我正在尝试编译我的文档，但是我反复收到如下错误：
Package listofitems Error: Empty list ignored, nothing to do. \end{align}
与该错误相对应的行是：
\begin{align}
\mathbf{A}^{n+1}=\underbrace{\left[\begin{array}{c}\mathbf{\bar{Z}} \\ \hline \mathbf{0}\end{array}\right]}_{n}
\setstackgap{L}{1.2 \normalbaselineskip}
\vcenter{\hbox{\stackunder[0.1pt]{\left.{\Centerstack{}}\right\}\scriptstyle n}{\left.{\Centerstack{}}\right\}\scriptstyle m}}}
\end{align}
我不知道这有什么问题。需要的输出如下：
相同的文件在我的另一台计算机上运行良好，这就是我卸载并重新安装 latex 的原因，但这个错误一直出现。我主文件的序言如下所示：
\documentclass[twoside]{iitbreport}
%% Default spacing: 1.5
%% Default font size: 12pt
%% Default font: txfonts (similar to times new roman)
%% Selectively comment out sections that you want to be left out but
%% maintaining the page numbers and other \ref
\includeonly{%
intro/introduction_v2,
lit/literature,
chapter/chapter1v1,
chapter/chapter2v1,
chapter/chapter3v1,
chapter/chapter4v1,
chapter/conclusion,
rnd/results,
dec,abs,pub,ack
}
%%% Some commonly used packages (make sure your LaTeX installation
%%% contains these packages, if not ask your senior to help installing
%%% the packages)
\usepackage{booktabs}
\graphicspath{{chapter/}}
\usepackage{chemformula} % Formula subscripts using \ch{}
\usepackage[T1]{fontenc} % Use modern font encodings
%\usepackage{amsthm}
\usepackage{blindtext}
\usepackage{graphicx}
\usepackage{chemist}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{epstopdf}
\epstopdfsetup{update} % only regenerate pdf files when eps file is newer
\usepackage{amsfonts}
\usepackage{float}
\usepackage{mathtools}
\usepackage{enumerate}
\usepackage{color}
\usepackage{subfig}
\usepackage{algorithm}
\usepackage{algpseudocode}
\usepackage{pifont}
\usepackage{mathrsfs}
\usepackage{arydshln}
\usepackage{amsbsy}
%\usepackage{booktabs}
\usepackage{dsfont}
\usepackage{graphicx}
\usepackage{url}
\usepackage{xcolor}
\usepackage{wasysym}
%\usepackage{geometry}
\usepackage{easybmat}
\usepackage{multirow,bigdelim}
\usepackage[usestackEOL]{stackengine}
\usepackage{listofitems}
\usepackage{tabstackengine}
\stackMath
\usepackage[utf8]{inputenc}
\allowdisplaybreaks
%%% Macro definitions for Commonly used symbols
\newcommand{\Rey}{\ensuremath{\mathrm{Re}}}
\newcommand{\avg}[1]{\ensuremath{\overline{#1}}}
\newcommand{\tenpow}[1]{\ensuremath{\times 10^{#1}}}
\newcommand{\pder}[2]{\ensuremath{\frac{\partial#1}{\partial#2}}}
\newtheorem{lemma}{Lemma}
\numberwithin{lemma}{section}
\newtheorem{remark}{Remark}
\numberwithin{remark}{section}
\newtheorem{theorem}{Theorem}
\numberwithin{theorem}{section}
\newtheorem{proof}{Proof}
\numberwithin{proof}{section}
%\theoremstyle{plain}
\newtheorem{assumption}{Assumption}
\numberwithin{assumption}{section}
% Referencing macros
\newcommand{\Eqref}[1]{Equation~\eqref{#1}}
\newcommand{\Tabref}[1]{Table~\ref{#1}}
\newcommand{\Figref}[1]{Figure~\ref{#1}}
\newcommand{\Appref}[1]{Appendix~\ref{#1}}
对应错误的行上方的行如下：
\section{Householder Transformation for Square root filtering}
\label{ap:householder}
In this section we briefly discuss the Householder transformation technique to implement square root filtering in EGPOF (Section \ref{sc:online_algo_EGPOF}) for improved numerical precision. Detailed discussion for the same can be found in \cite{Maybeck}.
