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\begin{document}
\newtheorem{dingli}{Theorem}[section]
\newtheorem{dingyi}[dingli]{Definition}
\newtheorem{tuilun}[dingli]{Corollary}
\newtheorem{zhuyi}[dingli]{Remark}
\newtheorem{yinli}[dingli]{Lemma}
\title{ The absolute values and support projections for a class of operator matrices
involving idempotents
\thanks{ This work was supported by NSF of
China (Nos: 11671242, 11571211) and the
Fundamental Research Funds for the Central Universities
(GK201801011).}}
\author{  Yuan Li\thanks{E-mail address:
 liyuan0401@aliyun.com}, \ Xiaomei Cai, \ Shuaijie Wang }
\date{} \maketitle\begin{center}
\begin{minipage}{16cm}
{ \small $$    \ \ \   \ College \ of \ Mathematics \ and \
Information \ Science,\ Shaanxi \ Normal \ University, $$ $$ Xi'an,
710062, People's \ Republic \ of\ China. $$ }
\end{minipage}
\end{center}
 \vspace{0.05cm}
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\begin{minipage}{16cm}
\ {\small {\bf Abstract }
 Let $\lambda\in \mathbb{R},$ $\mu\in \mathbb{R}$ and $B$ be a linear bounded operator from a Hilbert space $\mathcal{K}$ into another Hilbert space $\mathcal{H}.$ In this paper, we consider the formulas of the absolute value $|Q_{\lambda,\mu}|,$ where $Q_{\lambda,\mu}$ with respect to the decomposition  $\mathcal{H}\oplus\mathcal{K}$ have the operator matrix
  form $Q_{\lambda,\mu}:=\left(\begin{array}{cc}\lambda I&B\\B^*&\mu I\end{array}\right).$ Then the positive part and the support projection of  $Q_{\lambda,0}$  are obtained. Also, we characterize the symmetry $J$ such that a projection $E$ is the $J$-projection.
In particular, the minimal element of the set of all symmetries $J$ with the property $JE\geqslant0$ is described.
\\}
\endabstract
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\begin{minipage}{16cm}
{\bf  Keywords}: The positive part, Support projection, $J$-projection \\
{\bf  Mathematics Subject Classification}: 47A05,47B65,46C20\\
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\end{center}

\section{ Introduction}
Let $\mathcal{H}$ and $\mathcal{K}$ be complex Hilbert spaces,
$\mathcal{B(K,H)}$ be the set of all bounded linear operators from $\mathcal{K}$ into $\mathcal{H}.$
For an operator $A\in \mathcal{B(H,K)},$ we denote by $|A|$ the absolute value of operator $A,$
that is $|A|=(A^{\ast} A)^{\frac{1}{2}},$ where $A^*$ is the adjoint operator of $A.$
We write $A\geqslant0$ if $A$ is a positive operator, meaning $\langle Ax,x\rangle \geqslant 0,$
where we denote by $\langle,\rangle$ the inner product of $\mathcal{H}.$  Also, denote by $\mathcal{B({H})}^{+}$ the set of all positive bounded linear operators on $\mathcal{H}.$ If $A \in\mathcal{B({H})}^{+},$
then $A^{\frac{1}{2}} $ denotes the positive square root of $A.$
For an operator $T\in \mathcal{B(H,K)},$ we use $N(T),$ $R(T)$ and $\overline{R(T)}$ to
 denote the null
 space, the range of $T,$ and the closure of $R(T),$  respectively.
 If $\mathcal{M}$ is a closed subspace of $\mathcal{H},$ then
 $P_{\mathcal{M}}$ denotes the orthogonal projection onto $\mathcal{M}.$
Particularly, we use $P_{A}$ to denote the orthogonal projection onto
$\overline{R(A)}.$ Also, $P_{A}$ is said to be the support projection of $A$ and
 $E_A(\cdot)$ denotes the spectral measure of $A$
 for a self-adjoint operator $A\in\mathcal{B(H)}.$

As usually, the operator order (Loewner partial order) relation $A\geqslant B$ between two bounded self-adjoint operators is defined as $A-
B\geqslant 0.$
Let $A^+:=\frac{|A|+A}{2}$ and $A^-:=\frac{|A|-A}{2}$ be the positive and negative parts of a
self-adjoint operator $A\in\mathcal{B(H)}.$ Then the self-adjoint operator
$A$ can be written as $A=A^+-A^-$ with $A^+A^-=0.$

It is well known that every operator $T\in \mathcal{B(H)}$ has a (unique) polar decomposition
 $T = U(T^*T)^\frac{1}{2}$
where $U$ is a partial isometry from $N(T)^\perp$ onto $\overline{R(T)}$
with kernel space $N(T)$(See [5]).
 Also, we write $A\simeq B$ to mean that the operator $A$ and $B$ are unitarily
equivalent $(A=UBU^*)$ and $\mathcal{M}\simeq \mathcal{N}$ denotes
$\mathcal{M}=U\mathcal{N}$ for some unitary operator $U,$ where
$\mathcal{M}$ and $\mathcal{N}$ are closed subspaces of $\mathcal{H}.$

An operator $J\in \mathcal{B(H)}$ is said to be a symmetry if $J=J^*=J^{-1}.$
In this case, the operators $J^+:=\frac{I+J}{2}$ and $J^-:=\frac{I-J}{2}$ are
mutually annihilating orthogonal projections. Thus there is a bijection between
the set of all orthogonal projections and all symmetries. If $J$ is a non-scalar symmetry, then
 the form \[[x,y]:=\langle Jx, y\rangle \qquad(x,y\in \mathcal{H})\] is called an indefinite metric and
 $\mathcal{H}$ with $[.,.]$ is said to be a Krein space.

Let us denote by $\mathcal{B(H)}^{Id}$ the set of all projections (idempotents) on $\mathcal{B(H)}.$
 If $E\in \mathcal{B(H)}^{Id},$ then $E$ can be written as a $2\times2$ operator matrix: \begin{equation} E=\left(\begin{array}{cc}I&E_1\\0&0\end{array}\right): R(E)\oplus R(E)^{\perp}\rightarrow R(E)\oplus R(E)^{\perp}\end{equation}
where $E_1\in {\mathcal{B}}(R(E)^{\perp}, R(E)).$ Thus  \begin{equation} E+E^*=\left(\begin{array}{cc}2I&E_1\\E_1^*&0\end{array}\right).\end{equation}
Also, a projection $E$ is said to be $J$-projection if $E=JE^{\ast}J.$
Recently, some interesting decomposition properties of projections and $J$-projections were studied in [3,6,10,12]. An exposition of the properties of the Krein spaces can be found in the book by T. Ya. Azizov and I.S. Iokhvidov [1]. In particular, the
problem of the existence of $J$-projections and its properties are studied in [7-9,11].

In general, with respect to the decomposition $ \mathcal{H}\oplus\mathcal{K},$
 a self-adjoint operator $Q\in \mathcal{B(H\oplus K)}$ has the following matrix representation
$$Q=\left(\begin{array}{cc}Q_1&Q_2\\Q_2^*&Q_3\end{array}\right),$$ where $Q_1$ and $Q_3$ are self-adjoint operators and $Q_2\in \mathcal{B(K,H)}.$
However, the expression of $|Q|$ (in terms of some formulas involving $Q_1,$ $Q_2$ and $Q_3$) seems difficult.
In section 2, we mianly consider how to give the expression of $|Q|,$ on the assumpton of
$Q_1=\lambda I$ and $Q_3=\mu I,$ where $\lambda\in\mathbb{R}$ and $\mu\in\mathbb{R}$ are real numbers.
That is, we study the concrete form of $|Q_{\lambda,\mu}|,$
according to the formulas of
$\lambda,\mu,$ and $B,$ where $$Q_{\lambda,\mu}:=\left(\begin{array}{cc}\lambda I&B\\B^*&\mu I\end{array}\right):\mathcal{H}\oplus\mathcal{K}\rightarrow\mathcal{H}\oplus\mathcal{K}.$$
 The motivation of our research $|Q_{\lambda,\mu}|$  is based on calculating $|E+E^*|$ in (1.2).

In section 3, we mainly give the formulas of the absolute values and support projections
for operator matrices, which have analogous forms as (1.2). In section 4,
 we mainly characterize the symmetry $J$ such that a projection $E$ is the $J$-projection.
In particular, the minimal element of the set of all symmetries $J$ with $JE\geqslant0$ is given. That is
$$ min\{J: JE\geqslant0,\ J=J^{\ast}=J^{-1}\}=2P_{(E+E^{\ast})^{+}}-I.$$
Furthermore, using some new methods and techniques,  we recover some theorems
and corollaries of [2, 11].

\section{ Absolute values for a class of operator matrices }

The following lemma is a direct calculation from diagonalization of a self-adjoint matrix.

 {\bf Lemma 1.}  Let $b>0$ and $\mu\in\mathbb{R}.$ Then

$$\left(\begin{array}{cc}
 1+b&(1+\mu)\sqrt{b}\\\sqrt{b}(1+\mu)&\mu^2+b\end{array}\right)^{\frac{1}{2}}=\begin{cases}\left(\begin{array}{cc}
 1&\sqrt{b}\\\sqrt{b}&\mu\end{array}\right),\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \qquad \hbox{ if }  \mu\geqslant b & \\
\dfrac{1}{t}\left(\begin{array}{cc}
 2b-\mu+1&(1+\mu)\sqrt{b}\\ \sqrt{b}(1+\mu)&\mu^2-\mu+2b\end{array}\right), \ \ \ \hbox{ if }\mu< b&
\end{cases}$$ where $t=(\mu^2-2\mu+4b+1)^{\frac{1}{2}}.$

We get the following results inspired by Lemma 1 above.

 {\bf Lemma 2.}  Let $A\in \mathcal{B(H)}^{+}, \mu\in \mathbb{R}.$
 Suppose that $M\in\mathcal{B(H\oplus H)}$ has the operator matrix form
$M:=\left(\begin{array}{cc}
I+A&(1+\mu) A^\frac{1}{2}\\(1+\mu) A^\frac{1}{2}&\mu^2I+A\end{array}\right):\mathcal{H\oplus H}\rightarrow \mathcal{H\oplus H}.$

(i) If $\mu\geqslant\|A\|,$ then
$$ M^{\frac{1}{2}} = \left(\begin{array}{cc}
I& A^{\frac{1}{2}}\\ A^{\frac{1}{2}}&\mu I\end{array}\right):\mathcal{H\oplus H}\rightarrow \mathcal{H\oplus H}.$$

(ii) If $ \mu\leqslant0,$ then
$$ M^{\frac{1}{2}}= \left(\begin{array}{cc}
T^{-1}(2A-\mu I+I)&(1+\mu)T^{-1}A^\frac{1}{2}\\(1+\mu)T^{-1}A^\frac{1}{2}&T^{-1}(\mu^2I-\mu I+2A)\end{array}\right):\mathcal{H\oplus H}\rightarrow \mathcal{H\oplus H},$$
where $T=[(\mu-1)^2I+4A]^{\frac{1}{2}}.$

(iii) If $0<\mu<\|A\|$ and $\mathcal{H}_1:=E_A[0,\mu],$ $\mathcal{H}_2:=E_A(\mu,\|A\|],$ then
with respect to the decomposition $ \mathcal{H}\oplus\mathcal{H}=\mathcal{H}_1\oplus\mathcal{H}_2\oplus\mathcal{H}_1\oplus\mathcal{H}_2\simeq
(\mathcal{H}_1\oplus\mathcal{H}_1)\oplus(\mathcal{H}_2\oplus\mathcal{H}_2),$
$$ \begin{array}{rl}M^{\frac{1}{2}}&=\left(\begin{array}{cccc}
I_1&0&A_1^\frac{1}{2}&0\\0&T_2^{-1}(2A_2-\mu I_{2}+I_2)&0&(1+\mu)T_2^{-1}A_2^\frac{1}{2}\\ A_1^\frac{1}{2}&0&\mu I_1&0\\0&(1+\mu)T_2^{-1} A_2 ^\frac{1}{2}&0&T_2^{-1}(\mu^2I_2-\mu I_2+2A_2)
\end{array}\right)\\&\simeq\left(\begin{array}{cc}
I_1&A_1^\frac{1}{2}\\A_1^\frac{1}{2}&\mu I_1\end{array}\right) \oplus  \left(\begin{array}{cc}
T_2^{-1}(2A_2-\mu I_2+I_2)&(1+\mu)T_2^{-1}A_2^\frac{1}{2}\\(1+\mu)T_2^{-1}A_2^\frac{1}{2}&T_2^{-1}(\mu^2I_2-\mu I_2+2A_2)\end{array}\right),\end{array}$$
where $A_1:=A\mid_{\mathcal{H}_1},$ $A_2:=A\mid_{\mathcal{H}_2}$ and $T_2:=[(\mu-1)^2I_2+4A_2]^\frac{1}{2},$
where $I_i$ are identity operators on the subspace of $H_i$ for $i=1,2.$

{\bf Proof.} (i) and (ii) are easy to verify.

(iii) If $0<\mu<\|A\|$ and $\mathcal{H}_1=E_A[0,\mu],$ $\mathcal{H}_2=E_A(\mu,\|A\|],$
then $A_1=A\mid_{\mathcal{H}_1}\leqslant\mu I_1$ and  $A_2=A\mid_{\mathcal{H}_2}\geqslant\mu I_2,$ so
$\|A_1\|\leqslant\mu$ and $T_2=[(\mu-1)^2I_2+4A_2]^\frac{1}{2}$ is invertible.
It is easy to see that
$$\begin{array}{rl}&(2A_2-\mu I_2+I_2)(\mu^2I_2-\mu I_2+2A_2)-(1+\mu)^2A_2\\=&4A_2^2-6\mu A_2+\mu^2A_2+A_2-\mu(\mu-1)^2I_2\\ \geqslant &(\mu-1)^2(A_2-\mu I_2)\\ \geqslant&0.\end{array}$$
Then the positivity of the Schur complement of the following operator matrix implies
$$\left(\begin{array}{cc}
T_2^{-1}(2A_2-\mu I_2+I_2)&(1+\mu)T_2^{-1}A_2^\frac{1}{2}\\(1+\mu)T_2^{-1} A_2 ^\frac{1}{2}&T_2^{-1}(\mu^2I_2-\mu I_2+2A_2)
\end{array}\right)\geqslant0.$$

Obviously, with respect to the decomposition $H=\mathcal{H}_1\oplus\mathcal{H}_2\oplus\mathcal{H}_1\oplus\mathcal{H}_2,$
the block matrix representation of $M$ is
\begin{equation}M=\left(\begin{array}{cccc}
I_1+A_1&0&(1+\mu)A_1^\frac{1}{2}&0\\0&I_2+A_2&0&(1+\mu)A_2^\frac{1}{2}\\ (1+\mu)A_1^\frac{1}{2}&0&\mu^2I_1+A_1&0\\0&(1+\mu) A_2^\frac{1}{2} &0&\mu^2I_2+A_2
\end{array}\right).\end{equation}
 Then Lemma 1 and a direct calculation imply
$$M^{\frac{1}{2}}=\left(\begin{array}{cccc}
I_1&0&A_1^\frac{1}{2}&0\\0&T_2^{-1}(2A_2-\mu I_2+I_2)&0&(1+\mu)T_2^{-1}A_2^\frac{1}{2}\\ A_1^\frac{1}{2}&0&\mu I_1&0\\0&(1+\mu)T_2^{-1} A_2 ^\frac{1}{2}&0&T_2^{-1}(\mu^2I_2-\mu I_2+2A_2)
\end{array}\right),$$
so with respect to the decomposition $H=(\mathcal{H}_1\oplus\mathcal{H}_1)\oplus(\mathcal{H}_2\oplus\mathcal{H}_2),$ $$M^{\frac{1}{2}}=\left(\begin{array}{cc}
I_1&A_1^\frac{1}{2}\\A_1^\frac{1}{2}&\mu I_1\end{array}\right) \oplus  \left(\begin{array}{cc}
T_2^{-1}(2A_2-\mu I_2+I_2)&(1+\mu)T_2^{-1}A_2^\frac{1}{2}
\\(1+\mu)T_2^{-1}A_2^\frac{1}{2}&T_2^{-1}(\mu^2I_2-\mu I_2+2A_2)\end{array}\right).$$
\qquad $\square$

 Let $D\in \mathcal{B(K,H)}$ with $\dim N(D)=\dim N(D^*)=0.$
 Then the polar decomposition theorem implies that $D^*=V(DD^*)^{\frac{1}{2}},$ where
$V$ is a unitary operator.
Denote $\mathcal{H}_1:=E_{DD^*}[0,\mu]$ and $\mathcal{H}_2:=E_{DD^{\ast}}(\mu,\|D\|^{2}].$
If $\mathcal{K}_1:=V\mathcal{H}_1,$ and $\mathcal{K}_2:=V\mathcal{H}_2,$ then $V$ has the operator matrix form
 \begin{equation} V=\left(\begin{array}{cc}V_{11}&0\\0&V_{22}\end{array}\right):
\mathcal{H}_1\oplus\mathcal{H}_2\longrightarrow\mathcal{K}_1\oplus\mathcal{K}_2. \end{equation}
Also, define $A_1:=DD^{\ast}\mid _{\mathcal{H}_1}$ and $A_2:=DD^{\ast}\mid _{\mathcal{H}_2}.$ Then $D_1^{\ast}:=V_{11}A_1^\frac{1}{2}$ and $D_2^{\ast}:=V_{22} A_2^\frac{1}{2}$ satisfy
  that $D_1^{\ast}\in {\mathcal{B}}({\mathcal{H}_1,\mathcal{K}_1}),$ $D_2^{\ast}\in {\mathcal{B}}({\mathcal{H}_2,\mathcal{K}_2})$ and
 \begin{equation}D^{\ast}=V(DD^{\ast})^{\frac{1}{2}}=\left(\begin{array}{cc}
V_{11}&0\\0&V_{22}\end{array}\right)\left(\begin{array}{cc}
 A_1^\frac{1}{2}&0\\0& A_2^\frac{1}{2}\end{array}\right)=D_1^*\oplus D_2^{\ast}. \end{equation}

