text:
\begin{document}

\title{Full-Duplex Relay based on Zero-Forcing Beamforming}
\author{Jong-Ho~Lee~and~Oh-Soon~Shin
\thanks{J.-H. Lee is with the Division of Electrical, Electronic \& Control Engineering, Kongju National University,
Korea (e-mail: jongholee@kongju.ac.kr).}
\thanks{O.-S. Shin is with the School of Electronic Engineering, Soongsil University, Seoul 156-743, Korea (e-mail: osshin@ssu.ac.kr).}

}

\maketitle

\begin{abstract}
In this paper, we propose a full-duplex relay (FDR) based on a zero-forcing beamforming (ZFBF) for a multiuser MIMO relay system. The ZFBF is employed at the base station to suppress both the self-interference of the relay and the multiuser interference at the same time. Numerical results show that the proposed FDR can enhance the sum rate performance as compared to the half-duplex relay (HDR), if sufficient isolation between the transmit and receive antennas is ensured at the relay.
\end{abstract}

\begin{IEEEkeywords}
Decode-and-forward, full-duplex relay (FDR), zero-forcing beamforming (ZFBF).
\end{IEEEkeywords}

\section{Introduction} \label{sec1}
\IEEEPARstart{R}{elay} transmission technology has attracted growing attention due to its promising capability of extending cell coverage and increasing cell capacity of wireless communication systems. So far, most relay technologies have been developed under the half-duplex constraint \cite{Laneman04} that prevents relays from transmitting and receiving at the same time. Although this form of relay, called half-duplex relay (HDR), is easy to implement, it requires partitioning of resource for transmission and reception, reducing the whole system capacity.

In order to overcome the drawback of HDR, we consider a full-duplex relay (FDR) that is able to transmit and receive on the same frequency at the same time \cite{Wang05}, where it is crucial to reduce the effect of possibly strong self-interference caused by its own transmitter to its own receiver. In order to solve the self-interference problem, \cite{Slingsby95} and \cite{Anderson04} have studied the physical isolation between the transmit and receive antennas at the RS. Since the antenna isolation may not be enough for stable FDR operation, \cite{Bliss07}-\cite{Ju09} have suggested signal processing techniques to suppress the residual self-interference. In particular, \cite{Bliss07} and \cite{Riihonen09} suggested spatial interference suppression schemes that exploit multiple transmit and receive antennas at the RS. However, the need for multiple antennas at the relay increases the complexity and cost of the relay, which may become a huge barrier for deploying relay stations in wireless networks.

We propose an FDR that works under a multiuser MIMO scenario, where we need to suppress the multiuser interference as well as the self-interference in the FDR. We exploit a zero-forcing beamforming (ZFBF) in \cite{Yoo05} using multiple antennas at the base station (BS) rather than at the relay station (RS). It should be noted that the ZFBF is applied to cancel the self-interference of the FDR as well as to support multiuser MIMO. Moreover, the need for multiple antennas at the RS is eliminated by letting the BS perform the self-interference cancellation. The performance of the proposed FDR will be compared with that of an HDR with ZFBF and an ordinary multiuser MIMO without RS.

\section{System Model} \label{sec2}
We consider a downlink scenario where the BS has $L$ transmit antennas and a RS has one receive antenna and one transmit antenna. The RS is assumed to employ the decode-and-forward protocol \cite{Nabar04}, where the RS decodes all the signals from the BS and forwards them to the MS. We categorize mobile stations (MS's), each assumed to be equipped with one receive antenna, into two groups according to the reachability of the signals from the BS: 1) MS's in the first group can receive the signals from both the BS and RS. 2) MS's in the second group can receive only the signal from the RS. For notational convenience, we will refer to the set of $N_i$ MS's in the $i$-th group as MS-$i$, where $N_i$ denotes the number of MS's in the MS-$i$. Moreover, we assume that the BS selects the MS's from each MS group so that the sum rate is maximized \cite{Yoo05}.

The received signal $y_{i,j}$ at the $j$-th MS in the MS-$i$, $i = 1, 2$, is respectively given as
\begin{equation} \label{eq:received_signal}
\begin{array}{lcl}
{y}_{1,j} &=& \mathbf{h}_{1,j}^{\textrm{BS}}\mathbf{s}^{\textrm{BS}} + {h}_{1,j}^{\textrm{RS}}{s}^{\textrm{RS}} + {z}_{1,j}, \\
{y}_{2,j} &=& {h}_{2,j}^{\textrm{RS}}{s}^{\textrm{RS}} + {z}_{2,j},
\end{array}
\end{equation}
where $\mathbf{h}_{1,j}^{\textrm{BS}}$ denotes a $1 \times L$ channel vector between the BS and the $j$-th MS in MS-1. Moreover, ${h}_{i,j}^{\textrm{RS}}$ is a channel coefficient between the RS and the $j$-th MS in MS-$i$, $\mathbf{s}^{\textrm{BS}}$ denotes the $L \times 1$ transmit signal vector from the BS, ${s}^{\textrm{RS}}$ denotes the transmit signal from the RS, and ${z}_{i,j}$ denotes the zero-mean additive white Gaussian noise (AWGN) with unit variance. Note that $\mathbf{s}^{\textrm{BS}}$ and $s^{\textrm{RS}}$ are determined according to the MS selection. When generating $\mathbf{s}^{\textrm{BS}}$ at the BS and $s^{\textrm{RS}}$ at the RS, the total available transmit power is assumed to be ${P}_{T}^{\textrm{BS}}$ and ${P}_{T}^{\textrm{RS}}$, respectively. The signal transmission at the BS and RS is assumed to occur every time slot, the duration of which is assumed to be normalized to the unity.

\section{Relay Transmission Schemes} \label{sec3}
Although our main interest is in the FDR with ZFBF, the HDR transmission can also accommodate a ZFBF. For fair comparison between the HDR and FDR, we first develop a ZFBF for the HDR and the derived ZFBF will be incorporated into the FDR. In this section, we assume that all the exact channel state informations (CSI's) are available at the BS as in \cite{Marzetta06}. The impact of CSI feedback errors will be discussed in Section \ref{sec4}.

\subsection{Half-Duplex Relay (HDR)} \label{sec3a}
An HDR transmission scheme can be realized by time sharing of each slot between the BS and RS. Specifically, in the first $(1-t)$ fraction of each time slot, the BS can serve the RS as well as MS's in the MS-1 using a ZFBF scheme. The received signal at the RS during this period can be expressed as
\begin{equation} \label{eq:received_rs_hdr}
{y}_{\textrm{RS}} = \mathbf{h}_{\textrm{RS}}^{\textrm{BS}}\mathbf{s}^{\textrm{BS}} + {z}_{\textrm{RS}},
\end{equation}
where ${z}_{\textrm{RS}}$ also has unit variance. Let $\Gamma \subset \{ 1, 2, \cdots , N_{\textrm{1}} \}$ with $|\Gamma| < L$ be the set of MS's served by the BS, which is a subset of indices for the MS's in the MS-1. With $\Gamma=\{ \gamma_1, \gamma_2, \cdots, \gamma_{|\Gamma|} \}$, the received signals at the MS's associated with $\Gamma$, $\mathbf{y}_{\Gamma}=\left[ y_{1,\gamma_1}~y_{1,\gamma_2}~\cdots~y_{1,\gamma_{|\Gamma|}} \right]^{T}$, can be given as
\begin{equation} \label{eq:received_ms1_hdr}
\mathbf{y}_{\Gamma} = \mathbf{H}\left(\Gamma \right)\mathbf{s}^{\textrm{BS}} + \mathbf{z}_{\Gamma},
\end{equation}
where the $|\Gamma| \times L $ channel matrix $\mathbf{H}\left(\Gamma \right)$ is given as
\begin{equation}
\mathbf{H}\left( \Gamma \right) = \left[ \left(\mathbf{h}^{\textrm{BS}}_{1,\gamma_1}\right)^{T}~ \left(\mathbf{h}^{\textrm{BS}}_{1,\gamma_2} \right)^{T}~ \cdots ~ \left(\mathbf{h}^{\textrm{BS}}_{1,\gamma_{|\Gamma|}}\right)^{T} \right]^{T}. \label{eq:channel_stack_Gamma}
\end{equation}
From (\ref{eq:received_rs_hdr}) and (\ref{eq:received_ms1_hdr}), we have
\begin{equation}\left[
  \begin{array}{c}
    \mathbf{y}_{\Gamma} \\
    {y}_{\textrm{RS}}
  \end{array}
\right] = \mathbf{H}_{\Gamma}
\mathbf{s}^{\textrm{BS}}+
\left[
  \begin{array}{c}
    \mathbf{z}_{\Gamma} \\
    {z}_{\textrm{RS}}
  \end{array}
\right], \label{eq:received_hdr_zf}
\end{equation}
where the $(|\Gamma|+1) \times L$ channel matrix $\mathbf{H}_{\Gamma}$ is defined as
\begin{equation}
\mathbf{H}_{\Gamma} \triangleq \left[
  \begin{array}{c}
    \mathbf{H}\left(\Gamma \right) \\
    \mathbf{h}_{\textrm{RS}}^{\textrm{BS}} \\
  \end{array}
\right], \label{eq:channel_hdr_zf}
\end{equation}
and the $L \times 1$ BS transmit signal vector $\mathbf{s}^{\textrm{BS}}_{\Gamma}$ can be expressed as
\begin{equation}
\mathbf{s}^{\textrm{BS}} \triangleq \mathbf{W}_{\Gamma} \mathbf{P}_{\Gamma} \left[
  \begin{array}{c}
    \mathbf{x}_{\Gamma} \\
    {x}_{\textrm{RS}} \\
  \end{array}
\right]. \label{eq:tx_signal_hdr_zf}
\end{equation}
In (\ref{eq:tx_signal_hdr_zf}), the $|\Gamma| \times 1$ transmit symbol vector $\mathbf{x}_{\Gamma}$ is defined as $\mathbf{x}_{\Gamma} = \left[ x_{1,\gamma_1}~x_{1,\gamma_2}~\cdots~x_{1,\gamma_{|\Gamma|}} \right]^{T}$, where $x_{1,\gamma_m}$ is the data symbol destined for the $\gamma_m$-th MS in the MS-1. ${x}_{\textrm{RS}}$ is the transmit symbol destined for the RS, which the RS decodes and forwards to an MS in the MS-2. The $L \times (|\Gamma|+1)$ ZFBF precoder $\mathbf{W}_{\Gamma}$ and the $(|\Gamma|+1) \times (|\Gamma|+1)$ power allocation matrix $\mathbf{P}_{\Gamma}$ are defined as
\begin{eqnarray}
\mathbf{W}_{\Gamma} &\triangleq& \mathbf{H}_{\Gamma}^{\dag}\left(\mathbf{H}_{\Gamma}\mathbf{H}_{\Gamma}^{\dag} \right)^{-1} \nonumber \\
&=&\left[
  \begin{array}{ccccc}
    \mathbf{w}^{\textrm{BS}}_{1,\gamma_1} & \mathbf{w}^{\textrm{BS}}_{1,\gamma_2} & \cdots & \mathbf{w}^{\textrm{BS}}_{1,\gamma_{|\Gamma|}} & \mathbf{w}^{\textrm{BS}}_{\textrm{RS}}
  \end{array}
\right] \label{eq:bf_vector}
\end{eqnarray}
and
\begin{equation}
{\mathbf{P}}_{\Gamma} = \mbox{diag} \left[\sqrt{{P}_{1,\gamma_1}}, \sqrt{{P}_{1,\gamma_2}}, \cdots, \sqrt{{P}_{1,\gamma_{|\Gamma|}}}, \sqrt{{P}_{\textrm{RS}}} \right] \label{eq:power_hdr1}
\end{equation}
with $(\cdot)^{\dag}$ denoting the conjugate transpose. The $L \times 1$ vectors $\mathbf{w}^{\textrm{BS}}_{1,\gamma_m}$ and $\mathbf{w}^{\textrm{BS}}_{\textrm{RS}}$ in (\ref{eq:bf_vector}) denote the ZFBF vectors at the BS corresponding to the $\gamma_m$-th MS in the MS-1 and to the RS, respectively. Then, the sum rate for each link during the $(1-t)$ fraction of a time slot is given as
\begin{equation}
\begin{array}{lcl}
{R}_{\Gamma} &=& \sum_{m=1}^{|\Gamma|} {\log}_{2} \left( 1+{P}_{1,\gamma_m} \right), \\
{R}_{\textrm{RS}} &=& {\log}_{2} \left( 1+{P}_{\textrm{RS}} \right)
\end{array}\label{eq:cap_hdr_zf1}
\end{equation}
where the transmit power ${P}_{1,\gamma_m}$ allocated to the $\gamma_m$-th MS in the MS-1 and $P_{\textrm{RS}}$ allocated to the RS should satisfy the following power constraint,
\begin{equation}
\sum_{m =1}^{|\Gamma|} ||\mathbf{w}^{\textrm{BS}}_{1,\gamma_m}||^2 {P}_{1,\gamma_m} + ||\mathbf{w}^{\textrm{BS}}_{\textrm{RS}}||^2 {P}_{\textrm{RS},\Gamma} = {P}_{T}^{\textrm{BS}}. \label{eq:bs_pw_hdr_zf}
\end{equation}
Here, we define $\tau_{1,\gamma_m}=\frac{1}{||\mathbf{w}^{\textrm{BS}}_{1,\gamma_m}||^2}$ and $\tau_{\textrm{RS}}=\frac{1}{||\mathbf{w}^{\textrm{BS}}_{\textrm{RS}}||^2}$, and perform a water-filling as in \cite{Yoo05}. Then, ${P}_{1,\gamma_m}$ and $P_{\textrm{RS}}$ that maximize the sum rate ${R}_{\Gamma}+{R}_{\textrm{RS}}$ are given as
\begin{equation}
\begin{array}{lcl}
{P}_{1,\gamma_m} &=& \left(\mu \tau_{1,\gamma_m} -1 \right)^{+}, \\
{P}_{\textrm{RS}} &=& \left(\mu \tau_{\textrm{RS}} -1 \right)^{+}
\end{array}\label{eq:water_filling_hdr_zf1}
\end{equation}
where $(x)^{+}$ denotes $\max\{x,0 \}$ and
\begin{equation}
\sum_{m =1}^{|\Gamma|} \left(\mu - \frac{1}{\tau_{1,\gamma_m}} \right)^{+} + \left(\mu - \frac{1}{ \tau_{\textrm{RS}}} \right)^{+} = {P}_{T}^{\textrm{BS}}.
\end{equation}

