert  \xi^* A \xi \vert^2 \ge t   \right)  \ge \frac{\left( \mathbb{E} \vert \xi^* A \xi \vert^2  - t  \right)^2}{\mathbb{E}  \vert \xi^* A \xi \vert^4 }.
	\end{equation}
	In particular, setting $t=\mathbb{E} \vert \xi^* A \xi \vert^2/2 $ yields that
	\begin{equation}\label{equ:paleyzygmund_internwef}
	\mathbb{P} \left(  \vert  \xi^* A \xi \vert^2 \ge \frac{\mathbb{E} \vert \xi^* A \xi \vert^2}{2}  \right) \ge \frac{\left( \mathbb{E} \vert \xi^* A \xi \vert^2   \right)^2}{4 \mathbb{E}  \vert \xi^* A \xi \vert^4 }.
	\end{equation}

	To estimate $\mathbb{E}  \vert \xi^* A \xi \vert^4$ from above we note that the triangle inequality yields that
	\begin{equation}\label{ineq:interna1}
	\begin{split}
	\mathbb{E}  \vert \xi^* A \xi \vert^4 &\lesssim  \mathbb{E} \left[ \big\vert \xi^* A \xi  - \mathbb{E} \left[  \xi^* A \xi \right]  \big\vert^4 \right] + \vert \mathbb{E} \xi^* A \xi \vert^4\\
	&=  \mathbb{E} \left[ \big\vert \xi^* A \xi  - \mathbb{E} \left[  \xi^* A \xi \right]  \big\vert^4 \right] +  \left( \text{trace}\left(A\right) \right)^4.
	\end{split}
	\end{equation}
	In order to estimate the first summand we will use that $ \big\vert \xi^* A \xi  - \mathbb{E} \left[  \xi^* A \xi \right]  \big\vert$ has a mixed subgaussian/subexponential tail.
	We can bound the tail probability using the Hanson-Wright inequality (in the version of \cite{rudelson_hansonwright}), which states that there is a numerical constant $c>0$ such that for all $t>0$ it holds that
	\begin{equation}
	\mathbb{P} \left(  \vert  \xi^* A \xi - \mathbb{E} \left[ \xi^* A \xi \right] \vert > t   \right) \le 2   \exp \left( -c \ \text{min} \left\{ \frac{t^2}{K^4 \Vert A \Vert^4_{HS}}  , \frac{t}{K^2 \Vert A \Vert}  \right\} \right).
	\end{equation}
	This yields that
	\begin{align}
	\mathbb{E}  \left[ \big\vert \xi^* A \xi - \mathbb{E} \left[ \xi^* A \xi \right]  \big\vert^4 \right] &= 4 \int_{0}^{\infty} t^3 \ \mathbb{P} \left( \big\vert \xi^* A \xi - \mathbb{E} \left[  \xi^* A \xi \right] \big\vert > t  \right)  dt \\
	 &\le 8 \left(     \int_{0}^{\infty} t^3 \  \exp \left( -c  \frac{t^2}{K^4 \Vert A \Vert^2_{HS}} \right) dt     +  \int_{0}^{\infty} t^3 \  \exp \left( -c  \frac{t}{K^2 \Vert A \Vert} \right) dt   \right)\\
	& = 8 \left( K^8 \Vert A \Vert^4_{HS}  \int_{0}^{\infty} u \  \exp \left( -cu^2 \right) du  + K^8 \Vert A \Vert^4    \int_{0}^{\infty} u^3 \  \exp \left( -cu \right) du    \right) \\
	&\lesssim  K^8 \Vert A \Vert^4_{HS},
	\end{align}
	where the third line follows from a change of variables. Combining this inequality chain with (\ref{ineq:interna1}) we obtain that
	\begin{equation}
	\mathbb{E}  \vert \xi^* A \xi \vert^4  \lesssim K^8 \Vert A \Vert^4_{HS} +   \left( \text{trace}\left(A\right) \right)^4  .
	\end{equation}
	Inserting this into (\ref{equ:paleyzygmund_internwef}) finishes the proof.
\end{proof}
In order to apply Lemma \ref{lemma:PRsmallballprobabilities} we need a lower bound for $\mathbb{E} \left[ \vert \xi^* A \xi \vert^2 \right]$. The next lemma computes this quantity.
\begin{lemma}\label{lemma:PRsecondmoments}
	Let $ \xi = \left( \xi_1, \ldots, \xi_n \right) $ be a random vector with independent and identically distributed entries $ \xi_i  $ taking values in $ \mathbb{C} $ such that $ \mathbb{E} \xi_i = 0 $ and $ \mathbb{E} \vert \xi_i \vert^2 =1 $. Then for all matrices  $ A \in \mathcal{S}^n $ it holds that
	\begin{equation}\label{equ:secondmoment}
	\begin{split}
	&\mathbb{E} \left[ \vert \xi^* A \xi \vert^2 \right]\\
	=& \left( \text{trace} A \right)^2 + \left(  \mathbb{E} \vert \xi_i \vert^4   -1 \right) \sum_{i=1}^{n} A^2_{i,i} + \left( 1 + \big\vert \mathbb{E} \left[ \xi^2_1 \right] \big\vert^2  \right) \sum_{i \ne j} \text{Re} \left( A_{i,j} \right)^2\\
	 &+ \left( 1- \big\vert \mathbb{E}  \left[ \xi^2_1  \right] \big\vert^2  \right) \sum_{i \ne j} \text{Im} \left( A_{i,j} \right)^2.