 {\bf Lemma 3.} Let $\mu\in \mathbb{R}$ and $D\in \mathcal{B(K,H)}$ with $\dim N(D)=\dim N(D^{\ast})=0.$ If $0<\mu<\|D\|^{2}$ and if $\mathcal{H}_i,$ $\mathcal{K}_i$ and $D_i$ are the same as in (2.2) and (2.3) for $i=1,2,$ then with respect to the space decomposition
$\mathcal{H}\oplus\mathcal{K}=(\oplus_{i=1}^2{\mathcal{H}_i})\bigoplus({\oplus_{i=1}^2} {\mathcal{K}}_i)\simeq\bigoplus_{i=1}^2({\mathcal{H}_i}{\oplus}{\mathcal{K}_i)},$

$$\begin{array}{rl}&\begin{vmatrix}\left(\begin{array}{cc} I & D \\ D^{\ast}&\mu I\end{array}\right)\end{vmatrix}
\\=&\left(\begin{array}{cccc}I_1&0&D_1&0\\0&T_2^{-1}(2D_2D_2^{\ast}-\mu I_2+I_2)&0&(1+\mu)T_2^{-1}D_2\\D_1^{\ast}&0&\mu I_3&0
\\0&(1+\mu)D_2^{\ast}T_2^{-1}&0&(\mu^2 I_4-\mu I_4+2D_2^{\ast}D_2)
S_2^{-1}\end{array}\right)
\\ \simeq&\left(\begin{array}{cc} I_1 & D_1 \\ D^{\ast}_1&\mu I_3\end{array}\right) \oplus \left(\begin{array}{cc} T_2^{-1}(2D_2D_2^{\ast}-\mu I_2+I_2)&(1+\mu)T_2^{-1}D_2\\(1+\mu)D_2^{\ast}T_2^{-1}
&(\mu^2 I_4-\mu I_4+2D_2^{\ast}D_2)
S_2^{-1} \end{array}\right).\end{array}$$
where  $I_i$ (for $i=1,2)$ are identity operators on the subspace of ${\mathcal{H}_i},$
$I_i$ (for $i=3,4)$ are identity operators on the subspace of ${\mathcal{K}}_{i-2},$ $T_2=[(\mu-1)^2I_2+4D_2D_2^{\ast}]^{\frac{1}{2}}$
and $S_2=[(\mu-1)^2I_4+4D_2^{\ast}D_2]^{\frac{1}{2}}.$

{\bf Proof.}
  Define an operator $X$ from $\mathcal{H}\oplus\mathcal{H}=
(\oplus_{i=1}^2{\mathcal{H}_i})\oplus(\oplus_{i=1}^2{\mathcal{H}_i})$ into $\mathcal{H}\oplus\mathcal{K}=(\oplus_{i=1}^2{\mathcal{H}_i})\oplus({\oplus_{i=1}^2} {\mathcal{K}}_i)$ by  $$X:=diag(I_1,I_2,V_{11},V_{22}),$$
where $V_{ii}$ is the same as in (2.2) for $i=1,2.$ Then $X$ is a unitary operator and
$$\left(\begin{array}{cc}I&D\\D^{\ast}&\mu I\end{array}\right)
=X\left(\begin{array}{cccc}I_1&0&A_1^\frac{1}{2}&0\\0&I_2&0& A_2^\frac{1}{2}\\A_1^\frac{1}{2}&0&\mu I_3&0\\0& A_2^\frac{1}{2}&0&\mu I_4\end{array}\right)X^\ast,$$ where
$A_i:=DD^{\ast}\mid _{{\mathcal{H}}_i}$ for $i=1,2,$ which induces
$$\left(\begin{array}{cc}I&D\\D^{\ast}&\mu I\end{array}\right)^{2}=
X\left(\begin{array}{cccc}
I_1+A_1&0&(1+\mu) A_1^\frac{1}{2}&0\\0&I_2+A_2&0&(1+\mu) A_2^\frac{1}{2}\\(1+\mu) A_1^\frac{1}{2}&0&\mu^2I_3+A_1&0\\0&(1+\mu) A_2^\frac{1}{2}&0&\mu^2I_4+A_2\end{array}\right)X^\ast.$$

On the other hand, we have $$T_2^{2}D_2=[(\mu-1)^2I_2+4D_2D_2^{\ast}]D_2=D_2[(\mu-1)^2I_2+4D_2^{\ast}D_2]=D_2S_2^2,$$ which implies $T_2D_2=D_2S_2,$ so $T_2^{-1}D_2=D_2S_2^{-1}.$ Then
$$\begin{array}{rl}&V_{22}T_2^{-1}(\mu^2I_2-\mu I_2+2D_2D_2^{\ast})V_{22}^*
\\=&V_{22}(D_2D_2^*)^\frac{1}{2}(D_2D_2^*)^{-1}T_2^{-1}(\mu^2I_2-\mu I_2+2D_2D_2^{\ast})(D_2D_2^*)^\frac{1}{2}V_{22}^*\\=&D_2^{\ast}(D_2D_2^{\ast})^{-1}(\mu^2I_2-\mu I_2+2D_2D_2^{\ast})T_2^{-1}D_2\\=&
D_2^{-1}(\mu^2I_2-\mu I_2+2 D_2 D_2^{\ast})D_2S_2^{-1}\\=&
D_2^{-1}D_2(\mu^2I_4-\mu I_4+2 D_2^{\ast} D_2)S_2^{-1}\\=&
(\mu^2I_4-\mu I_4+2D_2^{\ast}D_2)S_2^{-1}.\end{array}$$
Thus by Lemma 2 (iii) and a direct calculation, we get
$$\begin{array}{rl}&\begin{vmatrix}\left(\begin{array}{cc}
I&D\\D^{\ast}&\mu I\end{array}\right)\end{vmatrix}=X\left(\begin{array}{cccc}
I_1+A_1&0&(1+\mu) A_1^\frac{1}{2}&0\\0&I_2+A_2&0&(1+\mu) A_2^\frac{1}{2}\\(1+\mu) A_1^\frac{1}{2}&0&\mu^2I_1+A_1&0\\0&(1+\mu) A_2^\frac{1}{2}&0&\mu^2I_2+A_2\end{array}\right)^\frac{1}{2}X^\ast\\=&\left(\begin{array}{cccc}
I_1&0&D_1&0\\0&T_2^{-1}(2D_2D_2^{\ast}-\mu I_2+I_2)&0&(1+\mu)T_2^{-1}D_2\\D_1^{\ast}&0&\mu I_3&0
\\0&(1+\mu)D_2^{\ast}T_2^{-1}&0&V_{22}T_2^{-1}(\mu^2I_2-\mu I_2+2D_2D_2^{\ast})V_{22}^*\end{array}\right)\\=&\left(\begin{array}{cccc}
I_1&0&D_1&0\\0&T_2^{-1}(2D_2D_2^{\ast}-\mu I_2+I_2)&0&(1+\mu)T_2^{-1}D_2\\D_1^{\ast}&0&\mu I_3&0
\\0&(1+\mu)D_2^{\ast}T_2^{-1}&0&(\mu^2I_4-\mu I_4+2D_2^{\ast}D_2)S_2^{-1} \end{array}\right)
 \\ \simeq & \begin{pmatrix} I_1 & D_1 \\ D^{\ast}_1&\mu I_3\end{pmatrix}
\oplus \begin{pmatrix} T_2^{-1}(2D_2D_2^{\ast}-\mu I_2+I_2)&(1+\mu)T_2^{-1}D_2\\
(1+\mu)D_2^{\ast}T_2^{-1}&(\mu^2I_4-\mu I_4+2D_2^{\ast}D_2)S_2^{-1} \end{pmatrix}.\end{array}$$ \ \ \ \ \ \ $\square$

Let $B\in \mathcal{B(K,H)}.$ It is clear that $B$  has the operator matrix form
$$ B=\left(\begin{array}{cc}
\widetilde{B}&0\\0&0\end{array}\right):
N(B)^{\perp}\oplus N(B)\rightarrow\overline{R(B)}\oplus R(B)^{\perp},$$
where $ \widetilde{B}\in B(N(B)^{\perp},\ \overline{R(B)})$ is injective and has dense range.
Then with respect to the space decomposition $\overline{R(B)}\oplus R(B)^{\perp}\oplus N(B)^{\perp}\oplus N(B),$
$$\left(\begin{array}{cc}I&B\\B^{\ast}&\mu I\end{array}\right)=\left(\begin{array}{cccc}
I_1&0& \widetilde{B}&0\\0&I_2&0&0\\ \widetilde{B}^{\ast}&0&\mu I_3&0\\0&0&0&\mu I_4\end{array}\right)
 ,$$
which implies
\begin{equation}\begin{vmatrix}\left(\begin{array}{cc}I&B\\B^{\ast}&\mu I\end{array}\right)\end{vmatrix}
\simeq\begin{vmatrix}\left(\begin{array}{cc}
I_1&\widetilde{B}\\\widetilde{B}^{\ast}&\mu I_3\end{array}\right)\end{vmatrix}
\oplus  I_2 \oplus  |\mu| I_4,\end{equation}
where $I_i$ (for $i=1,2,3,4$) are identity operators on the subspaces of $\overline{R(B)},$ $R(B)^{\perp},$
$N(B)^{\perp}$ and $N(B),$ respectively.

{\bf Remark.} Let $0<\mu<\|B\|^{2}$ and $\widetilde{B}^*=\widetilde{V}(\widetilde{B}\widetilde{B}^*)^\frac{1}{2}$
 be the polar decomposition of $\widetilde{B}^*,$
 where $\widetilde{V}$ is the unique unitary operator from $\overline{R(B)}$ onto $N(B)^{\perp}.$
 In a similar way to (2.2) and (2.3), we denote $\widetilde{\mathcal{H}_1}:=E_{\widetilde{B}\widetilde{B}^*}[0,\mu]$ and $\widetilde{\mathcal{H}_2}:=E_{\widetilde{B}\widetilde{B}^{\ast}}(\mu,\|D\|^{2}].$
 Thus $\overline{R(B)}=\widetilde{\mathcal{H}_1}\oplus\widetilde{\mathcal{H}_2}.$

If $\widetilde{\mathcal{K}_1}:=\widetilde{V}\widetilde{\mathcal{H}_1}$ and $\widetilde{\mathcal{K}_2}:=\widetilde{V}\widetilde{\mathcal{H}_2},$ then $N(B)^{\perp}=\widetilde{\mathcal{K}_1}\oplus\widetilde{\mathcal{K}_2}$ and
$\widetilde{V}$ has the operator matrix form
 \begin{equation} \widetilde{V}=\left(\begin{array}{cc}\widetilde{V}_{11}&0\\0&\widetilde{V}_{22}\end{array}\right):
\widetilde{\mathcal{H}_1}\oplus\widetilde{\mathcal{H}_2}\longrightarrow
\widetilde{\mathcal{K}_1}\oplus\widetilde{\mathcal{K}_2}. \end{equation}
Analogously, define $\widetilde{A_1}:=\widetilde{B}\widetilde{B}^{\ast}\mid _{\widetilde{\mathcal{H}_1}}$
and $\widetilde{A_2}:=\widetilde{B}\widetilde{B}^{\ast}\mid _{\widetilde{\mathcal{H}_2}}.$ Then $\widetilde{
B_1}^{\ast}:=\widetilde{V}_{11}\widetilde{A_1}^\frac{1}{2}$ and $\widetilde{B_2}^{\ast}:=\widetilde{V}_{22}\widetilde{A_2}^\frac{1}{2}$ satisfy
  that $\widetilde{B_1}^{\ast}\in {\mathcal{B}}({\widetilde{\mathcal{H}_1},\widetilde{\mathcal{K}_1}}),$ $\widetilde{B_2}^{\ast}\in {\mathcal{B}}(\widetilde{{\mathcal{H}_2}},\widetilde{\mathcal{K}_2})$ and
 $\widetilde{B}^{\ast}=\widetilde{B_1}^*\oplus \widetilde{B_2}^{\ast}.$
 Also, $I_{1i}$ and $I_{3i}$ (for $i=1,2$) are identity operators on the subspaces of
 $\widetilde{\mathcal{H}_i}$ and $\widetilde{\mathcal{K}_i},$
  respectively.

{\bf Lemma 4.} Let $B\in \mathcal{B(K,H)}$ and $\mu\in \mathbb{R}.$ Then

(i) If $\mu\geqslant \|B\|^{2},$ then
$$\begin{vmatrix}\left(\begin{array}{cc}
I&B\\B^{\ast}&\mu I\end{array}\right)\end{vmatrix}=\left(\begin{array}{cc}
I&B\\B^{\ast}&\mu I\end{array}\right):\mathcal{H}\oplus\mathcal{K}\rightarrow
\mathcal{H}\oplus\mathcal{K}.$$

(ii) If $\mu\leqslant0,$ then with respect to the decomposition $\mathcal{H}\oplus\mathcal{K},$ we have

$$\begin{vmatrix}\left(\begin{array}{cc}
I&B\\B^{\ast}&\mu I\end{array}\right)\end{vmatrix}
=\left(\begin{array}{cc}
 T^{-1}(2BB^{\ast}-\mu I+I)&(1+\mu)T^{-1}B\\(1+\mu)B^{\ast} T^{-1}&S^{-1}[(\mu^2-\mu)I+2B^*B]\end{array}\right),$$
where $T:=[(\mu-1)^2I+4BB^{\ast}]^{\frac{1}{2}}$ and $S:=[(\mu-1)^2I+4B^*B]^{\frac{1}{2}}.$

(iii) If $0<\mu<\|B\|^{2},$ then
with respect to the decomposition $\mathcal{H}\oplus\mathcal{K}\simeq(\widetilde{\mathcal{H}_2}\oplus\widetilde{\mathcal{K}_2})\oplus
(\widetilde{\mathcal{H}_1}\oplus\widetilde{\mathcal{K}_1})
\oplus R(B)^\perp\oplus N(B),$ we have

$$\begin{array}{rl}\begin{vmatrix}\left(\begin{array}{cc}
I&B\\B^{\ast}&\mu I\end{array}\right)\end{vmatrix}
\simeq&\left(\begin{array}{cc}
\widetilde{T_2}^{-1}(2\widetilde{B_2}\widetilde{B_2}^{\ast}-\mu I_{12}+I_{12})&(1+\mu)\widetilde{T_2}^{-1}\widetilde{B_2}
\\(1+\mu)\widetilde{B_2}^{\ast}\widetilde{T_2}^{-1}&(\mu^2I_{32}-\mu I_{32}+2\widetilde{B_2}^{\ast}\widetilde{B_2})
\widetilde {S_2}^{-1}\end{array}\right)
 \\ \oplus&\left(\begin{array}{cc}
I_{11}&\widetilde{B_1}\\ \widetilde{B_1}^{\ast}&\mu I_{31}\end{array}\right)\oplus I_2\oplus|\mu|I_4,\end{array}$$
 where $\widetilde{B_i},$ $I_{1i}$ and $I_{3i}$ (for $i=1,2$) are as above Remark,  $\widetilde {T_2}=[(\mu-1)^2I_{12}+4\widetilde{B_2}\widetilde{B_2}^{\ast}]^{\frac{1}{2}}$ and
 $\widetilde {S_2}=[(\mu-1)^2I_{32}+4\widetilde{B_2}^{\ast}\widetilde{B_2}]^{\frac{1}{2}}$

{\bf Proof.} (i) is clear.