In the remaining $t$ fraction of the time slot, we consider two possible scenarios: 1) the BS serves MS's in the MS-1 and the RS forwards the decoded data to an MS in the MS-2 at the same time. 2) the BS serves MS's in the MS-1 and the RS forwards the decoded data to another MS in the MS-1 at the same time. For the first scenario, the RS's signal destined for an MS in the MS-2 becomes the multiuser interference to the MS's in the MS-1. In the following, we cancel this interference by using a ZFBF, which will be incorporated into the FDR in the next subsection. The received signals at the MS's in the MS-1 and an MS in the MS-2 can be expressed as
\begin{equation}\left[
  \begin{array}{c}
    {\mathbf{y}}_{\check{\Gamma}} \\
    {y}_{2,j} \\
  \end{array}
\right] = \mathbf{H}_{\check{\Gamma},j}\left[
  \begin{array}{c}
    \mathbf{s}^{\textrm{BS}}\\
    {s}^{\textrm{RS}}
  \end{array}
\right]
+
\left[
  \begin{array}{c}
    {\mathbf{z}}_{\check{\Gamma}} \\
    {z}_{2,j} \\
  \end{array}
\right]. \label{eq:received_hdr_zf2}
\end{equation}
The $(|\check{\Gamma}|+1) \times (L+1)$ channel matrix ${\mathbf{H}}_{\check{\Gamma}}$ is given as
\begin{equation}
{\mathbf{H}}_{\check{\Gamma},j}=\left[
  \begin{array}{cc}
    \mathbf{H}\left( \check{\Gamma} \right) & \mathbf{h}\left(\check{\Gamma} \right) \\
    \mathbf{0}_{1 \times L} & {h}_{2,j}^{\textrm{RS}} \\
  \end{array}
\right], \label{eq:channel_hdr_zf2}
\end{equation}
where $\mathbf{H}\left( \check{\Gamma} \right)$ can be defined as in (\ref{eq:channel_stack_Gamma}) and $\mathbf{h}\left(\check{\Gamma} \right)=\left[{h}_{1,\check{\gamma}_1}^{\textrm{RS}}~{h}_{1,\check{\gamma}_2}^{\textrm{RS}}~\cdots~{h}_{1,\check{\gamma}_{|\check{\Gamma}|}}^{\textrm{RS}} \right]^{T}$. The $(L+1) \times 1$ transmit signal vector can be expressed in a matrix form
\begin{equation}
\left[
  \begin{array}{c}
    \mathbf{s}^{\textrm{BS}}\\
    {s}^{\textrm{RS}}
  \end{array}
\right] = {\mathbf{W}}_{\check{\Gamma},j}
{\mathbf{P}}_{\check{\Gamma},j}
\left[
  \begin{array}{c}
    \mathbf{x}_{\check{\Gamma}} \\
    \tilde{x}_{2,j} \\
  \end{array}
\right] \label{eq:transmit_hdr_zf2}
\end{equation}
with $\mathbf{x}_{\check{\Gamma}} = \left[ x_{1,\check{\gamma}_1}~x_{1,\check{\gamma}_2}~\cdots~x_{1,\gamma_{|\check{\gamma}|}} \right]^{T}$, $\check{\Gamma}=\{ \check{\gamma}_1~\check{\gamma}_2~\cdots~\check{\gamma}_{|\check{\Gamma}|} \}$ and $|\check{\Gamma}| \leq L$. In (\ref{eq:transmit_hdr_zf2}), the $(L+1) \times (|\check{\Gamma}|+1)$ ZFBF precoder and the $(|\check{\Gamma}|+1) \times (|\check{\Gamma}|+1)$ power allocation matrix are given as
\begin{equation}
{\mathbf{W}}_{\check{\Gamma},j} = {\mathbf{H}}_{\check{\Gamma},j}^{\dag}\left({\mathbf{H}}_{\check{\Gamma},j}{\mathbf{H}}_{\check{\Gamma},j}^{\dag} \right)^{-1} \label{eq:precode_hdr_zf2}
\end{equation}
and
\begin{equation}
{\mathbf{P}}_{\check{\Gamma},j} = \mbox{diag} \left[\sqrt{{P}_{1,\check{\gamma}_1}}, \sqrt{{P}_{1,\check{\gamma}_2}}, \cdots, \sqrt{{P}_{1,\check{\gamma}_{|\check{\Gamma}|}}},\sqrt{{P}_{2,j}} \right], \label{eq:power_hdr2}
\end{equation}
respectively. Note that $\tilde{x}_{2,j}$ denotes a delayed version of ${x}_{2,j}$ destined for the $j$-th MS in the MS-2 with a delay corresponding to the decoding delay at the RS. ${\mathbf{W}}_{\check{\Gamma}}$ in (\ref{eq:precode_hdr_zf2}) can be rewritten as \cite{Frank90}
\begin{equation}
{\mathbf{W}}_{\check{\Gamma},j} = \left[
  \begin{array}{cc}
    {\mathbf{W}}_{\check{\Gamma}}^{\textrm{BS}} & {\mathbf{w}}_{2,j}^{\textrm{BS}}  \\
    \mathbf{0}_{1 \times |\check{\Gamma}|} & {{w}}_{2,j}^{\textrm{RS}}
  \end{array}
\right] \label{eq:precode1_hdr_zf2}
\end{equation}
where ${{w}}_{2,j}^{\textrm{RS}}$ is the beamforming weight at the RS and ${\mathbf{w}}_{2,j}^{\textrm{BS}}$ denotes the $L \times 1$ ZFBF vectors at the BS corresponding to the $j$-th MS in the MS-2. The $L \times |\check{\Gamma}|$ beamforming matrix ${\mathbf{W}}_{\check{\Gamma}}^{\textrm{BS}}$ is defined as
\begin{equation}
{\mathbf{W}}_{\check{\Gamma}}^{\textrm{BS}} = \left[
  \begin{array}{cccc}
    {\mathbf{w}}_{1,\check{\gamma}_1}^{\textrm{BS}} & {\mathbf{w}}_{1,\check{\gamma}_2}^{\textrm{BS}} & \cdots & {\mathbf{w}}_{1,\check{\gamma}_{|\check{\Gamma}|}}^{\textrm{BS}}
  \end{array}
\right].
\end{equation}
Then, the transmit signals in (\ref{eq:transmit_hdr_zf2}) can be rewritten as
\begin{eqnarray}
\mathbf{s}^{\textrm{BS}} &=& \sum_{m=1}^{|\check{\Gamma}|} {\mathbf{w}}_{1,{\check{\gamma}}_m}^{\textrm{BS}}\sqrt{{P}_{1,\check{\gamma}_m}}{x}_{1,\check{\gamma}_m}+\mathbf{w}_{2,j}^{\textrm{BS}}\sqrt{{P}_{2,j}} \tilde{x}_{2,j}, \nonumber \\
{s}^{\textrm{RS}} &=& {w}_{2,j}^{\textrm{RS}} \sqrt{{P}_{2,j}} \tilde{x}_{2,j}.
\label{eq:transmit_hdr_zf2_2}
\end{eqnarray}
Note that the BS transmit signal $\mathbf{s}^{\textrm{BS}}$ contains not only the beamformed signal of $\mathbf{x}_{\check{\Gamma}}$ but also the beamformed signal of $\tilde{x}_{2,j}$. It should be noted that the BS can transmit the beamformed signal of $\tilde{x}_{2,j}$, since $\tilde{x}_{2,j}$ is a delayed version of ${x}_{2,j}$ that is originated from the BS. It is also seen that the RS transmits only the beamformed signal of $\tilde{x}_{2,j}$, due to the form of the precoding matrix ${\mathbf{W}}_{\check{\Gamma},j}$ in (\ref{eq:precode1_hdr_zf2}). It is reasonable because the RS can know only $\tilde{x}_{2,j}$ after decoding ${x}_{2,j}$.

The sum rate served to MS's is given as
\begin{equation}
\begin{array}{lcl}
{R}_{\check{\Gamma}} &=& \sum_{m=1}^{|\check{\Gamma}|}\log_{2} \left( 1+{P}_{1,\check{\gamma}_m} \right), \\
{R}_{2,j} &=& \log_{2} \left( 1+{P}_{2,j} \right). \label{eq:cap_hdr_zf2}
\end{array}
\end{equation}
From (\ref{eq:transmit_hdr_zf2_2}), it is obvious that the BS power constraint is given as
\begin{equation}
\sum_{m=1}^{|\check{\Gamma}|}|| {\mathbf{w}}_{1,\check{\gamma}_m}^{\textrm{BS}} ||^2 {P}_{1,\check{\gamma}_m} + || {\mathbf{w}}_{2,j}^{\textrm{BS}} ||^2  {P}_{2,j} = {P}_{T}^{\textrm{BS}}. \label{eq:bs_pw_hdr}
\end{equation}
As in (\ref{eq:water_filling_hdr_zf1}), the water-filling provides the available power $\bar{P}_{1,\check{\gamma}_m}^{\textrm{BS}}$ and $\bar{P}_{2,j}^{\textrm{BS}}$ at the BS. From the RS power constraint, the available power at the RS can be also computed as
\begin{equation}
\bar{P}_{2,j}^{\textrm{RS}}  = \frac{{P}_{T}^{\textrm{RS}}}{\left| {{w}}_{2,j}^{\textrm{RS}} \right|^{2}}.
\end{equation}
Now, we determine final transmit power ${P}_{1,\check{\gamma}_m}^{\star}$ and ${P}_{2,j}^{\star}$ using $\bar{P}_{1,\check{\gamma}_m}^{\textrm{BS}}$, $\bar{P}_{2,j}^{\textrm{BS}}$ and $\bar{P}_{2,j}^{\textrm{RS}}$. In the case of $\bar{P}_{2,j}^{\textrm{RS}} > \bar{P}_{2,j}^{\textrm{BS}}$, we simply set ${P}_{1,\check{\gamma}_m}^{\star} = \bar{P}_{1,\check{\gamma}_m}^{\textrm{BS}}$ and ${P}_{2,j}^{\star} = \bar{P}_{2,j}^{\textrm{BS}}$. Otherwise, we set ${P}_{2,j}^{\star} = \bar{P}_{2,j}^{RS}$. Then, we have
\begin{equation}
\sum_{m=1}^{|\check{\Gamma}|}|| {\mathbf{w}}_{1,\check{\gamma}_m}^{\textrm{BS}} ||^2 {P}_{1,\check{\gamma}_k}  = {P}_{T}^{\textrm{BS}}- || {\mathbf{w}}_{2,j}^{\textrm{BS}} ||^2  {P}_{2,j}^{\star}
\end{equation}
and perform the water-filling again to determine ${P}_{1,\check{\gamma}_m}^{\star}$. Then, ${P}_{1,\check{\gamma}_m}^{\star}$ and ${P}_{2,j}^{\star}$ are used to compute (\ref{eq:cap_hdr_zf2}). Assuming that all the received signals in the $(1-t)$ time fraction are decoded and forwarded by the RS during the remaining $t$ time fraction, $t$ is chosen such that $(1-t){R}_{\textrm{RS}} = t{R}_{2,j}$ \cite{Chae08}. This implies that the RS can forward only the information that has been received from the BS. Then, the overall sum rate in the HDR is given as
\begin{eqnarray}
{R}_{\textrm{HDR}} = \max_{\begin{smallmatrix} {\Gamma}, \check{\Gamma} \subset \{ 1, 2, \cdots , N_{1} \} \\ j \in \{ 1, 2, \cdots , N_2 \} \end{smallmatrix}} \frac{{R}_{\Gamma} {R}_{2,j} + R_{\textrm{RS}}({R}_{\check{\Gamma}}+{R}_{2,j})}{{R}_{2,j}+R_{\textrm{RS}}}. \label{eq:cap_hdr_zf}
\end{eqnarray}

For the second scenario, the RS receives ${x}_{1,j}$ from the BS and forwards $\tilde{x}_{1,j}$ to the $j$-th MS in the MS-1. ${\mathbf{H}}_{\check{\Gamma},j}$ can be given as
\begin{equation}
{\mathbf{H}}_{\check{\Gamma},j}=\left[
  \begin{array}{cc}
    \mathbf{H}\left( \check{\Gamma} \right) & \mathbf{h}\left(\check{\Gamma} \right) \\
    \mathbf{h}^{\textrm{BS}}_{1,j} & {h}_{1,j}^{\textrm{RS}} \\
  \end{array}
\right]. \label{eq:channel_hdr_zf2_na}
\end{equation}
Here, the MS-1 receives the signals from both the BS and RS such that $\mathbf{h}^{\textrm{BS}}_{1,j}$ is not a zero vector. The pseudo-inverse of (\ref{eq:channel_hdr_zf2_na}) yields
\begin{equation}
{\mathbf{W}}_{\check{\Gamma},j} = \left[
  \begin{array}{cc}
    {\mathbf{W}}_{\check{\Gamma}}^{\textrm{BS}} & {\mathbf{w}}_{1,j}^{\textrm{BS}}  \\
    {\mathbf{w}}_{1,j}^{\textrm{RS}} & {{w}}_{1,j}^{\textrm{RS}}
  \end{array}
\right], \label{eq:precode1_hdr_zf2_na}
\end{equation}
where the $1 \times |\check{\Gamma}|$ vector ${\mathbf{w}}_{1,j}^{\textrm{RS}}$ is also not a zero vector unlike in (\ref{eq:precode1_hdr_zf2}). From (\ref{eq:transmit_hdr_zf2}), we obtain the RS transmit signal shown as
\begin{eqnarray}
{s}^{\textrm{RS}} &=& {\mathbf{w}}_{1,j}^{\textrm{RS}}\mbox{diag} \left[\sqrt{{P}_{1,\check{\gamma}_1}}, \sqrt{{P}_{1,\check{\gamma}_2}}, \cdots, \sqrt{{P}_{1,\check{\gamma}_{|\check{\Gamma}|}}} \right]\mathbf{x}_{\check{\Gamma}} \nonumber\\ &&+{{w}}_{1,j}^{\textrm{RS}}\sqrt{{P}_{1,j}}\tilde{x}_{1,j}. \label{eq:transmit_hdr_zf2_na}
\end{eqnarray}
Note that the RS has no information on $\mathbf{x}_{\check{\Gamma}}$ since it received and decoded only ${x}_{1,j}$ in the $(1-t)$ time fraction. Therefore, it is obvious that the RS cannot compose (\ref{eq:transmit_hdr_zf2_na}), and thus we do not consider this transmission scenario. Due to the same reason, the scenario cannot be realized in the FDR as well, which will be described in the next subsection.

\subsection{Full-Duplex Relay (FDR)} \label{sec3c}
In the FDR transmission, the received signal at the RS and at an MS in each group can be expressed as
\begin{equation}\label{eq:received_rs_fdr}
\begin{array}{lcl}
{y}_{1,j} &=& \mathbf{h}_{1,j}^{\textrm{BS}}\mathbf{s}^{\textrm{BS}} + {h}_{1,j}^{\textrm{RS}}{s}^{\textrm{RS}} + {z}_{1,j}, \\
{y}_{2,j} &=& {h}_{2,j}^{\textrm{RS}}{s}^{\textrm{RS}} + {z}_{2,j}, \\
{y}_{\textrm{RS}} &=& \mathbf{h}_{\textrm{RS}}^{\textrm{BS}}\mathbf{s}^{\textrm{BS}}+{h}_{\textrm{RS}}^{\textrm{RS}}{s}^{\textrm{RS}}+{z}_{\textrm{RS}},
\end{array}
\end{equation}
where the second term in ${y}_{\textrm{RS}}$ represents the self-interference due to simultaneous transmission and reception. During an entire time slot, the BS serves MS's in the MS-1 as well as the RS, while the RS serves an MS in the MS-2 at the same time. In the proposed FDR, ${h}_{\textrm{RS}}^{\textrm{RS}}$ should be available at the BS. Here, we assume that the RS transmits ${h}_{\textrm{RS}}^{\textrm{RS}}$ to the BS by using an appropriate method, such as the analog linear modulation in \cite{Marzetta06}.

With $\Gamma \subset \{ 1, 2, \cdots , N_{1} \}$ and $|\Gamma| < L$, the received signals can be arranged in a matrix form as
\begin{equation}
\left[
  \begin{array}{c}
    \mathbf{y}_{\Gamma} \\
    {y}_{\textrm{RS}} \\
    {y}_{2,j} \\
  \end{array}
\right] = {\mathbf{H}}_{\Gamma,j}
\left[
  \begin{array}{c}
    \mathbf{s}^{\textrm{BS}}\\
    {s}^{\textrm{RS}}
  \end{array}
\right]
+
\left[
  \begin{array}{c}
    \mathbf{z}_{\Gamma} \\
    {z}_{\textrm{RS}} \\
    {z}_{2,j} \\
  \end{array}
\right], \label{eq:received_fdr_zf}
\end{equation}
where the $(|\Gamma|+2) \times (L+1)$ channel matrix ${\mathbf{H}}_{\Gamma,j}$ is given as
\begin{equation}
{\mathbf{H}}_{\Gamma,j} = \left[
  \begin{array}{cc}
    \mathbf{H}\left(\Gamma \right) & \mathbf{h}\left(\Gamma \right)  \\
    \mathbf{h}_{\textrm{RS}}^{\textrm{BS}} & {h}_{\textrm{RS}}^{\textrm{RS}} \\
    \mathbf{0}_{1 \times L} & {h}_{2,j}^{\textrm{RS}} \\
  \end{array}
\right] \label{eq:channel_fdr_zf}
\end{equation}
where $\mathbf{H}\left(\Gamma \right)$ is defined in (\ref{eq:channel_stack_Gamma}) and $\mathbf{h}\left(\Gamma \right)=\left[{h}_{1,\gamma_1}^{\textrm{RS}}~{h}_{1,\gamma_2}^{\textrm{RS}}~\cdots~{h}_{1,\gamma_{|\Gamma|}}^{\textrm{RS}} \right]^{T}$. The $L \times 1$ BS transmit signal vector $\mathbf{s}^{\textrm{BS}}$ and the RS transmit signal ${s}^{\textrm{RS}}$ are expressed in a matrix form as
\begin{equation}
\left[
  \begin{array}{c}
    \mathbf{s}^{\textrm{BS}}\\
    {s}^{\textrm{RS}}
  \end{array}
\right]=\mathbf{W}_{\Gamma,j}
\mathbf{P}_{\Gamma,j}
\left[
  \begin{array}{c}
    \mathbf{x}_{\Gamma} \\
    {x}_{2,j} \\
    \tilde{x}_{2,j} \\
  \end{array}
\right]. \label{eq:tx_signal_fdr_zf}
\end{equation}
In (\ref{eq:tx_signal_fdr_zf}), $\mathbf{x}_{\Gamma}$ and ${x}_{2,j}$ denote transmit symbols of the BS destined for the MS's in the MS-1 and for the $j$-th MS in the MS-2, respectively, and $\tilde{x}_{2,j}$ denotes transmit symbol of the RS. Moreover, the $(L+1) \times (|\Gamma|+2)$ ZFBF precoder and the $(|\Gamma|+2) \times (|\Gamma|+2)$ power allocation matrix in (\ref{eq:tx_signal_fdr_zf}) are defined as
\begin{equation}
\mathbf{W}_{\Gamma,j} = \mathbf{H}_{\Gamma,j}^{\dag}\left(\mathbf{H}_{\Gamma,j}\mathbf{H}_{\Gamma,j}^{\dag} \right)^{-1} \label{eq:precode_fdr_zf}
\end{equation}
and
\begin{equation}
\mathbf{P}_{\Gamma,j} = \mbox{diag} \left[\sqrt{{P}_{1,\gamma_1}}, \sqrt{{P}_{1,\gamma_2}}, \cdots , \sqrt{{P}_{1,\gamma_{|\Gamma|}}}, \sqrt{{P}_{\textrm{RS}}}, \sqrt{{P}_{2,j}} \right], \label{eq:power_fdr}
\end{equation}
respectively. In (\ref{eq:power_fdr}), $P_{1,\gamma_m}$ is the BS power allocated to the $\gamma_m$-th MS in the MS-1, $P_{\textrm{RS}}$ is the BS power allocated to the RS, and ${P}_{2,j}$ is the RS power allocated to the $j$-th MS in the MS-2.