	\end{split}
	\end{equation}
\end{lemma}

\begin{proof}
	First, we observe that
	\begin{align}
	\mathbb{E} \left[ \vert \xi^* A \xi \vert^2  \right] &= \mathbb{E} \left[   \left( \sum_{i,j}  A_{i,j} \overline{\xi_i} \xi_j \right) \left(  \sum_{i',j'} \overline{ A_{i',j'}} \xi_{i'} \overline{ \xi_{j'}} \right)   \right] \\
	&= \sum_{i,i',j,j'}  \mathbb{E} \left[ \left(  A_{i,j} \overline{\xi_i} \xi_j \right) \left(   \overline{ A_{i',j'}} \xi_{i'} \overline{ \xi_{j'}} \right)    \right]     \\
	& = \sum_{i,j} \mathbb{E} \left[  \left(  A_{i,i} \vert \xi_i \vert^2  \right)  \left(  A_{j,j} \vert \xi_j \vert^2  \right)   \right] +  \sum_{i\ne j ,i' \ne j'} \mathbb{E} \left[   \left(  A_{i,j} \overline{\xi_i} \xi_j \right) \left(  \overline{ A_{i',j'}} \xi_{i'} \overline{ \xi_{j'}} \right)   \right] \\
	& = \left(I\right)+ \left(II\right),
	\end{align}
	where in the third line we used that $ \mathbb{E} \left[ \xi_i \right] = 0 $ and that the entries of $\xi$ are independent, which implies that there are no summands where one index appears exactly three times. The first summand can be computed by
	\begin{align}
	\left(I\right) &= \sum_i A^2_{i,i} \mathbb{E} \left[ \vert \xi_i \vert^4 \right] + \sum_{i \ne j} A_{i,i} A_{j,j} \mathbb{E} \left[ \vert \xi_i \vert^2 \right] \mathbb{E} \left[ \vert \xi_j \vert^2  \right]\\
	&= \sum^n_{i=1} A^2_{i,i} \mathbb{E} \left[ \vert \xi_i \vert^4 \right] + \sum_{i \ne j} A_{i,i} A_{j,j}\\
	&= \left( \text{trace} A \right)^2 + \left(  \mathbb{E} \left[ \vert \xi_i \vert^4 \right]   -1 \right) \sum_{i=1}^{n} A^2_{i,i}  ,
	\end{align}
	where we have used that $A_{i,i} = \overline{ A_{i,i} } $ for all $ i \in \left[n\right] $ and $ \mathbb{E} \left[ \vert \xi_i \vert^2 \right]  =1$. The second summand can be computed by
	\begin{align}
	(II)&= \sum_{i \ne j, i' \ne j'} A_{i,j} \overline{ A_{i',j'}} \mathbb{E} \left[ \overline{ \xi_i} \xi_j \xi_{i'} \overline{\xi_{j'}}  \right]\\
	&= \sum_{i \ne j, i' \ne j'} A_{i,j}  A_{j',i'} \mathbb{E} \left[ \overline{ \xi_i} \xi_j \xi_{i'} \overline{\xi_{j'}}  \right]\\
	&=  \sum_{i \ne j} A^2_{i,j} \mathbb{E} \left[ \overline{ \xi_i}^2 \right] \mathbb{E} \left[ \xi^2_j  \right] + \sum_{i \ne j} A_{i,j} A_{j,i} \mathbb{E} \left[ \vert \xi_i \vert^2 \right] \mathbb{E} \left[ \vert \xi_j \vert^2  \right]\\
	&=  \sum_{i \ne j} A^2_{i,j}  \Big\vert \mathbb{E} \left[ \xi^2_i \right] \Big\vert^2  + \sum_{i \ne j} \vert A_{i,j} \vert^2\\
	&= \sum_{i \ne j} A^2_{i,j}  \vert \mathbb{E} \left[ \xi^2_i \right] \vert^2  + \sum_{i \ne j} \vert A_{i,j} \vert^2 \\
	&\overset{(a)}{=} \big\vert \mathbb{E} \left[ \xi^2_1 \right] \big\vert^2 \sum_{i \ne j} \left( \text{Re} \left(A_{i,j}\right)^2 - \text{Im} \left(A_{i,j}\right)^2   \right)   + \sum_{i \ne j} \vert A_{i,j} \vert^2\\
	&=  \left( 1 + \big\vert \mathbb{E} \left[ \xi^2_1 \right] \big\vert^2  \right) \sum_{i \ne j} \text{Re} \left( A_{i,j} \right)^2 + \left( 1- \big\vert \mathbb{E}  \left[ \xi^2_1  \right] \big\vert^2  \right) \sum_{i \ne j} \text{Im} \left( A_{i,j} \right)^2.
	\end{align}
	For equation $(a)$ we used the observation that
	\begin{equation}
	 A^2_{i,j}+A^2_{j,i} = A^2_{i,j} + \overline{ A_{i,j}}^2  = 2 \text{Re} \left(A_{i,j}\right)^2 - 2 \text{Im} \left(A_{i,j}\right)^2.
	 \end{equation}
	 By summing up $(I)$ and $(II)$ we obtain equality (\ref{equ:secondmoment}).

