text:
\section{Introduction}

Einstein, Podolsky, Rosen, Schrodinger, et al discovered quantum entanglement. Then it became a remarkable feature of quantum mechanics. It lies in the heart of quantum information theory \cite{2007Quantum,GUHNE20091}. In recent decades, entanglement has been recognized as a kind of valuable resource \cite{2007Quantum,2019Quantum,Patricia2019Resource}. It is widely used in various quantum information processing tasks such as quantum computing \cite{2005Experimental}, teleportation \cite{2004Deterministic}, dense coding \cite{2002Quantum}, cryptography \cite{2020Entanglement}, and quantum key distribution \cite{Xu2020}.

Although several useful separability criteria such as positive-partial-transpose (PPT) criterion \cite{Peres1996,horodecki1997}, range criterion, and realignment criterion \cite{Rudolph2002Some} were developed, all of them cannot strictly distinguish between the set of entangled states and that of separable ones. According to PPT criterion, any state with non positive partial transpose (NPT) must be entangled. Nevertheless,  the converse only holds for two-qubit and qubit-qutrit  systems. It has been shown  that determining whether a bipartite state is entangled is an NP-hard problem \cite{2003Proceedings}. In $2000$, Terhal first introduced the term entanglement witness (EW) by indicating that a violation of a Bell inequality can be expressed as a witness for entanglement \cite{2000Entanglement}. Nowadays, EWs are a fundamental tool for the detection of entanglement both theoretically and experimentally. Actually, many EWs have been implemented with local measurements \cite{2000Entanglement,2019Design, 2020Measurement}.

As we know, the partial transpose of NPT state $\rho$ is an EW. The negative eigenvalues of $\rho^\Gamma$ are a signature of entanglement. They are closely related to other problems in entanglement theory. The negativity  is a well-known computable entanglement measure \cite{2002Computable}. At the same time, it is the sum of the absolute values of negative eigenvalues of  $\rho^\Gamma$. Also, by the definition of $1$-distillable state \cite{Micha1998Mixed}, when $\rho^\Gamma$ has more negative eigenvalues, $\rho$ is  more likely to be $1$-distillable. Thus, it is important to explore the negative eigenvalues of $\rho^\Gamma$. The problem of determining how many negative eigenvalues the partial transpose of NPT state could contain has attracted great interest \cite{2008Universal,2012Qubit,2013Negative,2013Non}. It was first specified in \cite{2008Universal} that $\rho^\Gamma$ has one negative eigenvalue and three positive eigenvalues for any two-qubit entangled state $\rho$. Next the inertias of the partial transpose of $2\times n$ NPT states has been investigated in \cite{2020Inertias}. As far as we know, many quantum information queestions turn into intractable when $2\times n$ systems are replaced by $3\times 3$ systems, such as distillability problem \cite{Micha1998Mixed,1999Evidence}, distinguishable subspace problem \cite{2007On}, and separability problem \cite{2002Distinguishing}.

Due to Sylvester theorem \cite{2011Matrix}, we introduce a tool to study the inertia of EW, namely the inertia of a Hermitian operator is invariant under SLOCC. Firstly, we introduce what the two-qutrit subspaces contain for subspaces with various dimensions  in Lemma \ref{le:9}. Then we explore a property of the inertia of the partial transpose of bipartite NPT states in Lemma \ref{le:add}. Further we investigate the number of positive eigenvalues of  bipartite EWs in Lemma \ref{le:c=2}. Then we give some observations on the relation between the number of positive eigenvalues and that of negative eigenvalues in Lemma \ref{le:negative}.  Then we propose two observations on the partial transpose of the two-qutrit  NPT states in Lemmas \ref{le:12} and \ref{le:13}, demonstrated by Example \ref{ex:11}. Then we discuss all cases in the partial transpose of the two-qutrit bipartite NPT state in Theorem \ref{th:NPT2}. We draw a conclusion about the inertias of (the partial transpose) of the two-qutrit bipartite NPT states in Theorem \ref{th:N_33}. Next we propose Example \ref{ex:13arrays} to illustrate Theorem \ref{th:N_33}(i). We propose Example \ref{ex:3arrays} to show the relationship between the inertias of $2\times 3$ states and those of $3\times 3$ states.  Finally, we extend some conclusions on the inertias  from  $2\times n$ states to $3\times n$ states in Lemma \ref{le:3n}.

The remainder of this paper is organized as follows. In section \ref{sec:p} we introduce the preliminaries by clarifying the notations and presenting necessary definitions and results. In section \ref{sec:I_1} we focus on the bipartite EWs constructed by (the partial transpose) of NPT state, and determine intertias of such EWs. In section \ref{sec:I_2} we investigate the inertias of the partial transpose of the two-qutrit bipartite NPT states. Finally, we conclude in section \ref{sec:con}. In section \ref{sec:app} we find out the relationship between the inertias of $2\times 3$ states and those of $2\times 3$ states and  extend some conclusions on the inertias  from  $2\times n$ states to $3\times n$ states. We also partially test the existence of two unverified inertias using python program.

\section{Preliminaries}
\label{sec:p}
 In this section we introduce the facts used in this paper.
 We refer to $\mathbb{C}^{d}$ as the $d$-dimensional Hilbert space. We define $I_n$ as the  identity matrix of order $n$. We denote $\otimes^n_{i=1}\mathbb{C}^{d_i}:= \otimes^n_{i=1}\mathcal{H}_{A_i}$ as the n-partite Hilbert space, where $d_i$'s are local dimensions. Let $\mathbb{M}_n$ be the set of $n\times n$ matrices. If $\rho\in \mathcal{B}(\otimes^n_{i=1}\mathbb{C}^{d_i})$ is a positive semidefinite matrix of trace one, then $\rho$ is an $n$-partite quantum state. The partial transpose w.r.t. system A of a bipartite matrix $\rho \in \mathcal{B}(\mathcal{H}_A\otimes \mathcal{H}_B)$ is definied as,
  $\rho^{\Gamma}:=\sum_{i,j}\langle c_i|_A\rho |c_j \rangle \otimes |c_j \rangle \langle c_i|_A$,
 where the set of $|c_i \rangle$ is an arbitrary orthonormal basis in $\mathcal{H}_A$. If $\rho^{\Gamma}$ has at least one negative eigenvalue then $\rho$ has non-positive partial transpose (NPT).  If $\rho^{\Gamma}$ does not have negative eigenvalues then $\rho$ has positive partial transpose (PPT).  We refer to an $M\times N$ state as a bipartite $\rho$ such that rank $\rho_A=M$ and rank $\rho_B=N$. Given the Schmidt decomposition $|\psi\rangle =\sum_i\sqrt{c_i}|a_i,b_i\rangle$, we refer to the number of nonzero $c_i$ as the Schmidt rank of $\psi$. We denote the  number as SR($\psi$). There is an essential proposition for the matrix inertia, namely Sylvester theorem \cite{Horn1985}. It states that Hermitian matrices $A, B \in \bbM^n$
 have the same inertia if and only if there is a non-singular matrix $S$ such that $B = SAS^\dagger$.

 In the following we review the entanglement witness (EW), inertia and stochastic local operations and classical communications (SLOCC) equivalence.

 \begin{definition}
 	\label{de:EW}
 	Suppose $W\in \mathcal{B}(\otimes^n_{i=1}\mathbb{C}^{d_i})$ is Hermitian. We say that W is an $n$-partite EW if

 	(i) it is non-positive semidefinite,

 	(ii) $\langle\psi|W|\psi\rangle\geq 0$ for any product vector $|\psi\rangle=\otimes^n_{i=1}|a_i\rangle$ with $|a_i\rangle \in \mathcal{H}_i$.
 \end{definition}

 \begin{definition}
 	\label{de:in}
 	Let $A \in \mathbb{M}_n$ be Hermitian. The inertia of $A$, denoted by $\ln (A)$, is defined as the following sequence

 	\begin{eqnarray}
 	\label{eq:1}
 		\ln(A):=(v_-,v_0,v_+),
 	\end{eqnarray}
 	where $v_-,v_0$ and $v_+$ are respectively the numbers of negative, zero and positive eigenvalues of $A$.