The ZFBF precoder in (\ref{eq:precode_fdr_zf}) can be rewritten as \cite{Frank90}
\begin{equation}
\mathbf{W}_{\Gamma,j} = \left[
  \begin{array}{ccc}
    \mathbf{W}_{\Gamma}^{\textrm{BS}} & \mathbf{w}_{\textrm{RS}}^{\textrm{BS}} & \mathbf{w}_{2,j}^{\textrm{BS}} \\
    \mathbf{0}_{1 \times |\Gamma|} & 0 & {w}_{2,j}^{\textrm{RS}} \\
  \end{array}
\right], \label{eq:precode1_fdr_zf}
\end{equation}
where $\mathbf{w}_{\textrm{RS}}^{\textrm{BS}}$ and $\mathbf{w}_{2,j}^{\textrm{BS}}$ denote the $L \times 1$ ZFBF vectors at the BS, ${w}_{2,j}^{\textrm{RS}}$ is the beamforming weight at the RS and
\begin{equation}
\mathbf{W}_{\Gamma}^{\textrm{BS}} = \left[
  \begin{array}{cccc}
    {\mathbf{w}}_{1,{\gamma}_1}^{\textrm{BS}} & {\mathbf{w}}_{1,{\gamma}_2}^{\textrm{BS}} & \cdots & {\mathbf{w}}_{1,{\gamma}_{|\Gamma|}}^{\textrm{BS}}
  \end{array}
\right].
\end{equation}
Then, the transmit signals in (\ref{eq:tx_signal_fdr_zf}) can be rewritten as
\begin{eqnarray}
\mathbf{s}^{\textrm{BS}} &=& \sum_{m=1}^{|\Gamma|} {\mathbf{w}}_{1,{\gamma}_m}^{\textrm{BS}}\sqrt{{P}_{1,\gamma_m}}{x}_{1,\gamma_m}+\mathbf{w}_{\textrm{RS}}^{\textrm{BS}}\sqrt{{P}_{\textrm{RS}}} {x}_{2,j}\nonumber \\ &&+\mathbf{w}_{2,j}^{\textrm{BS}}\sqrt{{P}_{2,j}} \tilde{x}_{2,j}, \nonumber\\
{s}^{\textrm{RS}} &=& {w}_{2,j}^{\textrm{RS}} \sqrt{{P}_{2,j}} \tilde{x}_{2,j}.
\end{eqnarray}
As described in Section \ref{sec3a}, it is reasonable that $\mathbf{s}^{\textrm{BS}}$ contains not only the beamformed signal of $\mathbf{x}_{\Gamma}$ and $x_{2,j}$ but also the beamformed signal of $\tilde{x}_{2,j}$. Moreover, the RS transmits only the beamformed signal of $\tilde{x}_{2,j}$.

Substituting (\ref{eq:tx_signal_fdr_zf})-(\ref{eq:power_fdr}) into (\ref{eq:received_fdr_zf}), we get
\begin{equation}
{y}_{1,\gamma_m}=\sqrt{{P}_{1,\gamma_m}} {x}_{1,\gamma_m}+{z}_{1,\gamma_m} \label{eq:afterzfbf_fdr_zf1}
\end{equation}
with $m=1,2,\cdots,|\Gamma|$ and
\begin{equation}
\left[
  \begin{array}{c}
    {y}_{\textrm{RS}} \\
    {y}_{2,j} \\
  \end{array}
\right] =
\left[
  \begin{array}{c}
    \sqrt{{P}_{\textrm{RS}}} {x}_{2,j} \\
    \sqrt{{P}_{2,j}} \tilde{x}_{2,j} \\
  \end{array}
\right]
+
\left[
  \begin{array}{c}
    {z}_{\textrm{RS}} \\
    {z}_{2,j} \\
  \end{array}
\right]. \label{eq:afterzfbf_fdr_zf2}
\end{equation}
In (\ref{eq:afterzfbf_fdr_zf2}), it is seen that the ZFBF precoder $\mathbf{W}_{\Gamma,j}$ makes ${y}_{\textrm{RS}}$ free from the self-interference. Thus, the RS can decode ${x}_{2,j}$ and forward $\tilde{x}_{2,j}$ to the $j$-th MS in the MS-2 without interference. It is also obvious that we can set ${P}_{\textrm{RS}} = {P}_{2,j}$, since ${x}_{2,j}$ and $\tilde{x}_{2,j}$ are associated with the same information rate.

The sum rate served to MS's is found as
\begin{equation}
{R}_{\textrm{FDR}} = \max_{\begin{smallmatrix} \Gamma \subset \{1, 2, \cdots , N_{1} \} \\ j \in {1, 2, \cdots, N_2} \end{smallmatrix}} \sum_{m=1}^{|\Gamma|}\log_{2} \left( 1+{P}_{1,\gamma_m} \right) + \log_{2} \left( 1+{P}_{\textrm{RS}} \right) \label{eq:cap_fdr_zf}
\end{equation}
with the BS power constraint
\begin{equation}
\sum_{m=1}^{|\Gamma|} || \mathbf{w}_{1,{\gamma}_m}^{\textrm{BS}} ||^2 {P}_{1,{\gamma}_m} + \left( || \mathbf{w}_{\textrm{RS}}^{\textrm{BS}} ||^2 + || \mathbf{w}_{2,j}^{\textrm{BS}} ||^2 \right) {P}_{\textrm{RS}} = {P}_{T}^{\textrm{BS}}. \label{eq:bs_pw_fdr}
\end{equation}
After water-filling with (\ref{eq:cap_fdr_zf}) and (\ref{eq:bs_pw_fdr}) as in (\ref{eq:water_filling_hdr_zf1}), the available power $\bar{P}_{1,{\gamma}_m}$ and $\bar{P}_{\textrm{RS}}$ at the BS can be obtained. Similarly, the available power $\bar{P}_{2,j}$ at the RS can be computed as
\begin{equation}
\bar{P}_{2,j}  = \frac{{P}_{T}^{\textrm{RS}}}{\left|w_{2,j}^{\textrm{RS}} \right|^2}. \label{eq:rs_pw_fdr}
\end{equation}
Using $\bar{P}_{1,{\gamma}_m}$, $\bar{P}_{\textrm{RS}}$, and $\bar{P}_{2,j}$ obtained from (\ref{eq:bs_pw_fdr}) and (\ref{eq:rs_pw_fdr}), we determine final transmit power ${P}_{1,{\gamma}_m}^{\star}$ and ${P}_{\textrm{RS}}^{\star}$ of the BS. In the case of $\bar{P}_{\textrm{RS}} < \bar{P}_{2,j}$, we simply get ${P}_{1,{\gamma}_m}^{\star} = \bar{P}_{1,{\gamma}_m}$ and ${P}_{\textrm{RS}}^{\star} = \bar{P}_{\textrm{RS}}$. Otherwise, we set ${P}_{\textrm{RS}}^{\star} = \bar{P}_{2,j}$. Then we have
\begin{equation}
\sum_{m=1}^{|\Gamma|} || \mathbf{w}_{1,{\gamma}_m}^{\textrm{BS}} ||^2 {P}_{1,{\gamma}_m}  = {P}_{T}^{\textrm{BS}}- \left( || \mathbf{w}_{\textrm{RS}}^{\textrm{BS}} ||^2 + || \mathbf{w}_{2,j}^{\textrm{BS}} ||^2 \right) {P}_{\textrm{RS}}^{\star} \label{eq:bs_pw_fdr2}
\end{equation}
and ${P}_{1,{\gamma}_m}^{\star}$ is computed by the water-filling with (\ref{eq:bs_pw_fdr2}). Then, ${P}_{1,{\gamma}_m}^{\star}$ and ${P}_{\textrm{RS}}^{\star}$ are used to compute the sum rate of the proposed FDR in (\ref{eq:cap_fdr_zf}). In order to realize the FDR with ZFBF, the CSI between the RS's transmit antenna and receive antenna, $h_{\textrm{RS}}^{\textrm{RS}}$, is required at the BS. Moreover, the BS should inform the RS of $w_{2,j}^{\textrm{RS}} \sqrt{{P}_{\textrm{RS}}^{\star}}$ after constructing $\mathbf{W}_{\Gamma,j}$, ${P}_{1,{\gamma}_m}^{\star}$ and ${P}_{\textrm{RS}}^{\star}$.

\begin{figure*}[t]
 \centering \subfigure[]
  {\includegraphics[height=3in]{figure1a.jpg}} \label{fig1a}\\
 \centering\subfigure[]
  {\includegraphics[height=3in]{figure1b.jpg}} \label{fig1b}
 \caption{FDR signal model and its equivalent model ($L=2$ and $|\Gamma|=1$). (a) Signal model for the proposed FDR. (b) The equivalent multiuser MIMO transmission model.}\label{fig1}
\end{figure*}

We can establish an interesting connection between the proposed FDR and multiuser MIMO transmission. The signal model for the proposed FDR is illustrated in Fig. 1(a), where the number of BS transmit antennas is assumed to be $L=2$. If we regard the RS transmit antenna as a virtual BS transmit antenna in Fig. 1(a), the FDR signal model can equivalently be translated into a multiuser MIMO transmission model in Fig. 1(b), where the restriction on the received signals at the MS-2 is due to our assumption of the signal reachability in Section \ref{sec2}. Comparing Figs. 1(a) and 1(b), we see that the self-interference in the FDR corresponds to multiuser interference in the equivalent multiuser MIMO model. This justifies our approach of eliminating both the self-interference and the multiuser interference simultaneously, since the ZFBF was adopted to suppress the multiuser interference in the multiuser MIMO system \cite{Yoo05}.

It is worth discussing a practical issue to be considered in deploying the FDR. The self-interference signal might be much stronger than the desired signal at the receiver front-end of the RS. In order to avoid possible saturation in this case, it is important to make sure that the received signals at the RS fall within the dynamic range of the receiver. This can be assured through the use of an analog-to-digital converter (ADC) with high resolution and/or sufficient isolation between the transmit and receive antennas \cite{Slingsby95}, \cite{Anderson04}.

\section{Impact of CSI Feedback Errors} \label{sec4}
In order to investigate the impact of CSI feedback errors on the sum rate performance, we adopt the CSI error model in \cite{Yoo06}. According to the model, the CSI's at the BS can be represented as
\begin{equation}
\begin{array}{lcl}
\hat{\mathbf{h}}_{1,j}^{\textrm{BS}} &=& \mathbf{h}_{1,j}^{\textrm{BS}} + {e}_{1,j}^{\textrm{BS}}, \\
\hat{h}_{1,j}^{\textrm{RS}} &=& {h}_{1,j}^{\textrm{RS}}+{e}_{1,j}^{\textrm{RS}},\\
\hat{\mathbf{h}}_{\textrm{RS}}^{\textrm{BS}} &=& \mathbf{h}_{\textrm{RS}}^{\textrm{BS}}+{e}_{\textrm{RS}}^{\textrm{BS}},\\
\hat{h}_{\textrm{RS}}^{\textrm{RS}} &=& {h}_{\textrm{RS}}^{\textrm{RS}}+{e}_{\textrm{RS}}^{\textrm{RS}}, \\
\hat{h}_{2,j}^{\textrm{RS}} &=& {h}_{2,j}^{\textrm{RS}} + {e}_{2,j}^{\textrm{RS}},
\end{array}\label{eq:channel_fdr_zf_err}
\end{equation}
where ${e}_{1,j}^{\textrm{BS}}$, ${e}_{1,j}^{\textrm{RS}}$, ${e}_{2,j}^{\textrm{RS}}$, ${e}_{\textrm{RS}}^{\textrm{BS}}$, and ${e}_{\textrm{RS}}^{\textrm{RS}}$ denote the feedback errors, which are assumed to follow independent and identically distributed (i.i.d.) complex Gaussian distribution with zero mean and variance of $\sigma_{E}^{2}$.

As in (\ref{eq:channel_fdr_zf}), we can compose $\hat{\mathbf{H}}_{\Gamma,j}$ using (\ref{eq:channel_fdr_zf_err}) and the FDR ZFBF precoder with the feedback errors, $\hat{\mathbf{W}}_{\Gamma,j} = \hat{\mathbf{H}}_{\Gamma,j}^{\dag}\left(\hat{\mathbf{H}}_{\Gamma,j}\hat{\mathbf{H}}_{\Gamma,j}^{\dag} \right)^{-1}$, can be written as
\begin{equation}
\hat{\mathbf{W}}_{\Gamma,j} = \left[
  \begin{array}{ccccc}
    \hat{\mathbf{w}}_{1,{\gamma}_1}^{\textrm{BS}} & \cdots & \hat{\mathbf{w}}_{1,{\gamma}_{|\Gamma|}}^{\textrm{BS}} & \hat{\mathbf{w}}_{\textrm{RS}}^{\textrm{BS}} & \hat{\mathbf{w}}_{2,j}^{\textrm{BS}} \\
    0 & \cdots & 0 & 0 & \hat{w}_{2,j}^{\textrm{RS}} \\
  \end{array}
\right]. \label{eq:precode1_fdr_zf_err}
\end{equation}
Then, the SINR's at the MS's and RS are computed as
\begin{eqnarray}
\textrm{SINR}_{1,\gamma_m} &=& \frac{\left|\mathbf{h}^{\textrm{BS}}_{1,\gamma_m} \hat{\mathbf{w}}_{1,{\gamma}_m}^{\textrm{BS}}\right|^2{P}_{1,\gamma_m}}{1+I^{(1)}_{1,\gamma_m}+I^{(2)}_{1,\gamma_m}}, \nonumber \\
\textrm{SINR}_{\textrm{RS}} &=& \frac{ \left| \mathbf{h}^{\textrm{BS}}_{\textrm{RS}} \hat{\mathbf{w}}_{\textrm{RS}}^{\textrm{BS}}\right|^2 {P}_{\textrm{RS}}} {1+I^{(1)}_{RS}+ I^{(2)}_{RS}}, \nonumber \\
\textrm{SINR}_{2,j} &=& \left|{h}_{2,j}^{\textrm{RS}}\hat{w}_{2,j}^{\textrm{RS}} \right|^2 {P}_{2,j},
\label{eq:afterzfbf_fdr_zf_err}
\end{eqnarray}
where
\begin{eqnarray}
I^{(1)}_{1,\gamma_m} &=& \sum_{k \neq m} \left|\mathbf{h}^{\textrm{BS}}_{1,\gamma_m} \hat{\mathbf{w}}_{1,{\gamma}_k}^{\textrm{BS}}\right|^2{P}_{1,\gamma_k}+\left|\mathbf{h}^{\textrm{BS}}_{1,\gamma_m}\hat{\mathbf{w}}_{\textrm{RS}}^{\textrm{BS}}\right|^2{P}_{\textrm{RS}}, \nonumber \\
I^{(2)}_{1,\gamma_m} &=& \left| \mathbf{h}^{\textrm{BS}}_{1,\gamma_m} \hat{\mathbf{w}}_{2,j}^{\textrm{BS}} + {h}_{1,\gamma_m}^{\textrm{RS}}\hat{w}_{2,j}^{\textrm{RS}} \right|^2 {P}_{\textrm{RS}}, \nonumber \\
I^{(1)}_{RS} &=& \sum_{k} \left| \mathbf{h}^{\textrm{BS}}_{\textrm{RS}} \hat{\mathbf{w}}_{1,{\gamma}_k}^{\textrm{BS}} \right|^2 {P}_{1,\gamma_k}, \nonumber \\
I^{(2)}_{RS} &=& \left|\mathbf{h}^{\textrm{BS}}_{\textrm{RS}}\hat{\mathbf{w}}_{2,j}^{\textrm{BS}}+{h}_{\textrm{RS}}^{\textrm{RS}}\hat{w}_{2,j}^{\textrm{RS}} \right|^2 {P}_{\textrm{RS}}.
\label{eq:afterzfbf_interference}
\end{eqnarray}
Note that we have assumed that ${P}_{\textrm{RS}} = {P}_{2,j}$ in (\ref{eq:afterzfbf_fdr_zf2}), since the incoming information rate and outgoing information rate are the same at the RS. However, it is obvious that $\log_2(1+\textrm{SINR}_{\textrm{RS}}) \neq \log_2 (1+\textrm{SINR}_{2,j})$ with CSI feedback errors, even though the actual transmit power satisfies ${P}_{\textrm{RS}} = {P}_{2,j}$. Therefore, in this case, it may be reasonable to modify the sum rate in (\ref{eq:cap_fdr_zf}) for the FDR with ZFBF as
\begin{eqnarray}
\hat{R}_{\textrm{FDR}} &=& \max_{\begin{smallmatrix} \Gamma \subset \{1, 2, \cdots , N_{1} \} \\ j \in {1, 2, \cdots, N_2} \end{smallmatrix}} \sum_{m=1}^{|\Gamma|}\log_{2} \left( 1+\textrm{SINR}_{1,\gamma_m} \right)\nonumber\\ &+& \min \left\{ \log_2(1+\textrm{SINR}_{\textrm{RS}}), \log_2 (1+\textrm{SINR}_{2,j}) \right\}, \label{eq:cap_fdr_zf_err}
\end{eqnarray}
For the HDR with ZFBF, $\hat{R}_{\textrm{HDR}}$ can also be computed with the CSI feedback errors similarly to the procedure described above.