(ii) Obviously, $\mu\leqslant0$ implies that $T=[(\mu-1)^2I+4BB^{\ast}]^{\frac{1}{2}}$ and $S=[(\mu-1)^2I+4B^*B]^{\frac{1}{2}}$ are invertible.
Let $U$ be the unique partial isometry such that
$B^{\ast}=U(BB^{\ast})^{\frac{1}{2}},$ $R(U)=\overline{R(B^*)}$ and $R(U^*)=\overline{R(B)}.$
By a direct calculation, we get that   $$\begin{array}{rl}&\left(\begin{array}{cc}
I&B\\B^{\ast}&\mu I\end{array}\right)^2
=\left(\begin{array}{cc}I+BB^{\ast}&(1+\mu)B\\(1+\mu)B^{\ast}&\mu^2I+B^{\ast}B\end{array}\right)
\\=&\left(\begin{array}{cc}I&0\\0&U\end{array}\right)
\left(\begin{array}{cc}I+BB^{\ast}&(1+\mu)|B^{\ast}|\\(1+\mu)|B^{\ast}|&\mu^2I+BB^{\ast}\end{array}\right)
\left(\begin{array}{cc}I&0\\0&U^{\ast}\end{array}\right)+
\left(\begin{array}{cc}0&0\\0&\mu^2P_{B^*}^\perp\end{array}\right)\\=&\left(\begin{array}{cc}I&0\\0&UP_{B}\end{array}\right)
\left(\begin{array}{cc}I+BB^{\ast}&(1+\mu)|B^{\ast}|\\(1+\mu)|B^{\ast}|&\mu^2I+BB^{\ast}\end{array}\right)
\left(\begin{array}{cc}I&0\\0&P_{B}U^{\ast}\end{array}\right)+
\left(\begin{array}{cc}0&0\\0&\mu^2P_{B^*}^\perp\end{array}\right)
\\=&\left(\begin{array}{cc}I&0\\0&U\end{array}\right)
\left(\begin{array}{cc}I+BB^{\ast}&(1+\mu)|B^{\ast}|\\(1+\mu)|B^{\ast}|&\mu^2P_{B}+BB^{\ast}\end{array}\right)
\left(\begin{array}{cc}I&0\\0&U^{\ast}\end{array}\right)+
\left(\begin{array}{cc}0&0\\0&\mu^2P_{B^*}^\perp\end{array}\right),\end{array}$$
since $UU^*=P_{B^*},$ $U(BB^*)U^*=B^{\ast}B$ and $U P_{B}=U.$  Clearly,
$$\left(\begin{array}{cc}I+BB^{\ast}&(1+\mu)|B^{\ast}|
\\(1+\mu)|B^{\ast}|&\mu^2P_{B}+BB^{\ast}\end{array}\right)^{\frac{1}{2}}=
\left(\begin{array}{cc}I&0\\0&P_{B}\end{array}\right)
\left(\begin{array}{cc}I+BB^{\ast}&(1+\mu)|B^{\ast}|\\(1+\mu)|
B^{\ast}|&\mu^2I+BB^{\ast}\end{array}\right)^{\frac{1}{2}},$$
$P_{B^*}^\perp B^*=0$ and $BP_{B^*}^\perp =0.$
Setting $$\widetilde{U}=\left(\begin{array}{cc}I&0\\0&U\end{array}\right):
\mathcal{H\oplus H}\rightarrow\mathcal{H\oplus K},$$ we conclude from Lemma 2 (ii) that
$$\begin{array}{rl}&\begin{vmatrix}\left(\begin{array}{cc}I&B\\B^{\ast}&\mu I\end{array}\right)\end{vmatrix}
\\=&\widetilde{U}
\left(\begin{array}{cc}I+BB^{\ast}&(1+\mu)|B^{\ast}|\\(1+\mu)|B^{\ast}|&\mu^2P_{B}+BB^{\ast}
\end{array}\right)^{\frac{1}{2}}
\widetilde{U}^*+
\left(\begin{array}{cc}0&0\\0&\mu^2P_{B^*}^\perp\end{array}\right)^\frac{1}{2}
\\=&\left(\begin{array}{cc}I&0\\0&UP_{B}\end{array}\right)
\left(\begin{array}{cc}I+BB^{\ast}&(1+\mu)|B^{\ast}|\\(1+\mu)|B^{\ast}|&\mu^2I+BB^{\ast}
\end{array}\right)^{\frac{1}{2}}
\left(\begin{array}{cc}I&0\\0&U^{\ast}\end{array}\right)+
\left(\begin{array}{cc}0&0\\0&|\mu|P_{B^*}^\perp\end{array}\right)
\\=&\widetilde{U}
\left(\begin{array}{cc}
T^{-1}(2BB^{\ast}-\mu I+I)&(1+\mu)T^{-1}|B^{\ast}|\\(1+\mu)|B^{\ast}|T^{-1}&T^{-1}(\mu^2I-\mu I+2BB^{\ast})\end{array}\right)
\widetilde{U}^*+
diag(0,|\mu|P_{B^*}^\perp)
\\=&\left(\begin{array}{cc}
T^{-1}(2BB^{\ast}-\mu I+I)&(1+\mu)T^{-1}B\\(1+\mu)B^{\ast}T^{-1}&UT^{-1}(\mu^2I-\mu I+2BB^{\ast})U^{\ast}+|\mu|P_{B^*}^\perp\end{array}\right).\end{array}$$

Moreover, $B^{\ast}=U(BB^{\ast})^{\frac{1}{2}},$ implies
$(B^{\ast}B)^{\frac{1}{2}}=U(BB^{\ast})^{\frac{1}{2}}U^*,$  so $$(B^{\ast}B)^{\frac{1}{2}}U=U(BB^{\ast})^{\frac{1}{2}}U^*U=U(BB^{\ast})^{\frac{1}{2}}=B^*,$$ which yields
$$\begin{array}{rcl}U[(\mu-1)^2I+4BB^{\ast}]&=&(\mu-1)^2 U+4B^*(BB^{\ast})^{\frac{1}{2}}
\\&=&(\mu-1)^2 U+4(B^*B)^{\frac{1}{2}}B^{\ast}\\&=&
[(\mu-1)^2 +4B^*B]U.\end{array}$$ Thus $$UT^{-1}=U[(\mu-1)^2I+4BB^{\ast}]^{-{\frac{1}{2}}}=[(\mu-1)^2 +4B^*B]^{-{\frac{1}{2}}}U=S^{-1}U,$$  which induces
$$\begin{array}{rl}&UT^{-1}(\mu^2I-\mu I+2BB^{\ast})U^{\ast}+|\mu|P_{B^*}^\perp\\=&
[(\mu-1)^2 +4B^*B]^{-\frac{1}{2}}[(\mu^2-\mu)P_{B^*}+2B^*B]-\mu P_{B^*}^\perp\\=&
S^{-1}[(\mu^2-\mu)P_{B^*}+2B^*B-\mu(1-\mu)P_{B^*}^\perp]\\=&
S^{-1}[(\mu^2-\mu)I+2B^*B],\end{array}$$ since $P_{B^*}^\perp S^2=P_{B^*}^\perp[(\mu-1)^2 +4B^*B]=(1-\mu)^2P_{B^*}^\perp$ implies $ SP_{B^*}^\perp=(1-\mu)P_{B^*}^\perp.$

 (iii) According to equations (2.4),(2.5) and Lemma 3, we know that $$\begin{array}{rcl}\begin{vmatrix}\left(\begin{array}{cc}I&B\\B^{\ast}&\mu I\end{array}\right)\end{vmatrix}
&\simeq& \begin{vmatrix}\left(\begin{array}{cc}
I_1&\widetilde{B}\\ \widetilde{B}^{\ast}&\mu I_3\end{array}\right)\end{vmatrix}
\oplus  I_2 \oplus |\mu| I_4\\&=&\left(\begin{array}{cc}
\widetilde{T_2}^{-1}(2\widetilde{B_2}\widetilde{B_2}^{\ast}-\mu I_{12}+I_{12})&(1+\mu)\widetilde{T_2}^{-1}\widetilde{B_2}
\\(1+\mu)\widetilde{B_2}^{\ast}
\widetilde{T_2}^{-1}&(\mu^2I_{32}-\mu I_{32}+2\widetilde{B_2}\widetilde{B_2}^{\ast})
\widetilde {S_2}^{-1}\end{array}\right)
 \\& \oplus&\left(\begin{array}{cc}
I_{11}&\widetilde{B_1}\\ \widetilde{B_1}^{\ast}&\mu I_{31}\end{array}\right)\oplus I_2\oplus|\mu|I_4. \end{array}$$
 \qquad $\square$

{\bf Theorem 5.} Let $\lambda\in \mathbb{R}, \mu\in  \mathbb{R}$ and  $B\in \mathcal{B(K,H)}.$

 (i) If $\lambda=\mu=0,$ then

$$\begin{vmatrix}\left(\begin{array}{cc}
\lambda I&B\\B^{\ast}&\mu I\end{array}\right)\end{vmatrix}=\left(\begin{array}{cc}
|B^{\ast}|&0\\0&|B|\end{array}\right):\mathcal{H}\oplus\mathcal{K}\rightarrow
\mathcal{H}\oplus\mathcal{K}.$$

(ii) If $\lambda \neq 0$ and $\mu=0,$ then

 $$\begin{vmatrix}\left(\begin{array}{cc}
\lambda I&B\\B^{\ast}&0\end{array}\right)\end{vmatrix}=
\left(\begin{array}{cc}
\frac{T+\lambda^{2} T^{-1}}{2}&\lambda T^{-1}B
\\ \lambda B^{\ast}T^{-1} &2B^{\ast}T^{-1}B \end{array}\right):\mathcal{H}\oplus\mathcal{K}\rightarrow
\mathcal{H}\oplus\mathcal{K},$$
where $T=(\lambda^{2}I+4BB^{\ast})^{\frac{1}{2}}.$

(iii) If $\mu\lambda\geqslant\|B\|^{2}>0,$ then

$$ \begin{vmatrix}\left(\begin{array}{cc}
\lambda I&B\\B^{\ast}&\mu I\end{array}\right)\end{vmatrix}=\frac{|\lambda|}
{\lambda}\left(\begin{array}{cc}
\lambda I&B \\ B^{\ast}&\mu I\end{array}\right):\mathcal{H}\oplus\mathcal{K}\rightarrow
\mathcal{H}\oplus\mathcal{K}.$$

(iv) If $\mu\lambda<0,$ then with respect to the decomposition $\mathcal{H}\oplus\mathcal{K}:$
$$\begin{vmatrix}\left(\begin{array}{cc}
\lambda I&B\\B^{\ast}&\mu I\end{array}\right)\end{vmatrix}
=\left(\begin{array}{cc}
T^{-1}(2BB^{\ast}-\lambda \mu I+\lambda^{2}I)
&(\lambda+\mu)T^{-1}B
\\ (\lambda+\mu)B^*T^{-1} &S^{-1}(\mu^2I-\lambda \mu I+2B^{\ast}B)\end{array}\right),$$
where $T=[(\mu-\lambda)^2 I+4BB^{\ast}]^\frac{1}{2}$ and $S=[(\mu-\lambda)^2 I+4B^{\ast}B]^\frac{1}{2}.$

(v) If $0<\mu\lambda<\|B\|^{2},$ then
with respect to the decomposition $\mathcal{H}\oplus\mathcal{K}\simeq R(B)^\perp\oplus N(B)\oplus(\widetilde{\mathcal{H}_1}\oplus\widetilde{\mathcal{K}_1})
\oplus(\widetilde{\mathcal{H}_2}\oplus\widetilde{\mathcal{K}_2}),$ we have
$$\begin{array}{rl}\begin{vmatrix}\left(\begin{array}{cc}
\lambda I&B\\B^{\ast}&\mu I\end{array}\right)\end{vmatrix}&\simeq |\lambda| I_2\oplus|\mu|I_4\oplus|\lambda|\left(\begin{array}{cc}
I_{11}&\frac{\widetilde{B_1}}{\lambda}\\ \frac{\widetilde{B_1}^{\ast}}{\lambda}&\frac{\mu}{\lambda}I_{31}\end{array}\right)
\\& \oplus \left(\begin{array}{cc}
\widetilde{T}^{-1}(2\widetilde{B_2} \widetilde{B_2}^{\ast}-{\lambda}\mu I
+{\lambda^{2}}I)&
({\lambda}+\mu)\widetilde{T}^{-1}\widetilde{B_2} \\
({\lambda}+\mu)\widetilde{B_2}^{\ast} \widetilde{T}^{-1}&(\mu^2I_{32}-\lambda\mu I_{32}+2\widetilde{B_2}^{\ast}\widetilde{B_2})
\widetilde {S_2}^{-1}\end{array}\right),\end{array}$$
where $\widetilde{B_i}^{\ast}$ is the same to that of Lemma 4 (for $i=1,2$), $\widetilde{T}=[(\mu-\lambda)^2 I_{12}+4\widetilde{B_2}\widetilde{B_2}^{\ast}]^\frac{1}{2}$ and $\widetilde{S}=[(\mu-\lambda)^2 I_{32}+4\widetilde{B_2}^{\ast}\widetilde{B_2}]^\frac{1}{2}.$

(vi) If $\lambda=0$ and $\mu\neq 0,$ then

 $$\begin{vmatrix}\left(\begin{array}{cc}
0&B\\B^{\ast}&\mu I\end{array}\right)\end{vmatrix}=
\left(\begin{array}{cc}
2BS^{-1}B^{\ast}&{\mu}B S^{-1}
\\ {\mu}S^{-1}B^{\ast} &\frac{S+{\mu}^{2}S^{-1}}{2}\end{array}\right):\mathcal{H}\oplus\mathcal{K}\rightarrow
\mathcal{H}\oplus\mathcal{K},$$
where $S=(\mu^{2}I+4B^*B)^{\frac{1}{2}}.$

{\bf Proof.} (i) and (iii) are direct calculations. It is easy to verify that Lemma 4 (ii) implies (ii).
(iv) follows from Lemma 4 (ii) and the fact that if $\lambda\neq 0,$ then \begin{equation}\begin{vmatrix}\left(\begin{array}{cc}
\lambda I&B\\B^{\ast}&{\mu}I \end{array}\right)\end{vmatrix}=|\lambda|
\begin{vmatrix}\left(\begin{array}{cc}
I&\frac{B}{\lambda }\\ \frac{B^{\ast}}{\lambda}&\frac{{\mu}}{\lambda}I\end{array}\right)\end{vmatrix}.\end{equation}
Using equation (2.6) and Lemma 4 (iii), we get (v).
 (vi) is obvious from (ii) and the relation $\begin{vmatrix}\left(\begin{array}{cc}
0&B\\B^{\ast}&\mu I\end{array}\right)\end{vmatrix}\simeq \begin{vmatrix}\left(\begin{array}{cc}
 \mu I&B^{\ast}\\B&0\end{array}\right)\end{vmatrix}.$ \qquad $\square$

\section{Support projections for a class of operator matrices}

In this section, we will give the expressions of the positive parts and support projections for operator matrices \begin{equation}S_\lambda:=Q_{\lambda,0}=\left(\begin{array}{cc}\lambda I&B\\B^{\ast}&0\end{array}\right):
\mathcal{H}\oplus\mathcal{K}\rightarrow
\mathcal{H}\oplus\mathcal{K},\end{equation}  where $\lambda\in\mathbb{R}$ and $B\in \mathcal{B(K,H)}.$ The following lemma
follows from diagonalization of a self-adjoint matrix ([4]).

{\bf Lemma 6.} Let $a>0.$ Then $C:=\left(\begin{array}{cc}1&2\sqrt{a}
\\2\sqrt{a}&4a\end{array}\right)$ is positive
and the support projection of $C$ is
$$P_{C}= \left(\begin{array}{cc}\dfrac{1}{1+4a}&\dfrac{2\sqrt{a}}{1+4a}
\\\\ \dfrac{2\sqrt{a}}{1+4a}&\dfrac{4a}{1+4a}\end{array}\right).$$

Using Lemma 6,  we get some hints for the following results.

{\bf Lemma 7.}  Let $A\in \mathcal{B(H)}^{+}.$
 If $\widetilde{A}\in B(\mathcal{H}\oplus\mathcal{H})$ has the operator matrix form  $\widetilde{A}=\left(\begin{array}{cc}
 I&2A^{\frac{1}{2}}\\2A^{\frac{1}{2}}&
 4A\end{array}\right),$
then $\widetilde{A}\in B(\mathcal{H}\oplus\mathcal{H})^+$ and the support projection of $\widetilde{A}$ is
$$P_{\widetilde{A}}:=\left(\begin{array}{cc} (I+4A)^{-1}&2A^{\frac{1}{2}}(I+4A)^{-1}\\2A^{\frac{1}{2}}(I+4A)^{-1}&4A(I+4A)^{-1}\end{array}\right).$$

{\bf Proof.} Clearly, $\widetilde{A}\geqslant0.$
It is easy to verify that \begin{equation}P_{\widetilde{A}}^{2}=P_{\widetilde{A}}\geqslant0\ \ \ \ \ \ \ \ \hbox{and} \ \ \ \ \ \ \ \ P_{\widetilde{A}}\widetilde{A}=\widetilde{A}.\end{equation}
Let $x \in\mathcal{H}$ and $y\in\mathcal{H}$ satisfy $\widetilde{A}(x,y)^{t}=0.$ That is
$$\begin{cases}x+2A^{\frac{1}{2}}y
=0&\\2A^{\frac{1}{2}}x+4Ay=0,
\end{cases}$$
then $x=-2A^{\frac{1}{2}}y,$ so
$$P_{\widetilde{A}}(x,y)^{t}=0.$$ Thus
$N(\widetilde{A})\subseteq N(P_{\widetilde{A}})$ yields $\overline{R (\widetilde{A})}\supseteq R(P_{\widetilde{A}}).$
Then by equation (3.2), $\overline{R(\widetilde{A})}= R(P_{\widetilde{A}}),$ which says that
$P_{\widetilde{A}}$ is the support projection of $\widetilde{A}. \ \ \ \ \  \ \ \ \ \ \ \ \square $

{\bf Lemma 8.} Let $A,B\in \mathcal{B(H)}^{+}.$ If $B$ is invertible and $AB=BA,$ then the support projection of $\widetilde{S}:=\left(\begin{array}{cc}
B^2&2A^{\frac{1}{2}}B\\2A^{\frac{1}{2}}B&4A
\end{array}\right):\mathcal{H}\oplus\mathcal{H}
\rightarrow\mathcal{H}\oplus\mathcal{H}$ is $$P_{\widetilde{S}}=\left(\begin{array}{cc}B^2(B^2+4A)^{-1}&2A^{\frac{1}{2}}B(B^2+4A)^{-1}
\\2A^{\frac{1}{2}}B(B^2+4A)^{-1}&4A(B^2+4A)^{-1}
\end{array}\right):\mathcal{H}\oplus\mathcal{H}
\rightarrow\mathcal{H}\oplus\mathcal{H}.$$

{\bf Proof.} Obviously, $AB=BA$ yields $AB^{-1}=B^{-1}A$ and $A^{\frac{1}{2}}B^{-1}=B^{-1}A^{\frac{1}{2}},$ so with respect to the decomposition $\mathcal{H}\oplus\mathcal{H}$
$$S:=\left(\begin{array}{cc}I&2A^{\frac{1}{2}}B^{-1}\\2A^{\frac{1}{2}}B^{-1}&4AB^{-2}\end{array}\right)\geqslant 0.$$
It is easy to see that $\widetilde{S}$ can be written as
$$\widetilde{S}=S\left(\begin{array}{cc}B^2&0\\0&B^2\end{array}\right)
=\left(\begin{array}{cc}B^2&0\\0&B^2\end{array}\right)S,$$ and $diag(B^2,B^2)$ is invertible on the space $\mathcal{H}\oplus \mathcal{H}.$
Then $R(\widetilde{S})=R(S),$ so Lemma 7 implies that with respect to the decomposition $\mathcal{H}\oplus\mathcal{H}$
 $$\begin{array}{rl}P_{\widetilde{S}}=P_{S}&=
\left(\begin{array}{cc}(I+4AB^{-2})^{-1}&2A^{\frac{1}{2}}B^{-1}(I+4AB^{-2})^{-1}
\\2A^{\frac{1}{2}}B^{-1}(I+4AB^{-2})^{-1}&4AB^{-2}(I+4AB^{-2})^{-1}\end{array}\right)
\\\\&=\left(\begin{array}{cc}B^2(B^2+4A)^{-1}&2A^{\frac{1}{2}}B(B^2+4A)^{-1}
\\2A^{\frac{1}{2}}B(B^2+4A)^{-1}&4A(B^2+4A)^{-1}\end{array}\right).\end{array}$$
\qquad $\square$

{\bf Lemma 9.} Let $F\in \mathcal{B(K)}$ and $G\in \mathcal{B(H)}$ be self-adjoint operators.
If there exists a partial isometry operator $U\in \mathcal{B(H,K)}$
such that $F=UGU^{\ast}$ and $U^{\ast}UG=G,$ then $P_{F}=UP_{G}U^{\ast}.$

{\bf Proof.} Denote $Q:=UP_{G}U^{\ast}.$ Then $$Q^2=UP_{G}U^{\ast}UP_{G}U^{\ast}
=UP_{G}U^{\ast}=Q,$$ so $Q$ is an orthogonal projection. Taking $x\in \mathcal{H},$
we get that  $$ Fx=UGU^{\ast}x=UP_{G}U^{\ast}UGU^{\ast}x=QUGU^{\ast}x=QFx,$$ which yields
$R(F)\subseteq R(Q),$ so $ \overline{R(F)}\subseteq R(Q).$

On the other hand, if $Fx=0,$ then $$GU^{\ast}x=U^{\ast}UGU^{\ast}x=U^*Fx=0,$$
which implies that $U^{\ast}x\in N(G)=R(G)^{\perp},$ so $ P_{G}U^{\ast}x=0.$
Hence, $$Qx=UP_{G}U^{\ast}x=0,$$ which induces
$N(F)\subseteq N(Q),$ that is $\overline{R(F)}\supseteq R(Q).$
Thus $\overline{R(F)}=R(Q),$ which says  $P_{F}=Q=UP_{G}U^{\ast}.$  \ \ \ \ \ $ \square$

The following theorem is the main result of this section.