The a prior covariance matrix for EGPOF is $\mathbf{P}(k+1|k)$, and its square root is $\mathbf{P_{sq}}(k+1|k)$ such that $\mathbf{P}(k+1|k)=\mathbf{P_{sq}}(k+1|k)[\mathbf{P_{sq}}(k+1|k)]^T$. The aposterior error covariance matrix is $\mathbf{P}(k+1|k+1)$, and its square root is $\mathbf{P_{sq}}(k+1|k+1)$. Also, square root of $\left[\mathbf{\bar{Q}}-\mathbf{S}[\mathbf{R}]^{-1}[\mathbf{S}]^T\right]$ is given by $\mathbf{Q_{sq}}$ such that $\left[\mathbf{\bar{Q}}-\mathbf{S}[\mathbf{R}]^{-1}[\mathbf{S}]^T\right]=\mathbf{Q_{sq}}[\mathbf{Q_{sq}}]^T$. Then using Eq. \eqref{eq:pred_cov_lin}, the predicted covariance can be written as:
\begin{align}
\label{eq:pred_cov_household}
\mathbf{P}(k+1|k) &=\boldsymbol{\Phi}(k)\mathbf{P_{sq}}(k|k)[\mathbf{P_{sq}}(k|k)]^T\boldsymbol{\Phi}(k)^T+\boldsymbol{\Gamma}(k)\mathbf{Q_{sq}}[\mathbf{Q_{sq}}]^T[\boldsymbol{\Gamma}(k)]^T
\end{align}
Now it is desired to find the propagation equation for the square root of $\mathbf{P}(k+1|k)$; to find the relation to yield $\mathbf{P_{sq}}(k+1|k)$ such that $\mathbf{P_{sq}}(k+1|k)[\mathbf{P_{sq}}(k+1|k)]^T$ is equal to the right hand side of Eq. \eqref{eq:pred_cov_household}. Suppose that we can find an orthogonal $(n+m)\times (n+m)$ matrix $\mathbf{\tilde{T}}$ i.e. $\mathbf{\tilde{T}}[\mathbf{\tilde{T}}]^T=\mathbf{I}$ such that
\begin{align}
\left[\begin{array}{c:c}
\boldsymbol{\Phi}(k)\mathbf{P_{sq}}(k|k) & \boldsymbol{\Gamma}(k)\mathbf{Q_{sq}}
\end{array}\right]\mathbf{\tilde{T}}&=\left. \left[\begin{array}{c:c}\underbrace{\mathbf{P_{sq}}(k+1|k)}_{\text{n columns}} & \underbrace{\mathbf{0}}_{\text{m columns}} \end{array}\right]\right\}\text{n rows}
\end{align}
then in fact an $n \times n$ square root matrix $\mathbf{P_{sq}}(k+1|k)$ will have been found which satisfies the desired relationship. Further, note that the same procedure could also be applied to:
\begin{align}
\mathbf{P}(k+1|k+1)=\big[\mathbf{I}-\mathbf{L}(k+1)\mathbf{C}(k+1)\big] \mathbf{P}(k+1|k)
\big[\mathbf{I}-\mathbf{L}(k+1)\mathbf{C}(k+1)\big]^T + \mathbf{L}(k+1) \mathbf{R} [\mathbf{L}(k+1)]^T
\end{align}
where, we find an orthogonal $(n+r)\times (n+r)$ matrix $\mathbf{\tilde{T}}$ i.e. $\mathbf{\tilde{T}}[\mathbf{\tilde{T}}]^T=\mathbf{I}$ such that
\begin{align}
\left[\begin{array}{c:c}
\big[\mathbf{I}-\mathbf{L}(k+1)\mathbf{C}(k+1)\big]\mathbf{P_{sq}}(k+1|k) & \mathbf{L}(k+1) [\mathbf{R}]^{1/2}