 	Note that if $\rho$ is PPT, then $v_-=0$.
 \end{definition}

 \begin{definition}
 	\label{de:SL}
 	Two $n$-partite pure state $|\alpha \rangle$, $|\beta \rangle$ are SLOCC equivalent if there exists a product invertible operation $Y=Y_1\otimes...\otimes Y_n$ such that $|\alpha \rangle =Y|\beta \rangle$.
 \end{definition}

 Then we present several results for the state $\rho$ and the EWs \cite{2013Non,2020Inertias,Chen2013,3tensors1983}.
 \begin{lemma} \cite{3tensors1983}
 	\label{le:mn-2}
 	Suppose $\mathcal{V} \subseteq \bbC^m\otimes\bbC^n$ is a bipartite Hilbert subspace.

 	(i) If $\mathop{\rm Dim}(\mathcal{V})=mn-1$, then $\mathcal{V}$ is spanned by product vectors.

 	(ii) If $\mathop{\rm Dim}(\mathcal{V})=mn-2$, then $\mathcal{V}$ is either spanned by product vectors or up to SLOCC equivalence spanned by $\{\ket{0,0}+\ket{1,1},\ket{i,j}:(i,j)\neq(0,0),(0,1),(1,1)\}$.
 \end{lemma}

 \begin{lemma} \cite{2013Non}
 	\label{le:pre}
 	Suppose $\rho$ is an $m\times n$ NPT state. If $\rho$ is a pure state with Schmidt rank $r$, then

 	\begin{eqnarray}
 	\ln (\rho^{\Gamma})=(\frac{r^2-r}{2},mn-r^2,\frac{r^2+r}{2}) .
 	\end{eqnarray}
 \end{lemma}

\begin{lemma}\cite[Proposition $6$]{Chen2013}
	\label{le:2}
	Suppose $\mathcal{V} \subseteq \bbC^m \otimes \bbC^n$. If $\mathop{\rm Dim}(\mathcal{V}) >(m-1)(n-1)$ then $\mathcal{V}$ contains at least one product
	vector. Furthermore, if $\mathop{\rm Dim}(\mathcal{V}) >(m-1)(n-1)+1$ then $\mathcal{V}$ has infinitely many product vectors.
\end{lemma}

\begin{lemma} \cite[Lemma $5$]{2020Inertias}
	\label{le:mn}
	Suppose $W$ is an EW on $\mathbb{C}^m \otimes \mathbb{C}^n$.

	(i) Let $A$ be the non-positive eigen-space of $W$, i.e., the sum of negative and zero eigen-spaces
	of W. Then the product vectors in $A$ all belong to the zero eigen-space of W. In particular,
	every vector in the negative eigen-space of W is a pure entangled state.

	(ii) The number of negative eigenvalues of W is in $[1, (m-1)(n-1)]$. The decomposable EW
	containing exactly $(m-1)(n-1)$ negative eigenvalues exists.

	(iii) The number of positive eigenvalues of W is in $[2, mn-1]$.
\end{lemma}

 We will apply Lemma \ref{le:mn-2} to prove Lemma \ref{le:c=2}. We will use Lemmas \ref{le:pre}, \ref{le:2} and \ref{le:mn} to prove Lemma \ref{th:N_33}.
Then we denote the inertia set, $\mathcal{N}_{m,n}:=\{\ln (\rho^{\Gamma})|\rho\quad is \quad an\quad m\times n\quad NPT\quad state.\}$. We introduce some observations on the inertias.
\begin{lemma} \cite[Lemma $8$]{2020Inertias}
	\label{le:rel}
	(i) Suppose $\rho$ is an $m\times n$ NPT state and $\rho^{\Gamma}$ has the inertia $(a,b,c)$. Then there is a small enough $x>0$ and NPT state $\sigma:=\rho+xI_{mn}$, such that $\ln(\sigma^{\Gamma})=(a,0,b+c)$.

	(ii) Suppose $m_1\leq m_2$ and $n_1\leq n_2$. If $(a_1,b_1,c_1)\in \mathcal{N}_{m_1,n_1}$, with $a_1+b_1+c_1=m_1n_1$, then for any  $l\in[0,m_2n_2-m_1n_1]$ we have $(a_1, m_2n_2-m_1n_1-l,b_1+c_1+l)\in \mathcal{N}_{m_2,n_2}$.
\end{lemma}

Based on the above preliminary knowledge we are ready to study the inertia of the partial transposes of NPT states.

\section{Inertias of the partial transposes of NPT states}
\label{sec:I_1}
In this section we focus on the bipartite EWs constructed by the partial transpose of NPT state, and determine intertias of such EWs.

At first,   we  introduce what the two-qutrit subspaces contain for subspaces with various dimensions.
\begin{lemma}
	\label{le:9}

	Suppose $\mathcal{V} \subseteq \mathbb{C}^3 \otimes \mathbb{C}^3 $ has infinitely many pairwise linearly independent product vectors.

	(i) If the dimension of $\mathcal{V}$ is two,  then $\mathcal{V}$  up to SLOCC equivalence and system permutation contains $\{\ket{0,0},\ket{0,1}\}$.

	(ii) If the dimension of $\mathcal{V}$ is three, then $\mathcal{V}$ up to SLOCC equivalence and system permutation contains $\{\ket{0,0},\ket{0,1}\}$ or  $\{\ket{0,0},\ket{1,1}, (\ket{0}+\ket{1})(\ket{0}+\ket{1})\}$.

	(iii) If the dimension of $\mathcal{V}$ is four, then $\mathcal{V}$ up to SLOCC equivalence and system permutation contains $\{\ket{0,0},\ket{0,1}\}$ or $\{\ket{0,0},\ket{1,1}, (\ket{0}+\ket{1})(\ket{0}+\ket{1})\}$ or $\{\ket{0,0}, \ket{1,1}, \ket{2}(\ket{0}+\ket{1}), (a\ket{0}+b\ket{1}+\ket{2})(\ket{0}+f\ket{1})\}$ satisfying the conditions $a,b\neq 0$  and $f\neq 0,1$.

	(iv)  If the dimension of $\mathcal{V}$ is five, then $\mathcal{V}$ up to SLOCC equivalence and system permutation contains  $\{\ket{0,0},\ket{0,1}\}$ or $\{\ket{0,0},\ket{1,1}, (\ket{0}+\ket{1})(\ket{0}+\ket{1})\}$ or $\{\ket{0,0}, \ket{1,1}, \ket{2}(\ket{0}+\ket{1}), (a\ket{0}+b\ket{1}+\ket{2})(\ket{0}+f\ket{1})\}$, $\{\ket{0,0}, \ket{1,1}, \ket{2,2}, \}$
\end{lemma}
\begin{proof}

	(i) If the dimension of $\mathcal{V}$  is two, then we obtain that one product vector is SLOCC equivalent to $\ket{0,0}$. One can verify that the other product vector can only be SLOCC equivalent or system permutated to $\ket{0,1}$.