{\bf Theorem 10.} Let $B\in \mathcal{B(K,H)}$ and $S_\lambda$ is the same as (3.1) for $\lambda\in\mathbb{R}.$   Then

(i) If $\lambda\neq0,$ then $$S_{\lambda}^{+}=\frac{1}{2} \left(\begin{array}{cc}
\frac{T_\lambda+\lambda^2T_\lambda^{-1}
+2\lambda I}{2}&B+\lambda T_\lambda^{-1}B\\ B^{\ast}+ \lambda B^*T_\lambda^{-1}&
2B^{\ast}T_{\lambda}^{-1}B\end{array}\right):
\mathcal{H}\oplus\mathcal{K}\rightarrow \mathcal{H}\oplus\mathcal{K},$$ where $T_\lambda=(\lambda^2 I+4BB^{\ast})^{\frac{1}{2}}.$

(ii) $$P_{S_1^+}=\left(\begin{array}{cc}\frac{I+T^{-1}}{2}&T^{-1}B\\B^{\ast}T^{-1}&2B^{\ast}T^{-1}(I+T)^{-1}B\end{array}
\right):
\mathcal{H}\oplus\mathcal{K}\rightarrow \mathcal{H}\oplus\mathcal{K},$$
where $T:=T_1=(I+4BB^{\ast})^{\frac{1}{2}}.$

(iii) If $\lambda>0,$ then $$P_{S_\lambda^+}=\left(\begin{array}{cc}
\frac{I+\lambda T_\lambda^{-1}}{2}
&T_\lambda^{-1}B\\ B^{\ast}T_\lambda^{-1}&
2B^{\ast}T_\lambda^{-1}(\lambda I+T_\lambda)^{-1}B\end{array}
\right):
\mathcal{H}\oplus\mathcal{K}\rightarrow \mathcal{H}\oplus\mathcal{K},$$
where $T_\lambda=(\lambda^2 I+4BB^{\ast})^{\frac{1}{2}}.$

(iv) If $\lambda<0,$ then $$P_{S_\lambda^+}=\left(\begin{array}{cc}
\frac{I+\lambda T_\lambda^{-1}}{2}
&T_\lambda^{-1}B\\ B^{\ast}T_\lambda^{-1}
&\frac{V(I-\lambda T_\lambda^{-1})V^*}{2}\end{array}
\right):
\mathcal{H}\oplus\mathcal{K}\rightarrow \mathcal{H}\oplus\mathcal{K},$$ where $T_\lambda=(\lambda^2 I+4BB^{\ast})^{\frac{1}{2}}$ as above and $V$ is the unique partial isometry such that
$B^{\ast}=V(BB^{\ast})^{\frac{1}{2}}$ with $R(V)=\overline{R(B^*)}$ and $R(V^*)=\overline{R(B)}.$

(v) $$P_{S_0^+}=\frac{1}{2}\left(\begin{array}{cc}
P_B&V^*\\V&P_{B^*}\end{array}
\right):
\mathcal{H}\oplus\mathcal{K}\rightarrow \mathcal{H}\oplus\mathcal{K},$$ where $V$ is the same to (iv).

{\bf Proof.} (i) is immediate from the fact $S_\lambda^{+}=\dfrac{S_\lambda+|S_\lambda|}{2}$ and Theorem 5 (ii).

(ii) Let $B^{\ast}=V(BB^{\ast})^{\frac{1}{2}}$ be the polar decomposition of $B^*,$ where $V$ is a partial isometry.
Setting $$\widetilde{Q}:=\frac{1}{2}\left(\begin{array}{cc}
\frac{T+T^{-1}+2I}{2}&(I+T^{-1})|B^{\ast}|\\ |B^{\ast}|(I+T^{-1})&2BB^{\ast}T^{-1}
\end{array}\right):
\mathcal{H}\oplus\mathcal{H}\rightarrow \mathcal{H}\oplus\mathcal{H},$$ and
$$\widetilde{X}:=\left(\begin{array}{cc}
(I+T)^{2}&2(I+T)|B^{\ast}|
\\2(I+T)|B^{\ast}|&4BB^{\ast}\end{array}\right):
\mathcal{H}\oplus\mathcal{H}\rightarrow \mathcal{H}\oplus\mathcal{H}
,$$
we get that
\begin{equation}S_1^{+}=\left(\begin{array}{cc}I&0\\0&V\end{array}\right)
\widetilde{Q}
\left(\begin{array}{cc}I&0\\0&V^{\ast}\end{array}\right),\end{equation}
and \begin{equation}\left(\begin{array}{cc}I&0\\0&V^{\ast}\end{array}\right)
\left(\begin{array}{cc}I&0\\0&V\end{array}\right)\widetilde{Q}=\widetilde{Q}.\end{equation}
On the other hand, it is clear that $T^{-1}|B^{\ast}|=|B^{\ast}|T^{-1}$ and $$\widetilde{Q}=\left(\begin{array}{cc}I+T^{2}+2T&2(I+T)|B^{\ast}|\\2(I+T)|B^{\ast}|&4BB^{\ast}\end{array}\right)
\left(\begin{array}{cc}\frac{1}{4}T^{-1}&0\\0&\frac{1}{4}T^{-1}\end{array}\right)=\widetilde{X}
\left(\begin{array}{cc}\frac{1}{4}T^{-1}&0\\0&\frac{1}{4}T^{-1}\end{array}\right),$$
which implies $R(\widetilde{Q})=R(\widetilde{X}).$
Thus Lemma 8 yields that
$$P_{\widetilde{Q}}=P_{\widetilde{X}}=\left(\begin{array}{cc}\frac{I+T^{-1}}{2}&T^{-1}|B^{\ast}|
\\|B^{\ast}|T^{-1}&2BB^{\ast}T^{-1}(I+T)^{-1}\end{array}\right).$$
Combining Lemma 9, equations (3.3) and (3.4), we have $$
\begin{array}{rl}P_{S_1^+}&=
\left(\begin{array}{cc}I&0\\0&V\end{array}\right)P_{\widetilde{Q}}
\left(\begin{array}{cc}I&0\\0&V^{\ast}\end{array}\right)\\\\&=\left(\begin{array}{cc}
\frac{I+T^{-1}}{2}&T^{-1}B\\B^{\ast}T^{-1}&2B^{\ast}T^{-1}(I+T)^{-1}B\end{array}\right). \end{array}$$

(iii) If $\lambda>0,$ then it is clear that $$S_{\lambda}^{+}
=\lambda\left(\begin{array}{cc}I&\frac{B}{\lambda}\\ \frac{B^{\ast}}{\lambda}&0\end{array}\right)^+=\lambda\left(\begin{array}{cc}I&B_0\\ B_0^{\ast}&0\end{array}\right)^+,$$
where $B_0=\frac{B}{\lambda}.$
 Thus (ii) implies that
$$T=(I+4B_0B_0^{\ast})^{\frac{1}{2}}=(I+\frac{4BB^{\ast}}{\lambda^2})^{\frac{1}{2}}
=\frac{(\lambda^2 I+4BB^{\ast})^{\frac{1}{2}}}{\lambda}$$ and $$P_{S_\lambda^+}=\left(\begin{array}{cc}\frac{I+T^{-1}}{2}&T^{-1}B_0\\B_0^{\ast}T^{-1}
 &2B_0^{\ast}T^{-1}(I+T)^{-1}B_0\end{array}
\right)=\left(\begin{array}{cc}
\frac{I+\lambda T_\lambda^{-1}}{2}
&T_\lambda^{-1}B\\ B^{\ast}T_\lambda^{-1}&
2B^{\ast}T_\lambda^{-1}(\lambda I+T_\lambda)^{-1}B\end{array}
\right).$$

(iv) If $\lambda<0,$ then $$S_\lambda^{+}=\dfrac{S_\lambda+|S_\lambda|}{2}
=\dfrac{\lambda}{2}[\left(\begin{array}{cc}I&\frac{B}{\lambda}\\ \frac{B^{\ast}}{\lambda}&0\end{array}\right)-\begin{vmatrix}\left(\begin{array}{cc}I&\frac{B}{\lambda}\\ \frac{B^{\ast}}{\lambda}&0\end{array}\right)\end{vmatrix}]=-\lambda\left(\begin{array}{cc}I&\frac{B}{\lambda}\\ \frac{B^{\ast}}{\lambda}&0\end{array}\right)^-.$$
Let $B_0=\frac{B}{\lambda}$ and $V_0$ is the unique partial isometry such that
$B_0^*=V_0(B_0B_0^{\ast})^{\frac{1}{2}}$ with $R(V_0)=\overline{R(B_0^*)}=\overline{R(B^*)}$ and $R(V_0^*)=\overline{R(B_0)}=\overline{R(B)}.$ Denoting
 $$S':=\left(\begin{array}{cc}I&\frac{B}{\lambda}\\ \frac{B^{\ast}}{\lambda}&0\end{array}\right)^-=\left(\begin{array}{cc}I&B_0\\ B_0^*&0\end{array}\right)^-,$$
 we conclude from the following Proposition 11 that
$$P_{S'}=\left(\begin{array}{cc}\frac{I-T^{-1}}{2}&-T^{-1}B_0\\-B_0^{\ast}T^{-1}
 &\frac{V_0(T^{-1}+I)V_0^*}{2}\end{array}
\right):\mathcal{H}\oplus\mathcal{K}\rightarrow \mathcal{H}\oplus\mathcal{K},$$  where $T=(I+4B_0B_0^{\ast})^{\frac{1}{2}}=
\frac{(\lambda^2 I+4BB^{\ast})^{\frac{1}{2}}}{-\lambda}.$
Clearly, $V=-V_0,$ then $$P_{S_\lambda^{+}}=P_{S'}=\left(\begin{array}{cc}\frac{I-T^{-1}}{2}&-T^{-1}B_0\\-B_0^{\ast}T^{-1}
 &\frac{V_0(T^{-1}+I)V_0^*}{2}\end{array}
\right)=\left(\begin{array}{cc}
\frac{I+\lambda T_\lambda^{-1}}{2}
&T_\lambda^{-1}B\\ B^{\ast}T_\lambda^{-1}
&\frac{V(I-\lambda T_\lambda^{-1})V^*}{2}\end{array}
\right).$$

(v) It is clear that with respect to the decomposition $\mathcal{H}\oplus\mathcal{H}:$ $\widetilde{C}:=\dfrac{1}{2}\left(\begin{array}{cc}
|B^*|&|B^*|\\ |B^*|&|B^*|\end{array}\right)\geqslant 0.$
By a direct calculation, we get that $$S_0^+=\frac{1}{2}\left(\begin{array}{cc} |B^*|&B\\ B^*&|B|\end{array}
\right)=\left(\begin{array}{cc}I&0\\0&V\end{array}\right)
 \widetilde{C}
\left(\begin{array}{cc}I&0\\0&V^{\ast}\end{array}\right)\ \ \hbox{ and }\ \ \left(\begin{array}{cc}I&0\\0&V^{\ast}V\end{array}\right)
\widetilde{C}
= \widetilde{C}.$$ Thus Lemma 9 implies $$\begin{array}{rcl}P_{S_0^+}&=&\left(\begin{array}{cc}I&0\\0&V\end{array}\right)
P_{\widetilde{C}}\left(\begin{array}{cc}I&0\\0&V^*\end{array}\right)
\\&=&\dfrac{1}{2}\left(\begin{array}{cc}I&0\\0&V\end{array}\right)
\left(\begin{array}{cc}P_{B}
&P_{B}\\ P_{B}&P_{B}\end{array}\right) \left(\begin{array}{cc}I&0\\0&V^*\end{array}\right)
\\&=&\dfrac{1}{2}\left(\begin{array}{cc}P_B&V^*\\
V&P_{B^*}\end{array}
\right).\end{array}$$

\ \ \ \ \ $\square$

{\bf Proposition 11.} Let $B\in \mathcal{B(K,H)}$ and $S\in \mathcal{B}(\mathcal{H}\oplus\mathcal{K})$ has the operator matrix form as $S_1$ in (3.1).  Then

(i) $$S^{-}= \left(\begin{array}{cc}\frac{T+T^{-1}-2I}{4}&\frac{(T^{-1}-I)B}{2}\\ \frac{B^{\ast}(T^{-1}-I)}{2}&B^{\ast}T^{-1}B\end{array}\right):
\mathcal{H}\oplus\mathcal{K}\rightarrow \mathcal{H}\oplus\mathcal{K},$$
where $T=(I+4BB^{\ast})^{\frac{1}{2}}.$

(ii) $P_{S}=diag(I, P_{B^*}):
\mathcal{H}\oplus\mathcal{K}\rightarrow \mathcal{H}\oplus\mathcal{K}.$

(iii)
$$P_{S^-}=\left(\begin{array}{cc}\frac{I-T^{-1}}{2}&-T^{-1}B
\\-B^{\ast}T^{-1}& \frac{V(T^{-1}+I)V^*}{2}\end{array}\right):
\mathcal{H}\oplus\mathcal{K}\rightarrow \mathcal{H}\oplus\mathcal{K},$$ where $V$ is the unique partial isometry such that
$B^{\ast}=V(BB^{\ast})^{\frac{1}{2}}$ with $R(V)=\overline{R(B^*)}$ and $R(V^*)=\overline{R(B)}.$

{\bf Proof.} (i) is immediate from the fact $S^{-}=S^{+}-S.$

(ii) Let $V$ be the unique partial isometry such that
$B^{\ast}=V(BB^{\ast})^{\frac{1}{2}},$ $R(V)=\overline{R(B^*)}$ and $R(V^*)=\overline{R(B)}.$ Setting $$\widetilde{V}=diag(I,V):\mathcal{H}\oplus\mathcal{K}\rightarrow \mathcal{H}\oplus\mathcal{K}$$
and
 $$Q=\left(\begin{array}{cc}I&|B^*|\\|B^{\ast}|& 0\end{array}\right):\mathcal{H}\oplus\mathcal{H}\rightarrow \mathcal{H}\oplus\mathcal{H},$$ we claim that
$N(Q)=\{0\}\oplus N(B^*).$ Indeed, suppose that
$x\in \mathcal{H}$ and $y\in \mathcal{H}$ satisfy $Q(x\oplus y)=0.$ Then
$$x+|B^*|y=0 \ \ \ \ \hbox{ and } \ \ \ \ |B^*|x=0,$$ so $BB^*y=|B^*|(x+|B^*|y)=0,$ which implies
$y\in N(B^*)$ and $x=0.$
Thus $$P_{Q}=diag(I, P_{B}):\mathcal{H}\oplus\mathcal{H}\rightarrow \mathcal{H}\oplus\mathcal{H}.$$
Clearly, $S=\widetilde{V}Q\widetilde{V}^*$ and $\widetilde{V}^*\widetilde{V}Q=Q,$
so Lemma 9 yields that  $$P_{S}=\widetilde{V}P_{Q}\widetilde{V}^*=diag(I, VP_{B}V^*)=diag(I, VV^*)=diag(I, P_{B^*}).$$

(iii) It is easy to verify that
$$\begin{array}{rcl}I-2BB^*T^{-1}(I+T)^{-1}&=&(T+T^2)^{-1}(T+T^2-2BB^*)
\\&=&\dfrac{(T+T^2)^{-1}(2I+4BB^*+2T)}{2}\\&=&\dfrac{(T+T^2)^{-1}(I+T)^2}{2}\\&=&\dfrac{T^{-1}+I}{2}.\end{array}$$ Hence
$$P_{B^*}-2B^{\ast}T^{-1}(I+T)^{-1}B=VV^*-2VBB^*T^{-1}(I+T)^{-1}V^*=\frac{V(T^{-1}+I)V^*}{2},$$ which implies
$$P_{S^-}=P_{S}-P_{S^+}=\left(\begin{array}{cc}\frac{I-T^{-1}}{2}&-T^{-1}B
\\-B^{\ast}T^{-1}& \frac{V(T^{-1}+I)V^*}{2}\end{array}\right):
\mathcal{H}\oplus\mathcal{K}\rightarrow \mathcal{H}\oplus\mathcal{K}.$$\ \ \ \ \ $ \square$

The following is a direct corollary from Theorem 10.