\end{proof}
The lemmas above would allows us to find a lower bound for $Q_{\mathcal{Z}} \left(2u\right) $ in Lemma \ref{thm:smallballmethod}. We still need an upper bound for the Rademacher complexity $\mathbb{E} \left[ \underset{Z \in  \mathcal{Z} }{\sup}  \Big\vert \sum_{i=1}^{m}\varepsilon_i  \innerproduct{  \xi_i \xi^*_i, Z } \Big\vert   \right]$. The next lemma provides such a bound. In \cite{kueng2017low} a version of this lemma has already been presented. Nevertheless, we include a proof for completeness.
\begin{lemma}\label{lemma:PRmeanwidth} Assume that $m \ge C_1n$. Let $\alpha >0 $, $0 < \mu \le 1 $ and set $\mathcal{Z}:= \mathcal{M}_{1,\mu, \alpha}  \cap \left\{ Z \in \mathcal{S}^n: \ \Vert Z \Vert_{HS}=1  \right\}$. Then we have that
	\begin{equation}
	\mathbb{E} \left[ \underset{Z \in  \mathcal{Z} }{\sup}  \Big\vert  \innerproduct{Z , \sum_{i=1}^{m} \varepsilon_i \xi_i \xi^*_i } \Big\vert   \right] \le C_2 \left(1+\frac{1}{\alpha}\right) K^2  \sqrt{mn}.
	\end{equation}
	$C_1$ and $C_2$ are absolute constants.
\end{lemma}
\begin{proof}
	First, we note that by Hoelder's inequality and Lemma \ref{lemma:key1}  we obtain that
	\begin{equation}\label{ineq:interno1}
	\begin{split}
	\mathbb{E} \left[ \underset{Z \in  \mathcal{Z}}{\sup} \Big\vert  \innerproduct{Z , \sum_{i=1}^{m} \varepsilon_i  \xi^{\left(i\right)} \left(  \xi^{\left(i\right)}\right)^* }  \Big\vert  \right] & \le \left(   \underset{Z \in \mathcal{Z}}{\sup}  \Vert Z \Vert_1 \right)      \mathbb{E} \left[ \Big\Vert \sum_{i=1}^{m} \varepsilon_i  \xi^{\left(i\right)} \left(  \xi^{\left(i\right)}\right)^*  \Big\Vert \right]  \\
	&\le \left(1 + \frac{1}{\alpha} \right) \mathbb{E} \left[ \Big\Vert \sum_{i=1}^{m} \varepsilon_i  \xi^{\left(i\right)} \left(  \xi^{\left(i\right)}\right)^* \Big\Vert \right].
	\end{split}
	\end{equation}
	To bound $ \mathbb{E} \left[ \Big\Vert  \sum^m_{i=1} \varepsilon_i \xi^{\left(i\right)} \left(  \xi^{\left(i\right)}\right)^*   \Big\Vert  \right] $ let $ \mathcal{N} $ be an $\frac{1}{4}$-covering of the unit sphere $S^{n-1} \subset \mathbb{R}^n$ with respect to the Euclidean norm such that
	\begin{equation}\label{ineq:unionbound}
	\vert \mathcal{N} \vert \le 12^n.
	\end{equation}
	By \cite[Lemma 5.4]{vershynin2010introduction} we have that
	\begin{equation}\label{ineq:PRepsnet}
	\Big\Vert  \sum_{i=1}^{m} \varepsilon_i  \xi^{\left(i\right)} \left(  \xi^{\left(i\right)}\right)^*   \Big\Vert \le 2 \underset{x \in \mathcal{N}}{\sup} \Big\vert  \langle  x, \sum_{i=1}^{m} \varepsilon_i  \xi^{\left(i\right)} \left(  \xi^{\left(i\right)}\right)^*  x \rangle \Big\vert.
	\end{equation}
	Fix $ x \in \mathcal{N} $ and observe that
	\begin{equation}
	\langle x, \sum_{i=1}^{m} \varepsilon_i  \xi^{\left(i\right)} \left(  \xi^{\left(i\right)}\right)^*  x \rangle = \sum_{i=1}^{m} \varepsilon_i \vert \langle  \xi^{\left(i\right)} ,x \rangle \vert^2 = \sum_{i=1}^{m} z_i,
	\end{equation}
	where we have set $z_i :=  \varepsilon_i  \vert \langle \xi_i ,x \rangle  \vert^2 $. We observe that $ \mathbb{E} \left[z_i\right] = 0 $ and, moreover,
	\begin{equation*}
	\Vert z_i  \Vert_{\psi_1} = \Vert  \langle  \xi^{\left(i\right)} ,x \rangle^2  \Vert_{\psi_1} \lesssim  \Vert  \langle  \xi^{\left(i\right)} ,x \rangle  \Vert^2_{\psi_2} \lesssim K^2 ,
	\end{equation*}
	where the first equality follows directly from the definition of the $\Vert \cdot \Vert_{\psi_1} $-norm. \footnote{For the definition of the $\Vert \cdot \Vert_{\psi_1}$-norm see, e.g., \cite[Section 5.2.4]{vershynin2010introduction}.} The first inequality can be seen using \cite[Lemma  2.7.6]{vershynin2016high} and the second one using \cite[Lemma 3.4.2]{vershynin2016high}. By the Bernstein inequality (see, e.g., \cite[Theorem 2.8.1]{vershynin2016high}) we obtain that
	\begin{equation}\label{ineq:intern5}
	\mathbb{P} \left(  \Big\vert  \langle x, \sum_{i=1}^{m} \varepsilon_i \xi^{\left(i\right)} \left(  \xi^{\left(i\right)}\right)^*  x \rangle  \Big\vert   \ge t \right) = \mathbb{P} \left(  \Big\vert \sum_{i=1}^{m} z_i \Big\vert   \ge t \right)  \le 2 \exp \left( -c \min \left\{ \frac{t^2}{mK^4}; \frac{t}{K^2} \right\}  \right) ,
	\end{equation}
	where $ c>0 $ is some numerical constant. It follows from (\ref{ineq:unionbound}), (\ref{ineq:PRepsnet}), (\ref{ineq:intern5}), and a union bound that
	\begin{align}
	\mathbb{P} \left(  \Big\Vert  \sum_{i=1}^{m} \varepsilon_i  \xi^{\left(i\right)} \left(  \xi^{\left(i\right)}\right)^*   \Big\Vert   \ge t \right) & \overset{(\ref{ineq:PRepsnet})}{\le}   	\mathbb{P} \left(   2 \underset{x \in \mathcal{N}}{\sup} \Big\vert  \langle  x, \sum_{i=1}^{m} \varepsilon_i  \xi^{\left(i\right)} \left(  \xi^{\left(i\right)}\right)^*   x \rangle \Big\vert  \ge t \right)   \\
	& \le  \sum_{x\in \mathcal{N}}    	\mathbb{P} \left(   2 \Big\vert  \langle  x, \sum_{i=1}^{m} \varepsilon_i  \xi^{\left(i\right)} \left(  \xi^{\left(i\right)}\right)^*   x \rangle \Big\vert  \ge t \right)    \\
	& \overset{(\ref{ineq:unionbound}), (\ref{ineq:intern5})}{\le} 2  \cdot 12^{n}  \exp \left( -c' \min \left\{ \frac{t^2}{mK^4}; \frac{t}{K^2} \right\}  \right)\\
	&   = 2 \exp \left( \tilde{c} n -c' \min \left\{ \frac{t^2}{mK^4} ;  \frac{t}{K^2}  \right\}   \right),
	\end{align}
	where $\tilde{c} = \log 12 $. Then, whenever $ m \ge \frac{\tilde{c}}{c} n $, we obtain that
	\begin{align}
	&\mathbb{E}  \left[  \Big\Vert  \sum_{i=1}^{m} \varepsilon_i  \xi^{\left(i\right)} \xi^*_i  \Big\Vert \right]\\
	=& \int_{0}^{\infty} \mathbb{P} \left(  \Big\Vert  \sum_{i=1}^{m} \varepsilon_i  \xi^{\left(i\right)} \left(  \xi^{\left(i\right)}\right)^*  \Big\Vert   \ge t \right)  dt \\
	\le&   \int_{0}^{   K^2 \sqrt{\frac{\tilde{c}}{c'} nm }  } 1 dt  + 2 \int_{ K^2 \sqrt{\frac{\tilde{c}}{c'} nm }   }^{ mK^2  }  \exp \left( \tilde{c} n -  \frac{c' t^2}{mK^4} \right)  dt + 2 \int_{ mK^2 }^{ \infty}   \exp \left( \tilde{c} n - c' \frac{t}{K^2}  \right)  dt\\
	\lesssim & K^2 \sqrt{nm} + \exp \left(\tilde{c} n\right) \left(  \int_{ K^2 \sqrt{\frac{\tilde{c}}{c'} nm }   }^{  \infty }  \exp \left(  -  \frac{c't^2}{mK^4} \right)  dt +    \int_{ mK^2 }^{ \infty}   \exp \left( - \frac{c't}{K^2}  \right)  dt    \right).
	\end{align}
	In order to finish we need to estimate the two integrals.  By a change of variables and  \cite[Lemma C.7]{foucart2013mathematical} we obtain that
	\begin{align}
	\int_{ K \sqrt{\frac{\tilde{c}}{c} nm }   }^{  \infty }  \exp \left(  -  \frac{c' t^2}{mK^4} \right)  dt &= \frac{\sqrt{m}K^2}{\sqrt{2c'}} \int_{\sqrt{2 \tilde{c} n}}^{\infty}  \exp \left(  \frac{-t^2}{2} \right) dt \lesssim \sqrt{m} K^2 \exp \left( -\tilde{c} n \right),\\
	\int_{mK^2 }^{ \infty}   \exp \left( - \frac{c't}{K^2}  \right)  dt   &= \frac{K^2}{c'} \exp \left( -c'm\right).
	\end{align}
	Inserting this in the inequality chain above yields that
	\begin{equation}
	\mathbb{E}  \left[  \Big\Vert  \sum_{i=1}^{m} \varepsilon_i \xi^{(i)} \left(  \xi^{\left(i\right)}\right)^*   \Big\Vert \right] \lesssim K^2 \sqrt{nm}.
	\end{equation}
	Combined with inequality (\ref{ineq:interno1}) this finishes the proof.
\end{proof}
Now we have gathered all the ingredients to complete the proof.
\begin{proof}[Proof of Lemma \ref{lemma:smallballapplied} and Lemma \ref{lemma:smallballapplied_variant}]
We will start by showing that $\mathbb{E} \left[ \vert \innerproduct{\xi \xi^*, Z }   \vert^2 \right] \gtrsim \beta \Vert Z \Vert^2_{HS} $ for all $ Z \in \mathcal{M}_{2,\mu, \alpha} $ in the case of Lemma \ref{lemma:smallballapplied}, or all $ Z \in \mathcal{M}_{2,\mu, \alpha}  \cap \mathbb{R}^{n \times n} $ in the case of Lemma \ref{lemma:smallballapplied_variant}, respectively.