\section{Numerical Results} \label{sec5}

\begin{figure}[!t]
\centering
\includegraphics[height=2.5in]{figure2}
\caption{Sum rate of relay transmission schemes vs. $\sigma_{E}^{2}$ ($L=2$, $N=4$, $G=20$, $I=10$ and $Q=6$).} \label{fig2}
\end{figure}

\begin{figure}[!t]
\centering
\includegraphics[height=2.5in]{figure3}
\caption{Sum rate of relay transmission schemes vs. $Q$ ($L=2$, $N=4$, $G=20$, $I=10$ and $\sigma_{E}^{2}=0.01$).} \label{fig3}
\end{figure}

\begin{figure}[!t]
\centering
\includegraphics[height=2.5in]{figure4}
\caption{Sum rate of relay transmission schemes vs. $G$ ($L=2$, $N=4$, $I=10$, and $Q=6$).} \label{fig4}
\end{figure}

\begin{figure}[!t]
\centering
\includegraphics[height=2.5in]{figure5}
\caption{Sum rate of relay transmission schemes vs. $I$ ($L=2$, $N=4$, $G=20$, $Q=6$ and $\sigma_{E}^{2}=0.01$).} \label{fig5}
\end{figure}

We evaluate the ergodic sum rate performance of the HDR and FDR schemes through Monte Carlo simulations; 10,000 independent channel realizations are used to obtain the ergodic sum rate of each scheme. We also present the performance of an ordinary multiuser MIMO system without RS \cite{Yoo05} as a baseline. We assume ${P}_{T}^{\textrm{BS}} = 100$ and ${P}_{T}^{\textrm{RS}} = 50$. Moreover, for every MS in the first or second group, channel coefficient of the BS-to-MS link is assumed to follow i.i.d. complex Gaussian distribution with zero mean and unit variance. Similarly, channel coefficient of the RS-to-MS link is assumed to follow the same distribution with zero mean and the variance of $Q$ dB. These assumptions imply that the signal-to-noise ratio (SNR) of the BS-to-MS link and RS-to-MS link is set to 20dB and $(Q+17)$dB, respectively. It is also assumed that the average channel gain of each entry in $\mathbf{h}_{\textrm{RS}}^{\textrm{BS}}$ and that of ${h}_{\textrm{RS}}^{\textrm{RS}}$ are, respectively, $G$-dB and $I$-dB larger than that of the BS-to-MS link. Assuming that the number of MS's in each group is the same, we denote the number of MS's in each group as $N$ (i.e., $N_i = N$ with $i=1,2$).

In Fig. \ref{fig2}, we presnet the variation of ${\mathcal{E}}\left\{\hat{R}_{\textrm{HDR}}\right\}$ and ${\mathcal{E}}\left\{\hat{R}_{\textrm{FDR}}\right\}$ with $\sigma_{E}^{2}$, when $L=2$, $N=4$, $G=20$, $I=10$, and $Q=6$. The ergodic sum rate of ZFBF without RS \cite{Yoo05} is also presented. The sum rate performance for all transmission schemes is shown to decrease steeply when $\sigma_{E}^{2}>10^{-3}$. Moreover, it is seen that the proposed FDR with ZFBF outperforms the HDR with ZFBF and the ordinary ZFBF without RS, even though the CSI's at the BS are erroneous.

Fig. \ref{fig3} shows the variation of the ergodic sum rate performance with $Q$, when $L=2$, $N=4$, $G=20$, $I=10$, and $\sigma_{E}^{2}=0.01$. As shown in Fig. \ref{fig3}, even though the RS-to-MS link becomes more reliable, the HDR sum rate is almost constant due to the time sharing. However, the sum rate of the proposed FDR increases steeply compared with the HDR. Fig. \ref{fig4} presents how the ergodic sum rate performance varies with $G$, when $L=2$, $N=4$, $I=10$, and $Q=6$. In Fig. \ref{fig4}, $\sigma_{E}^{2}=0$ indicates that the perfect CSI's are available at the BS. Note that the larger value of $G$ makes the BS-to-RS link more reliable. The FDR with ZFBF is shown to outperform the HDR when $G>8$.

Fig. \ref{fig5} shows the variation of the ergodic sum rates with $I$, when $L=2$, $N=4$, $G=20$, $Q=6$, and $\sigma_{E}^{2}=0.01$. Note that the larger $I$ causes the stronger self-interference to the RS receiver in the case of the FDR with ZFBF. The HDR scheme must not depend on $I$. For the FDR to outperform the HDR, $I$ must be less than 21. This suggests that sufficient isolation between the transmit and receive antennas at the RS is necessary for the proposed FDR.

\section{Conclusion} \label{sec6}
We have proposed an FDR transmission scheme for the downlink of a cellular system. The FDR has been realized through the use of a ZFBF scheme based on multiple antennas at the base station. Numerical results have verified that the proposed FDR provides the performance improvement over the HDR when the antenna isolation is sufficient for the proposed FDR to operate within a tolerable self-interference range. We have also investigated the impact of CSI errors on the sum rate performance.


TITLE: Let $R$ be an Euclidean domain. Let $p$ be irreducible. Let $a\ge1$. Show that every non-zero submodule of $R/Rp^a$ contains $Rp^{a-1}/Rp^a.$
QUESTION [2 upvotes]: Let $R$ be a Euclidean domain. Let $p$ be irreducible. Let $a\geq 1$. Show that every non-zero submodule of $R/Rp^a$ contains $Rp^{a-1}/Rp^a$.

I bet the answer is stupidly obvious, but I just can't see it. So do I just consider a non-trivial submodule of $R/Rp^a$. It has a non-zero element (say) $u+Rp^a$. Where do I go from here? Do I use the division algorithm? What's $p$ being irreducible or $a\gneq 1$ got to do with it?
Thanks for any assistance!

REPLY [1 votes]: Let $K$ be a non-zero submodule of $R/Rp^a$. $K$ is of the form $I/Rp^a$ with $I$ a submodule of $R$ (that is, an ideal of $R$) with $I\supseteq Rp^a$. Since $R$ is euclidean (in fact it's enough to be a PID) we have $I=Rr$ for some $r\in R$. The inclusion $Rr\supseteq Rp^a$ shows that $r\mid p^a$, so $r=p^b$ with $0\le b\le a$. If $b=a$ then $I=Rp^a$ and therefore $K=0$, a contradiction. Thus $b\le a-1$ and $$K=I/Rp^a=Rp^b/Rp^a\supseteq Rp^{a-1}/Rp^a.$$


TITLE: Eigenvalues and Eigenvectors in Quantum Mechanics
QUESTION [1 upvotes]: I can calculate them mathematically but I don't understand the physical meaning of these concepts. My knowledge  is purely algorithmic and lacks actual insight, how do I picture eigenvalues/eigenvectors in my mind.

REPLY [2 votes]: It's easy to understand eigenvalues and eigenvectors in the context of matrices. Any n by n matrix can transform n-D space in some form. To make it easy to understand let's work with a 2by2 matrix
\begin{equation}
A=\begin{bmatrix}
1 & 1 \\
0 & 1
\end{bmatrix}
\end{equation}
This $A$ can transform the entire 2D space. Let's check it for one vector in 2d space:
\begin{bmatrix}
0  \\
1
\end{bmatrix} $\quad$This is nothing but $\hat{j}$ (read the column vector as $0\hat{i}+1\hat{j}$).
When $A$ acts on 2D cartesian space, every vector in the space including $\begin{bmatrix}
  0 \\
  1
\end{bmatrix}$ transforms as:
\begin{equation}
\begin{bmatrix}
  1 & 1 \\
  0 &1
\end{bmatrix} \begin{bmatrix}
  0 \\
  1
\end{bmatrix}= \begin{bmatrix}
  1  \\
  1
\end{bmatrix}
\end{equation}
So $A$ transforms $\hat{j}$ into $\hat{i}+\hat{j}$ ($=\hat{j'}$ for the transformed space). You can also check that it transforms $\hat{i}$ to $\hat{i}$ itself.
Check for any general vector and you'll see that $A$ shears the entire x-y plane. The coordinate axes in the transformed space are at $45^\circ$.
Now, $A$ is analogous to an operator in quantum mechanics and the column vectors are analogous to the wavefunction (usually called state vector in matrix language of Quantum Mechanics).
If $A$ acting on any vector gives the same vector but scaled by some factor, then this particular vector is called the eigenvector of $A$ (analogous to eigenfunction of operator $A$):
\begin{equation}
A\begin{bmatrix}
   a  \\
  b
\end{bmatrix}
=\lambda \begin{bmatrix}
   a  \\
  b
\end{bmatrix} \quad ...(*)\end{equation}
. In other words, If a vector in the original space retains it's direction in the transformed space then it is an eigenvector. In case of $A$ you can now tell that $\hat{i}$ is an eigenvector, so is any scaled version of $\hat{i}$, e.g,  $\begin{bmatrix}
   5  \\
  0
\end{bmatrix}$$=5$$\begin{bmatrix}
   1  \\
  0
\end{bmatrix}$ $=5\hat{i}$ , which is $\hat{i}$ scaled by 5, is also an eigenvector.
The scalar that scales the length of eigenvector in the transformed space is known as the eigenvalue($\lambda$, see $(*)$) corresponding to that particular eigenvector. e.g, in case of $\hat{i}$, $A\hat{i}=\begin{bmatrix}
1 & 1 \\
0 & 1
\end{bmatrix} \begin{bmatrix}
1  \\
0
\end{bmatrix}=\begin{bmatrix}
1  \\
0
\end{bmatrix}=1.\begin{bmatrix}
1  \\
0
\end{bmatrix} $, compare this to $(*)$ and you'll see that the eigenvalue corresponding to the eigenvector $\hat{i}$ is $1$.
To give an example in Quantum Mechanics(in Heisenberg's matrix mechanics form): If you operate the Hamiltonian matrix($H$, a 3by3 matrix) on an eigenvector of $H$ ($\psi$, a 3by1 column vector), then $H\psi=E\psi$, where $E$ (=Energy) is the eigenvalue of the Hamiltonian operator corresponding to the eigenvector $\psi$.


TITLE: Injective homomorphism between two finite groups with the same order
QUESTION [0 upvotes]: I want to show given two groups $G,H$ and the homomorphism $f:G\rightarrow H$ injective with $\text{ord}(G)=\text{ord}(H)<\infty$, then $f$ is bijective.
Thank you very much in advance.

REPLY [5 votes]: Hint: Since $f$ is injective, we know that $H$ contains a "copy" of $G$. Namely, the image $\text{im}(f) \le H$ is a subgroup of $H$ isomorphic to $G$ (why?). But if $G$ and $H$ are the same order, what can we conclude about $\text{im}(f)$ and $H$?


TITLE: How do I find all $n × n$ matrices $C$ such that $CD = DC.$
QUESTION [0 upvotes]: Two $n × n$ matrices $A$ and $B$ are said to be simultaneously diagonalizable if there exists an invertible $n × n$ matrix $S$ such that $S^{−1}AS$ and $S^{−1}BS$ are both diagonal.

Let $D$ be a diagonal $n × n$ matrix with distinct entries on the diagonal. How would I find all $n × n$ matrices $C$ such that $CD = DC$?

Isn't it just the $0$ matrix, because there exist no such matrices?

REPLY [2 votes]: It is useful to remember (and easy to show) that if $D$ is diagonal and $A$ is any matrix of the same size, then $DA$ is obtained from $A$ by multiplying every entry by the diagonal entry of$~D$ corresponding to its row, while $AD$ is obtained from $A$ by multiplying every entry by the diagonal entry of$~D$ corresponding to its column. (More concisely, left-multiplication by a diagonal matrix multiplies each row by a scalar, while right-multiplication does so with every column). Now since the diagonal entries of $D$ are assumed to be all different, it follows that as soon as $A$ has any off-diagonal entry that is nonzero, then $DA$ and $AD$ will differ in that position. On the other hand if all off-diagonal entries of $A$ are zero (that is, if $A$ is a diagonal matrix), then clearly $DA=AD$.
So the answer to your question is precisely the set of diagonal matrices.


TITLE: Topological Quantum Computing beyond Anyonic Braiding
QUESTION [3 upvotes]: In materials such as those that exhibit fractional quantum hall states, the ground-state topological degeneracy is known to be robust against external perturbations. This ultimately tells us that we might be able to use such states as a fault-tolerant quantum computer. One example is anyonic braiding, where we braid anyons to form qubits and perform calculations.
My question is the following: how can we utilize robust topological states beyond anyonic braiding to form a fault-tolerant quantum computer? Is there a more physically realizable model of a topological quantum computer? I feel like we can utilize topological order to a greater extent than just braiding anyons. Any possible systems/papers would be greatly appreciated.

REPLY [5 votes]: The first question is in a system supporting anyons, what are the universal/topologically invariant physical quantities? Our current theoretical understanding is that all such quantities are related in one way or another to anyon braiding, to be more precise, the $S$ and $T$ matrices. Here $T$ matrix is a diagonal matrix whose diagonal entries are the topological twist of anyons (roughly speaking the exchange statistics), and the element of $S$ matrix represents mutual braiding of anyons. All the other invariant quantities, such as fusion rules and quantum dimensions can be derived from $S$ and $T$. For example, one can obtain fusion rules by applying Verlinde's formula from the $S$ matrix.
This does not mean that braiding is the only thing one can do for quantum computing using anyons. One can also fuse anyons and measure the outcome, or do inteferometry measurement. For example, just using measurement one can perform "braiding" without actually moving anyons physically, as explained in https://arxiv.org/abs/0808.1933. In some (potentially physically relevant) anyon models, braiding alone is not sufficient for universal quantum computation, and by adding measurement one can make it universal, for example https://arxiv.org/abs/1504.02098 and https://arxiv.org/abs/1405.7778 for the example of $SU(2)_4$ anyons.
Another interesting idea is to seriously make use of the topology of the manifold. $S$ and $T$ matrices are in fact related to modular transformations of the manifold. Roughly speaking, it is almost impossible to directly access the $T$ matrix of anyons with just braiding, while the so-called "Dehn twist" of the manifold can do so. This forms the basis of a proposal https://arxiv.org/abs/cond-mat/0512066 that attempted to make the Ising anyon model computationally universal. Of course, dynamically changing the topology is very difficult in reality, but still a quite beautiful idea. A recent refinement is to introduce geometric defects into the system, which serves as "branch cut" for anyons and effectively changes the topology. Braiding such defects therefore realizes Dehn twist, see https://arxiv.org/abs/1208.4834. A broader context is that one can have various symmetry defects in an anyon system, which have distinct and possiblly richer braiding statistics. For example, defects in an Abelian anyon system can behave like non-Abelian anyons. For a general discussion see https://arxiv.org/abs/1410.4540, and for Abelian phases https://arxiv.org/abs/1305.7203.


TITLE: Result of increasing the radius of earth?
QUESTION [2 upvotes]: How can increasing the radius of earth may cause an impact on the solar system ? Like, would earth may start making a bigger orbit (due to increase in size and wait) or vice versa ? or else ?
PS: The base of my question arises from the point of view of "Growing Population", and managing it by "artificially increasing the size of earth" (Instead of say, finding possibilities to live on other planets like Mars). So, if the radius increases, naturally the circumference of the sphere, and hence surface to live on will increase.
I understand however that probably exploring newer planets might be more feasible solution. But just for curiosity I shared my query. :)

REPLY [0 votes]: Increasing the radius does not require new matter.
See Menger sponge as a conceptual model; and assume that when pieces of Earth are removed from its surface (excavation for raw materials) and then placed in a vertical order (buildings), Earth's surface takes a less dense form.
If Earth's surface were hypothetically covered by tall buildings. Earth's radius would have increased by the height of the buildings, but its surface mass would also be less dense.
Therefore now the question becomes: How much of Earth's mass can be tampered with by reducing its density, before Earth's ecosystem becomes too risky for its habitants?
Maybe that would help estimate a more realistic maximum sustainable human population.
*As per the Menger sponge, less density may translate into more surface area for habitants.


text:
\begin{document}

\maketitle

Institute of Mathematics, Silesian University of Technology\\
Kaszubska 23, 44-101 Gliwice, Poland

\begin{abstract}
We describe the lattice of ideals  and derivations of the Lie algebra of infinite column-finite matrices. As a corollary we obtain a new uncountably dimensional simple Lie algebra.
\end{abstract}

Keywords: infinite-dimensional Lie algebra, infinite matrices, ideals, derivations \\

2000 MSC: 17B20, 17B65

\section{Introduction}
The works of S.~Lie, W.~Killing and E. Cartan were the starting points for systematic development of the theory of  finite-dimensional Lie algebras. We mention here the classification of finite dimensional simple Lie algebras over the algebraically closed fields (for fields of characteristic $0$ due to E.~Cartan and W.~Killing and for characteristic $p>3$ given in works of R.~E.~Block, R.~L.~Wilson,  H.~Strade, A.~Premet) and the representation theory (the highest weight classification of irreducible modules of general linear Lie algebras). The nice exposition can be found in \cite{Bou1},  \cite{Bou2},  \cite{Bou3}, \cite{Jac62}, \cite{Hum72}, \cite{Str13}.