	(ii) Suppose the dimension of $\mathcal{V}$ is three. Since $\mathcal{V}$ has infinitely many product vectors, it may contain the set $\{\ket{0,0},\ket{0,1}\}$ from (i). Suppose $\mathcal{V}$ does not contain the set. It contains product vectors $\{\ket{a_0,b_0},\ket{a_1,b_1},\ket{a_2,b_2}\}$.  Up to SLOCC equivalence, we may assume that $\ket{a_0,b_0}=\ket{0,0}$ and $\ket{a_1,b_1}=\ket{1,1}$. Suppose the third product vector is $(a\ket{0}+b\ket{1}+c\ket{2})(d\ket{0}+f\ket{1}+g\ket{2})$. Recall that $\mathcal{V}$ has infinitely many pairwise linearly independent product vectors. So we have that $x\ket{0,0}+y\ket{1,1}+(a\ket{0}+b\ket{1}+c\ket{2})(d\ket{0}+f\ket{1}+g\ket{2})$ is a product vector and $x,y\neq 0$. Then we obtain that the  rank of  $x\proj{0}+y\proj{1}+(a\ket{0}+b\ket{1}+c\ket{2})(d\bra{0}+f\bra{1}+g\bra{2})= \begin{bmatrix} x+ad&af &ag\\ bd&y+bf &bg\\ cd &cf &cg\end{bmatrix}$ is one. Because these three row vectors are pairwise linearly dependent and $x,y\neq 0$, we have $c=f=0$. Then we obtain that $(a\ket{0}+b\ket{1})(d\ket{0}+f\ket{1})$ is SLOCC equivalent to $(\ket{0}+\ket{1})(\ket{0}+\ket{1})$. Then $\mathcal{V}$ up to SLOCC equivalence and system permutation contains $\{\ket{0,0},\ket{1,1}, (\ket{0}+\ket{1})(\ket{0}+\ket{1})\}$.

	(iii) Suppose the dimension of $\mathcal{V}$  is four.   Since $\mathcal{V}$ has infinitely many product vectors, it may contain the sets  $\{\ket{0,0},\ket{0,1}\}$ or  $\{\ket{0,0},\ket{1,1}, (\ket{0}+\ket{1})(\ket{0}+\ket{1})\}$ from (ii). Suppose $\mathcal{V}$ does not contain the sets. That is, we hypothesize that the space spanned by any three product vectors of the set does not generate infinitely many product vectors. The sets contains product vectors $\{\ket{a_0,b_0},\ket{a_1,b_1},\ket{a_2,b_2},\ket{a_3,b_3}\}$.  Up to SLOCC equivalence, we may assume that $\ket{a_0,b_0}=\ket{0,0}$ and $\ket{a_1,b_1}=\ket{1,1}$. Then we discuss three cases (iii.A)-(iii.C) in terms of the third product vector $\ket{a_2,b_2}=\ket{\alpha}$ and the fouth product vector $\ket{a_3,b_3}=\ket{\beta}$.

	Because of the hypothesis, $\ket{\alpha}$ can only be  $\ket{2}(a\ket{0}+b\ket{1}+c\ket{2})$. We will divide $\ket{\alpha}$ into two cases  up to SLOCC equivalence and permutation. One is $\ket{2,2}$, the other is $\ket{2}(m\ket{0}+n\ket{1})$.  When $\alpha$ is $\ket{2,2}$, we will prove it is impossible  that $\ket{\alpha}$ is $\ket{2,2}$ and $\ket{\beta}$ is $(a\ket{0}+b\ket{1}+c\ket{2})(d\ket{0}+f\ket{1}+g\ket{2})$ in (iii.A). Then we obtain that $\ket{\alpha}$ and $\ket{\beta}$ are not up to SLOCC equivalent to $\ket{2,2}$. Then we consider the other case for  $\ket{\alpha}=\ket{2}(m\ket{0}+n\ket{1})$. If $m=0$ or $n=0$, one can prove that $\{\ket{0,0},\ket{1,1},\ket{\alpha}\}$ generates infinitely many product vectors. We draw a contradiction. So $m,n\neq 0$. We obtain that $\ket{\alpha}$ is SLOCC equivalent to $\ket{2}(\ket{0}+\ket{1})$ in this case. Because $\ket{\beta}$ are not up to SLOCC equivalent to $\ket{2,2}$, we will divide $\ket{\beta}$ into two cases when $\ket{\alpha}=\ket{2}(\ket{0}+\ket{1})$.  Then we will dicuss the case that $\ket{\alpha}$ is $\ket{2}(\ket{0}+\ket{1})$ and $\ket{\beta}$ is $(a\ket{0}+b\ket{1})(d\ket{0}+f\ket{1}+\ket{2})$ in (iii.B), and the case that $\ket{\alpha}$ is $\ket{2}(\ket{0}+\ket{1})$ and $\ket{\beta}$ is $(a\ket{0}+b\ket{1}+\ket{2})(d\ket{0}+f\ket{1})$ in (iii.C). Suppose $d=0$ or $f=0$. $\mathcal{V}$ up to SLOCC equivalence and system permutation contains $\{\ket{0,0},\ket{1,1},(a\ket{0}+b\ket{1}+\ket{2})(d\ket{0}+f\ket{1})\}$. We get that  $\mathcal{V}$ up to SLOCC equivalence and system permutation contains $\{\ket{0,0},\ket{0,1}\}$. We draw a contradiction. Then we have $d,f\neq 0$. We will suppose $d=1$ and $f\neq 0$ in (iii.C).

	(iii.A) Suppose $\ket{\alpha}$ is $\ket{2,2}$ and $\ket{\beta}$ is $(a\ket{0}+b\ket{1}+c\ket{2})(d\ket{0}+f\ket{1}+g\ket{2})$.  $\mathcal{V}$ has infinitely many pairwise linearly independent product vectors. Then we obtain that the  rank of  $x\proj{0}+y\proj{1}+z\proj{2}+(a\ket{0}+b\ket{1}+c\ket{2})(d\bra{0}+f\bra{1}+g\bra{2})= \begin{bmatrix} x+ad&af &ag\\ bd&y+bf &bg\\ cd &cf &z+cg\end{bmatrix}$ is one. These three row vectors are pairwise linearly dependent.  Because of the hypothesis, we get that $x,y,z\neq 0$. Then one can verify that the rank of the above matrix is more than one. Then we draw a contradiction.

	(iii.B) Suppose $\ket{\alpha}$ is $\ket{2}(\ket{0}+\ket{1})$, $\ket{\beta}$ is $(a\ket{0}+b\ket{1})(d\ket{0}+f\ket{1}+\ket{2})$. Then we obtain that the  rank of  $x\proj{0}+y\proj{1}+z\ket{2}(\bra{0}+\bra{1})+(a\ket{0}+b\ket{1})(d\bra{0}+f\bra{1}+\bra{2})= \begin{bmatrix} x+ad&af &a\\ bd&y+bf &b\\zeta &z &0\end{bmatrix}$ is one. One can verify that if the rank of the  matrix is one, then $a=b=x=y=0$. We draw a contradiction.

	(iii.C) Suppose $\ket{\alpha}$ is $\ket{2}(\ket{0}+\ket{1})$, $\ket{\beta}$ is $(a\ket{0}+b\ket{1}+\ket{2})(\ket{0}+f\ket{1})$ and $f\neq 0$. Then we obtain that the  rank of  $x\proj{0}+y\proj{1}+z\ket{2}(\bra{0}+\bra{1})+(a\ket{0}+b\ket{1}+\ket{2})(\bra{0}+f\bra{1})= \begin{bmatrix} x+a&af &0\\ b&y+f &0\\ z+1 &z+f &0\end{bmatrix}$ is one. Because of the hypothesis, we get that $x,y,z\neq0$. If $f=1$, then by (ii) we have that $\{\ket{0,0}, \ket{2}(\ket{0}+\ket{1}), (a\ket{0}+b\ket{1}+\ket{2})(\ket{0}+\ket{1})\}$ can generate infinitely many product vectors when $b=0$.  We draw a contradiction. We obtain that $f\neq 1$. Then we consider the values of $a$ and $b$.  Suppose $a=0$. We have that $\{\ket{1}\bra{1},\ket{2}(\bra{0}+\bra{1}),(b\ket{1}+\ket{2})(\bra{0}+f\bra{1})\}$ is a $3$-dimensional subspace. We draw a contradiction. So we have that $a\neq0$. Similarly, we get that $b\neq 0$.  Suppose $a,b\neq 0$ and $f\neq0,1$. We define $n=\dfrac{x+a}{af}$. $\dfrac{x+a}{af}=\dfrac{b}{y+f}=\dfrac{z+1}{z+f}=n$, if and only if the rank of $\begin{bmatrix} x+a&af &0\\ b&y+f &0\\ z+1 &z+f &0\end{bmatrix}$ is one. We obtain that $x=afn-a$, $y=\dfrac{b-fn}{n}$  and $z=\dfrac{1-fn}{1-n}$. Because the value of $n$ is arbitrary, there are various combinations of values of $x,y,z\neq 0$.  So we get that $\mathcal{V}$ up to SLOCC equivalence and system permutation contains  $\{\ket{0,0}, \ket{1,1}, \ket{2}(\ket{0}+\ket{1}), (a\ket{0}+b\ket{1}+\ket{2})(d\ket{0}+f\ket{1})\}$ and $a,b\neq 0$ and $f\neq 0,1$.