\noindent We first consider the second case and assume that the condition $\mathbb{E} \left[ \vert \xi_i \vert^4 \right] \ge 1+ \beta $ is satisfied for some $ \beta>0 $. By Lemma \ref{lemma:PRsecondmoments} we obtain that for all $Z \in \mathcal{M}_{2,\mu,\alpha} $ under the conditions of Lemma \ref{lemma:smallballapplied}
\begin{equation}\label{ineq:chain3}
\begin{split}
&\mathbb{E} \left[ \vert  \innerproduct{\xi \xi^*, Z }   \vert^2 \right]\\
=& \left( \text{trace} Z\right)^2 + \left(  \mathbb{E} \vert \xi_i \vert^4   -1 \right) \sum_{i=1}^{n} Z^2_{i,i} + \left( 1 + \big\vert \mathbb{E} \left[ \xi^2_1 \right] \big\vert^2  \right) \sum_{i \ne j} \text{Re} \left( Z_{i,j} \right)^2\\
&+ \left( 1- \big\vert \mathbb{E}  \left[ \xi^2_1  \right] \big\vert^2  \right) \sum_{i \ne j} \text{Im} \left( Z_{i,j} \right)^2\\
\ge& \beta \sum_{i=1}^{n} Z^2_{i,i}   + \beta \Vert Z-\diagonal{Z} \Vert^2_{HS}\\
=& \beta \Vert Z \Vert^2_{HS}.
\end{split}
\end{equation}
Under the assumptions of Lemma \ref{lemma:smallballapplied_variant} we observe that $\sum_{i \ne j}  \text{Im} \left( Z_{i,j} \right)^2=0  $ and $\sum_{i \ne j}  \text{Re} \left( Z_{i,j} \right)^2 = \Vert Z - \diagonal{Z} \Vert^2_{HS} $. Hence, a similar argument as before also leads to
\begin{equation}
\mathbb{E} \left[ \vert  \innerproduct{\xi \xi^*, Z }   \vert^2 \right] \ge  \beta \Vert Z \Vert^2_{HS}.
\end{equation}
 Under the first assumption  we obtain by Lemma \ref{lemma:PRsecondmoments} that for all $Z \in \mathcal{M}_{2,\mu, \alpha}  $
\begin{equation}\label{ineq:chain1}
\begin{split}
&\mathbb{E} \left[ \vert  \innerproduct{\xi \xi^*, Z }   \vert^2 \right]\\
=& \left( \text{trace} Z\right)^2 + \left(  \mathbb{E} \vert \xi_i \vert^4   -1 \right) \sum_{i=1}^{n} Z^2_{i,i} + \left( 1 + \big\vert \mathbb{E} \left[ \xi^2_1 \right] \big\vert^2  \right) \sum_{i \ne j} \text{Re} \left( Z_{i,j} \right)^2\\
&+ \left( 1- \big\vert \mathbb{E}  \left[ \xi^2_1  \right] \big\vert^2  \right) \sum_{i \ne j} \text{Im} \left( Z_{i,j} \right)^2\\
\ge&  \beta  \sum_{i \ne j} \text{Re} \left( Z_{i,j} \right)^2 + \beta  \sum_{i \ne j} \text{Im} \left( Z_{i,j} \right)^2  \\
\ge& \beta \Vert Z - \diagonal{Z} \Vert^2_{HS}.
\end{split}
\end{equation}
Similarly, under the assumptions of Lemma \ref{lemma:smallballapplied_variant} we can again use that $\sum_{i \ne j}  \text{Im} \left( Z_{i,j} \right)^2=0  $ and $\sum_{i \ne j}  \text{Re} \left( Z_{i,j} \right)^2 = \Vert Z - \diagonal{Z} \Vert^2_{HS} $ to obtain by an analogous argument that
\begin{equation}
\mathbb{E} \left[ \vert  \innerproduct{\xi \xi^*, Z }   \vert^2 \right] \ge  \beta \Vert Z - \diagonal{Z} \Vert^2_{HS}.
\end{equation}
The remainder of the proof will be the same for Lemma \ref{lemma:smallballapplied} and Lemma \ref{lemma:smallballapplied_variant}. By Lemma \ref{lemma:key1} we have that
\begin{align}
\Vert \diagonal{Z} \Vert_{HS} &\le \left(  \sqrt{1-\frac{1}{\left(1+ \alpha^{-1}\right)^2 }}   + 3 \mu   \right) \Vert Z \Vert_{HS}\\
&\le \left(  0.9  + \frac{1}{27}  \right) \Vert Z \Vert_{HS}\\
&\le 0.99 \Vert Z \Vert_{HS}.
\end{align}
By the triangle inequality it follows that
\begin{align}
\Vert Z -\diagonal{Z} \Vert_{HS} &\ge \Vert Z \Vert_{HS} - \Vert \diagonal{Z} \Vert_{HS}\\
&\ge \frac{1}{\tilde{C}} \Vert Z \Vert_{HS}
\end{align}
for $\tilde{C}=100$. Inserting into (\ref{ineq:chain1}) one obtains that
\begin{equation}
\mathbb{E} \left[ \vert \innerproduct{\xi \xi^*, Z }   \vert^2 \right] \ge \frac{\beta}{\tilde{C}^2} \Vert Z \Vert^2_{HS}.
\end{equation}
Hence, we have shown in all cases that $\mathbb{E} \left[ \vert  \innerproduct{\xi \xi^*, Z }   \vert^2 \right] \ge \frac{\beta}{\tilde{C}^2} \Vert Z \Vert^2_{HS} $.\\