However at the present time there is no general theory of the infinite-dimensional Lie algebras, they have been investigated only in special settings. There are few classes of   infinite-dimensional Lie algebras that were
more or less intensively studied from the geometric point of view.
In preface to his monography \cite{Kac90}  V.~Kac consider as the first class the Lie algebras of vector fields which have many geometric applications (cohomology theory  \cite{FT83}, \cite{Fuks86}.)
The second class  consists of Lie algebras of smooth mappings of a given manifold into a finite-dimensional Lie algebra (some central extensions of these algebras are referred as current algebras).
The third class consists of the classical Lie algebras of operators in a Hillbert or Banach space \cite{delaHar72}.
Nowadays, there are studied intensively the  Kac-Moody algebras defined by generalized Cartan matrices (the fourth class), which have many applications in modern physics \cite{Kac90}, \cite{Bau90}, \cite{Bau97}.

Algebraic point of view was used in investigations of free Lie algebras and graded Lie algebras. In the book by Amayo and Stewart \cite{AS74} the results on the lattice structure of infinite-dimensional Lie algebras are surveyed. They translate some group theoretic definitions and results on infinite discrete groups into the language of Lie algebras. They provide results on soluble ideals, locally nilpotent radicals, Lie algebras with chain or other finitness restrictions, Frattini theory, Engel conditions, etc.. This shows the deep parallel of both theories however the proofs may differ, because Lie algebras depend on underlying field. These results are interesting for group theorists, however for Lie theory there are more limited \cite{Hum74}.

 In many papers appear, as examples, the Lie algebras of infinite matrices.
 We mention here the Lie algebra $\mathfrak{g}\mathfrak{l}_{\infty}$  of $\mathbb{Z} \times \mathbb{Z}$ infinite matrices over $\mathbb{C}$ which have only finite number of nonzero entries:
 $$\mathfrak{g}\mathfrak{l}_{\infty}= \{ A=(a_{ij})_{i,j \in \mathbb{Z}} :  \,\,{\rm all \,\,but\,\, the\,\, finite\,\, number \,\,of}\,\, a_{ij} \,\,\ {\rm are} \,\,0\} $$

 \noindent It has a basis consisting of matrices $E_{ij}$ which have $1$ in $(i,j)$-entry and zeros elsewhere and so it is countably dimensional. $\mathfrak{g}\mathfrak{l}_{\infty}$ is a subalgebra of $\mathfrak{g}\mathfrak{l}_{J}$ --  the Lie algebra of generalized Jacobian matrices, i.e. infinite matrices having nonzero entries in finite number of diagonals, introduced in \cite{KP81},
 \cite{DJKM81}:
 \begin{displaymath}
 \mathfrak{g}\mathfrak{l}_{J}= \{ A=(a_{ij})_{i,j \in \mathbb{Z}} :  \,\,\ a_{ij} = 0 \,\,\,\ {\rm for\,\, all} \,\,\,| i-j| >> 0 \}
\end{displaymath}

\noindent    The last algebra is denoted in \cite{KacB13} as $\bar{a}_{\infty}$ and is uncountably dimensional.   The representations of $\mathfrak{g}\mathfrak{l}_{\infty}$ play important role in soliton's theory . A central extension $a_{\infty}$ of $\bar{a}_{\infty}$ has also many applications in physics. It contains for example any affine Kac-Moody algebra and Virasoro algebra \cite{KacB13}.

 However there is no systematic study of  Lie algebras of infinite matrices. In this paper we shall consider the Lie algebra of column-finite infinite matrices indexed by positive integers $\mathbb{N}$,  describe the lattice of its ideals and describe its derivations.

\bigskip

Let $R$ be a commutative, unital ring and let $M_{cf}(\mathbb{N},R)$ denote the set of all $\mathbb{N} \times \mathbb{N}$  column-finite matrices over $R$, i.e. infinite matrices which have only finite number of nonzero entries in each column.
It forms an associative $R-$algebra. In case $R=K$  -- a field, the column-finite matrices correspond to the linear endomorphisms of the vector space $K^{(\mathbb{N})}$ with respect to the canonical basis
 (elements of  $K^{(\mathbb{N})}$ are infinite columns $(a_1, \ldots , a_n, \ldots)^T$ with only finite number of nonzero entries). It is clear that  $M_{cf}(\mathbb{N},K)$ is not countably dimensional as a vector space and it is imposssible to find reasonable basis for it.

 $M_{cf}(\mathbb{N},R)$ with respect to Lie product $[A,B]=AB-BA$ forms a Lie algebra  $\mathfrak{g}\mathfrak{l}_{cf}(\mathbb{N}, R)$. By  $\mathfrak{g}\mathfrak{l}_{rcf}(\mathbb{N}, R)$ we denote its Lie subalgebra
of all matrices which are simultaneously row-finite and column-finite. The additional condition of row-finitness means that also the transpose of the  matrix defines an endomorphism of  $K^{(\mathbb{N})}$,
 i.e. the adjoint endomorphism of the dual space  $K^{\mathbb{N}}$ preserves the subspace  $K^{(\mathbb{N})}$   ($K^{\mathbb{N}}$ consists of all infinite columns  $(a_1, \ldots , a_n, \ldots)^T$).

\smallskip

By  $\mathfrak{g}\mathfrak{l}_{fr}(\mathbb{N}, R)$ we denote the Lie subalgebra of  $\mathfrak{g}\mathfrak{l}_{cf}(\mathbb{N}, R)$ consisting of all matrices which have nonzero entries in only finite number of rows. It is clear that
$$\mathfrak{g}\mathfrak{l}_{fr}(\mathbb{N}, R) = \cup_{n>0}\mathfrak{g}\mathfrak{l}(n, \mathbb{N}, R),$$
where  $\mathfrak{g}\mathfrak{l}(n,\mathbb{N}, R)$ is a Lie subalgebra of matrices which have nonzero entries in the first $n$ rows.
For any matrix $A$ from  $\mathfrak{g}\mathfrak{l}(n,\mathbb{N}, R)$ we can define a trace $\textrm{Tr}(A)$ as a sum of first $n$  entries on the main diagonal. In fact, we can extend $\textrm{Tr}(A)$ onto  $\mathfrak{g}\mathfrak{l}_{fr}(\mathbb{N}, R)$. One can check that $\textrm{Tr}$ is a Lie homomorphism
$$\textrm{Tr}:  \mathfrak{g}\mathfrak{l}_{fr}(\mathbb{N}, R) \to R$$
onto a commutative Lie algebra $R$. Its kernel  $\mathfrak{s}\mathfrak{l}_{fr}(\mathbb{N}, R)$ is an ideal in  $\mathfrak{g}\mathfrak{l}_{fr}(\mathbb{N}, R)$ and consists of all matrices with trace $0$.

By  $\mathfrak{d}_{sc}(\mathbb{N}, R)$ we denote the Lie subalgebra of scalar matrices $\alpha \cdot E$, where $\alpha \in R$ and
$E$ is the infinite identity matrix. Of course, $\mathfrak{d}_{sc}(\mathbb{N}, R)$ is contained in the center of
 $\mathfrak{g}\mathfrak{l}_{cf}(\mathbb{N}, R)$.

The direct limit $\mathfrak{g}\mathfrak{l}_{\infty}(R)$ of Lie  algebras $\mathfrak{g}\mathfrak{l}_n(R)$ of $n \times n$ matrices under natural embeddings $\mathfrak{g}\mathfrak{l}_n(R) \to \mathfrak{g}\mathfrak{l}_{n+1}(R)$, given by:
$$A \rightarrow \left(\begin{array}{c|c} A&0\cr \cline{1-2}
0&0\cr
\end{array}\right)$$
is the Lie algebra of countable dimension. It can be viewed as  the Lie algebra of infinite $\mathbb{N} \times \mathbb{N}$ matrices $A$, which have only finite number of nonzero entries. Similarly, we obtain a Lie subalgebra  $\mathfrak{s}\mathfrak{l}_{\infty}(R)$ of matrices $A$ having $\Tr (A)=0$. By results of Ado and Iwasawa any   finite-dimensional Lie algebra over a field $K$  can be embedded into $\mathfrak{g}\mathfrak{l}_n(K)$  for some natural $n$ and so into $\mathfrak{g}\mathfrak{l}_{\infty}(K)$.

So it is a quite natural to ask:
\begin{quote}
Which infinite-dimensional Lie algebras can be embedded into  $\mathfrak{g}\mathfrak{l}_{cf}(\mathbb{N}, K)$?
\end{quote}

\noindent K.R. Goodearl, P. Menal and J. Moncasi proved in \cite{GMM93} that any countably dimensional associative $K-$ algebra can be embedded into the
algebra of row-finite and column-finite infinite $\mathbb{N} \times \mathbb{N}$  matrices. It means that any countably dimensional Lie algebra which arise from such algebra by taking a Lie product can be embedded into
 $\mathfrak{g}\mathfrak{l}_{rcf}(\mathbb{N}, K)$.

In \cite{Noe05} it was shown that the Lie algebra of derivations of $\mathfrak{s}\mathfrak{l}_{\infty}(K)$ ($K$ a field) is isomorphic to
factor algebra $\mathfrak{g}\mathfrak{l}_{rcf}(\mathbb{N}, K)/ \mathfrak{d}_{sc}(\mathbb{N}, K)$. In \cite{HKZ17} the derivations of the Lie algebra of strictly uppertriangular matrices
 $\mathfrak{n}(\mathbb{N}, R)$ were described.
The simplicity of $\mathfrak{s}\mathfrak{l}_{fr}(\mathbb{N}, K)$ for any field $K$ was proved in \cite{Hol16}, \cite{HZ16}. These results give motivation for a systematic study of $\mathfrak{g}\mathfrak{l}_{cf} (\mathbb{N}, R)$.

We will use the direct sum decomposition
$\mathfrak{g} \oplus \mathfrak{h} $ of two Lie algebras $\mathfrak{g}$, $\mathfrak{h} $ in usual sense \cite{Hen12}. Our main result is the following

\begin{thm} For any field $K$ the following seven subalgebras $\{0\}$,
$\mathfrak{d}_{sc}(\mathbb{N}, K)$,
  $\mathfrak{s}\mathfrak{l}_{fr}(\mathbb{N}, K)$,
 $\mathfrak{g}\mathfrak{l}_{fr}(\mathbb{N}, K)$,
 $\mathfrak{d}_{sc}(\mathbb{N}, K) \oplus\mathfrak{s}\mathfrak{l}_{fr}(\mathbb{N}, K)$,
 $\mathfrak{d}_{sc}(\mathbb{N}, K)  \oplus\mathfrak{g}\mathfrak{l}_{fr}(\mathbb{N}, K)$,
$\mathfrak{g}\mathfrak{l}_{cf}(\mathbb{N}, K)$
are the only ideals of  $\mathfrak{g}\mathfrak{l}_{cf}(\mathbb{N}, K)$.
\end{thm}
The lattice of ideals of $\mathfrak{g}\mathfrak{l}_{cf}(\mathbb{N}, K)$  is shown in the  figure below (we abbreviate notation for more convenience, see the beginning of Section $2$.)
\begin{center}
\begin{tikzpicture}
 \draw (0,0)--(-2,1)--(0,2.5)--(0,3.5)--(0,05);
 \draw (0,0)--(3,1.5)--(3,2.5)--(0,3.5);
 \draw (3,1.5)--(0,2.5);
 \fill (0,0) node [fill=white]{\{0\}};
 \fill (-2,1) node [fill=white]{$\mathfrak{d}_{sc}$};
 \fill (3,1.5) node [fill=white]{$\mathfrak{sl}_{fr}$};
 \fill (3,2.5) node [fill=white]{$\mathfrak{gl}_{fr}$};
 \fill (0,2.5) node [fill=white]{$\mathfrak{d}_{sc} \oplus \mathfrak{sl}_{fr} $};
 \fill (0,3.5) node [fill=white]{$\mathfrak{d}_{sc}\oplus \mathfrak{gl}_{fr}$};
 \fill (0,5) node [fill=white]{$\mathfrak{gl}_{cf}$};
\end{tikzpicture}
\end{center}

We note  that in finite-dimensional case the Lie algebra $\mathfrak{gl}(n, \mathbb{C})$ has only four ideals
$\{0\}$, $\mathfrak{d}(n, \mathbb{C})$, $\mathfrak{sl}(n,\mathbb{C})$ and $\mathfrak{gl}(n, \mathbb{C})$ \cite{Hen12} and has the following lattice of ideals:

\begin{center}
\begin{tikzpicture}
 \draw (0,0)--(-2,2)--(0,4)--(2,2)--(0,0);
 \fill (0,0) node [fill=white]{$\{0\}$};
 \fill (-2,2) node [fill=white]{$\mathfrak{d}_{sc}(n, \mathbb{C})$};
 \fill (2,2) node [fill=white]{$\mathfrak{sl}(n, \mathbb{C})$};
 \fill (0,4) node [fill=white]{$\mathfrak{gl}(n, \mathbb{C})$};
\end{tikzpicture}
\end{center}

 As a conseqence of Theorem 1.1.   we obtain

\begin{Cor}
The Lie algebra $\mathfrak{s}\mathfrak{l}_{cf}(\mathbb{N}, K)$ and the factor Lie algebra
$\mathfrak{gl}_{cf}(\mathbb{N}, K)/(\mathfrak{d}_{sc}(\mathbb{N}, K) \oplus \mathfrak{gl}_{fr}(\mathbb{N}, K))$  are  simple.
\end{Cor}

Our proof of Theorem 1.1. gives a different proof of simplicity of $\mathfrak{s}\mathfrak{l}_{cf}(\mathbb{N}, K)$  than it is given in \cite{Hol16}, \cite{HZ16}.
We note that  $\mathfrak{gl}_{cf}(\mathbb{N}, K)/(\mathfrak{d}_{sc}(\mathbb{N}, K) \oplus \mathfrak{gl}_{fr}(\mathbb{N}, K))$
is a new example of uncountably dimensional Lie algebra.
In forthcomming paper the autors will study their properties.

Direct consequence of the Theorem 1.1.  is the following
\begin{Cor}
The derived algebra of $\mathfrak{g}\mathfrak{l}_{cf}(\mathbb{N}, K)$  coincides with
$\mathfrak{g}\mathfrak{l}_{cf}(\mathbb{N}, K)$, i.e. it is perfect.
The ideal $\mathfrak{d}_{sc}(\mathbb{N}, K)$, is the center of $\mathfrak{g}\mathfrak{l}_{cf}(\mathbb{N}, K)$.
\end{Cor}

The second main result describes derivations.
\begin{thm}
Let $R$ be any commutative unital ring. Every derivation of $\mathfrak{g}\mathfrak{l}_{cf}(\mathbb{N}, R)$  is a sum of inner and central derivations.
\end{thm}
We refer to section 3 for definitions. The derivations of $\mathfrak{gl}(n, R)$ were described in \cite{WY06}.

Our results are proved for $\mathfrak{g}\mathfrak{l}_{cf}(\mathbb{N}, R)$, but  the Proposition 3.4. says that $\mathfrak{g}\mathfrak{l}_{cf}(\mathbb{N}, R)$  and $\mathfrak{g}\mathfrak{l}_{cf}(\mathbb{Z}, R)$ - the Lie algebra of
$\mathbb{Z} \times \mathbb{Z}$ column-finite infinite matrices are isomorphic.
So $\mathfrak{gl}_{cf}(\mathbb{Z}, R)$ has the same lattice of ideals (if $R=K$) and similar description of derivations.
The proofs for $\mathbb{N} \times  \mathbb{N}$ are technically simpler.
We note that $\mathfrak{g}\mathfrak{l}_{cf}(\mathbb{Z}, \mathbb{C})$ contains a Lie algebra of generalized Jacobian matrices $\mathfrak{gl}_J$.
Recently, A. Fialowski and K. Iohara in \cite{FI17}, \cite{FI18} studied homology of $\mathfrak{gl}_J$ and its subalgebras. They remarked that there is a little known about algebraic properties of $\mathfrak{gl}_J$. The authors plan to fill this gap in forthcoming paper. The following problem seems to be interesting:
\begin{quote}
Describe the lattice of ideals of Lie algebra of generalized Jacobian matrices.
\end{quote}

In section $2$ we give the proofs of Theorem 1.1. and corollaries, in section $3$ we prove Theorem 1.4. and Proposition 3.4.

\section{Ideals of the Lie algebra of column-finite infinite matrices}

In this section we describe ideals of  the Lie algebra $\mathfrak{gl}_{cf}(\mathbb{N},K)$ over a field $K$.
We recall that a subspace $\mathfrak{I}$ is an ideal of Lie algebra $\mathfrak{L}$, if it  satisfies the condition $[\mathfrak{L},\mathfrak{I}]\subseteq \mathfrak{I}$, we denote it $\mathfrak{I}\triangleleft \mathfrak{L}$. This condition is equivalent to $[\mathfrak{I},\mathfrak{L}]\subseteq \mathfrak{I}$ (it follows from anticommutativity and bilinearity of Lie product).

By $\langle \mathfrak{g}_{1}, \mathfrak{g}_{2} \rangle$ we denote the interval of ideals of $\mathfrak{gl}_{cf}(\mathbb{N},\mathbb{K})$ between
 $\mathfrak{g}_{1}$ and  $\mathfrak{g}_{2}$, i.e.the set of  all $\mathfrak{I}\triangleleft \mathfrak{gl}_{cf}(\mathbb{N},\mathbb{K})$ such that
 $\mathfrak{g}_{1} \subseteq \mathfrak{I} \subseteq \mathfrak{g}_{2}$.  The ideal $\mathfrak{I}$ is called proper if $\mathfrak{I} \not=\mathfrak{g}_{1}$ and  $\mathfrak{I} \not=\mathfrak{g}_{2}$.