{\bf Corollary 12.} Let $E\in {\mathcal{B(H)}}^{Id}$ have the form (1.1). Then with respect to the decomposition $\mathcal{H}=R(E)\oplus R(E)^{\perp},$

(i)$$(E+E^{\ast})^{+}=\dfrac{1}{2}\left(\begin{array}{cc}T+T^{-1}+2I&(I+T^{-1})E_1
\\E_1^{\ast}(I+T^{-1})&E_1^{\ast}T^{-1}E_1\end{array}\right),$$

(ii)$$P_{(E+E^{\ast})^+}=\dfrac{1}{2}\left(\begin{array}{cc}I+T^{-1}&T^{-1}E_1
\\E_1^{\ast}T^{-1}&E_1^{\ast}(T+T^{2})^{-1}E_1\end{array}\right),$$
where $T=(I+E_1E_1^{\ast})^{\frac{1}{2}}.$

\section{$J$-projections}

A projection $E\in \mathcal{B(H)}^{Id}$ is said to be $J$-projection if $E=JE^{\ast}J,$
which is equivalent to say that $JE$ is self-adjoint.
A $J$-projection $E$ is called to be positive (negative) if $JE\geqslant0 (JE\leqslant0$).
In this section,
 we mainly characterize the symmetry $J$ such that a projection $E$ is the $J$-projection.
In particular, the minimal element of the set of all symmetries $J$ with $JE\geqslant 0$ is given.
The following lemma is needed.

{\bf Lemma 13.} Let $E\in \mathcal{B(H)}^{Id}$ have the form (1.1). If $J$ is a symmetry,
 then $E$ is a $J$-projection if and only if $J$ has the operator matrix form
\begin{equation}J=\left(\begin{array}{cc}J_1(I+E_1E_1^{\ast})^{-\frac{1}{2}}&J_1(I+E_1E_1^{\ast})^{-\frac{1}{2}}E_1\\
E_1^{\ast}(I+E_1E_1^{\ast})^{-\frac{1}{2}}
J_1&J_2(I+E_1^{\ast}E_1)^{-\frac{1}{2}}
\end{array}\right):R(E)\oplus R(E)^{\perp}\rightarrow R(E)\oplus R(E)^{\perp},\end{equation}
 where $J_1$ and $J_2$ are  symmetries on the subspaces
 $R(E)$ and $R(E)^{\perp}$ respectively, satisfying $J_1E_1+E_1J_2=0.$

{\bf Proof.} Sufficiency. If $J_1E_1+E_1J_2=0,$ then $J_1E_1=-E_1J_2,$ which induces
\begin{equation}J_1E_1E_1^{\ast}=-E_1J_2E_1^{\ast}=-E_1(-E_1^{\ast}J_1)=E_1E_1^{\ast}J_1.\end{equation}
Similarly, we get that \begin{equation}J_2E_1^{\ast}E_1=E_1^{\ast}E_1J_2.\end{equation}
 Combining equations (1.1) and (4.1)-(4.3), we easily verify the equation $JE^{\ast}J=E.$

Necessity. Without loss of generality, we may assume
\begin{equation}J=\left(\begin{array}{cc}
Q_{11}&Q_{12}\\Q_{12}^{\ast}&Q_{22}
\end{array}\right):R(E)\oplus R(E)^{\perp}\rightarrow R(E)\oplus R(E)^{\perp},\end{equation}
where $Q_{11}$ and $Q_{22}$ are self-adjoint operators.
It follows from the fact $J^2=I$  that \begin{equation}\begin{cases}Q_{11}^2+
Q_{12}Q_{12}^{\ast}=I\ \qquad\qquad\textcircled{1}
&\\Q_{11}Q_{12}+Q_{12}Q_{22}=0\ \ \qquad \textcircled{2}&\\Q_{12}^{\ast}Q_{12}+Q_{22}^2=I. \qquad\qquad\textcircled{3}\end{cases}\end{equation}

On the other hand, $E=JE^{\ast}J$ implies that $JE$ is self-adjoint, so
\begin{equation}Q_{12}=Q_{11}E_1.\end{equation}
Using equations $ \textcircled{1}$ of (4.5) and (4.6), we have $$Q_{11}(I+E_1E_1^{\ast})Q_{11}=I,$$
which yields $Q_{11}$ is invertible on the subspace $R(E)$ and $Q_{11}^2=(I+E_1E_1^{\ast})^{-1}.$
Thus $$|Q_{11}|=(I+E_1E_1^{\ast})^{-\frac{1}{2}}$$ follows from the fact that  $Q_{11}$ is self-adjoint.
Let $Q_{11}=V|Q_{11}|$ be the polar decomposition of $Q_{11}.$
 Then $V=Q_{11}|Q_{11}|^{-1}$ satisfies
$V=V^{\ast}=V^{-1},$  so $J_1:=V$ is a symmetry on the subspace $R(E)$ and \begin{equation}Q_{11}=J_1|Q_{11}|=J_1(I+E_1E_1^{\ast})^{-\frac{1}{2}}.\end{equation}
Similarly,  equations $ \textcircled{3}$ of (4.5) and (4.6) imply
$$Q_{22}^2=I-E_1^*Q_{11}^2E_1=I-E_1^*(I+E_1E_1^{\ast})^{-1}E_1=
I-(I+E_1^*E_1)^{-1}E_1^{\ast}E_1=(I+E_1^{\ast}E_1)^{-1},$$ since
$(I+E_1^*E_1)^{-1}E_1^{\ast}=E_1^{\ast}(I+E_1E_1^*)^{-1}.$  Thus
\begin{equation}|Q_{22}|=(I+E_1^{\ast}E_1)^{-\frac{1}{2}}.\end{equation}
Setting $J_2:=Q_{22}|Q_{22}|^{-1},$  we easily verify  that $J_2=J_2^{\ast}=J_2^{-1}$  and
\begin{equation}Q_{22}=J_2|Q_{22}|=J_2(I+E_1^{\ast}E_1)^{-\frac{1}{2}}.\end{equation}
Moreover, by equations $ \textcircled{2}$ of (4.5) and (4.7)-(4.9), we get that $$(J_1E_1+E_1J_2)(I+E_1^{\ast}E_1)^{-\frac{1}{2}}=Q_{11}E_1+E_1Q_{22}=0,$$
which induces $J_1E_1+E_1J_2=0.$  \qquad $\square$

In the following, we give some new characterizations for the positivity of the $J$-projections.

{\bf Corollary 14.} Let $E\in \mathcal{B(H)}^{Id}$ have the form (1.1). Then $JE\geqslant0$
if and only if $$J=\left(\begin{array}{cc}(I+E_1E_1^{\ast})^{-\frac{1}{2}}&(I+E_1E_1^{\ast})^{-\frac{1}{2}}E_1\\
E_1^{\ast}(I+E_1E_1^{\ast})^{-\frac{1}{2}}
&J_2(I+E_1^{\ast}E_1)^{-\frac{1}{2}}
\end{array}\right):R(E)\oplus R(E)^{\perp}\rightarrow R(E)\oplus R(E)^{\perp},$$
 where $J_2$ is a symmetry on the subspace
 $R(E)^{\perp}$  with $E_1=-E_1J_2.$

{\bf Proof.} Sufficiency. With respect to the decomposition $R(E)\oplus R(E)^{\perp},$ it is clear that $$JE=\left(\begin{array}{cc}(I+E_1E_1^{\ast})^{-\frac{1}{2}}&(I+E_1E_1^{\ast})^{-\frac{1}{2}}E_1\\
E_1^{\ast}(I+E_1E_1^{\ast})^{-\frac{1}{2}}&
E_1^{\ast}(I+E_1E_1^{\ast})^{-\frac{1}{2}}
E_1\end{array}\right).$$
Calculating the Schur complement of $JE,$ we know $JE\geqslant0.$

 Necessity. By Lemma 13, we only need to show $J_1=I.$
 It is easy to see that $JE\geqslant0$ implies
 $$Q_{11}=J_1|Q_{11}|=J_1(I+E_1E_1^{\ast})^{-\frac{1}{2}}\geqslant0,$$ so
  $$J_1(I+E_1E_1^{\ast})^{-\frac{1}{2}}=Q_{11}=|Q_{11}|=(I+E_1E_1^{\ast})^{-\frac{1}{2}},$$
which yields $J_1=I$ as desired. \qquad $\square$

{\bf Theorem 15.} Let $E\in \mathcal{B(H)}^{Id}.$ Then  \begin{equation}min\{J: JE\geqslant0,\ J=J^{\ast}=J^{-1}\}=2P_{(E+E^{\ast})^{+}}-I,\end{equation}
where the ``min" is in the sense of Loewner partial order.

{\bf Proof.} Suppose that $E\in \mathcal{B(H)}^{Id}$ has the form (1.1). Then by Corollary 12,
 we get that $$2P_{(E+E^{\ast})^{+}}-I=
\left(\begin{array}{cc}T^{-1}&T^{-1}E_1
\\E_1^{\ast}T^{-1}&E_1^{\ast}(T+T^2)^{-1}
E_1-I\end{array}\right):R(E)\oplus R(E)^{\perp}\rightarrow R(E)\oplus R(E)^{\perp},$$
where $T=(I+E_1E_1^{\ast})^{\frac{1}{2}}.$

It is easy to see that  $$\begin{array}{rl}E_1^{\ast}(T+T^2)^{-1}E_1-I&=E_1^{\ast}[(I+E_1E_1^{\ast})^{\frac{1}{2}}+I+E_1E_1^{\ast}]^{-1}E_1-I
\\&=[(I+E_1^{\ast}E_1)^{\frac{1}{2}}+I+E_1^{\ast}E_1]^{-1}E_1^{\ast}E_1-I
\\&=-[(I+E_1^{\ast}E_1)^{\frac{1}{2}}+I+E_1^{\ast}E_1]^{-1}[I+(I+E_1^{\ast}E_1)^{\frac{1}{2}}]
\\&=-(I+E_1^{\ast}E_1)^{-\frac{1}{2}}.\end{array}$$

Setting $J_2=-I$ in proof of Corollary 14, we get $(2P_{(E+E^{\ast})^{+}}-I)E\geqslant0,$
so the left hand is no more than the right hand of equation (4.10).

 If $JE\geqslant0,$ then by Corollary 14, we have $$J=\left(\begin{array}{cc}(I+E_1E_1^{\ast})^{-\frac{1}{2}}&(I+E_1E_1^{\ast})^{-\frac{1}{2}}E_1\\
E_1^{\ast}(I+E_1E_1^{\ast})^{-\frac{1}{2}}&
J_2(I+E_1^{\ast}E_1)^{-\frac{1}{2}}\end{array}
\right):R(E)\oplus R(E)^{\perp}\rightarrow R(E)\oplus R(E)^{\perp}.$$  Obviously, $J_2E_1^*E_1=E_1^*E_1J_2$ implies $$J_2(I+E_1^{\ast}E_1)^{-\frac{1}{2}}-[-(I+E_1^{\ast}E_1)^{-\frac{1}{2}}]=(I+J_2)(I+E_1^{\ast}E_1)^{-\frac{1}{2}}\geqslant0,$$
which yields $J\geqslant2P_{(E+E^{\ast})^{+}}-I$ as desired.\qquad $\square$

{\bf Proposition 16.}([11, Proposition 4]) Let $E\in \mathcal{B(H)}^{Id}$ and $J$ be a symmetry.  If $E=JE^{\ast}J,$ then

(a) $JE\geqslant0$ if and only if $J\geqslant2P_{(E+E^{\ast})^{+}}-I.$

(b) $JE\geqslant0$ if and only if $(J+I)E(J+I)\geqslant0  \ \ \ \hbox{ and } \ \ \ R(E)\cap R(I-J)=\{0\}.$

{\bf Proof.} (a) Necessity is clear by Theorem 15.

Sufficiency.  Since $E=JE^{\ast}J,$  it follows from Lemma 13 that $$J=\left(\begin{array}{cc}J_1(I+E_1E_1^{\ast})^{-\frac{1}{2}}&J_1(I+E_1E_1^{\ast})^{-\frac{1}{2}}E_1\\
E_1^{\ast}(I+E_1E_1^{\ast})^{-\frac{1}{2}}J_1
&J_2(I+E_1^{\ast}E_1)^{-\frac{1}{2}}
\end{array}\right):R(E)\oplus R(E)^{\perp}\rightarrow R(E)\oplus R(E)^{\perp}.$$ Then $J\geqslant2P_{(E+E^{\ast})^{+}}-I$ yields
$J_1(I+E_1E_1^{\ast})^{-\frac{1}{2}}\geqslant(I+E_1E_1^{\ast})^{-\frac{1}{2}},$ so
$J_1\geqslant I$ follows from the fact $J_1E_1E_1^*=E_1E_1^*J_1.$
Thus $J_1=I,$ as $J_1=J_1^*=J_1^{-1}.$  Then Corollary 14 induces $JE\geqslant0$ as desired.

 (b)  Necessity. If $JE\geqslant0,$ then $$(J+I)E(J+I)=(J+I)JE(J+I)\geqslant0.$$
Let $x\in R(E)\cap R(I-J).$ Then there exists $y\in \mathcal{H}$ such that $x=Ex=(I-J)y,$  so $$Jx=JEx=J(I-J)y=(J-I)y=-x,$$
which yields $$0\leqslant\langle JEx,x\rangle=\langle -x,x\rangle=-\|x\|^{2}.$$
 Thus $x=0.$

Sufficiency. By Corollary 14, we need to prove that $J_1=I.$

Since $J_1=J_1^*=J_1^{-1},$ then without loss of generality,
we assume \begin{equation}J_1=\left(\begin{array}{cc}
I&0\\0&-I\end{array}\right):\mathcal{M}\oplus \mathcal{M}^{\perp}\rightarrow \mathcal{M}\oplus \mathcal{M}^{\perp},\end{equation}
 where $\mathcal{M}\subseteq R(E)$ and $\mathcal{M}^{\perp}=R(E)\ominus \mathcal{M}.$
Then $J_1E_1E_1^{\ast}=E_1E_1^{\ast}J_1$ implies that $\mathcal{M}$ is a reduced subspace of $E_1E_1^{\ast},$
so \begin{equation}(I+E_1E_1^{\ast})^{-\frac{1}{2}}
=\left(\begin{array}{cc}Q_1&0\\0&Q_2
\end{array}\right):\mathcal{M}\oplus \mathcal{M}^{\perp}\rightarrow \mathcal{M}\oplus \mathcal{M}^{\perp},\end{equation}
 where $Q_1$ and $Q_2$ are positive and invertible.
Using equation (4.7), we have \begin{equation}Q_{11}=J_1(I+E_1E_1^{\ast})^{-\frac{1}{2}}
=\left(\begin{array}{cc}Q_1&0\\0&-Q_2
\end{array}\right):\mathcal{M}\oplus \mathcal{M}^{\perp}\rightarrow \mathcal{M}\oplus \mathcal{M}^{\perp}.\end{equation}
  By equations (1.1),(4.1) and a direct calculation, we know that $(J+I)E(J+I)\geqslant0$ implies $$(Q_{11}+I)^2+(Q_{11}+I)E_1E_1^{\ast}Q_{11}\geqslant0,$$
so $$Q_{11}(I+E_1E_1^*)Q_{11}+Q_{11}+(I+E_1E_1^*)Q_{11}+I=(Q_{11}+I)^2+(Q_{11}+I)E_1E_1^{\ast}Q_{11}\geqslant0.$$ Using again equation (4.7), we get that  $$Q_{11}(I+E_1E_1^*)Q_{11}=I  \hbox{   }\hbox{ and }\hbox{   }
(I+E_1E_1^*)Q_{11}=Q_{11}^{-1},$$ so $$Q_{11}+2I+Q_{11}^{-1}
=Q_{11}(I+E_1E_1^*)Q_{11}+Q_{11}+
(I+E_1E_1^*)Q_{11}+I\geqslant0.$$
Then
$$\left(\begin{array}{cc}Q_1+2I+Q_1^{-1}&0\\0&-Q_2+2I-Q_2^{-1}\end{array}\right)\geqslant0$$ follows from equation (4.13),
which yields $$-(Q_2^\frac{1}{2}-Q_2^{-\frac{1}{2}})^2=-Q_2+2I-Q_2^{-1}\geqslant0,$$
so $Q_2^\frac{1}{2}-Q_2^{-\frac{1}{2}}=0.$ Thus $Q_2=I,$ which implies
Thus \begin{equation} (I+E_1E_1^{\ast})^{-\frac{1}{2}}=\left(\begin{array}{cc}Q_1&0\\0&I\end{array}\right) \ \ \
\hbox{ and } \ \ \ \ Q_{11}=\left(\begin{array}{cc}Q_1&0\\0&-I\end{array}\right).\end{equation}

Let $x\in \mathcal{M}^{\perp}.$ Then $(I+E_1E_1^{\ast})^{-\frac{1}{2}}x=x,$
which yields $(I+E_1E_1^{\ast})x=x,$ so $E_1E_1^{\ast}x=0.$
Hence, \begin{equation} x\in N(E_1^{\ast})\subseteq R(E).\end{equation}

Moreover, combining equations (4.1) and (4.14), we conclude that
$$Jx=\left(\begin{array}{cc}Q_{11}x\\E_1^{\ast}Q_{11}x\end{array}\right)
=\left(\begin{array}{cc}-x\\-E_1^{\ast}x\end{array}\right)
=\left(\begin{array}{cc}-x\\0\end{array}\right),$$ which induces $(I-J)x=x-Jx=2x,$
so \begin{equation}x=(I-J)\dfrac{x}{2}\in R(I-J).\end{equation}
Thus equations (4.15) and (4.16) imply $$x\in R(I-J)\cap R(E)=\{0\}.$$
 Then $\mathcal{M}^{\perp}=0,$ so $ J_1=I$ as desired.
 \ \ \ \ \ $\square$

{\bf Proposition 17.} ([2, Theorem 2.1]) Let $\mathcal{M}$ be a non-trivial closed subspace of $\mathcal{H}$ and $J$ be a symmetry.
Then there exists $E\in\mathcal{B(H)}^{Id}$ such that $R(E)=\mathcal{M}$ and
$JE^{\ast}J=E$ if and only if $P_{\mathcal{M}}J\mid_{\mathcal{M}}$
is invertible on the subspace $\mathcal{M}.$
In this case, $E$ is unique.