\noindent By Lemma \ref{lemma:PRsmallballprobabilities} it follows that for all $Z\in \mathcal{M}_{2,\mu, \alpha}  $
\begin{align}
\mathbb{P} \left( \big\vert \innerproduct{\xi \xi^*, Z }   \big\vert \ge \frac{ \sqrt{ \beta}  \Vert Z \Vert_{HS}}{\sqrt{2}\tilde{C} }  \right)  &\ge \mathbb{P} \left( \big\vert \innerproduct{\xi \xi^*, Z }   \big\vert^2 \ge \frac{ \mathbb{E} \left[ \vert \innerproduct{\xi \xi^* , Z} \vert^2 \right]  }{2  }  \right)\\
&\ge \frac{\beta^2 \Vert Z \Vert_{HS}^4}{ \tilde{C}^4 C \left( K^8 \Vert Z \Vert^4_{HS}  + \left( \text{trace} \left( Z \right)  \right)^4  \right)}.
\end{align}
Note that for all $Z \in \mathcal{M}_{2,\mu, \alpha}  $
\begin{align}
 \vert \text{trace} \left(Z\right) \vert^4 \le& \Vert Z \Vert^4_1 \overset{ \left( \ref{ineq:lowrank}  \right)  }{\lesssim} \Vert Z \Vert^4_{HS},
\end{align}
where in the last inequality we also used that $ \alpha = \frac{4}{5} $. This shows that for all $Z \in \mathcal{M}_{2,\mu, \alpha}  $ it holds that
\begin{equation}
\mathbb{P} \left( \big\vert \innerproduct{\xi \xi^*, Z }   \big\vert \ge \frac{ \sqrt{\beta} \Vert Z \Vert_{HS}}{\sqrt{2}\tilde{C} }  \right)  \gtrsim \frac{\beta^2}{K^8},
\end{equation}
where we used that $ K \gtrsim 1 $ due to (\ref{lemma:Kbound}). Now recall that $ \mathcal{Z}:= \mathcal{M}_{2,\mu, \alpha}   \cap  \left\{ Z\in \mathcal{S}^n: \ \Vert Z \Vert_{HS} =1  \right\}  $. Thus we have shown that
\begin{equation}
 Q_{\mathcal{Z}} \left(2u\right) \ge \frac{\beta^2}{C''K^8},
\end{equation}
where $Q_{\mathcal{Z}} \left(\cdot \right)$ is defined in (\ref{definition:Q}). We used that $ u= \frac{ \sqrt{\beta}}{2\sqrt{2} \tilde{C}}  $ and $C''>0$ is a constant chosen large enough. From Lemma \ref{lemma:PRmeanwidth} it follows that
\begin{equation}
	\mathbb{E} \left[ \underset{Z \in  \mathcal{Z} }{\sup}  \Big\vert \frac{1}{m}  \innerproduct{Z , \sum_{i=1}^{m} \varepsilon_i \xi^{\left(i\right)} \left(  \xi^{\left(i\right)}\right)^*  } \Big\vert   \right] \lesssim  K^2  \sqrt{\frac{n}{m}}.
\end{equation}
Combining this inequality with our choice of $u$ and choosing the constant in assumption (\ref{ineq:condition2}) large enough it follows that
\begin{align}
\frac{1}{u}\mathbb{E} \left[ \underset{Z \in \mathcal{Z}}{\sup} \Big\vert  \frac{1}{m} \sum_{i=1}^{m} \varepsilon_i \innerproduct{ \xi^{\left(i\right)} \left(  \xi^{\left(i\right)}\right)^*  , Z }   \Big\vert \right] \le \frac{\beta^2}{8C''K^8}.
\end{align}
Applying Lemma \ref{thm:smallballmethod} yields that with probability at least $1-2\exp\left(-2t^2\right) $
\begin{align}
\underset{Z \in \mathcal{Z}}{\inf}   \left( \frac{1}{m} \sum_{i=1}^{m} \big\vert \innerproduct{\xi^{\left(i\right)} \left(  \xi^{\left(i\right)}\right)^* , Z} \big\vert \right) &\gtrsim  \sqrt{\beta} \left(  Q_{\mathcal{Z}} \left(2u\right)  - \frac{4}{u} \mathbb{E} \left[ \underset{Z \in \mathcal{Z}}{\sup} \Big\vert  \frac{1}{m} \sum_{i=1}^{m} \varepsilon_i \innerproduct{ \xi^{\left(i\right)}   \left(  \xi^{\left(i\right)}\right)^*    , Z }   \Big\vert \right]       -\frac{t}{\sqrt{m}}   \right)\\
&\ge \sqrt{\beta} \left(  \frac{\beta^2}{2C''K^8} -  \frac{t}{\sqrt{m}}   \right).
\end{align}
Setting $t= \frac{\sqrt{m} \beta^2 }{4 C'' K^8} $ it follows that with probability at least $1- 2 \exp \left( \frac{-m\beta^4}{8 (C'')^2 K^{16}}   \right) $ it holds that
\begin{equation}
\underset{Z \in \mathcal{Z}}{\inf}   \left( \frac{1}{m} \sum_{i=1}^{m} \big\vert \innerproduct{\xi^{\left(i\right)} \left(  \xi^{\left(i\right)}\right)^* , Z} \big\vert \right) \gtrsim \frac{\beta^{5/2}}{K^8}.
\end{equation}
Hence, by the definition of $\mathcal{A}$ and $\mathcal{Z}$ it follows that
\begin{equation}\label{ineq:intern78}
\underset{Z \in \mathcal{M}_{2,\mu, \alpha} }{\inf} \frac{\Vert \mathcal{A} \left(Z\right) \Vert_{\ell_1}}{\Vert Z \Vert_{HS}} \gtrsim  \frac{\beta^{5/2}}{K^8} m .
\end{equation}
Due to $\alpha=\frac{4}{5} $ and Lemma \ref{lemma:key1} we have that $\Vert Z \Vert_{1} \le \frac{9}{4} \Vert Z \Vert_{HS} $ for all $Z \in \mathcal{M}_{2,\mu, \alpha}  $. Combined with (\ref{ineq:intern78}) this shows (\ref{ineq:key22}).
\end{proof}