Since in this section $K$ is the fixed field, we will use abbreviation  $\mathfrak{gl}_{cf}$ for  $\mathfrak{gl}_{cf}(\mathbb{N},K)$. Similarly we use abbreviations
$\mathfrak{d}_{sc}$,
  $\mathfrak{s}\mathfrak{l}_{fr}$,
 $\mathfrak{g}\mathfrak{l}_{fr}$,
 $\mathfrak{d}_{sc} \oplus\mathfrak{s}\mathfrak{l}_{fr}$,
 $\mathfrak{d}_{sc} \oplus\mathfrak{g}\mathfrak{l}_{fr}$ for its corresponding  subalgebras.
 By $\{0\}$ we denote the  trivial algebra.

We denote by  $E$ -- the infinite identity matrix and by $E_{ij}$ - the matrix  which has $1$ in $(i,j)$-entry and zeros elsewhere.
For a matrix $A=(a_{ij}) \in \mathfrak{gl}_{cf}$ we will write formally $A= \sum _{ij}a_{ij} E_{ij}$.

\begin{lem}\label{idCF}
Lie subalgebras
$\{0\}$,
$\mathfrak{d}_{sc}$,
  $\mathfrak{s}\mathfrak{l}_{fr}$,
 $\mathfrak{g}\mathfrak{l}_{fr}$,
 $\mathfrak{d}_{sc} \oplus\mathfrak{s}\mathfrak{l}_{fr}$,
 $\mathfrak{d}_{sc} \oplus\mathfrak{g}\mathfrak{l}_{fr}$ and  $\mathfrak{gl}_{cf}$
 are ideals in the Lie algebra $\mathfrak{gl}_{cf}$.
\end{lem}
\begin{proof}
It is clear  that $\{0\}\triangleleft\mathfrak{gl}_{cf}$ and $\mathfrak{gl}_{cf}\triangleleft\mathfrak{gl}_{cf}$.
Any  $A\in\mathfrak{d}_{sc}$ commutes with every $B\in\mathfrak{gl}_{cf}$, which means that  $[A,B]=0$ and  so $\mathfrak{d}_{sc}\triangleleft\mathfrak{gl}_{cf}$.

Let $A\in\mathfrak{gl}_{fr}$ and $n$ be a number of the last nonzero row in the matrix $A$. For all $B\in\mathfrak{gl}_{cf}$ the  matrix $AB$ has at most $n$ nonzero rows.
Since  $B$ is column-finite, it  has only finite numbers of nonzero entries in the first $n$ columns so $BA\in\mathfrak{gl}_{fr}$. Finally
$[A,B]=AB-BA\in\mathfrak{gl}_{fr}$.

Let $A\in\mathfrak{sl}_{fr}$ and $B\in\mathfrak{gl}_{cf}$. We know that $AB\in\mathfrak{gl}_{fr}$ and $BA\in\mathfrak{gl}_{fr}$. Let $m$ be the number of the last
nonzero row in the matrix $AB$, $k$ be the number of the last nonzero row in the matrix $BA$ and $n=\max\{m,k\}$. Denote by $(AB)_n$ and $(BA)_n$ square  $n\times n$ matrices
from upper left corner of  $AB$ and $BA$ respectively. We have $\textrm{Tr }(AB-BA)=\textrm{Tr }((AB)_n-(BA)_n)=0$.\qed
\end{proof}

\begin{lem}\label{sl}
Let  $\mathfrak{I}\triangleleft \mathfrak{gl}_{cf}$. Then either $\mathfrak{I}\subset \mathfrak{d}_{sc}$ or $\mathfrak{sl}_{cf} \subset \mathfrak{I}$.
\end{lem}
\begin{proof}
Step 1: We show that if $A$ is diagonal matrix and $a_{ii}\not=a_{jj}$ then $E_{ij}\in\mathfrak{I}$.

\noindent We have $E_{ij}=(a_{ii}-a_{jj})(a_{ii}-a_{jj})^{-1}E_{ij}=[A,(a_{ii}-a_{jj})^{-1}E_{ij}]$.\par
Step 2: We show that if $A$ is not a diagonal matrix and $a_{ij}\not= 0$ then for some $k\not=i$ there  is $E_{ik}\in\mathfrak{I}$.\\
We have
$$[A,E_{ii}]=\sum_{n\in\mathbb{N}}a_{in}E_{in}-\sum_{m\in\mathbb{N}}a_{mi}E_{mi}\in\mathfrak{I}$$ so
$$[\sum_{n\in\mathbb{N}}a_{in}E_{in}-\sum_{m\in\mathbb{N}}a_{mi}E_{mi},E_{jj}]=a_{ij}E_{ij}-a_{ji}E_{ji}\in\mathfrak{I}.$$
If $a_{ji}=0$ we obtain $a_{ij}E_{ij}\in\mathfrak{I}$ (and $E_{ij}\in\mathfrak{I}$, because $\mathfrak{I}$ is the linear space). If $a_{ji}\not= 0$ we have
$$[a_{ij}E_{ij}-a_{ji}E_{ji},E_{ij}]=-a_{ji}E_{jj}+a_{ji}E_{ii}\in\mathfrak{I}$$
so $E_{ii}-E_{jj}\in\mathfrak{I}$. Using step $1$ we obtain $E_{ik}\in\mathfrak{I}$ for all $k\not= i$ if $\textrm{char }\mathbb{K}\not=2$ or
$E_{ik}\in\mathfrak{I}$ for all $k\notin \{i,j\}$ if $\textrm{char }\mathbb{K}=2$.\par
Step 3: We show that if for some $i\not= j$ $E_{ij}\in\mathfrak{I}$ then $\mathfrak{I}$ contains all matrices $B\in\mathfrak{sl}_{fr}$.\\
For every matrix $A\in\mathfrak{gl}_{cf}$ we have
$$[A,E_{ij}]=\sum_{m\in\mathbb{N}}a_{mi}E_{mj}-\sum_{n\in\mathbb{N}}a_{jn}E_{in}=\sum_{m=i\vee n=j}b_{mn}E_{mn},$$
where $b_{ij}=a_{ii}-a_{jj}$, $b_{in}=-a_{jn}$ for $n\notin\{i,j\}$, $b_{mj}=a_{mi}$ for $m\notin\{i,j\}$, $b_{ii}=-a_{ji}$ and $b_{jj}=a_{ji}$. Because $A$ is
an arbitrary matrix we can choose entries $b_{mn}$ in any way, with condition $b_{ii}=-b_{jj}$ (which means that the trace is equal to zero). In particular we obtain
that for all $m\not=j$
$E_{mj}\in\mathfrak{I}$ and for all $n\not=i$ $E_{in}\in\mathfrak{I}$. Repeating the argument from the beginning of the paragraph we obtain
$\sum_{m=i\vee n=j}b_{mn}E_{mn}\in\mathfrak{I}$ for all natural numbers $i\not=j$. Since $\mathfrak{I}$ is a linear space, finite sum of matrices of this type
also is in $\mathfrak{I}$ and the proof is complete.\qed
\end{proof}

\begin{lem}\label{gl}
Let  $\mathfrak{I}\triangleleft \mathfrak{gl}_{cf}$. Then either $\mathfrak{I}\subset \mathfrak{d}_{sc} \oplus \mathfrak{gl}_{fr}$ or $\mathfrak{I}=\mathfrak{gl}_{cf}$.
\end{lem}
\begin{proof}
Step 1: We show that if  $A\in\mathfrak{I}\setminus(\mathfrak{d}_{sc}\oplus\mathfrak{gl}_{fr})$ is not a diagonal matrix then $\mathfrak{I}$ contains a diagonal matrix $D$ for which the set
$\{i\in\mathbb{N}: d_{ii}\not=d_{i+1,i+1}\}$ is infinite.\\
Assume at first that $A$ has finite number of nonzero entries beyond diagonal. We know that for all natural numbers $i\not=j$ we have $E_{ij}\in\mathfrak{I}$ (lemma
\ref{sl}). Since $\sum_{i\not=j}a_{ij}E_{ij}$ is the finite sum,  $A-\sum_{i\not=j}a_{ij}E_{ij}$ is a diagonal matrix, $A-\sum_{i\not=j}a_{ij}E_{ij}\in\mathfrak{I}$
and $A-\sum_{i\not=j}a_{ij}E_{ij}\notin\mathfrak{gl}_{fr} \oplus \mathfrak{d}_{sc}$.\\
Assume now that $A$ has infinite number of nonzero entries beyond the diagonal.
Choose a sequence of nonzero entries $a_{i_1,j_1}$, $a_{i_2,j_2}$, $a_{i_3,j_3}$,
\dots in such a way that
for all $n\in\mathbb{N}$ $i_1>1$, $j_1>1$, $i_n\not=j_n$, $i_{n+1}>1+i_n$, $i_{n+1}>1+j_n$, $j_{n+1}>1+j_n$, $j_{n+1}>1+i_n$, $1+k_n<i_{n+1}$ and $1+k_n<j_{n+1}$
where $k_n$ is the  number of row with last nonzero entry in column $j_n$. We can assume also that $i_n>j_n$ for all $n$ or $i_n<j_n$ for all $n$.
We write the
proof for $i_n<j_n$,  a second case is similar (by transposing the corresponding matrices). We have
$$[\sum_{n\in\mathbb{N}}E_{i_n,i_n},A]=\sum_{n\in\mathbb{N}}\sum_{m\in\mathbb{N}}a_{i_n,m}E_{i_n,m}-\sum_{n\in\mathbb{N}}\sum_{m\in\mathbb{N}}a_{n,i_m}E_{n,i_m}$$
and
$$[\sum_{n\in\mathbb{N}}\sum_{m\in\mathbb{N}}a_{i_n,m}E_{i_n,m}-\sum_{n\in\mathbb{N}}\sum_{m\in\mathbb{N}}a_{n,i_m}E_{n,i_m},\sum_{n\in\mathbb{N}}E_{j_n,j_n}]=$$
$$=\sum_{n\in\mathbb{N}}\sum_{m\in\mathbb{N}}a_{i_n,j_m}E_{i_n,j_m}+\sum_{n\in\mathbb{N}}\sum_{m\in\mathbb{N}}a_{j_n,i_m}E_{j_n,i_m}\in\mathfrak{I}$$
so
\begin{displaymath}\begin{split}
&[\sum_{n\in\mathbb{N}}E_{i,i+1},\sum_{n\in\mathbb{N}}\sum_{m\in\mathbb{N}}a_{i_n,j_m}E_{i_n,j_m}+\sum_{n\in\mathbb{N}}\sum_{m\in\mathbb{N}}a_{j_n,i_m}E_{j_n,i_m}
]=\\
&=\sum_{n\in\mathbb{N}}\sum_{m\in\mathbb{N}}a_{i_n,j_m}E_{i_n-1,j_m}+\sum_{n\in\mathbb{N}}\sum_{m\in\mathbb{N}}a_{j_n,i_m}E_{j_n-1,i_m}-\\
&-\sum_{n\in\mathbb{N}}\sum_{m\in\mathbb{N}}a_{i_n,j_m}E_{i_n,j_m+1}-\sum_{n\in\mathbb{N}}\sum_{m\in\mathbb{N}}a_{j_n,i_m}E_{j_n,i_m+1}\in\mathfrak{I}.
\end{split}\end{displaymath}
We obtain
\begin{displaymath}\begin{split}
&[\sum_{n\in\mathbb{N}}E_{i_n-1,i_n-1},\sum_{n\in\mathbb{N}}\sum_{m\in\mathbb{N}}a_{i_n,j_m}E_{i_n-1,j_m}+\sum_{n\in\mathbb{N}}\sum_{m\in\mathbb{N}}
a_{j_n,i_m}E_{j_n-1,i_m}-\\
&-\sum_{n\in\mathbb{N}}\sum_{m\in\mathbb{N}}a_{i_n,j_m}E_{i_n,j_m+1}-\sum_{n\in\mathbb{N}}\sum_{m\in\mathbb{N}}a_{j_n,i_m}E_{j_n,i_m+1}]=\\
&=\sum_{n\in\mathbb{N}}\sum_{m\in\mathbb{N}}a_{i_n,j_m}E_{i_n-1,j_m}\in\mathfrak{I}.
\end{split}\end{displaymath}

We have
$$\sum_{n\in\mathbb{N}}\sum_{m\in\mathbb{N}}a_{i_n,j_m}E_{i_n-1,j_m}=\sum_{n\in\mathbb{N}}\sum_{m\geq n}a_{i_n,j_m}E_{i_n-1,j_m}$$
 because $1+k_n<i_{n+1}$.
Denote for simplicity $h_n=i_n-1$, $a_{i_n,j_m}=b_{h_n,j_m}$ so
$$\sum_{n\in\mathbb{N}}\sum_{m\geq n}a_{i_n,j_m}E_{i_n-1,j_m}=\sum_{n\in\mathbb{N}}
\sum_{m\geq n}b_{h_n,j_m}E_{h_n,j_m}.$$
We define
$$x_{h_n,h_m}=\begin{cases}
-b_{h_m,j_m}^{-1}b_{h_n,j_m}&\textrm{for }m=n+1,\\
-b_{h_m,j_m}^{-1}(b_{h_n,j_m}-\sum_{k=n+1}^{m-1}x_{h_n,h_k}b_{h_k,j_m}) &\textrm{otherwise}.
\end{cases} $$
We have
$$[\sum_{n\in\mathbb{N}}\sum_{m>n}x_{h_n,h_m}E_{h_n,h_m},\sum_{n\in\mathbb{N}}\sum_{m\geq n}b_{h_n,j_m}E_{h_n,j_m}]=\sum_{n\in\mathbb{N}}b_{h_n,j_n}E_{h_n,j_n}.$$
Finally
$$[\sum_{n\in\mathbb{N}}b_{h_n,j_n}E_{h_n,j_n},\sum_{n\in\mathbb{N}}E_{j_n,h_n}]=\sum_{n\in\mathbb{N}}b_{h_n,j_n}E_{h_n,h_n}-\sum_{n\in\mathbb{N}}
b_{h_n,j_n}E_{j_n,j_n}\in\mathfrak{I}$$
is the required diagonal matrix.
\par
Step 2: We show that if $A$ is a diagonal matrix and if the set $$\mathbf{H}=\{i\in\mathbb{N}: a_{ii}\not=a_{i+1,i+1}\}$$ is infinite, then
$\sum_{i\in\mathbf{H}}E_{i,i+1}\in\mathfrak{I}$.
We have
\begin{displaymath}\begin{split}& [A,\sum_{i\in\mathbf{H}}(a_{ii}-a_{i+1,i+1})^{-1}E_{i,i+1}]=\\
&=\sum_{i\in\mathbf{H}}a_{ii}(a_{ii}-a_{i+1,i+1})^{-1}E_{i,i+1}-\sum_{i\in\mathbf{H}}a_{i+1,i+1}(a_{ii}-a_{i+1,i+1})^{-1}E_{i,i+1}=\\
&=\sum_{i\in\mathbf{H}}E_{i,i+1}.
\end{split}\end{displaymath}
\par
Step 3: Let $\mathbf{H}$ be an infinite subset of $\mathbb{N}$. We show that if $\sum_{i\in\mathbf{H}}E_{i,i+1}\in\mathfrak{I}$ then
$\sum_{i\in\mathbf{G}}E_{i,i+1}\in\mathfrak{I}$, where $\mathbf{H}\subset\mathbf{G}\subset\mathbb{N}$ and $\mathbb{N}\setminus\mathbf{G}$ does not contain two
consecutive natural numbers.\\
If $\mathbf{H}$ satisfies the assumptions of this step we can take $\mathbf{H}=\mathbf{G}$. Otherwise
define the family of sets $\{\mathbf{Z}_n\}_{n\in\mathbb{N}}$ as $\mathbf{Z}_1=\emptyset$ and
$$\mathbf{Z}_n=\begin{cases}\{n\}\mbox{  }\,\,\textrm{if }\{n-1,n\}\cap\mathbf{H}=\emptyset \wedge \mathbf{Z}_{n-1}=\emptyset,\\
\emptyset\mbox{  } \,\,\,\,\,\,\,\,\,\textrm{otherwise.} \end{cases}$$
Let $\mathbf{Z}=\cup_{n=1}^{\infty}\mathbf{Z}_n$. Note that $\mathbf{Z}\cap\mathbf{H}=\emptyset$ and $\mathbb{N}\setminus (\mathbf{Z}\cup\mathbf{H})$ does not
contain two consecutive natural numbers. Denote $G:=\mathbf{Z}\cup\mathbf{H}$. Let $f:\mathbf{Z}\rightarrow\mathbf{H}$ be a strictly increasing function and
$f(n)>n$ for all $n\in\mathbf{Z}$. We have
$$[\sum_{i\in\mathbf{Z}}E_{i,f(i)},\sum_{j\in\mathbf{H}}E_{j,j+1}]=\sum_{i\in\mathbf{Z}}E_{i,f(i)+1}\in\mathfrak{I}$$
so
$$[\sum_{i\in\mathbf{Z}}E_{i,f(i)+1},\sum_{i\in\mathbf{Z}}E_{f(i)+1,i}]=\sum_{i\in\mathbf{Z}}E_{i,i+1}\in\mathfrak{I}.$$
Finally $$\sum_{i\in\mathbf{H}}E_{i,i+1}+\sum_{i\in\mathbf{Z}}E_{i,i+1}=\sum_{i\in\mathbf{G}}E_{i,i+1}\in\mathfrak{I}.$$\par
Step 4: We show that if $\sum_{i\in\mathbf{G}}E_{i,i+1}\in\mathfrak{I}$ and $\mathbb{N}\setminus\mathbf{G}$ does not contain two
consecutive natural numbers then $\sum_{n\in\mathbb{N}}E_{i,i+1}\in\mathfrak{I}$.\\
We can assume that $1\in\mathbf{G}$ because $E_{12}\in\mathbf{G}$ (from lemma \ref{sl}) and $\mathfrak{I}$ is a linear space. Denote
$\mathbb{N}\setminus\mathbf{G}=\{n_1,n_2,n_3,\dots\}$ and define sets $\mathbf{Z}_1=\{n_1,n_3,n_5,\dots\}$ and $\mathbf{Z}_2=\{n_2,n_4,n_6,\dots\}$. Observe that
for all $m,n\in\mathbf{Z}_1$ or $m,n\in\mathbf{Z}_2$ is satisfied inequality $|m-n|\geq 4$. We have
$$[\sum_{j\in\mathbf{Z}_1}E_{j,j-1},\sum_{i\in\mathbf{G}}E_{i,i+1}]=\sum_{j\in\mathbf{Z}_1}E_{jj}-\sum_{j\in\mathbf{Z}_1}E_{j-1,j-1}\in\mathfrak{I}$$
so
$$[\sum_{j\in\mathbf{Z}_1}E_{jj}-\sum_{j\in\mathbf{Z}_1}E_{j-1,j-1},\sum_{i\in\mathbf{Z}_1}E_{i,i+1}]=\sum_{j\in\mathbf{Z}_1}E_{j,j+1}\in\mathfrak{I}$$
and
$$\sum_{i\in\mathbf{G}}E_{i,i+1}+\sum_{j\in\mathbf{Z}_1}E_{j,j+1}=\sum_{i\in\mathbf{G}\cup\mathbf{Z}_1}E_{i,i+1}\in\mathfrak{I}.$$
Analogously
$$[\sum_{j\in\mathbf{Z}_2}E_{j,j-1},\sum_{i\in\mathbf{G}\cup\mathbf{Z}_1}E_{i,i+1}]=\sum_{j\in\mathbf{Z}_2}E_{jj}-\sum_{j\in\mathbf{Z}_2}E_{j-1,j-1}\in
\mathfrak{I}$$
so
$$[\sum_{j\in\mathbf{Z}_2}E_{jj}-\sum_{j\in\mathbf{Z}_2}E_{j-1,j-1},\sum_{i\in\mathbf{Z}_2}E_{i,i+1}]=\sum_{j\in\mathbf{Z}_2}E_{j,j+1}\in\mathfrak{I}$$
and finally
$$\sum_{i\in\mathbf{G}\cup\mathbf{Z}_1}E_{i,i+1}+\sum_{j\in\mathbf{Z}_2}E_{j,j+1}=\sum_{n\in\mathbb{N}}E_{n,n+1}\in\mathfrak{I}.$$
\par
Step 5: We show that if $\sum_{i\in\mathbb{N}}E_{i,i+1}\in\mathfrak{I}$ then $\mathfrak{I}=\mathfrak{gl}_{cf}$.\\
We show that equation $[X,\sum_{i\in\mathbb{N}}E_{i,i+1}]=A$ has a solution for all $A\in\mathfrak{gl}_{cf}$. This equation is equivalent to a system of equations:
$$\begin{cases}
x_{m,n-1}-x_{m+1,n}=a_{m,n}\\
x_{m+1,1}=a_{m,1}
\end{cases}\mbox{  } n,m\in\mathbb{N},1<n,$$
which has, for example, the solution
$$
\begin{cases}
x_{1n}=0\mbox{  } n\in\mathbb{N},\\
x_{m+1,1}=a_{m,1}\mbox{  } m\in\mathbb{N},\\
x_{m+1,n}=x_{m,n-1}-a_{mn}\mbox{  } m,n\in\mathbb{N},1<n.
\end{cases}
$$
\end{proof} \qed