\end{array}\right]\mathbf{\tilde{T}}&=\left. \left[\begin{array}{c:c}\underbrace{\mathbf{P_{sq}}(k+1|k+1)}_{\text{n columns}} & \underbrace{\mathbf{0}}_{\text{r columns}} \end{array}\right]\right\}\text{n rows}
\end{align}
One of the method of generating such a $\mathbf{P_{sq}}(k+1|k)$ and $\mathbf{P_{sq}}(k+1|k+1)$ is Householder transformation technique. Conceptually, it generated $\mathbf{\tilde{T}}$ as
\begin{align}
\mathbf{\tilde{T}}=\mathbf{\tilde{T}}^n\mathbf{\tilde{T}}^{(n-1)}\ldots \mathbf{\tilde{T}}^1
\end{align}
where $\mathbf{\tilde{T}}^k$ is generated recursively as
\begin{align}
\mathbf{\tilde{T}}^k=\mathbf{I}-d^k\mathbf{u}^k[\mathbf{u}^k]^T
\end{align}
with the scalar $d^k$ and the $(n+m)$-vector $\mathbf{u}^k$ defined in the following. However, the computational algorithm never calculates these $\mathbf{\tilde{T}}^k$'s or $\mathbf{\tilde{T}}$ explicitly. The initial condition on the $(n+m)\times n$ $\mathbf{A}^k$ is
\begin{align}
\mathbf{A}^1&=[\mathbf{P_{sq}}(k+1|k)]^T=\left[\begin{array}{c:c}
\boldsymbol{\Phi}(k)\mathbf{P_{sq}}(k|k) & \boldsymbol{\Gamma}(k)\mathbf{Q_{sq}}
\end{array}\right]^T
\end{align}
Now letting $\mathbf{A}^k_j$ represent the $j^{th}$ column of $\mathbf{A}^k$, perform the $n$-step recursion, for $k=1,2,\ldots,n$:
\begin{align}
a^k &=\sqrt{\sum_{j=k}^{n+m}[A^k_{jk}]^2}\text{sign}\{A^k_{kk}\} \nonumber \\
d^k&=\frac{1}{a^k(a^k+A^k_{kk})} \nonumber \\
u_j^k&=\begin{cases}
0& j< k\\
a^k+A^k_{kk}& j=k \\
A^k_{jk} & j=(k+1),\ldots ,(n+m)
\end{cases} \nonumber \\
y^k_j&=\begin{cases}
0 & j<k \\
1 & j=k \\
d^k[\mathbf{u}^{k}]^T\mathbf{A}^k_j & j=(k+1), \ldots , n
\end{cases} \nonumber\\
\mathbf{A}^{k+1}&=\mathbf{A}^k-\mathbf{u}^k [\mathbf{y}^{k}]^T
\label{eq:iter}
\end{align}
At stage $k$, the first $(k-1)$ columns of $\mathbf{A}^k$ are zero below the diagonal of the upper square partition, and $\mathbf{u}^k$ has been chosen so that the sub-diagonal elements of $\mathbf{A}^{k+1}_k$ will be zero. After the $n$ iterations of Eq. \eqref{eq:iter},
\begin{align}
\mathbf{A}^{n+1}=\underbrace{\left[\begin{array}{c}\mathbf{\bar{Z}} \\ \hline \mathbf{0}\end{array}\right]}_{n}
\setstackgap{L}{1.2 \normalbaselineskip}
\vcenter{\hbox{\stackunder[0.1pt]{\left.{\Centerstack{}}\right\}\scriptstyle n}{\left.{\Centerstack{}}\right\}\scriptstyle m}}}
\end{align}
and then $\mathbf{P_{sq}}(k+1|k)$ is generated as
\begin{align}
\mathbf{P_{sq}}(k+1|k)=\left[\mathbf{\bar{Z}}\right]^T
\end{align}