\end{proof}
Then we introduce a property of the inertia of an $n\times n$ bipartite NPT state.
\begin{lemma}
	\label{le:add}
	Suppose $\rho$ is an $n\times n$ bipartite NPT state, and $\rho^\Gamma=P-Q$ where $P,Q\ge0$ and $P\perp Q$. If $\ket{a,b}\in \cR(\rho^\Gamma)$, then $\rho^\Gamma+\proj{a,b}$ has the inertia $(\mathop{\rm rank} Q, n-\mathop{\rm rank} P-\mathop{\rm rank} Q, \mathop{\rm rank} P )$ or $(-1+\mathop{\rm rank} Q, n-\mathop{\rm rank} P-\mathop{\rm rank} Q, 1+\mathop{\rm rank} P )$ or $(-1+\mathop{\rm rank} Q,  n+1-\mathop{\rm rank} P-\mathop{\rm rank} Q, \mathop{\rm rank} P )$.
\end{lemma}
\begin{proof}
	Because $\ket{a,b}\in \cR(\rho^\Gamma)=\cR(P)+\cR(Q)$, we have $\ket{a,b}\in \cR(P)$, $\ket{a,b}\in \cR(Q)$, or $\ket{a,b}\in \{\ket{x},\ket{y}\}$ where $\ket{x}\in \cR(P)$ and $\ket{y}\in \cR(Q)$. If $\ket{a,b}\in \cR(P)$ or $\cR(Q)$ then by the definition of inertia, we obtain that $\rho^\Gamma+\proj{a,b}$ has inertia $(\mathop{\rm rank} Q, n-\mathop{\rm rank} P-\mathop{\rm rank} Q, \mathop{\rm rank} P)$. Next suppose $\ket{a,b}\in \{\ket{x},\ket{y}\}$. We can write up $P=P_1+\proj{x}$ and $Q=Q_1+\proj{y}$ where
	\begin{eqnarray}
	\label{eq:p1q1}
	&& P_1,Q_1\ge0,
	\notag\\&&
	\mathop{\rm rank} P=1+\mathop{\rm rank} P_1,
	\notag\\&&
	\mathop{\rm rank} Q=1+\mathop{\rm rank} Q_1.
	\end{eqnarray}
	We have
	\begin{eqnarray}
	&&\rho^\Gamma+\proj{a,b}
	\\=&&
	P-Q+\proj{a,b}
	\\=&&
	(P_1+\proj{x})-(Q_1+\proj{y})+\proj{a,b}
	\\=&&
	P_1-Q_1+\sigma,
	\end{eqnarray}
	where $\sigma=\proj{x}-\proj{y}+\proj{a,b}$ has rank one or two. Further, \eqref{eq:p1q1} implies that the nonzero vector in $\cR(\sigma)$ is linearly independent from any nonzero vector in $\cR(P_1)+\cR(Q_1)$. If $\mathop{\rm rank}\sigma=1$ then $\sigma$ is positive semidefinite. So $\rho^\Gamma+\proj{a,b}$ has the inertia $(-1+\mathop{\rm rank} Q, n+1-\mathop{\rm rank} P-\mathop{\rm rank} Q, \mathop{\rm rank} P )$. On the other hand if $\mathop{\rm rank}\sigma=2$ then $\sigma$ has at least one positive eigenvalue. So $\rho^\Gamma+\proj{a,b}$ has the inertia $(\mathop{\rm rank} Q, n-\mathop{\rm rank} P-\mathop{\rm rank} Q, \mathop{\rm rank} P )$ or $(-1+\mathop{\rm rank} Q, n-\mathop{\rm rank} P-\mathop{\rm rank} Q, 1+\mathop{\rm rank} P )$. We have proven the assertion.
\end{proof}

 Then we investigate the number of positive eigenvalues of  bipartite EWs.

\begin{lemma}
	\label{le:c=2}
	Every bipartite EW has at least three positive eigenvalues.
\end{lemma}

\begin{proof}
	It is known that every bipartite EW has at least two positive eigenvalues. Thus, we have to show that there is no EW with exact two positive eigenvalues.
	Assume that $W$ is an $m\times n$ EW which has inertia $(a,b,2)$. Then the spectral decomposition of $\rho^\Gamma$ reads as
	\begin{eqnarray}
	\label{eq:sdec-1}
	W=-\sum_{j=1}^a\proj{v_j}+0\cdot\sum_{j=a+1}^{a+b}\proj{v_j}+\proj{v_{mn-1}}+\proj{v_{mn}},
	\end{eqnarray}
	where $\{\ket{v_j}\}_{j=1}^{mn}$ are pairwisely orthogonal.
	It follows from Lemma \ref{le:mn-2} that the non-positive eigenspace of $W$ is either spanned by product vectors or up to SLOCC equivalence spanned by $\{\ket{0,0}+\ket{1,1},\ket{i,j}:(i,j)\neq(0,0),(0,1),(1,1)\}$.

	First, if $\mathop{\rm span}\{\ket{v_j}\}_{j=1}^{a+b}$ is spanned by product vectors, there exists a product vector $\ket{x,y}$ which is orthogonal to $\mathop{\rm span}\{\ket{v_{mn-1}},\ket{v_{mn}}\}$ but non-orthogonal to $\ket{v_1}$. Hence, we obtain that
	\begin{eqnarray}
	\label{eq:c=2-1}
	0\leq \bra{x,y}W\ket{x,y}\leq -\bra{x,y}(\proj{v_1})\ket{x,y}<0.
	\end{eqnarray}
	It is a contradiction.