\noindent {\sc Proof of Theorem 1.1.}

We have $\dim \{0\}=0$ and $\dim\mathfrak{d}_{sc}=1$ so there are no proper ideal in interval $\langle\{0\},\mathfrak{d}_{sc}\rangle$.
Similarly, since $\mathfrak{sl}_{fr}$ in $\mathfrak{d}_{sc} \oplus \mathfrak{sl}_{fr}$, $\mathfrak{gl}_{fr}$ in $\mathfrak{d}_{sc} \oplus \mathfrak{gl}_{fr}$ and
$\mathfrak{sl}_{fr}$ in $\mathfrak{gl}_{fr}$ have codimension $1$, it follows that intervals
 $\langle\mathfrak{sl}_{fr}, \mathfrak{d}_{sc} \oplus \mathfrak{sl}_{fr}\rangle$, $\langle\mathfrak{gl}_{fr}, \mathfrak{d}_{sc} \oplus \mathfrak{gl}_{fr}\rangle$ and
 $\langle\mathfrak{sl}_{fr}, \mathfrak{gl}_{fr}\rangle$
do not contain proper ideals. From Lemma $2.2$ and $2.3$  it follows that the same holds for intervals
$\langle\{0\}, \mathfrak{sl}_{fr}\rangle$, $\langle\mathfrak{d}_{sc} \oplus \mathfrak{gl}_{fr}, \mathfrak{gl}_{cf}\rangle$.
The theorem is proved.   \qed

\bigskip

The Corollary $1.2.$ follows from homomorphism theorems for Lie algebras. In [HKZ] it was shown that $\sum_{i=1}^{\infty} E_{i,i+2}$ belongs to the derived Lie subalgebra of $\mathfrak{n}(\mathbb{N}, R)$ and so to the derived Lie subalgebra of $\mathfrak{gl}_{cf}(\mathbb{N},R)$ which shows the first part of Corollary $1.3$. The second part is the obvious consequence of Theorem $1.1.$

\section{Derivations of column-finite infinite matrices}

In this section by $R$ we denote a commutative, associative unital ring. The $R$-linear  map
$$\varphi:  \mathfrak{gl}_{cf}(\mathbb{N}, R)  \rightarrow  \mathfrak{gl}_{cf}(\mathbb{N}, R)$$
is called a derivation if
$$\varphi ([X,Y])=[\varphi (X),Y]+[X,\varphi (Y)].$$

The set
$Der(\mathfrak{gl}_{cf}(\mathbb{N}, R))$ of all derivations of $\mathfrak{gl}_{cf}(\mathbb{N}, R)$
forms a Lie algebra.
For a fixed
$A \in \mathfrak{gl}_{cf}(\mathbb{N}, R)$
the map
$ \ad  A: X \rightarrow [A,X] $
is a derivation, called inner derivation. A derivation is called central  if $Im(\varphi)$ is a subset of the center.

By $\mathcal{ND}(\mathbb{N},R)$ we denote a subset  of $\mathfrak{gl}_{cf}(\mathbb{N}, R)$  of all matrices which have only zeros on the  main diagonal and by $\mathfrak{d}=\mathfrak{d}(\mathbb{N}, R)$ the Lie subalgebra of all diagonal matrices. Any matrix $X \in \mathfrak{gl}_{cf}(\mathbb{N}, R)$  is a sum of two matrices $X= D+N$ where $D$ is diagonal and $N \in\mathcal{ ND}(\mathbb{N}, R)$.

\begin{lem}
For each derivation $\varphi \in Der(\mathfrak{gl}_{cf}(\mathbb{N}, R))$ there exists  $A$ in $\mathcal{ND}(\mathbb{N}, R)$ such that $(\varphi -\ad A)(\mathfrak{d})\subseteq  \mathfrak{d}$.
\end{lem}
\begin{proof}

 We denote $\varphi(E_{kk})=[h^{(k)}_{ij}]$. Let $\varphi(D)=C$ where $D\in \mathfrak{d}$ and $C\in \mathfrak{gl}_{cf}(\mathbb{N}, R)$\\
Step 0: We prove that $h^{(k)}_{ij}$ can be different from zero only if $i=j$ or $i=k$ or $j=k$.\\
For any $d\in D$ we have $[d,E_{kk}]=0$ so
\begin{displaymath}\begin{split}
0=\varphi ([d,E_{kk}])&=[d,\varphi (E_{kk})]+[\varphi (d),E_{kk}]=\\
&=d\varphi (E_{kk})-\varphi (E_{kk})d+\varphi (d)E_{kk}-E_{kk}\varphi (d).
\end{split}\end{displaymath}
If $i\neq k$ and $j\neq k$ we have the equality $d_{ii}h^{(k)}_{ij}-h^{(k)}_{ij}d_{jj}=h^{(k)}_{ij}(d_{ii}-d_{jj})=0$. Matrix $d$ is arbitrary,  so we can assume
that $d_{ii}\neq d_{jj}$ if $i\neq j$. In this case we obtain $h^{(k)}_{ij}=0$.\par
Step 1: We prove that $c_{ij}=d_{ii}h^{(i)}_{ij}+d_{jj}h^{(j)}_{ij}$  if $i\not= j$.\\
Let $A=D-d_{ii}E_{ii}-d_{jj}E_{jj}$ and $\varphi (A)=B$.
We have $[E_{ii},A]=0$ so
\begin{displaymath}\begin{split}
0=\varphi ([E_{ii},a])&=[E_{ii},\varphi (a)]+[\varphi (E_{ii}),A]=\\
&=E_{ii}\varphi (A)-\varphi (A)E_{ii}+\varphi (E_{ii})A-A\varphi (E_{ii}).
\end{split}\end{displaymath}
We obtain equality $0=b_{ij}-0+0-0=b_{ij}$.
We have
$$C=\varphi (D)=\varphi (A+d_{ii}E_{ii}+d_{jj}E_{jj})=\varphi (A)+d_{ii}\varphi (E_{ii})+d_{jj}\varphi (E_{jj})$$ so
$c_{ij}=0+d_{ii}h^{(i)}_{ij}+d_{jj}h^{(j)}_{ij}$.\par
\smallskip
Step 2: We prove that $h^{(i)}_{ij}=-h^{(j)}_{ij}$ for all  $i\neq j$.\\

Since
\begin{displaymath}\begin{split}
0&=\varphi ([E_{ii},E_{jj}])=[E_{ii},\varphi (E_{jj})]+[\varphi (E_{ii}),E_{jj}]=\\
&=E_{ii}\varphi (E_{jj})-\varphi (E_{jj})E_{ii}+\varphi(E_{ii})E_{jj}-E_{jj}\varphi (E_{ii})
\end{split}\end{displaymath}
we get $0=h^{(j)}_{ij}+h^{(i)}_{ij}$.\par
\smallskip
Step 3: Denote $\varphi(D)=C=X+Y$ where $X\in \mathfrak{d}$ and $Y\in \mathcal{ND}(\mathbb{N}, R)$. From the previous steps it follows
$$y_{ij}=d_{ii}h^{(i)}_{ij}+d_{jj}h^{(j)}_{ij}=d_{ii}h^{(i)}_{ij}-d_{jj}h^{(i)}_{ij}.$$
Let $A\in \mathcal{ND}(\mathbb{N}, R)$, $a_{ij}=-h^{(i)}_{ij}$. We have
$$Y=D(-A)-(-A)D=AD-DA=[A,D],$$
and so $$\varphi(D)=X+Y=X+[A,D]=X+\ad A (D).$$
Finally
$$\varphi(D)-\ad A (D)=(\varphi-\ad A)(D)=X.$$ \qed
\end{proof}