\section*{Acknowledgements}

This work has been supported by the German Science Foundation (DFG) in the context of the joint project {\em Bilinear Compressed Sensing} (KR 4512/2-1) as part of the Priority Program 1798 as well as the Emmy Noether Junior Research Group {\em Randomized Sensing of Signals and Images} (KR 4512/1-1). Furthermore, the authors want to thank Peter Jung for inspiring discussions.

\bibliographystyle{abbrv}

redpajama_set_name: RedPajamaArXiv


text:
N.J. Marriage Equality Pursuants to Re-File Lawsuit Following DOMA Decision
New motion to be filed on the back of last week's Supreme Court decision overturning a key section of DOMA.
By Brandon Baker·	 7/3/2013, 9:37 a.m.
Call it counterintuitive, but Garden State Equality and six same-sex New Jersey couples are hopeful that the state's civil unions law will be declared unconstitutional.
The group, which in conjunction with Lambda Legal had previously filed a motion against the state, will file for a second time this morning in an effort to force New Jersey to recognize same-sex marriages from other states. The renewed motion is in response to a change of circumstance following last week's crucial SCOTUS decision on the Defense of Marriage Act, which overturned Section 3 of the law.
According to the Associated Press, the state has until Aug. 2 to respond to the motion, which would be followed by a response from Garden State Equality on Aug. 9 and eventuate in a hearing on Aug. 15 in Trenton, N.J.
Meanwhile, New Jersey Gov. Chris Christie, who you can read all about in the new issue of Philly Mag, maintained in a TownSquare Media "Ask the Governor" program last week that his position on same-sex marriage post-DOMA remains the same, and that he would veto a hypothetical same-sex marriage bill again if it came to his desk. ("Let the people decide," he says.) He is, however, in the process of considering a bill that would ban gay conversion therapy, which has successfully passed through the Senate by a vote of 28-9 since we last reported on the matter.
Follow G Philly on Twitter | Like G Philly on Facebook | Follow G Philly on Instagram
Have an event you'd like featured on G Philly? Send details to [email protected].

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Existing trails already span much of the distance from New York to D.C., but the first piece of funding to fill in the gaps has only just fallen into place suggesting a long road ahead for trail organizers.
Four days after the attacks of Sept. 11, 2001, David Brickley found himself at a conference in Arlington, Va.
The Mid-Atlantic Governors Conference on Greenways, Blueways and Green Infrastructure almost didn't happen — no-fly zones were still in place, making it difficult for people to travel as the nation was still reeling from the terrorist attacks just a few days earlier. Conference organizers had considered but decided against canceling the event, which 1,500 people were scheduled to attend.
Instead, Mr. Brickley, then director of Virginia's Department of Conservation and Recreation, sent Virginia park rangers to North Carolina to pick up hundreds of small American flags to hand out to the 700 people who showed up for the conference. The parks and recreation directors and government and conservation leaders from 13 states started planning something to commemorate the events that occurred just a few days prior, Mr. Brickley said.
"Let's see if there's some way we can pay homage and tribute to those heroes of 9/11, because connectivity is a major thing that we try to promote in trails and greenways across the country," Mr. Brickley said.
With that, the idea for connecting the three Sept. 11 sites with 1,100 miles of bike trails was born. And after more than a decade, the first piece of funding is in place to make the trail a reality.
Mr. Brickley is now the president of the nonprofit September 11th National Memorial Trail Alliance, which was formed in 2002 but didn't begin working on a plan for the trails until about four years ago, he said.
The goal is to form a triangle of bike paths among the three sites — the National September 11 Memorial & Museum in New York City, the National 9/11 Pentagon Memorial and the Flight 93 National Memorial in Somerset County.
Somerset County recently received a $50,000 matching grant from the Pennsylvania Department of Conservation and National Resources, which will be used to hire a consultant to flesh out how the Great Allegheny Passage will connect with the Flight 93 site and, the more complicated project, how to link the Flight 93 site with the Liberty Water Gap Trail, which begins at the Delaware Water Gap between Pennsylvania and New Jersey and runs to New York City.
The $50,000 in matching funds came from corporations, foundations, individual donations and in-kind contributions, he said.
Two of the legs are largely in place. The Great Allegheny Passage and the C&O Canal Towpath, which runs from Pittsburgh to Washington, D.C., will make up most of the leg from the Flight 93 site to the Pentagon memorial. Somerset County trail coordinator Brett Hollern said a 25-mile spur from the Great Allegheny Passage at Rockwood to the Flight 93 site makes the most sense because other routes would be very hilly.
Mr. Hollern said he's hesitant to give a concrete timetable for the project, but he said Somerset County plans to have a consultant begin working on the routes from the Flight 93 site by the end of the year.
That leg will be about 440 miles long and use existing roads or require new trails to be built.
(c)2013 the Pittsburgh Post-Gazette. Distributed by MCT Information Services.