	Second, if $\mathop{\rm span}\{\ket{v_j}\}_{j=1}^{a+b}$ is spanned by $\{\ket{0,0}+\ket{1,1},\ket{i,j}:(i,j)\neq(0,0),(0,1),(1,1)\}$ up to SLOCC equivalence. It follows that $\mathop{\rm span}\{\ket{v_{mn-1}},\ket{v_{mn}}\}$ is spanned by $\ket{0,0}-\ket{1,1},\ket{0,1}$. It is known that every two-qubit EW has inertia $(1,0,3)$. Hence, we assume one of $m,n$ is greater than two. Suppose one of the local dimensions is two. For example, we may assume $m=2$. Using the projector $(I_2\oplus 0_{n-2})\otimes (I_n-(I_2\oplus 0_{n-2}))$ we obtain that
	\begin{eqnarray}
	\label{eq:c=2-2}
	\begin{split}
	\tilde{W}&:=\big((I_2\oplus 0_{n-2})\otimes (I_n-(I_2\oplus 0_{n-2}))\big) W \big((I_2\oplus 0_{n-2})\otimes (I_n-(I_2\oplus 0_{n-2}))\big)\\
	&=-\big(I_2\otimes (I_n-(I_2\oplus 0_{n-2}))\big)(\sum_{j=1}^a\proj{v_j})\big((I_2\oplus 0_{n-2})\otimes (I_n-(I_2\oplus 0_{n-2}))\big).
	\end{split}
	\end{eqnarray}
	It follows that $\tilde{W}$ is negative semidefinite. It implies $\mathop{\rm Tr}(\tilde{W})<0$. However, we have
	\begin{eqnarray}
	\label{eq:c=2-3}
	\begin{split}
	\mathop{\rm Tr}(\tilde{W})&=\mathop{\rm Tr}\Big(\big((I_2\oplus 0_{n-2})\otimes (I_n-(I_2\oplus 0_{n-2}))\big) W \big((I_2\oplus 0_{n-2})\otimes (I_n-(I_2\oplus 0_{n-2}))\big)\Big)\\
	&=\sum_{i,j}\bra{i,j}\big((I_2\oplus 0_{n-2})\otimes (I_n-(I_2\oplus 0_{n-2}))\big) W \big((I_2\oplus 0_{n-2})\otimes \\&(I_n-(I_2\oplus 0_{n-2}))\big)\ket{i,j}\\
	&\geq 0.
	\end{split}
	\end{eqnarray}
	The last inequality follows from $W$ is an EW. Thus, we derive a contradiction. Finally, suppose $m,n>2$. Using the projector $(I_m-(I_2\oplus 0_{m-2}))\otimes (I_n-(I_2\oplus 0_{n-2}))$ we similarly obtain that $\tilde{W}=(I_m-(I_2\oplus 0_{m-2}))\otimes (I_n-(I_2\oplus 0_{n-2}))W(I_m-(I_2\oplus 0_{m-2}))\otimes (I_n-(I_2\oplus 0_{m-2}))$ is negative semidefinite. It follows that $\mathop{\rm Tr}(\tilde{W})<0$. However, for the same reason \eqref{eq:c=2-3} we also have $\mathop{\rm Tr}(\tilde{W})\geq 0$. We derive a contradiction again. Therefore, such an EW which has exact two positive eigenvalues does not exist.

	This completes the proof.
\end{proof}

We give some observations on the relation between $v_-$ and $v_+$ on the inertia.

\begin{lemma}
\label{le:negative}
Suppose $W$ is an EW on $\bbC^m \otimes \bbC^n$. Suppose the negative space of $W$ is spanned by $\{\ket{u_j}\}$ and the positive space is spanned by $\{\ket{v_j}\}$. If $\{\ket{u_j}\}$ is supported in $\bbC^{m_1}\otimes \bbC^{n_1}$ and $\{\ket{v_j}\}$ is supported in $\bbC^{m_2}\otimes \bbC^{n_2}$ and $m_1>m_2$ and $n_1>n_2$,  then we can choose a projector $P=((I_{m_1}-(I_{m_2}\oplus 0_{m_1-m_2}))\oplus 0_{m-m_1})\otimes((I_{n_1}-(I_{n_2}\oplus 0_{n_1-n_2}))\oplus 0_{n-n_1})$ such that $V=PWP$ is negative semidefinite.
\end{lemma}

\section{Inertias of the partial transpose of the two-qutrit bipartite NPT states}
\label{sec:I_2}
	In this section we focus on the inertias of  the two-qutrit bipartite NPT states.  In Subsec. \ref{subsec:4_1},  we propose two observations on of the two-qutrit  NPT states in Lemmas \ref{le:13} and \ref{le:12} firstly.  For understanding the partial transpose of the two-qutrit bipartite NPT states in Example \ref{ex:11}. Then we discuss all cases in  the two-qutrit bipartite NPT state in detail in Theorem \ref{th:NPT2}. In Subsec. \ref{subsec:4_2}, we draw a conclusion about $\mathcal{N}_{3,3}$  in Theorem \ref{th:N_33}. Finally we propose Example \ref{ex:13arrays} to illustrate Theorem \ref{th:N_33}(i).
\subsection{Supporting lemmas}
\label{subsec:4_1}
At first, we propose some observations on the two-qutrit  NPT states.

\begin{lemma}
	\label{le:12}
	Let $\rho$ be a two-qutrit NPT state. Then we have,

	(i) if $|0,0\rangle$ and $|0,1\rangle$ are in $\ker \rho^{\Gamma}$, then $\ln (\rho^{\Gamma})$ is one of the $13$ arrays in Theorem \ref{th:N_33}(i).

	(ii) $(4,2,3)\notin \mathcal{N}_{3,3}$.

\end{lemma}
\begin{proof}
	(i) If $|0,0\rangle \in \ker \rho^{\Gamma}$, then we have $\rho^{\Gamma} |0,0\rangle =0$. Then we have that the first row and column of $\rho^{\Gamma}$ consist of zero entries. Similarly, if $|0,1\rangle \in \ker \rho^{\Gamma}$, then we have $\rho^{\Gamma} |0,1\rangle =0$. Then we have that the second row and column of $\rho^{\Gamma}$ consist of zero entries.

	So $\rho^{\Gamma}$ can be written as
	$	\begin{bmatrix}
	0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
	0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
	0 & 0 & * & * & * & * & * & * & * \\
	0 & 0 & * & * & * & * & * & * & * \\
	0 & 0 & * & * & * & * & * & * & * \\
	0 & 0 & * & * & * & * & * & * & * \\
	0 & 0 & * & * & * & * & * & * & * \\
	0 & 0 & * & * & * & * & * & * & * \\
	0 & 0 & * & * & * & * & * & * & *
	\end{bmatrix}.$
	Because $ |0,0\rangle ,|0,1\rangle \in \ker(\rho^{\Gamma})$, we have $ |0,0\rangle, |0,1\rangle \in \ker(\rho)$.
	Then $\rho$ can be written as
	$	\begin{bmatrix}
	0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
	0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
	0 & 0 & a & 0 & 0 & b & 0 & 0 & c \\
	0 & 0 & 0 & * & * & * & * & * & * \\
	0 & 0 & 0 & * & * & * & * & * & * \\
	0 & 0 & b^* & * & * & d & * & * & * \\
	0 & 0 & 0 & * & * & * & * & * & * \\
	0 & 0 & 0 & * & * & * & * & * & * \\
	0 & 0 & c^* & * & * & * & * & * & *
	\end{bmatrix}$.
	We have $\begin{bmatrix} a & b \\ b^* & d \end{bmatrix} \geq 0 $ by $\rho \geq 0$. If $a=0$, then $b=c=0$.

	If $a\neq 0$, then we have
	$\rho=	\begin{bmatrix}
	0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
	0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
	0 & 0 & a & 0 & 0 & b & 0 & 0 & c \\
	0 & 0 & 0 & * & * & * & * & * & * \\
	0 & 0 & 0 & * & * & * & * & * & * \\
	0 & 0 & b^* & * & * & d & * & * & * \\
	0 & 0 & 0 & * & * & * & * & * & * \\
	0 & 0 & 0 & * & * & * & * & * & * \\
	0 & 0 & c^* & * & * & * & * & * & *
	\end{bmatrix} \geq 0$. There exists a local invertible product operator $V=A\otimes B$ such that $V\rho V^+ = \begin{bmatrix} 0&0&0\\ 0&0&0\\ 0&0&a \end{bmatrix} \oplus \sigma$, where $\sigma$ is a $2\times 3$ state.

	We have known that $ \mathcal{N}_{2,3}=\{(1,2,3), (1,1,4),(1,0,5), (2,0,4)\}$ from \cite[Corollary $11(a)$]{2020Inertias}. Then one can verify that $\ln (\rho^{\Gamma})$ is one of the 13 arrays in Theorem \ref{th:N_33}(i).