\begin{lem}
For each derivation $\varphi \in Der(\mathfrak{gl}_{cf}(\mathbb{N}, R))$  there exists  $B\in \mathfrak{gl}_{cf}(\mathbb{N}, R)$ such that $(\varphi -\ad B)(\mathfrak{d})\subseteq \mathfrak{d}$ and $(\varphi -\ad B)(\mathcal{ND}(\mathbb{N}, R))\subseteq \{0\}$.
\end{lem}
\begin{proof}
For each derivation $\varphi$ we have $$(\varphi -\ad A)(\mathfrak{d})\subseteq \mathfrak{d}$$ for some $A\in \mathcal{ND}(\mathbb{N}, R)$. Denote $\psi =\varphi -\ad A$.\par
\smallskip
Step 1: We will prove that $\psi (E_{i, i+1})=x_iE_{i, i+1}$ where $x_i\in R$.\\
For any $D\in \mathfrak{d}$ we have:
\begin{displaymath}\begin{split}
\psi ([D,E_{i, i+1}])&=\psi (DE_{i,i+1}-E_{i, i+1}D)=\\
&=\psi (d_{ii}E_{i, i+1}-d_{i+1,i+1}E_{i, i+1})=\\
&=(d_{ii}-d_{i+1,i+1})\psi (E_{i, i+1}).
\end{split}\end{displaymath}
On the other hand
\begin{displaymath}\begin{split}
\psi ([D,E_{i, i+1}])&=[D,\psi (E_{i, i+1})]+[\psi (D),E_{i, i+1}]=\\
&=D\psi (E_{i, i+1})-\psi (E_{i, i+1})D+\psi (D)E_{i, i+1}-E_{i, i+1}\psi (D).
\end{split}\end{displaymath}
 Denote $\psi (E_{i, i+1})=C$. $\psi (D)$ is a diagonal matrix so $$\psi (D)E_{i, i+1}-E_{i, i+1}\psi (D)=rE_{i, i+1}$$ where $r\in R$.
For $(j,k)\neq (i,i+1)$ we obtain equality $$(d_{ii}-d_{i+1, i+1})c_{jk}=(d_{jj}-d_{kk})c_{jk}.$$ We can assume that $d_{ii}-d_{i+1,i+1}=0$ and $d_{jj}-d_{kk}=1$, then we have $c_{jk}=0$
for $j\not=k$. On the other hand if we take $D=E_{ii}$ we obtain $c_{jj}=0$ for all $j\in\mathbb{N}$.
\par
\smallskip
Step 2: We will prove that $\psi (E_{i j})=(x_i+x_{i+1}+...+x_{j-1})E_{i j}$ for $i<j$.\\
Denote $j=i+k$. We will use mathematical induction on $k$. From the previous step we get equality when $k=1$. We assume that this property is satisfied for some $k$.
For $k+1$ we have
\begin{displaymath}\begin{split}
\psi (E_{i,i+k+1})&=\psi ([E_{i,i+1},E_{i+1,i+k+1}])=\\
&=[E_{i,i+1},\psi (E_{i+1,i+k+1})]+[\psi (E_{i,i+1}),E_{i+1,i+k+1}]=\\
&=E_{i,i+1}\psi (E_{i+1,i+k+1})-\psi (E_{i+1,i+k+1})E_{i,i+1}+\\
&-\psi (E_{i,i+1})E_{i+1,i+k+1}-E_{i+1,i+k+1}\psi (E_{i,i+1})=\\
&=(x_{i+1}+...+x_{i+k}))E_{i,i+1}E_{i+1,i+k+1}-\\
&-(x_{i+1}+...+x_{i+k}))E_{i+1,i+k+1}E_{i,i+1}+\\
&+x_iE_{i,i+1}E_{i+1,i+k+1}-x_iE_{i+1,i+k+1}E_{i,i+1}=\\
&=(x_{i+1}+...+x_{i+k}))E_{i,i+k+1}-0+x_iE_{i,i+k+1}-0=\\
&=(x_i+x_{i+1}+...+x_{i+k})E_{i,i+k+1}.
\end{split}\end{displaymath}
\smallskip
Step 3: We will prove that $\psi (D)= r_D E$ for all $D\in \mathfrak{d}$ and $r_D\in R$.
Let  $\psi (D)=C$. For $i\in\mathbb{N}$ we have
$$\psi ([D,E_{i,i+1}])=\psi ((d_{ii}-d_{i+1,i+1})E_{i,i+1})=(d_{ii}-d_{i+1,i+1})x_i E_{i,i+1}.$$
On the other hand
\begin{displaymath}\begin{split}
\psi ([D,E_{i,i+1}])&=[D,\psi (E_{i,i+1})]+[\psi (D),E_{i,i+1}]=\\
&=D\psi (E_{i,i+1})-\psi (E_{i,i+1})D+\psi (D)E_{i,i+1}-E_{i,i+1}\psi (D)=\\
&=d_{ii}x_i E_{i,i+1}-d_{i+1,i+1}x_i E_{i,i+1}+c_{ii}E_{i,i+1}-c_{i+1,i+1}E_{i,i+1}=\\
&=(d_{ii}-d_{i+1,i+1})x_iE_{ii+1}+(c_{ii}-c_{i+1i+1})E_{ii+1}.
\end{split}\end{displaymath}
We get $c_{ii}-c_{i+1,i+1}=0$ for $i\in\mathbb{N}$, so $\psi (D)=r_D E$ where $r_D\in R$.\par
\smallskip
Step 4: We will prove that $\psi (E_{i j})=-\psi (E_{j i})$ for $i\not= j$.\\
Analogously to steps 1 and 2, we can show that $\psi (E_{j i})=rE_{ji}$ for $j>i$ and some $r\in R$. Denote $\psi (E_{i j})=x_{ij}E_{ij}$ for
$i\not= j$. We have
$$\psi ([E_{ij},E_{ji}])=\psi (E_{ii}-E_{jj})=\psi (E_{ii})-\psi(E_{jj})=r_iE-r_jE.$$
On the other hand
\begin{displaymath}\begin{split}
\psi ([E_{ij},E_{ji}])&=\psi (E_{ij})E_{ji}-E_{ji}\psi (E_{ij})+E_{ij}\psi(E_{ji})-\psi(E_{ji})E_{ij}=\\
&=x_{ij}E_{ii}-x_{ij}E_{jj}+x_{ji}E_{ii}-x_{ji}E_{jj}.
\end{split}\end{displaymath}
In this way we obtain that  $r_i=r_j$ and $$x_{ij}E_{ii}-x_{ij}E_{jj}+x_{ji}E_{ii}-x_{ji}E_{jj}=0.$$ So $x_{ij}+x_{ji}=0$ for $i\not= j$.
\par
\smallskip
Step 5: Let $j>i+5$ and $\beta ^{(i,j)}$ be a set of matrices which can have nonzero entries only in the $i$-th row  in the columns $k\geq j$. We will prove that
$\psi (\beta ^{(i,j)})\subset \beta ^{(i,j)}$.\\
Let $B^{(i,j)}\in \beta ^{(i,j)}$. We have
\begin{displaymath}\begin{split}
\psi(B^{(i,j)})&=\psi([E_{ii},B^{(i,j)}])=[\psi(E_{ii}),B^{(i,j)}]+[E_{ii},\psi(B^{(i,j)})]=\\
&=0+[E_{ii},\psi(B^{(i,j)})]=E_{ii}\psi(B^{(i,j)})-\psi(B^{(i,j)})E_{ii},
\end{split}\end{displaymath}
so $\psi(B^{(i,j)})$ can have nonzero entries only in the $i$-th row and the  $i$-th column. For $k<j$ and $k\not=i-1$ we have
$$\psi([B^{(i,j)},E_{k,k+1}])=\psi(B^{(i,j)}E_{k,k+1}-E_{k,k+1}B^{(i,j)})=\psi(0-0)=0.$$
On the other hand
\begin{displaymath}\begin{split}
&\psi([B^{(i,j)},E_{k,k+1}])=[\psi(B^{(i,j)}),E_{k,k+1}]+[B^{(i,j)},\psi(E_{k,k+1})]=\\
&=\psi(B^{(i,j)})E_{k,k+1}-E_{k,k+1}\psi(B^{(i,j)})+\\
&+x_{k,k+1}(B^{(i,j)}E_{k,k+1}-E_{k,k+1}B^{(i,j)})=\\
&=\psi(B^{(i,j)})E_{k,k+1}-E_{k,k+1}\psi(B^{(i,j)}),
\end{split}\end{displaymath}
so $k$-th column  of the matrix $\psi(B^{(i,j)})$ have only $0$'s in every row. We get $\psi (\beta ^{(1,j)})\subseteq \beta ^{(1,j)}$ and it is enough to show that
$\psi(B^{(i,j)})_{i,i-1}=0$ for $i>1$. We shall use induction to end the proof. We have
$$\psi([B^{(i,j)},E_{i-1,i}])=\psi(B^{(i,j)}E_{i-1,i}-E_{i-1,i}B^{(i,j)})=\psi(0-E_{i-1,i}B^{(i,j)})\subset\beta ^{(i-1,j)}$$
and
\begin{displaymath}\begin{split}
\psi([B^{(i,j)},E_{i-1,i}])=\psi(B^{(i,j)})E_{i-1,i}-E_{i-1,i}\psi(B^{(i,j)})+\\
+x_{i-1,i}(B^{(i,j)}E_{i-1,i}-E_{i-1,i}B^{(i,j)})=\\
\psi(B^{(i,j)})E_{i-1,i}-E_{i-1,i}\psi(B^{(i,j)})+x_{i-1,i}(0-E_{i-1,i}B^{(i,j)}).
\end{split}\end{displaymath}
Since $$-E_{i-1,i}\psi(B^{(i,j)})+x_{i-1,i}(-E_{i-1,i}B^{(i,j)})\subseteq \beta ^{(i-1,j)}$$
we obtain $\psi(B^{(i,j)})E_{i-1,i}=0$ and hence $\psi(B^{(i,j)})_{i,i-1}=0$.
\par
\smallskip
Step 6: Let $N\in \mathcal{ND}(\mathbb{N}, R)$ and $\psi(N)=M$. We will prove that if $n_{ij}=0$ then $m_{ij}=0$.\\
Denote $Y=\psi ([E_{ii},N])$. We have $Y=\psi ([E_{ii},N])=\psi (E_{ii}N-NE_{ii})$. Using step 5 and step 2 we get $$y_{ij}=n_{ij}(x_i+x_{i+1}+...+x_{j-1})=0$$ for $i<j$ or $$y_{ij}=-n_{ij}(x_i+x_{i+1}+...+x_{j-1})=0$$ for $i>j$. On the other hand
\begin{displaymath}\begin{split}
\psi ([E_{ii},N])&=[E_{ii},\psi (N)]+[\psi (E_{ii}),N]=\\&
E_{ii}\psi (N)-\psi (N)E_{ii}+\psi (E_{ii})N-N\psi (E_{ii}).
\end{split}\end{displaymath}
But $\psi (E_{ii})\in \mathfrak{d}$,  hence we have $y_{ij}=m_{ij}-0+0-0=m_{ij}$.\par
\smallskip
Step 7: Let $Y\in \mathcal{ND}(\mathbb{N}, R)$ and $\psi (Y)=Z$. For $i<j$ we have $z_{ij}=y_{ij}(x_i+x_{i+1}+...+x_{j-1})$ and $z_{ji}=-y_{ji}(x_i+x_{i+1}+...+x_{j-1})$.\\
Let $D\in \mathfrak{d}$ and $d_{11}=0$, $d_{ii}=-(x_1+x_2+...+x_{i-1})$ for $i>1$. We have
\begin{displaymath}\begin{split}
z_{ij}&=y_{ij}(x_i+x_{i+1}+...+x_{j-1})=\\
&=y_{ij}(x_1+x_2+...+x_{j-1})-y_{ij}(x_1+x_2+...+x_{i-1})=\\
&=-d_{jj}y_{ij}+d_{ii}y_{ij}=d_{ii}y_{ij}-d_{jj}y_{ij}
\end{split}\end{displaymath}
 and
\begin{displaymath}\begin{split}
z_{ji}&=-y_{ji}(x_i+x_{i+1}+...+x_{j-1})=\\
&=-y_{ji}(x_1+x_2+...+x_{j-1})+y_{ij}(x_1+x_2+...+x_{i-1})=\\
&=d_{jj}y_{ji}-d_{ii}y_{ji}
\end{split}\end{displaymath}
 so $\psi (Y)=Z=[D,Y]$.

This means that for all
$N\in \mathcal{ND}(\mathbb{N}, R)$ we have $\varphi (N)-[A,N]=[D,N]$ and  so $$\varphi (N)-[A,N]-[D,N]=\varphi (N)-[A+D,N]=(\varphi -ad B)(N)=0$$ where $B=A+D$. \qed
\end{proof}

\begin{lem}
If $\varphi$ is a derivation of $\mathfrak{gl}_{cf}(\mathbb{N}, R)$ such that   $\varphi(D)\subseteq D$ and $\varphi(\mathcal{ND}(\mathbb{N}, R))=\{0\}$ then $\varphi$ is a central derivation.
\end{lem}
\begin{proof}
Let  $\varphi (D)=C$. For $i\in\mathbb{N}$ we have
$$\varphi ([D,E_{ii+1}])=\varphi ((d_{ii}-d_{i+1,i+1})E_{ii+1})=0.$$
On the other hand
\begin{displaymath}\begin{split}
\varphi ([D,E_{i,i+1}])&=[D,\varphi (E_{i,i+1})]+[\varphi (D),E_{i,i+1}]=\\
&=D\varphi (E_{i,i+1})-\varphi (E_{i,i+1})D+\varphi (D)E_{i,i+1}-E_{i,i+1}\varphi (D)=\\
&=0-0+c_{ii}E_{i,i+1}-c_{i+1,i+1}E_{i,i+1}=(c_{ii}-c_{i+1,i+1})E_{i,i+1}.
\end{split}\end{displaymath}
We get $c_{ii}-c_{i+1,i+1}=0$ for $i\in\mathbb{N}$, so $\varphi (D)=r_D E$ where $r_D\in R$.\par
We have $\varphi (D)=r_D E$ where $r_D\in R$. This means that there  exists a homomorphism $\sigma : \mathfrak{d}\rightarrow R$ of
$R$-modules, defined by $\sigma (D)=r_D$. We have
$$\varphi(A)=\varphi (D+N)=\varphi (D)=\sigma (D)E$$
 for $A\in \mathfrak{gl}_{cf}(\mathbb{N}, R)$, $D\in \mathfrak{d}$, $N\in \mathcal{ND}(\mathbb{N}, R)$. Hence $\varphi $ is the
central derivation of
$\mathfrak{gl}_{cf}(\mathbb{N},R)$
induced by $\sigma $.
\end{proof}

\noindent {\sc Proof of Theorem 1.4.}

By previous Lemmas for each derivation $\varphi $ there exists $B \in \mathfrak{gl}_{cf}(\mathbb{N},R)$ such that $\varphi - \ad B=\hat{\varphi}$,
where $\hat{\varphi} $ is central derivation, so $\varphi = \ad B+\hat{\varphi} $.
\qed

\bigskip

Now we prove
\begin{prop}
Lie algebras $\mathfrak{gl}_{cf}(\mathbb{N},R)$ i $\mathfrak{gl}_{cf}(\mathbb{Z},R)$ are isomorphic.
\end{prop}
\begin{proof}
We recall that Lie algebra $\mathfrak{gl}_{cf}(\mathbb{Z},R)$ consists of column-finite infinite $\mathbb{Z} \times \mathbb{Z}$ matrices.

We define a function $\sigma :\mathbb{Z}\rightarrow \mathbb{N}$ as follows
$$\sigma (x)=\begin{cases} 2x+1\textrm{ dla } x\geq 0\\ -2x\textrm{\,\,\,\, \,\,dla } x< 0\end{cases}.$$
$\sigma$ is the bijection and induces the  bijection  $\phi : M_{cf}(\mathbb{Z},K)\rightarrow M_{cf}(\mathbb{N},K)$ given by
$$\phi (\sum\limits_{i,j\in\mathbb{Z}}a_{ij}E_{ij})=\sum\limits_{i,j\in\mathbb{Z}}a_{ij}E_{\sigma (i)\sigma (j)}.$$
We have
$$\phi(cA)=\sum\limits_{i,j\in\mathbb{Z}}ca_{ij}E_{\sigma (i)\sigma (j)}=c\sum\limits_{i,j\in\mathbb{Z}}a_{ij}E_{\sigma (i)\sigma (j)}=cf(A),$$
\begin{align*}
\phi(A+B)&=\sum\limits_{i,j\in\mathbb{Z}}(a_{ij}+b_{ij})E_{\sigma (i)\sigma (j)}=\\
&=\sum\limits_{i,j\in\mathbb{Z}}a_{ij}E_{\sigma (i)\sigma (j)}+\sum\limits_{i,j\in\mathbb{Z}}b_{ij}E_{\sigma (i)\sigma (j)}=\phi(A)+\phi(B),
\end{align*}
\begin{align*}
\phi(AB)&=\phi (\sum\limits_{i,j\in\mathbb{Z}}(\sum\limits_{n=-\infty}^\infty a_{in}b_{nj})E_{ij})=
\sum\limits_{i,j\in\mathbb{Z}}(\sum\limits_{n=-\infty}^\infty a_{in}b_{nj})E_{\sigma (i)\sigma (j)}=\\
&=\sum\limits_{k,m\in\mathbb{N}}(\sum\limits_{n=-\infty}^\infty a_{\sigma ^{-1}(k)n}b_{n\sigma ^{-1}(m)})E_{km}=\\
&=(\sum\limits_{k,m\in\mathbb{N}}a_{\sigma ^{-1}(k)\sigma ^{-1}(m)}E_{km})(\sum\limits_{k,m\in\mathbb{N}}b_{\sigma ^{-1}(k)\sigma ^{-1}(m)}E_{km})=\\
&=(\sum\limits_{i,j\in\mathbb{Z}}a_{ij}E_{\sigma (i)\sigma (j)})(\sum\limits_{i,j\in\mathbb{Z}}b_{ij}E_{\sigma (i)\sigma (j)})=\phi(A)\phi(B) ,
\end{align*}
and so
$$\phi([A,B])=[\phi(A),\phi(B)].$$
It means that  $\phi $ defines the isomorphism of Lie algebras $\mathfrak{gl}_{cf}(\mathbb{N},R)$ and $\mathfrak{gl}_{cf}(\mathbb{Z},R)$. \qed
\end{proof}

config: arxiv


TITLE: Change of Variables: which order of integration limits?
QUESTION [4 upvotes]: An exercise in Larson/Edwards Calculus (10th ed.) asks the reader to evaluate the double integral
$$\int_R\int 4(x^2+y^2)\,\mathrm{d}A$$
using a given change of variables
$$x=\frac{1}{2}(u+v)\mathrm{,}\quad y=\frac{1}{2}(u-v)$$
over the region pictured below.

Since, with the given change of variables, $\,v=x-y\,$ and $\,u=x+y$, it looks like my $v$ limits of integration are going to be $-1$ and $1$, and my $u$ limits of integration are going to be $-1$ and $1$ as well.  My question is: how do I know which number should be the upper limit of integration and which number should be the lower limit of integration? Going off the picture, it looks like I should be integrating from $v=1$ to $v=-1$, and from $u=-1$ to $u=1$, so my integral would look like this:
$$\int_{1}^{-1}\int_{-1}^{1}f(u,v)\,\mathrm{d}u\,\mathrm{d}v$$
This gives me the opposite of the correct answer (the correct answer multiplied by $-1$). I know I get the correct answer by integrating
$$\int_{-1}^{1}\int_{-1}^{1}f(u,v)\,\mathrm{d}u\,\mathrm{d}v$$
instead, but I don't see any reason why I ought to put the limits in that order. I suppose it might always be correct to choose the least of the two values to be the lower limit and the greatest of the two values to be the upper limit (thought I'm not sure), but even if that were so I still wouldn't really understand why that's the case.
Edit: I tried writing the original integral without changing the variables, like this:
$$\int_{-1}^{0}\int_{-x-1}^{x+1}g(x,y)\,\mathrm{d}y\,\mathrm{d}x+\int_0^1\int_{x-1}^{-x+1}g(x,y)\,\mathrm{d}y\,\mathrm{d}x$$
and then manipulating the inequalities
$$-x-1\leq y\leq x+1\quad\mathrm{and}\quad x-1\leq y \leq -x+1$$
to find $-1\leq x+y\mathrm{,}$ $-1\leq x-y\mathrm{,}$ $x-y\leq 1\mathrm{,}$ and $x+y\leq 1$. Substituting $v$ for $x-y$ and $u$ for $x+y$, I get $-1\leq v\leq 1$ and $-1\leq u\leq 1$. Basing my limits of integration off these inequalities seems to work, but I'm not sure this procedure for finding limits of integration works in general.

REPLY [1 votes]: No, you could not always just use the least value as the lower limit:  See what would happen if you had chosen the substitution $v = y-x$ instead of $x-y$.
The correct reasoning is the following:  You want the area "above" (in $u$) the line $x+y = -1$ yet "below" (in $u$)  the line $x+y = +1$.  Here "above" and "below" mean greater that $u$ is greater than (less than) $-1$ (or $+1$).  Now on  $x+y = -1$ we have $u=-1$ so the lower limit in $u$ is $-1$.  Similarly the upper limit is $+1$.
The best way to see that is that the area you want has $-1 \leq x+y \leq +1$ so the lower limit is $-1$ and the upper limit is $+1$.
Now consider $v = x-y$: You want the area with $-1 \leq x-y \leq +1$ so again the integral goes from $-1$ to $+1$.
To see if you understand it, try using a dfferent $v = y-x$ and see that you get the correct limits then.


TITLE: Find $f(x) = \lim_{n\to\infty}\frac{\lfloor x \rfloor + \cdots + \lfloor x^n \rfloor}{x^n}$
QUESTION [3 upvotes]: I found this cool problem in a textbook. I googled it and used MSE's search tool to check if it has been asked before or not, but it seems that it hasn't been asked before.
Find the function $f(x)$  that the following limit defines: $$f(x) = \lim_{n\to\infty}\frac{\lfloor x \rfloor + \cdots + \lfloor x^n \rfloor}{x^n}$$
I have already solved it and I have shared my solution as an answer. Other solutions are welcome too.

REPLY [2 votes]: Using $\lfloor t\rfloor=t-\{t\}$, the numerator is a geometric progression from $x$ to $x^n$, minus a number that does not exceed $n$.
Hence, for $|x|>1$,
$$\frac {x^{n+1}-x}{(x-1)x^n}\to\frac x{x-1}$$ while $$-\frac n{x^n}\to0.$$
For $x=1$, $\dfrac n1\to\infty$.
For $0<x<1$, $\dfrac{n\cdot0}{x^n}\to0$.
For $x=0$, not defined.
For $-1<x<0$, $-\dfrac n{2x^n}\to-\infty$.
For $x=-1$, alternatively $-\dfrac1{x^n}$ and $0$, undefined.