I Wish I Knew How To… Program the Canvas Control with Xojo Desktop will provide you with the ability to learn how to modify pictures, graphics, make animation, and how to make two games.
The focus of this book is to work with topics related to the Canvas Control. All of these examples are created with native Xojo code and does not use any third party programs. The examples have been tested on Windows 7, 8.1, OS X 10.9, and Ubuntu 14.04 Operating Systems.
By the time you finish the book you will be able to apply skills from these examples to create your own programs!
Examples were tested on Xojo 2014 r2.There is a newer book that has updated code for Xojo 2018 r4 that work with the changes in the Xojo Language. Version 1 of this book will remain availalble for those programming on Xojo 2014 r2.
There are 13 chapters and contains over 240 pages with over 100 example programs and subprograms.
Examples include files, text modification, a calculator, REALSQL Database, handling errors and more. Many screenshots have been added to show the results of the code with an index to help find topics quickly.
The 'I Wish I Knew' series contains technical data and advice that makes sense and contains practical and numerous examples with explanations to allow you to ease into the steep programming curve. You can create Real Studio programs today!
I Wish I Knew How To ... Begin Programming Xojo Web is an introductory book which will give you the tools and knowledge to begin programming a webpage for the internet. This book includes static and dynamic webpages which can also be integrated with a database.
This book introduces many of the fundamentals to create webpages which include CGI and Standalone applications. Both HTTP and HTTPS standalone examples are covered with a focus on HTTP CGI's. There are many different configurations for a complete HTTP and HTTPS CGI server and are beyond this introductory writing and are only lightly discussed. Working examples are for the Xojo Web class and does not cover socket examples, as this is a complete book on its own.
The 'I Wish I Knew' series contains technical data and advice that makes sense and contains practical and numerous examples with explanations to allow you to ease into the steep programming curve. You can create Microsoft Access database applications today!
This book "I Wish I Knew How to … Program Access with Real Studio Desktop Apps in Windows" delves into the mystery of programming a front end database.


Twenty one (21) trainees joined the 3-day Saori Weaving Workshop jointly held by the Natinal Council on Disability Affairs and the National Vocational Rehabilitation Center last December 12-14, 2011. Saori is an art of weaving by hand dedicated to free expression and self-development for everyone, regardless of physical or mental ability, age, or artistic aptitude. It is pure improvisation from the heart, with no premeditated pattern in mind. From the colourful woven material, clothing, bags, tapestries, and many useful items that can be appreciated by all, can be created.
Ms. Rebecca M. Santos. Founder and Executive Director of Saori VSA Arts Philippines, with trainors Mr. Jowee Martinez and Mr. Sandy Ching (deaf trainor), taught the group composed of 12 deaf, 4 with orthopaedic disabilities and 5 dependents. The enthusiastic group capped their training with a showcase of artistic talents expressed in their attractive woven products exhibited on the last day of the Workshop.
The NVRC officials thanked NCDA for choosing NVRC as project host which stands to usher in work opportunities for new Saori weavers. With assurance of a market in Japan, this widens the limited work opportunities for people with cross-disabilities including those with visual problems.
With the Workshop's success, NVRC Social Welfare Officer III Ms. Amor Macapanpan requested NCDA for a part 2. Deputy Executive Director Mateo A. Lee, Jr. said that it is just a beginning of new opportunities for PWD entrepreneurs.


"It was a unique experience for pupils to have the chance to use kit that they would not otherwise have access to, and learn from professionals. It built confidence and you could see a change in the children from the time that they walked in, to the time that they left the workshops".
"It was great for our younger students to learn about the Poppies: Wave through the use of digital media. The hands on approach gave them a tool to question and promote understanding about their local history and themes surrounding the First World War. It helps to establish links for when they re-visit these topics back in the classroom".
"The digital workshops were excellent; engaging local children with the significance of the Poppies sculpture and developing their understanding about its representation of the First World War both nationally and locally".


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Interest Rate in Uganda averaged percent from until , reaching an all time high of 23 percent in November of and a record low of 9 percent in February of In Uganda, interest rates decisions are taken by the Bank of Uganda. The Bank of Uganda official interest rate . Bank of Uganda has raised the Central Bank Rate by 1 percentage points to 17 percent in October , according to a Monetary Policy statement, issued by Governor Prof. Emmanuel Tumusiime -Mutebile.
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KAMPALA (Reuters) - Uganda's central bank raised its key lending rate to 10 percent on Wednesday, a one percentage point jump from August's percent, amid rising inflation. Uganda's central bank left its Central Bank Rate (CBR) at percent and adopted a neutral monetary policy stance as it expects inflation to remain around its.
Uganda's central bank cut its key lending rate by percentage points in an attempt to spur growth, as Africa's top coffee exporter continues to recover from the effects of a drought last year. Bank of Uganda (BoU) yesterday moved to provide a stimulus once again in the economy by reducing the Central Bank Rate (CBR) from 15 per cent to 14 per.
Uganda's central bank held its benchmark lending rate at 12 percent on Wednesday. DFCU's is currently at 24% from 25% in July when the central bank last published interest rates and bank charges charged by Uganda's 24 commercial banks. Orient Bank's prime lending rate has been reduced to % effective 1 October from the current %; it was 26% in July.


Underarm Masking Is Now A Real Thing!
Our jaws are still on the floor from all the lewks.
We all have goals we want to achieve in life, and we aren't even a little bit embarrassed to admit that, for us, reaching Sephora VIB Rouge status is one of them.
Whether you're off to school or work, you're going to want to make sure you've got some of these low-maintenance beauty items handy as you charge forth through your day.
We've said it before and we'll say it again, when fashion brands collaborate with beauty brands, perfection ensues.
We had a feeling that our beloved beauty brands would have some tricks up their sleeves in honor of National Lipstick Day on Saturday.
We don't really need an excuse to go shopping for beauty products but now that summer is in full swing we can't help but cater to the new season.
It goes without saying, but you should all be wearing sunscreen every single day, all year-round, even on cloudy days.
You can use a different one every night without having to do an elaborate makeup — and you'll look fabulous no matter what.
We can't wait to see some swatches!


The lighting designer and artist Tupac Martir (featured in Jocks&Nerds issue 18) has been appointed the creative director of the exhibition Sølv.
Meaning "silver" in Danish, Sølv features sculptural works from eight international artists. It is curated by the Gold-Smidt Assembly, a bespoke art dealer and creative consultancy.
Known as "Mr. London Fashion Week" for his lighting designs at runway shows, Martir's activities extend far beyond fashion.
Born in Britain to Mexican parents, and educated in Mexico, Italy and the US, his creative output includes painting, installations, performance art, theatre and graphic novels.
He has written and directed a multi-disciplinary opera, Nierka, and worked with Nederlands Dans Theater. In 2014, he curated the Looking Outside My Window festival at Rich Mix in Bethnal Green, and in 2015 he lit the façade of the British Museum for its Mexican Day of the Dead celebrations.