	(ii) Suppose $\ln(\rho^{\Gamma})=(4,2,3)$. Then we have $\mathop{\rm Dim}(\ker(\rho^{\Gamma}))=2$. Because the dimension of  non-positive eigen-space of $\rho$ is $6>(3-1)\times(3-1)+1$, we obtain that the non-positive eigen-space of $\rho$ has infinitely many product vectors from Lemma \ref{le:2}. From Lemma \ref{le:mn} we obtain that $\ker(\rho^{\Gamma})$ contains  infinitely many product vectors. We have $\ker(\rho^{\Gamma})=\mathrm{span}  \{\ket{a,b}, \ket{c,d}\}$. If  $\ket{a}$ and $\ket{c}$ are linearly dependent, then we can find an invertible operator $W=C\otimes D$ such that $W\ket{a,b}=\ket{0,0}$ and $W\ket{c,d}=\ket{0,1}$. Then we have $\ker((W^{-1})^{\dagger}\rho W^{-1})=\mathrm{span} \{\ket{0,0},\ket{0,1}\}$. Let $\sigma =(W^{-1})^{\dagger}\rho W^{-1}$. We obtain that $\ln(\sigma^{\Gamma})=\ln(\rho^{\Gamma})$ by Sylvester theorem.  Then $\ket{0,0}$ and $\ket{0,1}$ are in $\ker(\sigma^{\Gamma})$. From (i) we derive a contradiction.
	So we have $(4,2,3)\notin \mathcal{N}_{3,3}$.

\end{proof}

\begin{lemma}
	\label{le:13}
	Let $\rho$ be a two-qutrit NPT state. Then we have,

	(i) if $\ket{0,0}$, $\ket{1,1}$ and $(\ket{0}+\ket{1})(\ket{0}+\ket{1})$ are in $\ker \rho^{\Gamma}$, then $\ln (\rho^{\Gamma})$ is one of the $13$ arrays in Theorem \ref{th:N_33}(i).

	(ii) if $\ket{0,0}, \ket{1,1}, \ket{2}(\ket{0}+\ket{1}), (a\ket{0}+b\ket{1}+\ket{2})(\ket{0}+f\ket{1})$ satisfying the conditions $a,b\neq 0$  and $f\neq 0,1$ are in $\ker \rho^{\Gamma}$, then $\ln (\rho^{\Gamma})$ is one of the $12$ arrays in theorem \ref{th:N_33}(i).

	(iii) $(3,3,3), (2,4,3)\notin  \mathcal{N}_{3,3}$.
\end{lemma}
\begin{proof}
	(i) If $\ket{0,0} \in \ker \rho^{\Gamma}$, then we have $\rho^{\Gamma} \ket{0,0} =0$. Then we have that the first row and column of $\rho^{\Gamma}$ consist of zero entries. Similarly, if $\ket{1,1} \in \ker \rho^{\Gamma}$, then we have $\rho^{\Gamma} \ket{1,1} =0$. Then we have that the fifth row and column of $\rho^{\Gamma}$ consist of zero entries. We have  $(\ket{0}+\ket{1})(\ket{0}+\ket{1}) \in \ker \rho^{\Gamma}$.

	So $\rho^{\Gamma}$ can be written as
	$	\begin{bmatrix}
	0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
	0 & a & * & -a & 0 & * & * & * & * \\
	0 & b & * & -b & 0 & * & * & * & * \\
	0 & c & * & -c & 0 & * & * & * & * \\
	0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
	0 & d & * & -d & 0 & * & * & * & * \\
	0 & e & * & -e & 0 & * & * & * & * \\
	0 & f & * & -f & 0 & * & * & * & * \\
	0 & g & * & -g & 0 & * & * & * & *
	\end{bmatrix}.$ Because $ \ket{0,0}, \ket{1,1},  (\ket{0}+\ket{1})(\ket{0}+\ket{1}) \in \ker(\rho^{\Gamma})$, we have $ \ket{0,0}, \ket{1,1},  (\ket{0}+\ket{1})(\ket{0}+\ket{1})  \in \ker(\rho)$.
	Then $\rho$ can be written as
	$	\begin{bmatrix}
	0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
	0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & h^* \\
	0 & 0 & * & 0 & 0 & * & 0 & g & * \\
	0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -h^* \\
	0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
	0 & 0 & * & 0 & 0 & * & -g & 0 & * \\
	0 & 0 & 0 & 0 & 0 & -g^* & * & * & * \\
	0 & 0 & g^* & 0 & 0 & 0 & * & * & * \\
	0 & h & * & -h & 0 & * & * & * & *
	\end{bmatrix}.$

	Because $\rho$ is a  positive semi-definite matrix, then $\rho$ can be written as
	$	\begin{bmatrix}
	0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
	0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
	0 & 0 & * & 0 & 0 & * & 0 & g & * \\
	0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
	0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
	0 & 0 & * & 0 & 0 & * & -g & 0 & * \\
	0 & 0 & 0 & 0 & 0 & -g^* & * & * & * \\
	0 & 0 & g^* & 0 & 0 & 0 & * & * & * \\
	0 & 0 & * & 0 & 0 & * & * & * & *
	\end{bmatrix}.$

	Then we obtain that $\ket{0,0}$ and $\ket{0,1}$ are in the  $\ker (\rho^{\Gamma})$. We obtain that $\ln (\rho^{\Gamma})$ is one of the $13$ arrays in Theorem \ref{th:N_33}(i) from Lemma \ref{le:12}(i).

	(ii) If $\ket{0,0} \in \ker \rho^{\Gamma}$, then we have $\rho^{\Gamma} \ket{0,0} =0$. Then we have that the first row and column of $\rho^{\Gamma}$ consist of zero entries. Similarly, if $\ket{1,1} \in \ker \rho^{\Gamma}$, then we have $\rho^{\Gamma} \ket{1,1} =0$. Then we have that the fifth row and column of $\rho^{\Gamma}$ consist of zero entries. We have  $\ket{2}(\ket{0}+\ket{1}) \in \ker \rho^{\Gamma}$. Then we have that $\rho^{\Gamma}\ket{2}(\ket{0}+\ket{1})=0$.

	So $\rho^{\Gamma}$ can be written as
	$	\begin{bmatrix}
	0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
	0 & * & * & * & 0 & * & l & -l & * \\
	0 & * & * & * & * & * & m & -m & * \\
	0 & * & * & * & 0 & * & n & -n & * \\
	0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
	0 & * & * & * & 0 & * & o & -o & * \\
	0 & * & * & * & 0 & * & p & -p & * \\
	0 & * & * & * & 0 & * & q & -q & * \\
	0 & * & * & * & 0 & * & r & -r& *
	\end{bmatrix}.$
	Because $ \ket{0,0}, \ket{1,1},  \ket{2}(\ket{0}+\ket{1}), (a\ket{0}+b\ket{1}+\ket{2})(\ket{0}+f\ket{1}) \in \ker(\rho^{\Gamma})$, we have $ \ket{0,0}, \ket{1,1}, \ket{2}(\ket{0}+\ket{1}), (a\ket{0}+b\ket{1}+\ket{2})(\ket{0}+f\ket{1})  \in \ker(\rho)$. Then we have that $\rho \ket{0,0} =0$, $\rho \ket{1,1} =0$, $\rho \ket{2}(\ket{0}+\ket{1})=0$ and  $\rho (a\ket{0}+b\ket{1}+\ket{2})(\ket{0}+f\ket{1})=0$.
	Then $\rho$ can be written as
	$	\begin{bmatrix}
	0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
	0 & 0 & * & 0 & 0 & 0 & 0 & 0 & * \\
	0 & 0 & * & 0 & 0 & * & 0 & 0 & * \\
	0 & 0 & 0 & 0 & 0 & * & 0 & 0 & * \\
	0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
	0 & 0 & * & 0 & 0 & * & 0 & 0 & * \\
	0 & 0 & 0 & 0 & 0 & * & 0 & 0 & * \\
	0 & 0 & * & 0 & 0 & * & 0 & 0 & * \\
	0 & 0 & * & 0 & 0 & * & 0 & 0 & *
	\end{bmatrix}.$
	Because $\rho$ is a  positive semi-definite matrix and $\rho$ is a hermitian matrix, then $\rho$ can be written as
	$	\begin{bmatrix}
	0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
	0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
	0 & 0 & * & 0 & 0 & * & 0 & 0 & * \\
	0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
	0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
	0 & 0 & * & 0 & 0 & * & 0 & 0 & * \\
	0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
	0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
	0 & 0 & * & 0 & 0 & * & 0 & 0 & *
	\end{bmatrix}.$

	Then we obtain that $\ket{0,0}$ and $\ket{0,1}$ are in the  $\ker (\rho^{\Gamma})$. We obtain that $\ln (\rho^{\Gamma})$ is one of the $13$ arrays in Theorem \ref{th:N_33}(i) from Lemma \ref{le:12}(i).