{\bf Proof.} Let $$J=\left(\begin{array}{cc}Q_{11}&Q_{12}
\\Q_{12}^{\ast}&Q_{22}\end{array}\right):
\mathcal{M}\oplus \mathcal{M}^{\perp}\rightarrow \mathcal{M}\oplus \mathcal{M}^{\perp},$$
where $Q_{11}$ and $Q_{22}$ are self-adjoint operators.

Sufficiency. If $P_{\mathcal{M}}J\mid_{\mathcal{M}}$ is invertible on the subspace $\mathcal{M},$
then $Q_{11}$ is invertible.
Define an idempotent operator $E$ as $$E=\left(\begin{array}{cc}I& Q_{11}^{-1}Q_{12}\\0&0\end{array}\right):
\mathcal{M}\oplus \mathcal{M}^{\perp}\rightarrow\mathcal{M}\oplus \mathcal{M}^{\perp}.$$
It is easy to verify that $R(E)=\mathcal{M}$ and
$JE^{\ast}J=E.$

Necessity. Suppose that there exists $E\in\mathcal{B(H)}^{Id}$ such that $R(E)=\mathcal{M}$ and
$JE^{\ast}J=E.$ Then by Lemma 13 $$J=\left(\begin{array}{cc}
J_1(I+E_1E_1^{\ast})^{-\frac{1}{2}}
&J_1(I+E_1E_1^{\ast})^{-\frac{1}{2}}E_1\\
E_1^{\ast}(I+E_1E_1^{\ast})^{-\frac{1}{2}}
J_1&J_2(I+E_1^{\ast}E_1)^{-\frac{1}{2}}
\end{array}\right):\mathcal{M}\oplus \mathcal{M}^{\perp}\rightarrow \mathcal{M}\oplus \mathcal{M}^{\perp},$$ where $E$ has the form as (1.1). Thus $P_{\mathcal{M}}J\mid_{\mathcal{M}}=
J_1(I+E_1E_1^{\ast})^{-\frac{1}{2}}$
is invertible on the subspace $\mathcal{M}.$
Also, Lemma 13 (the above equation) implies $E_1=(P_{\mathcal{M}}J\mid_{\mathcal{M}})^{-1}
P_{\mathcal{M}}J\mid_{\mathcal{M}^{\perp}}$
is unique, which says the uniqueness of $E.$    \ \ \ \ \ $\square$

The following result is obtained in [2,7,11].
However, our methods are completely different from those of [2,7,11].

{\bf Corollary 18.} For a $J$-projection $E$, there exists uniquely a $J$-positive projection $Q,$
a $J$-negative projection $R$ such that $$E=Q+R,\ \ \  QR=RQ=0\ \ \ \hbox {and} \ \ \ QR^{\ast}=R^{\ast}Q=0.$$

{\bf Proof.} Suppose that $E$ has operator matrix form (1.1).
Using Lemma 13, we conclude that $JEJ=E^{\ast}$ is equivalent to that $J$ has the form (4.1).
With respect to the space decomposition $\mathcal{H}=R(E)\oplus R(E)^{\perp},$ we define $Q$ and $R$ as \begin{equation}Q:=\left(\begin{array}{cc}
\dfrac{I_1+J_1}{2}&\dfrac{I_1+J_1}{2}E_1
\\0&0\end{array}\right) \quad \hbox{ and }\  \  \
R:=\left(\begin{array}{cc}\dfrac{I_1-J_1}{2}&
\dfrac{I_1-J_1}{2}E_1\\\\0&0\end{array}\right),\end{equation}
where $I_1$ is the identity operator
on the subspace $R(E).$  Clearly, $Q$ and $R$ are projections with $E=Q+R.$
A direct calculation yields that $QR=RQ=0\ \hbox{and}\ QR^{\ast}=R^{\ast}Q=0.$
Moreover, combining equations (4.1) and (4.17), we get that $$JQ=\left(\begin{array}{cc}I_1&0\\0&E_1^{\ast}\end{array}\right)
\left(\begin{array}{cc}\dfrac{I_1+J_1}{2}T^{-1}&
\dfrac{I_1+J_1}{2}T^{-1}\\\\
\dfrac{I_1+J_1}{2}T^{-1}&\dfrac{I_1+J_1}{2}T^{-1}\end{array}\right)
\left(\begin{array}{cc}I_1&0\\0&E_1\end{array}\right)\geqslant0$$
and $$JR=-\left(\begin{array}{cc}I_1&0\\0&E_1^{\ast}\end{array}\right)
\left(\begin{array}{cc}\dfrac{I_1-J_1}{2}T^{-1}
&\dfrac{I_1-J_1}{2}T^{-1}\\\\
\dfrac{I_1-J_1}{2}T^{-1}&\dfrac{I_1-J_1}{2}T^{-1}\end{array}\right)
\left(\begin{array}{cc}I_1&0\\0&E_1\end{array}\right)\leqslant0,$$ since $\dfrac{I_1+J_1}{2}T^{-1}\geqslant0$ and $\dfrac{I_1-J_1}{2}T^{-1}\geqslant0,$ where $T=(I+E_1E_1^*)^{\frac{1}{2}}.$
Thus $Q$ is a $J$-positive projection and
 $R$ is a $J$-negative projection.

 To show uniqueness, we assume that $E=Q_1+Q_2$, where $Q_1$ and $Q_2$ satisfy $JQ_1\geqslant0,$\ $JQ_2\leqslant0$
and $Q_1Q_2^{\ast}=Q_2^{\ast}Q_1=0.$ It is obvious that $$JE=JQ_1+JQ_2 \ \
\hbox{ and }\ \ (JQ_2) (JQ_1)=(Q_2^{\ast}J)(JQ_1)=Q_2^{\ast}Q_1=0.$$
 Since $(JR)(JQ)=0$ and $JE=JQ+JR,$ then $JQ=JQ_1$ and $ JR=JQ_2$ follow from the uniqueness of the positive part and the negative part for a self-adjoint operator.
 Thus $Q=Q_1$ and $ R=Q_2.$ \qquad  $\square$

{\bf Remark 19.} In general, a $J$-projection $E$ is representable (not uniquely) as $E=E_++E_-,$
where $E_+$ and $E_-$ are positive and negative $J$-projection, respectively.


text:
\section{Numerical tests}
\label{sec:numerical}

In this section, we will demonstrate the performance of the IMEX$k$-DG$k$-S scheme, $k=1, 2, 3$. Two models will be considered. One is the telegraph equation with $\Omega_v=\{-1,1\}$ and $\lgl f \rgl = \frac{1}{2}(f|_{v=1}+f|_{v=-1})$, the other is the one-group transport equation in slab geometry with $\Omega_v=[-1,1]$ and $\lgl f\rgl = \frac{1}{2}\int_{-1}^1 fdv$. For the latter, we discretize the velocity space

with $16$ Gaussian quadrature points. The meshes in space are uniform unless otherwise specified.

Based on the energy and Fourier analysis in Section \ref{sec:stability}, for the one-group transport equation in slab geometry, the time step size $\Dt$ for the IMEX$k$-DG$k$-S scheme is chosen as $\Dt_{CFLk}$,
\begin{subequations}
\label{eqn:CFL}
\begin{align}
\mbox{IMEX1-DG1-S}: \hspace{0.2in} &  \Delta t_{CFL1}=\left\{
\begin{array}{ll} 0.75 h,& \vareps\leq  0.5h, \\
\min(0.75 h ,  \frac{\vareps^2 h }{\vareps-0.5h}),&  \vareps>  0.5h,
\end{array}
\right.\\
\mbox{IMEX2-DG2-S}: \hspace{0.2in} &  \Delta t_{CFL2}=\left\{
\begin{array}{ll} 0.75 h,& \vareps\leq  0.025h,\\
\min(0.75h,\frac{\vareps^2 h/\sqrt{10} }{\vareps-0.025h}), & \vareps\geq 0.025h,
\end{array}
\right.\label{eqn:CFL:2}\\
\mbox{IMEX3-DG3-S}: \hspace{0.2in} &  \Delta t_{CFL3}=\left\{
\begin{array}{ll} 0.75 h,& \vareps\leq  0.05h,\\
\min(0.75 h ,  \frac{0.1\vareps^2 h }{\vareps-0.05h}),&  \vareps> 0.05h.
\end{array}
\right.
\end{align}
\end{subequations}
When the schemes are unconditionally stable, $\Dt=0.75 h$ is used to ensure good resolution. The time step conditions in \eqref{eqn:CFL} also work well for the telegraph equation.
We want to mention that, when boundary conditions are Dirichlet (see next subsection), due to the numerical boundary treatment, a smaller time step size is taken for the second order IMEX2-DG2-S scheme in the diffusive regime.

For the linear solver, we apply the Schur complement discussed in Section \ref{sec:matrix-vector} and GMRES \cite{saad1986gmres} solver, which is implemented under the framework of C++ library PETSC \cite{balay2019petsc}.

\subsection{Numerical boundary condition}\label{sec:numerical_bc}

For some numerical tests, the following inflow (also Dirichlet) boundary conditions are given:
$$ f(x_L,v,t) = f_L(v,t),\;\;v\geq0\quad\text{and}\quad f(x_R,v,t)=f_R(v,t),\;\;v\leq 0.$$
These conditions are insufficient to define the boundary conditions for $\rho=\lgl f\rgl$ and $g$, hence numerical treatments are needed. We here adopt a close-loop strategy similar to \cite{jang2015high,peng2019stability}.

For simplicity, we present the strategy using the one-group transport equation in slab geometry with the velocity space being continuous. In implementation, we substitute integrals in $\Omega_v$ with their discrete counterparts.

\medskip
\noindent\textbf{Main idea:} Our numerical boundary treatment is based on the following idea. At the left boundary, we set
\begin{subequations}
\label{eq:close-loop}
\begin{align}
&\rho_L(t)+\vareps g_L(v, t)=f_L(v,t), \quad v\geq0 \quad\text{(inflow)},
\label{eq:close-loop1}\\
&\rho_L(t)+\vareps g_L(v,t)=\rho_h(x_{\frac{1}{2}}^{+}, t)+\vareps g_h(x_{\frac{1}{2} }^{+},v, t), \quad v<0\quad\text{(outflow)},
\label{eq:close-loop2}\\
&\lgl g_L(v, t)\rgl=0.
\label{eq:close-loop3}
\end{align}
\end{subequations}
Integrate \eqref{eq:close-loop1} in $v$ from $0$ to $1$ and \eqref{eq:close-loop2} from $-1$ to $0$, and sum them up, we get
\begin{subequations}
\begin{align}
&\rho_L=\rho_L(t) = \frac{1}{2}\Big(\rho_h(x_{\frac{1}{2}}^{+}, t)+\int_{0}^{1}f_L(v,t)dv+\vareps\int_{-1}^0 g_h(x_{\frac{1}{2} }^{+},v, t)dv\Big),\\
&g_L=g_L(v,t) = \begin{cases}
		\frac{1}{\vareps}( f_L(v,t) - \rho_L(t) ),\;\;v\geq 0,\\
		\frac{1}{\vareps}(\rho_h(x_{\frac{1}{2}}^{+}, t)+\vareps g_h(x_{\frac{1}{2} }^{+},v, t)-\rho_L(v,t) ),\;\;v<0.
		\end{cases}
\end{align}
\end{subequations}
With a similar idea, we obtain at  the right boundary
\begin{subequations}
\begin{align}
&\rho_R=\rho_R(t) = \frac{1}{2}\Big(\rho_h(x_{N+\frac{1}{2}}^{-}, t)+\int_{-1}^{0}f_R(v,t)dv+\vareps\int_{0}^1 g_h(x_{N+\frac{1}{2} }^{-},v, t)dv\Big),\\
&g_R=g_R(v,t) = \begin{cases}
		\frac{1}{\vareps}(\rho_h(x_{N+\frac{1}{2}}^{-}, t)+\vareps g_h(x_{N+\frac{1}{2} }^{-},v, t)-\rho_R(v,t) ),\;\;v\geq 0,\\
		\frac{1}{\vareps}( f_R(v,t) - \rho_R(t) ),\;\;v<0.
		\end{cases}
\end{align}
\end{subequations}

\noindent\textbf{Numerical flux:} The boundary strategies are imposed through numerical fluxes. On boundaries, we modify numerical fluxes in \eqref{eq:flux} to be
\begin{subequations}
\begin{align}
&(\breve{\rho_h})_{\frac{1}{2}}:=\rho_L,\;\;\widehat{\lgl vg_h\rgl}_\frac{1}{2}:= {\lgl vg_h\rgl}_{\frac{1}{2} }^+,\;\;\\
&(\breve{\rho_h})_{N+\frac{1}{2}}:=\rho_R,\;\;\widehat{\lgl vg_h\rgl}_{N+\frac{1}{2} }:= {\lgl vg_h\rgl}_{N+\frac{1}{2} }^-+c_R(\rho_R-\rho_{N+\frac{1}{2} }^-) ,
\\
&(\widetilde{vg_h})_\frac{1}{2}:=
\left\{
\begin{array}{ll}
{vg_L},&\mbox{if}\; v\geq0\\
({vg_h})_\frac{1}{2}^+,&\mbox{if}\; v<0
\end{array}
\right.
,\quad
(\widetilde{vg_h})_{N+\frac{1}{2} }:=
\left\{
\begin{array}{ll}
({vg_h})_{N+\frac{1}{2}}^-,&\mbox{if}\; v\geq0\\
vg_R,&\mbox{if}\; v<0
\end{array}
\right. .
\end{align}
\end{subequations}
The penalty term $c_R(\rho_R-\rho_{N+\frac{1}{2} }^-)$ is added to maintain the accuracy of the schemes, and one can refer to \cite{castillo2002optimal,li2017optimal} for details on the role of this penalty term. In our simulation, we take $c_R=1$.

We want to mention that, due to this numerical boundary treatment, the IMEX$2$-DG$2$-S scheme is no longer unconditionally stable in the diffusive regime. We modify the time step condition in \eqref{eqn:CFL} with  $\Dt_{CFL2}=0.1h$ when $\vareps\leq 0.025h$. Note that this time step condition can still be larger than
a parabolic time step condition $\Dt= O(h^2)$.

\subsection{Numerical examples}

\textbf{Example 1: smooth example \cite{peng2019stability}.}  We consider the one-group transport equation in slab geometry with a smooth example  on $\Omega_x=[0,2\pi]$. The initial conditions are
$$
\rho(x,0)=\sin(x),\qquad
g(x,v,0)=-v \cos(x),
$$
with periodic boundary conditions, and $\sigma_s=1$, $\sigma_a=0$. $\Omega_x$ is partitioned with a uniform mesh and we define $N=\frac{(x_R-x_L)}{h}$. Numerical errors in $L_\infty$-norm and convergence orders are obtained by Richardson extrapolation:
\begin{subequations}
\begin{align}
&E^{\rho}_{N}=|| \rho_h(x,T)-\rho_{\frac{h}{2}}(x,T) ||_{L_{\infty} (\Omega_x)},\quad\text{and},\quad O^{\rho}_{N}=log_{2}(E^{\rho}_N/E^{\rho}_{2N}),\\
&E^{g}_{N}=\max_{j=1,\dots N_v}|| g_h(x,v_j,T)-g_{\frac{h}{2}}(x,v_j,T) ||_{L_{\infty} (\Omega_x)},\quad\text{and},\quad O^{g}_{N}=log_{2}(E^{g}_N/E^{g}_{2N}).
\end{align}
\end{subequations}
Numerical results at $T=1$ with $\vareps=0.5$, $\vareps=10^{-2}$ and $\vareps=10^{-6}$ are shown in Tables \ref{tab:accuracy1}-\ref{tab:accuracy3}.