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News.com's Ina Fried has some of the inside story on how the Microsoft-Tellme deal went down on Superbowl Sunday.
It was only a few hours before kickoff on Super Bowl Sunday, but Microsoft CEO Steve Ballmer wasn't at home getting ready to watch the Bears take on the Colts.
Instead, the Microsoft chief executive was in his conference room inside Microsoft's building 34, the main executive offices, with a room full of colleagues and executives from speech recognition company Tellme Networks. Ballmer peppered Tellme executives with questions about their business, entering the answers into an Excel spreadsheet he had started from scratch.
Three-and-a-half hours later he had built a complete business model, and one thing was clear: an acquisition made a lot of sense.


The Aging Younger Show where the alternatives matter. It is important to understand that in the media, the term "stem cell" seems to be equivalent with "embryonic stem cell", and the term "stem cell research" seems to be equivalent to "embryonic stem cell "research" and although fueled by Big Pharma, it could not be further from the truth. While these terms are NOT equivalent, the confusion is not an accident. Those who control the media through the advertising dollars of big drug companies don't want you to know about adult stem cells, or umbilical cord stem cells, But tune in and you will find out for yourself.


text:
If you discover issues, have ideas for improvements or new features,
please report them to the [issue tracker][1] of the repository or
submit a pull request. Please, try to follow these guidelines when you
do so.

## Issue reporting

* Check that the issue has not already been reported.
* Check that the issue has not already been fixed in the latest code
  (a.k.a. `master`).
* Be clear, concise and precise in your description of the problem.
* Open an issue with a descriptive title and a summary in grammatically correct,
  complete sentences.
* Include Powerpack's version number.
* Include any relevant code to the issue summary.

## Pull requests

* Read [how to properly contribute to open source projects on Github][2].
* Fork the project.
* Use a topic/feature branch to easily amend a pull request later, if necessary.
* Write [good commit messages][3].
* Use the same coding conventions as the rest of the project.
* Commit and push until you are happy with your contribution.
* Make sure to add tests for it. This is important so I don't break it
  in a future version unintentionally.
* Add an entry to the [Changelog](CHANGELOG.md) accordingly. See [changelog entry format](#changelog-entry-format).
* Please try not to mess with the Rakefile, version, or history. If
  you want to have your own version, or is otherwise necessary, that
  is fine, but please isolate to its own commit so I can cherry-pick
  around it.
* Make sure the test suite is passing ([including rbx and jruby][7]) and the code you wrote doesn't produce
  RuboCop offenses.
* [Squash related commits together][5].
* Open a [pull request][4] that relates to *only* one subject with a clear title
  and description in grammatically correct, complete sentences.

### Changelog entry format

Here are a few examples:

```
* [#123](https://github.com/bbatsov/powerpack/issues/123): Fixed a bug in  `Something#something`. ([@bbatsov][])
* Added new method `Something#something`. ([@bbatsov][])
```

* Mark it up in [Markdown syntax][6].
* The entry line should start with `* ` (an asterisk and a space).
* If the change has a related GitHub issue (e.g. a bug fix for a reported issue), put a link to the issue as `[#123](https://github.com/bbatsov/powerpack/issues/123): `.
* Describe the brief of the change. The sentence should end with a punctuation.
* At the end of the entry, add an implicit link to your GitHub user page as `([@username][])`.
* If this is your first contribution to Powerpack project, add a link definition for the implicit link to the bottom of the changelog as `[@username]: https://github.com/username`.

[1]: https://github.com/bbatsov/powerpack/issues
[2]: http://gun.io/blog/how-to-github-fork-branch-and-pull-request
[3]: http://tbaggery.com/2008/04/19/a-note-about-git-commit-messages.html
[4]: https://help.github.com/articles/using-pull-requests
[5]: http://gitready.com/advanced/2009/02/10/squashing-commits-with-rebase.html
[6]: http://daringfireball.net/projects/markdown/syntax
[7]: http://blog.stwrt.ca/2013/09/06/installing-rubinius-with-rbenv

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Established in 1935, Link Engineering Company is the global leader in designing and manufacturing a complete range of test systems and in providing test services to the transportation industry in the areas of brake, transmission, driveline, steering, wheel, hub, tire, bearing, and electric motor applications. Headquartered in Plymouth, Michigan (suburb of Detroit), LINK operates facilities providing test services, test system sales, and customer support across the United States, Brazil, China, Japan, Germany, Korea, and India. LINK allows its global customers to realize their optimal test solutions through its in-depth industry knowledge, comprehensive test services and support, and state-of-the-art test systems.


text:
Q: Is "the" necessary in "the number of the nodes", "the list of the handles" and similar phrases? I wrote:

In this formula, the number of the nodes is multiplied by (H + P) because for each node, it is checked against the list of the handles, and if a match is found, a pattern is retrieved from the list of the patterns.

In this sentence, patterns, nodes, and handles are specific to the reader then I used "the" before them, I just want to double sure that the sentence above required the second "the" in of-phrases and it's not equivalent to:

In this formula, the number of nodes is multiplied by (H + P) because for each node, it is checked against the list of handles, and if a match is found, a pattern is retrieved from the list of patterns.

Are they equal in the context I described?
What about the following sentences:

the time complexity is O(N * (H + P)) where N is the number of the DOM-tree nodes, H is the number of handles and P is the number of the patterns. ...
These operations depend on the size of the handles, H, and the size of the patterns, P.

Is the second "the" extra in this sentence too? because I feel I need to emphasize the items, so I need "the".

A: You do not need to use the second "the" in your example.
The determiner "the" is used to signify a particular object, which in your case is either "number" or "list"

the number
  the list

If you had a case of nodes in a set A and nodes in a set B and only wanted to reference either A or B, you might write

the number of the A nodes
  the number of the nodes in A
  the number of nodes in A

In your second part

the time complexity is O(N * (H + P)) where N is the number of the DOM-tree nodes, H is the number of handles and P is the number of the patterns...
These operations depend on the size of the handles (H) and the size of the patterns (P).

Since you are using the "the +plural" pattern it is equivalent to "each +singular"

These operations depend on the size of each handle (H) and the size of each pattern (P).

unless it is the name of a group

My friend has two daughters and we call his wife and daughters: The Ladies.

which refers to all three (note capitalization), however

The Ladies in the band are all trained musicians.

refers to each woman assuming they are three other women.

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