	(iii) If $\ln (\rho^{\Gamma})= (3,3,3)$ or $(2,4,3)$, we obtain that the non-positive eigen-space of $\rho$ has infinitely many product vectors form Lemma \ref{le:2}. Then we obtain that the zero eigen-space of $\rho$ has infinitely many product vectors from Lemma \ref{le:mn}(i). If $\ln (\rho^{\Gamma})= (3,3,3)$, then the zero eigen-space of $\rho$ is up to
	SLOCC equivalence  spanned by $\{\ket{0,0},\ket{0,1}\}$ or  $\{\ket{0,0},\ket{1,1}, (\ket{0}+\ket{1})(\ket{0}+\ket{1})\}$ by Lemma \ref{le:9}(ii).  If $\ln (\rho^{\Gamma})= (2,4,3)$, then the zero eigen-space of $\rho$ is up to
	SLOCC equivalence  spanned by $\{\ket{0,0},\ket{0,1}\}$ or $\{\ket{0,0},\ket{1,1}, (\ket{0}+\ket{1})(\ket{0}+\ket{1})\}$ or $\{\ket{0,0}, \ket{1,1}, \ket{2}(\ket{0}+\ket{1}), (a\ket{0}+b\ket{1}+\ket{2})(\ket{0}+f\ket{1})\}$ satisfying the conditions $a,b\neq 0$  and $f\neq 0,1$ by Lemma \ref{le:9}(iii). Then in these two cases we obtain that $\ket{0,0}$ and $\ket{0,1}$ are in the  $\ker (\rho^{\Gamma})$. $\ln (\rho^{\Gamma})$ is one of the 12 arrays in Theorem \ref{th:N_33}(i) from (i) and (ii). Then we derive a contradiction. So we have  $(3,3,3), (2,4,3)\notin  \mathcal{N}_{3,3}$.
\end{proof}
We propose an example to understand the two-qutrit bipartite NPT states in Example \ref{ex:11}. Then we discussed all cases in the two-qutrit bipartite NPT states in detail in Theorem \ref{th:NPT2}.
\begin{example}
	\label{ex:11}
	Suppose $\rho =(\ket{0,0}+\ket{1,1}+\ket{2,2})(\bra{0,0}+\bra{1,1}+\bra{2,2})+(\ket{0}(a\ket{0}+b\ket{1}))(\bra{0}(a^*\bra{0}+b^*\bra{1}))$ is a two-qutrit bipartite NPT state.

	Then $\rho$ can be written as
	$	\begin{bmatrix}
	1+\left| a \right|^2 & ab^* & 0 & 0 & 1 & 0 & 0 & 0 & 1 \\
	a^*b & \left| b \right|^2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
	0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
	0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
	1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 \\
	0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
	0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
	0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
	1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 \\
	\end{bmatrix}$.
	Then $\rho^\Gamma$ can be written as
	$	\begin{bmatrix}
	1+\left| a \right|^2 & ab^* & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
	a^*b & \left| b \right|^2 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\
	0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\
	0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
	0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\
	0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\
	0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\
	0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\
	0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\
	\end{bmatrix}$.
	Then we obtain that $\det \rho^\Gamma=-1-\left| a \right|^2$. We get that $\rho^\Gamma$ has six constant eigenvalues, $-1,-1,1,1,1,1$. The other three eigenvalues of $\rho^\Gamma$ are the three roots of $x^3+(-1-\left| a \right|^2-\left| b \right|^2)x^2+(-1+\left| b \right|^2)x+1+\left| a \right|^2$. Suppose these three roots are $x_1>x_2>x_3$ respectively. We obtain that $x_1+x_2+x_3=1-\left| b \right|^2$ and $x_1x_2x_3=-1-\left| a \right|^2$. Then we obtain that three eigenvalues are $1$, $\frac{1}{2}(\left| a \right|^2+\left| b \right|^2-\sqrt{4+4\left| a \right|^2+\left| a \right|^4+2\left| a \right|^2\left| b \right|^2+\left| b \right|^4})$ and $\frac{1}{2}(\left| a \right|^2+\left| b \right|^2+\sqrt{4+4\left| a \right|^2+\left| a \right|^4+2\left| a \right|^2\left| b \right|^2+\left| b \right|^4})$. We get that $x_1,x_2>0$ and $x_3<0$. So nine eigenvalues of $\rho^\Gamma$ are $-1,-1,x_3,1,1,1,1,x_1,x_2$. The interia $\ln \rho^\Gamma=(3,0,6)$.

\end{example}

In this example, we investigate the interia of a two-qutrit bipartite NPT state $\rho=\proj{\alpha}+\proj{\beta}$ of rank two. We define the case of SR$(\ket{\alpha})=m$ and SR$(\beta)=n$ as a binary array $(m,n)$. We get that $\rho$ has six cases, $(1,1)$, $(2,1)$, $(2,2)$, $(3,1)$, $(3,2)$, $(3,3)$. In the case of $(1,1)$, $\rho$ is a separable state. We need to get rid of $(1,1)$.  On the other hand, the case $(3,3)$ is equivalent to the one of the remaining four cases. So it suffices to study four cases of two-qutrit bipartite NPT states of rank two as follows.

\begin{theorem}
	\label{th:NPT2}
	Suppose $\rho=\proj{\alpha}+\proj{\beta}$ is a two-qutrit bipartite NPT state.

	(i) If SR$(\ket{\alpha})=2$ and SR$(\ket{\beta})=1$, then we have $\ln \rho^\Gamma=(1,4,4)$.

	(ii) If SR$(\ket{\alpha})=2$ and SR$(\ket{\beta})=2$, then we have $\ln \rho^\Gamma= (2,2,5)$.

	(iii) If SR$(\ket{\alpha})=3$ and SR$(\ket{\beta})=1$, then we have $\ln \rho^\Gamma=(3,0,6)$.

	(iv) If SR$(\ket{\alpha})=3$ and SR$(\ket{\beta})=2$, then we have $\ln \rho^\Gamma=(2,2,5),(2,1,6),(2,0,7),(3,0,6),(3,1,5)$.

\end{theorem}

\begin{proof}
	Suppose $\rho=\proj{\alpha}+\proj{\beta}$ is a two-qutrit bipartite NPT state,
	\begin{eqnarray}
		\label{eq:NPT}
		\rho=\proj{\alpha}+\proj{\beta}
	\end{eqnarray}.

	(i) If SR$(\ket{\alpha})=2$ in Eq\eqref{eq:NPT}, then we have that $\ket{\alpha}$ is up to SLOCC equivalent to $\ket{0,0}+\ket{1,1}$.  If SR$(\ket{\beta})=1$, then we have that $\ket{\beta}$ is up to LOCC equivalent to $\ket{2,2}$. Then we have $\rho^\Gamma=\begin{bmatrix}
	1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
	0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\
	0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
	0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
	0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\
	0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
	0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
	0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
	0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\
	\end{bmatrix}$. Then we get that nine eigenvalues are $-1,0,0,0,0,1,1,1,1$. Then $\ln \rho^\Gamma=(1,4,4) $. In conclusion, we obtain that if SR$(\ket{\alpha})=2$ and SR$(\ket{\beta})=1$, then we have $\ln \rho^\Gamma=(1,4,4)$.