 We observe that the IMEX$k$-DG$k$-S scheme, $k=1, 2, 3$, has the optimal $k$-th order accuracy that seems to be uniform in
$\vareps$.

\begin{table}[htbp]
  \centering
  \caption{Errors and convergence orders for the example 1, IMEX1-DG1-S}
  \vspace{0.1in}
    \begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|}
    \hline
	$\vareps$& $N$ & $E^{\rho}_N$  & order  &$E_N^g$ &order \\
    \hline
     \multirow{5}{*}{0.5} & 10  &1.921E-02& - & 1.909E-02&- \\
     					       & 20 & 8.709E-02&1.14 & 8.390E-03 & 1.19\\
					       & 40 & 3.540E-02 & 1.30 & 3.699E-03& 1.18\\
     					       & 80 & 1.619E-03& 1.13 &  1.744E-03& 1.08\\
					       & 160 &7.737E-04& 1.07 & 8.469E-04&1.04\\ \hline
     \multirow{5}{*}{$10^{-2}$} & 10  &1.029E-02& - & 1.467E-02 &-\\
     					       & 20 & 4.371E-03& 1.23 & 6.742E-03 &1.12\\
					       & 40 & 2.014E-03 & 1.12 & 3.204E-03& 1.07\\
     					       & 80 & 9.648E-04& 1.06 & 1.553E-03& 1.04\\
					       & 160 &5.265E-04& 0.87 & 7.986E-04&0.96\\ \hline
     \multirow{5}{*}{$10^{-6}$} & 10  & 1.011E-02 & - &  1.459E-02& - \\
     					       & 20 & 4.306E-03 & 1.23 & 6.709E-03  & 1.12 \\
					       & 40 &  1.988E-03 & 1.12 &3.189E-03&  1.07 \\
     					       & 80 & 9.520E-04& 1.06 & 1.546E-03& 1.04\\
					       & 160 &4.657E-04& 1.03 & 7.618E-04 &1.02  \\ \hline

          \end{tabular}
          \label{tab:accuracy1}
\end{table}

\begin{table}[htbp]
  \centering
  \caption{Errors and convergence orders for the example 1, IMEX2-DG2-S}
    \vspace{0.1in}
    \begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|}
    \hline
	$\vareps$& $N$ & $E^{\rho}_N$  & order  &$E_N^g$ &order \\
    \hline
     \multirow{5}{*}{0.5} & 10  &3.505E-02& - & 3.911E-02 &- \\
     					       & 20 & 8.916E-03&1.97 & 9.991E-03 &1.97\\
					       & 40 & 2.205E-03 & 2.02 & 2.590E-03& 1.95\\
     					       & 80 & 5.479E-04& 2.01 &  6.563E-04& 1.98\\
					       & 160 &1.365E-04& 2.01 & 1.650E-04&1.99\\ \hline
     \multirow{5}{*}{$10^{-2}$} & 10  &3.519E-02& - & 4.215E-02 &-\\
     					       & 20 & 8.763E-03& 2.01 & 8.869E-03 &2.25\\
					       & 40 & 2.206E-03 &2.00 & 2.283E-03& 1.96\\
     					       & 80 & 5.523E-04& 2.00 & 5.906E-04& 1.95\\
					       & 160 &1.381E-04&2.00 & 1.536E-04&1.94\\ \hline
     \multirow{5}{*}{$10^{-6}$} & 10  & 3.518E-02 & - &  3.482E-02&- - \\
     					       & 20 & 8.726E-03 & 2.01 & 8.629E-03  & 2.01 \\
					       & 40 & 2.195E-03 & 1.99 &2.172E-03&  1.99 \\
     					       & 80 & 5.494E-04& 2.00 & 5.435E-04& 2.00\\
					       & 160 &1.374E-04& 2.00 & 1.360E-04 &2.00  \\ \hline

          \end{tabular}
          \label{tab:accuracy2}
\end{table}

\begin{table}[htbp]
  \centering
  \caption{Errors and convergence orders for the example 1, IMEX3-DG3-S}
    \vspace{0.1in}
    \begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|}
    \hline
	$\vareps$& $N$ & $E^{\rho}_N$  & order  &$E_N^g$ &order \\
    \hline
     \multirow{5}{*}{0.5} 	       & 10  &2.588E-03& - & 2.676E-03 &- \\
     					       & 20 & 3.215E-04&  3.01 & 4.103E-04 &2.71\\
					       & 40 & 4.028E-05 & 3.00 & 6.495E-05& 2.66\\
     					       & 80 & 5.036E-06&  3.00 & 9.198E-06& 2.82\\
					       & 160 &6.303E-07& 3.00 & 1.22E-06&   2.91\\ \hline
     \multirow{5}{*}{$10^{-2}$}   & 10  &2.510E-03& - & 2.543E-03 &-\\
     					       & 20 & 3.214E-04& 2.97 & 3.724E-04 &2.77\\
					       & 40 & 4.039E-05 & 2.99 & 1.109E-04& 1.75\\
     					       & 80 & 5.061E-06& 3.00 & 5.292E-06& 4.39\\
					       & 160 &6.328E-07& 3.00 & 6.659E-07&2.99\\
					       & 320 &7.910E-08& 3.00 & 8.355E-08&2.99\\
					       \hline
     \multirow{5}{*}{$10^{-6}$}   & 10  & 2.505E-03 & - &  2.554E-03&- \\
     					       & 20 & 3.211E-04 & 2.96 & 3.174E-04  & 3.01 \\
					       & 40 & 4.041E-05 & 2.99 &3.998E-05&  3.00 \\
     					       & 80 & 5.060E-06& 3.00 & 5.007E-06& 3.00\\
					       & 160 &6.327E-07& 3.00 &  6.269E-07 &3.00  \\ \hline

          \end{tabular}
          \label{tab:accuracy3}
\end{table}

\smallskip
\textbf{Example 2: two-material problem \cite{larsen1989asymptotic,lemou2008new}.} We consider a two-material problem on $\Omega_x=[0, 11]$ and $\Omega_v=[-1, 1]$ with isotropic inflow boundary conditions. The setup is as follows.
\begin{align}
&\sigma_s=0,\;\;\sigma_a=1,\;\;\text{if}\; x\in[0,1],\notag\\
&\sigma_s=100,\;\;\sigma_a=0,\;\;\text{if}\;x\in[1,11],\notag\\
&f_L(v,t) = 5,\;\; f_R(v,t)=0,\;\;f(x,v,0) = 0,\;\;
\end{align}
and $\vareps=1$.

We examine the numerical solutions at a shorter time
$T=1.5$ and also the steady state solution obtained at $T=20000$. This problem has a pure absorbing region with the length of one mean-free path on the left and a pure scattering region with the length of $1000$ mean-free path on the right. Subregiones with different scales coexist.
From the left boundary, an isotropic inflow enters the computational region, and it instantly becomes anisotropic. An interior layer is formed near the interface between the absorbing region and the scattering region.

We use a non-uniform mesh with $h=h^{(1)}=\frac{1}{20}$ on $[0,1]$ and $h=h^{(2)}=\frac{1}{2}$ on $[1,11]$, and the time step $\Dt$ is determined by  \eqref{eqn:CFL} using $h=h^{(1)}$. With this example, we also want to compare the performance of  the IMEX$k$-DG$k$-S schemes proposed here and the IMEX$k$-LDG$k$ schemes in \cite{peng2019stability} with the weight function $\omega = \exp(-\frac{\varepsilon}{100h})|_{h=h^{(1)}}$.
Numerical results are shown in Figure \ref{fig:two_material_short} and Figure \ref{fig:two_material_long}.  The reference solution is obtained by  the first order forward Euler upwind finite difference scheme applied to the original kinetic equation \eqref{eq:f_model} with $h=\frac{11}{20000}$ and $\Dt=10^{-5}$ for $T=1.5$, and with $h=\frac{11}{2000}$ and $\Dt=10^{-4}$ for $T=20000$.

At $T=1.5$, our proposed schemes capture the solutions very well. The third order IMEX3-LDG3-S scheme has the best result.  We observe that the IMEX$k$-DG$k$-S schemes outperform the IMEX$k$-LDG$k$ schemes with the chosen weight. Unlike for IMEX$k$-LDG$k$ schemes, one does not need to choose a weight function for our proposed methods for this example when both transport dominant and diffusion dominant regions coexist.

At $T=20000$ when the solution reaches its steady state, the numerical solutions by the IMEX$k$-DG$k$-S scheme, $k=1, 2, 3$ match the reference solutions well, and they are comparable with those in \cite{peng2019stability} by  IMEX$k$-LDG$k$ scheme. Higher order schemes lead to better resolution as expected.

\begin{figure}
  \centering
  \subfigure[$\rho$, IMEX$1$-DG$1$-S]
  {
     \includegraphics[width=0.31\linewidth]{picture/two_materials_short_time_rho_imex1.png}
    }
  \subfigure[$\rho$, IMEX$2$-DG$2$-S]
  {
        \includegraphics[width=0.31\linewidth] {picture/two_materials_short_time_rho_imex2.png}
    }
   \subfigure[$\rho$, IMEX$3$-DG$3$-S]
  {
        \includegraphics[width=0.31\linewidth] {picture/two_materials_short_time_rho_imex3.png}
    }
      \subfigure[$\rho$, IMEX$1$-LDG$1$]
  {
     \includegraphics[width=0.31\linewidth]{picture/two_materials_short_time_rho_imex1_old.png}
    }
  \subfigure[$\rho$, IMEX$2$-LDG$2$]
  {
        \includegraphics[width=0.31\linewidth] {picture/two_materials_short_time_rho_imex2_old.png}
    }
   \subfigure[$\rho$, IMEX$3$-LDG$3$]
  {
        \includegraphics[width=0.31\linewidth] {picture/two_materials_short_time_rho_imex3_old.png}
    }
  \caption{ Example 2: two-material problem

  $T=1.5$, zoomed in with $x\in[0,2]$.
  		}
  \label{fig:two_material_short}
\end{figure}

\begin{figure}
  \centering
  \subfigure[$\rho$, IMEX$1$-DG$1$-S]
  {
     \includegraphics[width=0.31\linewidth]{picture/two_materials_rho_imex1.png}
    }
  \subfigure[$\rho$, IMEX$2$-DG$2$-S]
  {
        \includegraphics[width=0.31\linewidth] {picture/two_materials_rho_imex2.png}
    }
   \subfigure[$\rho$, IMEX$3$-DG$3$-S]
  {
        \includegraphics[width=0.31\linewidth] {picture/two_materials_rho_imex3.png}
    }
  \caption{ Example 2: two-material problem,

   $T=20000$, $x\in[0,11]$.
  		}
  \label{fig:two_material_long}
\end{figure}

\smallskip
\textbf{Example 3: problem with varying scattering frequency and constant source term \cite{lemou2008new}.} We consider the one-group transport equation in slab geometry
with a source term $G$:
\begin{align}
\vareps \partial_t f + v\partial_x f = -\frac{\sigma_s}{\vareps}(\lgl f \rgl - f) + \vareps \sigma_a f+\vareps G.
\end{align}
The computational domain is $\Omega_x=[0,1]$, and
\begin{align}
&\sigma_s(x) = 1+100x^2,\;\;\sigma_a=0,\;\;G=1,\;\;
\notag\\
&f_L(v,t) = 0,\;\; f_R(v,t)=0,\;\; f(x,v,0)=0,\;\;\vareps = 10^{-2}.
\end{align}
The effective scaling is determined by $\frac{\vareps}{\sigma_s(x)}$, hence, it is varying in the computational domain.

We use a uniform mesh with $h=\frac{1}{40}$ and the source term $G$
 is treated explicitly. Numerical results for $T=0.4$ are presented in Figure \ref{fig:changing_scattering}. The reference solution is obtained by the first order forward Euler upwind finite difference scheme applied to \eqref{eq:f_model} with $h=\frac{1}{20000}$ and $\Dt=0.1\vareps h$. As the value of $\sigma_s(x)$ is larger on the right, the scattering effect is stronger on that side. As a result, sharp feature exists near the right boundary. All schemes match the reference solution well on this relatively coarse mesh, and high order schemes perform better, especially near the right boundary.

\begin{figure}
  \centering
  \subfigure[$\rho$, IMEX$1$-DG$1$-S]
  {
     \includegraphics[width=0.31\linewidth]{picture/change_scattering_rho_eps1e2_100x2_imex1.png}
    }
  \subfigure[$\rho$, IMEX$2$-DG$2$-S]
  {
        \includegraphics[width=0.31\linewidth] {picture/change_scattering_rho_eps1e2_100x2_imex2.png}
    }
   \subfigure[$\rho$, IMEX$3$-DG$3$-S]
  {
        \includegraphics[width=0.31\linewidth] {picture/change_scattering_rho_eps1e2_100x2_imex3.png}
    }
  \caption{ Example 3: changing scattering problem for one-group transport equation, $T=0.4$.
  		}
  \label{fig:changing_scattering}
\end{figure}

\smallskip
\textbf{Example 4: diffusive and kinetic regime with isotropic inflow Dirichlet boundary conditions \cite{boscarino2013implicit,lemou2008new}.} In this example, we consider the one-group transport equation in  slab geometry on

$\Omega_x=[0,1]$, and
\begin{align}
\sigma_s=1,\;\;\sigma_a=0,\;\;f_L(v,t)=1,\;\; f_R(v,t)=0,\;\;f(x,v,0)=0,
\end{align}
with $\vareps=1, 10^{-8}$.

In Figure \ref{fig:dirichlet}, we report numerical results on a uniform mesh with $h=\frac{1}{40}$. The reference solution for $\vareps=1$ is obtained by the first order forward Euler upwind finite difference scheme applied to \eqref{eq:f_model} with $h=\frac{1}{2000}$ and $\Dt = 0.5\varepsilon h$, while the reference solution for $\vareps=10^{-8}$ is obtained by a central difference scheme solving the diffusion limit \eqref{eq:diffusion_limit} with $h=\frac{1}{2000}$ and $\Dt = 0.25h^2$.
For comparison, we also include  in Figure \ref{fig:dirichlet} the numerical results  by
 the IMEX$k$-LDG$k$ schemes in \cite{peng2019stability}  with the weight function $\omega=\exp(-\frac{\varepsilon}{h})$ and  $\omega=1$, and when $\vareps=1$ .

When the problem is relatively kinetic with $\vareps=1$, it is observed that the numerical solutions by the proposed methods match the reference solutions well. The results are comparable with that by the IMEX$k$-LDG$k$ methods with the weight function $\omega=\exp({-\frac{\varepsilon}{h}})$, and both  are better than that by the IMEX$k$-LDG$k$ methods with the constant  weight function $\omega=1$.
Note that in this example, the initial and boundary conditions at $x=0$ are not compatible, and this introduces a Dirac delta structure in $\partial_x\rho$ at $t=0$ and subsequently sharper features in the solution form   near the left boundary. All these pose challenge to approximate the weighted diffusion term $\omega\partial_{xx}\rho$, unless $\omega$ is chosen to be small to balance the term $\partial_{xx}\rho$. This explains the IMEX$k$-LDG$k$ schemes with the weight function $\omega=\exp(-\frac{\varepsilon}{h})$ outperform that with $\omega=1$. Our IMEX-DG-S schemes on the other hand do not have a weight function to tune for this example.

When the problem is relatively diffusive with $\vareps=10^{-8}$,

 we take $\Dt=0.25h$ in the diffusive regime, instead of the original $\Dt=0.75h$ in \eqref{eqn:CFL} (still stable), for both the IMEX1-DG1-S and IMEX3-DG3-S schemes.
The numerical solutions by the IMEX$k$-DG$k$-S scheme, $k=1, 2, 3$, match the reference solutions well.
The higher order schemes lead to better resolution.

\begin{figure}
  \centering
    \subfigure[$\rho$ for $\vareps=1$, IMEX$1$-DG$1$-S]
  {
     \includegraphics[width=0.31\linewidth]{picture/inflow_bc_rho_eps1_imex1.png}
    }
  \subfigure[$\rho$ for $\vareps=1$, IMEX$2$-DG$2$-S]
  {
        \includegraphics[width=0.31\linewidth] {picture/inflow_bc_rho_eps1_imex2.png}
    }
   \subfigure[$\rho$ for $\vareps=1$, IMEX$3$-DG$3$-S]
  {
        \includegraphics[width=0.31\linewidth] {picture/inflow_bc_rho_eps1_imex3.png}
    }
    \\
    \subfigure[$\rho$ for $\vareps=1$, IMEX$1$-LDG$1$ with $\omega=\exp(-\frac{\varepsilon}{h})$]
  {
     \includegraphics[width=0.31\linewidth]{picture/inflow_bc_rho_eps1_imex1_old.png}
    }
  \subfigure[$\rho$ for $\vareps=1$, IMEX$2$-LDG$2$ with $\omega=\exp(-\frac{\varepsilon}{h})$]
  {
        \includegraphics[width=0.31\linewidth] {picture/inflow_bc_rho_eps1_imex2_old.png}
    }
   \subfigure[$\rho$ for $\vareps=1$, IMEX$3$-LDG$3$ with $\omega=\exp(-\frac{\varepsilon}{h})$]
  {
        \includegraphics[width=0.31\linewidth] {picture/inflow_bc_rho_eps1_imex3_old.png}
    }
    \\
      \subfigure[$\rho$ for $\vareps=1$, IMEX$1$-LDG$1$ with $\omega=1$]
  {
     \includegraphics[width=0.31\linewidth]{picture/inflow_bc_rho_eps1_imex1_old_weight1.png}
    }
  \subfigure[$\rho$ for $\vareps=1$, IMEX$2$-LDG$2$ with $\omega=1$]
  {
        \includegraphics[width=0.31\linewidth] {picture/inflow_bc_rho_eps1_imex2_old_weight1.png}
    }
   \subfigure[$\rho$ for $\vareps=1$, IMEX$3$-LDG$3$ with $\omega=1$]
  {
        \includegraphics[width=0.31\linewidth] {picture/inflow_bc_rho_eps1_imex3_old_weight1.png}
    }
    \\
     \subfigure[$\rho$ for $\vareps=10^{-8}$, IMEX$1$-DG$1$-S]
  {
     \includegraphics[width=0.31\linewidth]{picture/inflow_bc_rho_eps1e8_imex1.png}
    }
  \subfigure[$\rho$ for $\vareps=10^{-8}$, IMEX$2$-DG$2$-S]
  {
        \includegraphics[width=0.31\linewidth] {picture/inflow_bc_rho_eps1e8_imex2.png}
    }
   \subfigure[$\rho$ for $\vareps=10^{-8}$, IMEX$3$-DG$3$-S]
  {
        \includegraphics[width=0.31\linewidth] {picture/inflow_bc_rho_eps1e8_imex3.png}
    }
  \caption{ Example 4: diffusive and kinetic regime with isotropic inflow Dirichlet boundary conditions for one-group transport equation. Top three rows: $\rho$ for $\vareps=1$ and $T=0.1,\;0.4,\;1.0,\;1.6,\;4.0$; Bottom row: $\rho$ for $\vareps=10^{-8}$ and $T=0.15,\;0.25,\;2.0$.
  		}
  \label{fig:dirichlet}
\end{figure}

\smallskip
\textbf{Example 5: Riemann problem for telegraph equation \cite{boscarino2013implicit,jang2015high}.} We consider a Riemann problem with $\Omega_v=\{-1,1\}$, $\sigma_s=1$, $\sigma_a=0$ and the initial data
\begin{align}
\begin{cases}
	\rho(x,0)=2,\; g(x,v,0)=0,\; x\leq 0,\\
	\rho(x,0)=1,\; g(x,v,0)=0,\; x>0.
\end{cases}
\end{align}
Two different cases are considered: the more kinetic case with $\vareps=0.7$ and  $\Omega_x=[-1,1]$, and the more diffusive case with
$\vareps=10^{-6}$ and $\Omega_x=[-2,2]$. For both,  a uniform partition of $\Omega_x$ with $h=\frac{1}{40}$ is used, and the final time is set as $T=0.15$. Numerical results for $\rho$ and $j(x,t)=\lgl vg \rgl$ are presented in Figure \ref{fig:riemann1} and Figure \ref{fig:riemann2}. The reference solution for $\vareps=0.7$ is obtained by the first order forward Euler upwind finite difference scheme solving \eqref{eq:f_model} with a uniform mesh $h=\frac{1}{1000}$ and $\Dt=0.05\varepsilon h$. The reference solution for $\vareps=10^{-6}$ is calculated by a central difference scheme solving the diffusion limit \eqref{eq:diffusion_limit} with $h=\frac{1}{1000}$ and $\Dt = 0.25h^2$.

For the kinetic case

with $\vareps=0.7$, results from all schemes match the reference solution well.
Compared with the first order scheme,
 the second and the third order schemes give less dissipative results and capture the sharp features better. Small oscillation near discontinuity can be further reduced

 by  applying nonlinear limiters. With the discontinuity present in the solution, when IMEX-LDG schemes are applied to this example (see Section 6.1.2 in \cite{peng2019stability}), the quality of the computed solutions really depends on the choice of the weight function.  For the diffusive case with  $\vareps=10^{-6}$, all schemes capture the  solution well, and high order schemes show better resolutions.

\begin{figure}
  \centering
  \subfigure[$\rho$, IMEX$1$-DG$1$-S]
  {
     \includegraphics[width=0.31\linewidth]{picture/riemann_imex1_rho_07.png}
    }
  \subfigure[$\rho$, IMEX$2$-DG$2$-S]
  {
        \includegraphics[width=0.31\linewidth] {picture/riemann_imex2_rho_07.png}
    }
   \subfigure[$\rho$, IMEX$3$-DG$3$-S]
  {
        \includegraphics[width=0.31\linewidth] {picture/riemann_imex3_rho_07.png}
    }
    \\
      \subfigure[$j$, IMEX$1$-DG$1$-S]
  {
     \includegraphics[width=0.31\linewidth]{picture/riemann_imex1_j_07.png}
    }
  \subfigure[$j$, IMEX$2$-DG$2$-S]
  {
        \includegraphics[width=0.31\linewidth] {picture/riemann_imex2_j_07.png}
    }
   \subfigure[$j$, IMEX$3$-DG$3$-S]
  {
        \includegraphics[width=0.31\linewidth] {picture/riemann_imex3_j_07.png}
    }
  \caption{ Example 5: Riemann problem for the telegraph equation. $\vareps=0.7$ and $T=0.15$.
  		}
  \label{fig:riemann1}
\end{figure}

\begin{figure}
  \centering
  \subfigure[$\rho$, IMEX$1$-DG$1$-S]
  {
     \includegraphics[width=0.31\linewidth]{picture/riemann_imex1_rho_1e6.png}
    }
  \subfigure[$\rho$, IMEX$2$-DG$2$-S]
  {
        \includegraphics[width=0.31\linewidth] {picture/riemann_imex2_rho_1e6.png}
    }
   \subfigure[$\rho$, IMEX$3$-DG$3$-S]
  {
        \includegraphics[width=0.31\linewidth] {picture/riemann_imex3_rho_1e6.png}
    }
    \\
      \subfigure[$j$, IMEX$1$-DG$1$-S]
  {
     \includegraphics[width=0.31\linewidth]{picture/riemann_imex1_j_1e6.png}
    }
  \subfigure[$j$, IMEX$2$-DG$2$-S]
  {
        \includegraphics[width=0.31\linewidth] {picture/riemann_imex2_j_1e6.png}
    }
   \subfigure[$j$, IMEX$3$-DG$3$-S]
  {
        \includegraphics[width=0.31\linewidth] {picture/riemann_imex3_j_1e6.png}
    }
  \caption{ Example 5: Riemann problem for telegraph equation. $\vareps=10^{-6}$ and $T=0.15$.
  		}
  \label{fig:riemann2}
\end{figure}

config: arxiv


TITLE: How to rationalize $(\sqrt{4 + h} - 2) / h$?
QUESTION [3 upvotes]: How to rationalize this?
  $$\frac{\sqrt{4 + h} - 2} h$$

I tried multiplying it by $$\frac{\sqrt{4 + h} + 2}{\sqrt{4 + h} + 2}$$  but I could not get the correct answer.
The correct answer is $\frac{1}{\sqrt{4 + h} + 2}$.

REPLY [1 votes]: Doesn't rationalizing usually go in the reverse direction? Anyway, you get from point a to point b as follows:
$$
\frac{1}{\sqrt{4+h}+2}=\frac{1}{\sqrt{4+h}+2}\cdot\frac{\sqrt{4+h}-2}{\sqrt{4+h}-2}=\frac{\sqrt{4+h}-2}{h}
$$


TITLE: Give the Bayesian Posterior Mode
QUESTION [0 upvotes]: Suppose that $X_1, X_2, \ldots, X_n$ are IID Bernoulli random variables with success probability equal to an unknown parameter $\theta \in [0,1]$.
Let $A$ and $B$ be nonnegative constants. If we impose the prior $\pi(\theta) \propto (\theta^a)((1-\theta)^b)$, then what is the Bayesian posterior mode?

REPLY [1 votes]: To find the Posterior mode, we need to maximize the posterior distribution with respect to $\theta$. So first we find the joint density:
$\displaystyle f(X|\theta) = \prod_{i=1}^n \theta^{x_i}(1-\theta)^{1-x_i}=\theta^{\sum x_i}(1-\theta)^{n-\sum x_i}$
Then we can get the posterior distribution:
$\displaystyle \pi(\theta |x)\propto f(x|\theta)\pi(\theta)\propto \theta^{\sum x_i}(1-\theta)^{n-\sum x_i} \theta^a(1-\theta)^b=\theta^{\sum x_i+a}(1-\theta)^{n+b-\sum x_i}=\exp\left((n-n\bar{X}+b)\log(1-\theta)+(n\bar{X}+a)\log\theta\right)$
To find the posterior mode, we need to maximize the posterior with respect to $\theta$:
Let $g(\theta)=(n-n\bar{X}+b)\log(1-\theta)+(n\bar{X}+a)\log\theta$
$\displaystyle g^{\prime}(\theta)=-\frac{n-n\bar{X}+b}{1-\theta}+\frac{n\bar{X}+a}{\theta}=0$
Solving for $\theta$, we get $\displaystyle\hat{\theta}=\frac{n\bar{X}+a}{n+a+b}$, which you can check is the maximum by making sure $g^{\prime\prime}(\theta)>0$.


TITLE: Uniform Topology on $\mathbb{R}^{\Gamma} $ is different from Product Topology on $\mathbb{R}^{\Gamma}$
QUESTION [0 upvotes]: This question is from My topology assignment and i was unable to solve it and unfortunately I have no intution on how this question should be approached.

Question:If $\Gamma$ is an infinite set then prove that Uniform Topology on $\mathbb{R}^{\Gamma} $ is different from Product   Topology on $\mathbb{R}^{\Gamma}$.

Definition of uniform Topology: Let $\Gamma$ be a non empty set and let $\bar{d}$ be the standard bounded metric on $\mathbb{R}$ . Define a metric $\bar{\mathbb{\rho}}$ on $\mathbb{R}^{\Gamma}$ by  $\bar{\mathbb{\rho}}(x,y)$= {lub${\bar{d}(x_{\alpha} ,y_{\alpha})}: \alpha \in \Gamma$}.Then $ \bar{\mathbb{\rho}}$ is a metric called uniform metric and topology induced by $\bar{\mathbb{\rho}}$ is called uniform topology on $\mathbb{R}^{\Gamma}$.
Can you please tell me on how should this question should be solved.  Also, I am really sorry that I would not be able to Get any attemp as I don't have anything worthy to show.
Thank you!

REPLY [0 votes]: The "canonical" base (basis) $B_P$ for the Product Topology is the set of all $\prod_{g\in \Gamma} V_g$ such that (i) each $V_g$ is open in $\Bbb R,$ and (ii) $\{g\in \Gamma :V_g\ne \Bbb R\}$ is finite.
Let $U$ be an  open $\bar \rho$-ball of radius $1/3$ in the Uniform Topology. For some set $\{W_g: g\in \Gamma \}$ of non-empty subsets of $\Bbb R,$ we have $U\subseteq V=\prod_{g\in \Gamma}W_g$ where $\sup W_g -\inf W_g\le 2/3$ for each $g.$
No non-empty member of $B_P$ is a subset of $V.$ So $U$ is not a union of members of $B_P.$ So $U$ is not open in the Product Topology.


TITLE: Is the derivative operator Hadamard differentiable?
QUESTION [0 upvotes]: Is the derivative operator $f\mapsto f'$ Hadamard differentiable? If so, under which conditions (does the derivative need to be bounded maybe?). I searched for references but cannot find anything. I would love a step-by-step proof that I could study.

REPLY [1 votes]: The question is not well defined; to make sense of it you need to specify two Banach spaces $X$ and $Y$ such that $f'\in Y$ for every $f\in  X$. Having done that you can then ask about differentiability of $D:X\to Y$.
Once you do that it seems the answer must be yes, just because $D$ is linear: If $f,g\in X$ and $h$ is a non-zero scalar then $$\frac{D(f+hg)-D(f)}{h}=Dg$$and $Dg\to Dg$ as $h\to0$.


TITLE: Given integers $a \ge b > 0$ and a prime number $p$, prove that ${pa \choose pb} \equiv {a \choose b} \mod p$.
QUESTION [1 upvotes]: I've been grappling with this problem for a while but haven't solved it.
Given integers $a \ge b > 0$ and a prime number $p$, prove that ${pa \choose pb} \equiv {a \choose b} \mod p$.

REPLY [15 votes]: The following is a combinatorial argument.
Draw $a$ concentric circles and divide them radially into $p$ parts each.  There are then a total of $pa$ regions.  There are $\binom{pa}{pb}$ ways to select $pb$ of these regions.  Consider the action of rotation by $2\pi/p$ on these selections.  There are $\binom{a}{b}$ selections which are fixed by the rotation: these are the selections that consist of $b$ complete annuli.  All others fall into orbits of size $p$.  The desired conclusion $$\binom{pa}{pb} \equiv \binom{a}{b}\, (\text{mod}\,p)$$
follows.


TITLE: Identity groups where the square of each element is the neutral element
QUESTION [0 upvotes]: I need to identity all groups G where $\forall g \in G: g^2$ is the neutral element of G applies.
This does apply for all groups with 1 or 2 elements, but not for groups with 3 elements. It seems clear that each element (except for the neutral element) need to be of order 2. But how does this help me to identity all groups for which this applies?
[Edit]
It seems that those groups need to be abelian, and (what is trivial to show) all products of $\mathbb{Z}_2$. But are all those groups isomorphic to a product of $\mathbb{Z}_2$? How do I know if I got all matching groups?

REPLY [1 votes]: These groups are abelian, in fact $\mathbb{F}_2$-vector spaces and therefore $\cong \mathbb{F}_2^{\oplus S}$ for some set $S$. (They are not necessarily of the form $\mathbb{F}_2^S$ for some set $S$.)


TITLE: Lebesgue measure of set involving rationals, complement and closed set
QUESTION [0 upvotes]: I have the following problem:
Let rational numbers $\mathbb{Q} = \left \{q_1,q_2,q_3,...  \right \}$, $G=\bigcup_{n=1}^{\infty} (q_n-\frac{1}{n^2},q_n+\frac{1}{n^2})$.
Show that for every closed set F, either $m(G \setminus F) > 0$ or $m(F \setminus G) > 0$ where $m(A)$ is Lebesgue measure of set A.
In the previous part of this problem, I have already shown that G is measurable and $m(G) < \infty$. Also, if $\mathbb{Q} \subseteq F$ where F is a closed set, $F = \mathbb{R}.$ Sets involved are subsets of $\mathbb{R}$.
I tried to attempt this problem as follows:
Case 1: $F = \mathbb{R}$. $m(F \setminus G) = m(\mathbb{R} \setminus G) = m(\mathbb{R}) - m(G)$ = $\infty - m(G)$ > 0 as $m(G) < \infty$
Case 2: $F \neq \mathbb{R}$. As if $\mathbb{Q} \subseteq F$, $F = \mathbb{R}$. We have $F \neq \mathbb{R}$, so $\mathbb{Q} \nsubseteq F$, which is eqivalent to $\exists q_i \in \mathbb{Q}$ such that $q_i \notin F$. But, $q_i \in G$, $G \nsubseteq F$. We have the following two subcases:
Case 2.1: $F \subset G$. $m(G \setminus F) = m (G) - m(F)$ as if $F \subset G$, $m(F) \leq m(G) < \infty$. But, I can only get $m (G) - m(F) \geq 0$ in this case as I understand that $F \subset G$ does not mean $m (F) < m(G), m(G) - m(F) > 0$ from Lebesgue measure of proper subset
Case 2.2: Otherwise, $m(G \setminus F) = m(G) - m(G \cap F)$. As $G \cap F \subseteq G$, $m(G \cap F) \leq m(G)$, $m(G) - m(G \cap F) \geq 0$. I cannot get $m(G) - m(G \cap F) > 0$.
I want to ask if my approach is incorrect? How can I solve this problem?

REPLY [2 votes]: Note that $m(F \cup G) = m(F \setminus G) + m(F \cap G) + m(G \setminus F)$, as the latter three sets form a disjoint union of $F \cup G$.
So suppose both $m(F \setminus G)$ and $m(G \setminus F)$ are $0$, then $m(F \cup G) = m(F \cap G)$; can you rule this out?


TITLE: Projecting an ellipse defined on a sphere onto the XY plane
QUESTION [0 upvotes]: I want to find the equation for and the area of the ellipse defined by $\Theta_H$ and $\Psi_H$ when it gets projected onto the XY plane.
The following diagram shows these ellipses.

Let us suppose we know the radius $r$. I know that I can approximate the major and minor axes of the ellipse via:
$$ a = 2 r \text{tan}\left(\frac{\Theta_H}{2}\right) $$
$$ b = 2 r \text{tan}\left(\frac{\Psi_H}{2}\right) $$
If I could find the vector equation of the ellipse, then I could just set the z component to zero to project it onto the XY Plane.

Questions:

What is the vector equation for the ellipse? I.e. the 3D xyz equation.
Is there another way to do this projection rather than approximating the curve with a ellipse in a plane, finding the equation for that plane, and then setting the z component to zero?

REPLY [1 votes]: Let, for simplification purposes : $a:=\frac12 \Theta_H$ and $b:=\frac12 \Psi_H$.
Let $(\psi_0,\theta_0)$ be the center of the curve.
A first parameterization is
$$\begin{cases}\psi&=&\psi_0+a \cos(t)\\\theta&=&\theta_0+b \sin(t)\end{cases}$$
i.e., a "true ellipse" wrt coordinates $(\theta, \phi)$ parameterized by a 'non-physical" parameter $t$.
Then, in a second step, plug these expressions in the classical cartesian/spherical formulas:
$$\begin{cases}x&=&\cos \psi \cos \theta\\
y&=&\sin \psi \cos \theta\\
z&=&\sin \theta
\end{cases}$$
Suppressing parameter $z$ (I don't see another method) gives the projection onto the $x-y$ plane.


TITLE: Non-Linear Second Order Differential Equation Regarding Elasticity
QUESTION [0 upvotes]: What methods can I use to solve the following equation?
$$q''(p)=\frac{q'(p)^2}{q(p)}+\frac{q'(p)}{p}$$
I know from wolfram alpha that the solution is $q(p)=c_1p^{c_2}$.

REPLY [0 votes]: You can also divide through by $q'(p)$ to get
$$ \frac{q''}{q'} = \frac{q'}{q} + \frac{1}{p}$$
$$ \ln{q'} = \ln{q} + \ln{p} + c_1 $$
$$  q' = c_2qp $$
This gives
$$ q(p) = ae^{bp^2}$$


TITLE: Simplify the expression $^{66}C_{21}+\sum_{r=1} ^{15}\left( ^{81-r}C_{20}\right).$
QUESTION [0 upvotes]: Simplify the expression $^{66}C_{21}+\sum_{r=1} ^{15}\left( ^{81-r}C_{20}\right).$
The answer is $^{81} C_{21}$.
Maybe we can sum diagonals of Pascal's triangle, but I can't see the correct method.

REPLY [1 votes]: Are you familiar with the hockey stick identity? It says that
$$
\sum_{i=0}^k {}^{n+i}C_n= {}^{n+k+1}C_{n+1}
$$
(I highly advise you to look at the picture that link rather than my formula, and think about what your sum looks like in the same way. It is, in my opinion, a lot more insightful than the dry formulas.)
Coupled with the fact that, with some abuse of notation, $\sum_{r=1}^{15}=\sum_{r=1}^{61}-\sum_{r=16}^{61}$ and some re-indexing, you should have enough to conclude.


TITLE: Combination - Ordering Boy Scouts
QUESTION [1 upvotes]: In how many different ways can 9 distinct boy scouts be arranged in a 3 × 3 formation? In such a formation, there are 3 scouts in the ﬁrst row, 3 in the second, and 3 in the third. Two formations are the same if in every row, both formations contain the same three scouts in the same order.
I think it's 9 permute 3, 6 permute 3, 3 permute 3 multiplied together = 9! but I'm not really sure.

REPLY [1 votes]: One may arbitrarily define an ordering of the grid; something like:
 1 | 2 | 3
---+---+---
 4 | 5 | 6
---+---+---
 7 | 8 | 9

After an order is defined, you may rephrase the question like this: "in how many different ways can 9 distinct boy scouts be arranged in places 1,...,9?"
One thing about math is abstractions; for example, it wouldn't matter if you'd arrange boy scouts or girl scouts or dolphins, right? Boy scouts are just an example, a representative of the class of all things that can be distinguished. And one thing about combinatoric questions is finding a specific representative ("example") which hides the real (and usually simple) truth behind it.
In your case, the so-called grid just hides the simpler 9-places line. Similarly, one could ask "in how many different ways can 9 distinct boy scouts be arranged in a car with 2 front seats, 4 mid seats, and 3 back seats". Or whatever.