	(ii) If SR$(\ket{\alpha})=2$ in Eq\eqref{eq:NPT}, then we have that $\ket{\alpha}$ is up to SLOCC equivalent to $\ket{0,0}+\ket{1,1}$.  If SR$(\ket{\beta})=2$, then we have that $\ket{\beta}$ cotains $\ket{2,y_2}$ and $(a\ket{0}+b\ket{1})(\ket{y_2})$. Because $(a\ket{0}+b\ket{1})(\ket{y_0})$ is equivalent to $\ket{0,y_0}$, we have that $\ket{\beta}$ is up to SLOCC equivalent to $\ket{0,y_0}+\ket{2,y_2}$. We get that at least one of $\ket{y_0}$ and $\ket{y_2}$ contains $\ket{2}$. So $\ket{\beta}$ has two cases. One case is that $\ket{\beta}$ is up to SLOCC equivalent to $\ket{0}(a\ket{0}+b\ket{1})+\ket{2,2}$, the other case is  that $\ket{\beta}$ is up to SLOCC equivalent to $\ket{0,2}+\ket{2}(a\ket{0}+b\ket{1})$.

	(ii.a) Suppose $\ket{\beta}=\ket{0}(a\ket{0}+b\ket{1})+\ket{2,2}$ in Eq\eqref{eq:NPT}.
	Then we have $\rho^\Gamma=\begin{bmatrix}
	1+\left| a \right|^2 & ab^* & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
	ba^* & \left| b \right|^2 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\
	0 & 0 & 0 & 0 & 0 & 0 & a^* & b^* & 0 \\
	0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
	0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\
	0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
	0 & 0 & a & 0 & 0 & 0 & 0 & 0 & 0 \\
	0 & 0 & b & 0 & 0 & 0 & 0 & 0 & 0 \\
	0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\
	\end{bmatrix}$. Then we get that nine eigenvalues are $-(a^2+b^2)^{-\frac{1}{2}},\frac{1}{2}(a^2+b^2-(4+4a^2+a^4+2a^2b^2+b^4)^{-\frac{1}{2}},0,0,1,1,1,(a^2+b^2)^{-\frac{1}{2}},\frac{1}{2}(a^2+b^2-(4+4a^2+a^4+2a^2b^2+b^4)^{-\frac{1}{2}}$. Then $\ln \rho^\Gamma=(2,2,5)$.

	(ii.b) Suppose $\ket{\beta}=\ket{0,2}+\ket{2}(a\ket{0}+b\ket{1})$ in Eq\eqref{eq:NPT}.

	Then we have $\rho^\Gamma=\begin{bmatrix}
	1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & a \\
	0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & b \\
	0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\
	0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
	0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\
	0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
	0 & 0 & 0 & 0 & 0 & 0 & \left| a \right|^2 & ab^* & 0 \\
	0 & 0 & 0 & 0 & 0 & 0 & ba^* & \left| b \right|^2 & 0 \\
	a^* & b^* & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
	\end{bmatrix}$. Then we get that six eigenvalues are $0,0,1,1,1,a^2+b^2$. The other three eigenvalues are the roots of $x^3+(-1-a^2-b^2)x-a^2$. These three roots are two negative roots and one positive root. Then $\ln \rho^\Gamma=(2,2,5)$. In conclusion, we obtain that if SR$(\ket{\alpha})=2$ and SR$(\ket{\beta})=2$, then we have $\ln \rho^\Gamma=(2,2,5)$.

	(iii) If SR$(\ket{\alpha})=3$ in Eq\eqref{eq:NPT}, then we have that $\ket{\alpha}$ is up to SLOCC equivalent to $\ket{0,0}+\ket{1,1}+\ket{2,2}$.  If SR$(\ket{\beta})=1$, then we have that $\ket{\beta}$ is up to SLOCC equivalent to $\ket{0}(a\ket{0}+b\ket{1})$.
	From Example \ref{ex:11}, we get that $\ln \rho^\Gamma=(3,0,6)$.

	(iv) If SR$(\ket{\alpha})=3$ in Eq\eqref{eq:NPT}, then we have that $\ket{\alpha}$ is up to SLOCC equivalent to $\ket{0,0}+\ket{1,1}+\ket{2,2}$.  If SR$(\ket{\beta})=2$, then we have that $\ket{\beta}$ is up to SLOCC equivalent to $\ket{0}(a\ket{0}+b\ket{1}+c\ket{2})+\ket{1}(d\ket{0}+e\ket{1})$. Suppose $M$ is a two-qutrit invertible matrix. There exists a two-qutrit invertible matrix $M$, such that $(M(M^T)^{-1}\otimes I_3)(\ket{0,0}+\ket{1,1}+\ket{2,2})=(\ket{0,0}+\ket{1,1}+\ket{2,2})$ and $M(M^T)^{-1}(\ket{0}(a\ket{0}+b\ket{1}+c\ket{2})+\ket{1}(d\ket{0}+e\ket{1}))=\ket{0}(a'\ket{0}+b'\ket{1}+c'\ket{2})+e'\ket{1,1}$ or $\ket{0}(a'\ket{0}+b'\ket{1}+c'\ket{2})+d'\ket{1,0}$. Then $\ket{\beta}$ has two cases, $\ket{0}(a\ket{0}+b\ket{1}+c\ket{2})+e\ket{1,1}$ and $\ket{0}(a\ket{0}+b\ket{1}+c\ket{2})+d\ket{1,0}$. The case  $\ket{0}(a\ket{0}+b\ket{1}+c\ket{2})+d\ket{1,0}$ is SLOCC equivalent to the case $\ket{0}(a\ket{0}+b\ket{1}+c\ket{2})+\ket{1,0}$. There exists a two-qutrit invertible matrix $M_1$, such that $(M_1(M_1^T)^{-1}\otimes I_3)(\ket{0,0}+\ket{1,1}+\ket{2,2})=(\ket{0,0}+\ket{1,1}+\ket{2,2})$ and $(M_1(M_1^T)^{-1}\otimes I_3)(\ket{0}(a\ket{0}+b\ket{1}+c\ket{2})+d\ket{1,0})=\ket{0}(a''\ket{0}+b''\ket{1})+d''\ket{1,0}$ or $\ket{0}(a''\ket{0}+c''\ket{2})+d''\ket{1,0}$. The case  $\ket{0}(a\ket{0}+b\ket{1}+c\ket{2})+\ket{1,0}$ is divided into two cases $\ket{0}(a\ket{0}+b\ket{1})+\ket{1,0}$ and  $\ket{0}(a\ket{0}+c\ket{2})+\ket{1,0}$. Similarly, the case  $\ket{0}(a\ket{0}+b\ket{1}+c\ket{2})+e\ket{1,1}$ is divided into two cases $\ket{0}(a\ket{0}+b\ket{1})+e\ket{1,1}$ and  $\ket{0}(b\ket{1}+c\ket{2})+e\ket{1,1}$.  If SR$(\ket{\beta})=2$, then $\ket{\beta}$ has four cases $\ket{0}(a\ket{0}+b\ket{1})+\ket{1,0}$, $\ket{0}(a\ket{0}+c\ket{2})+\ket{1,0}$, $\ket{0}(a\ket{0}+b\ket{1})+e\ket{1,1}$, $\ket{0}(b\ket{1}+c\ket{2})+e\ket{1,1}$.

	(iv.a) Suppose $\ket{\beta}=\ket{0}(a\ket{0}+b\ket{1})+\ket{1,0}$ in Eq\eqref{eq:NPT}. Then we have $\rho^\Gamma=\begin{bmatrix}
	1+\left| a \right|^2 & ab^* & 0 & a^* & b^* & 0 & 0 & 0 & 0 \\
	ba^* & \left| b \right|^2 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\
	0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\
	a & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\
	b & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\
	0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\
	0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\
	0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\
	0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\
	\end{bmatrix}$. We know that $A\rho^\Gamma A^\dagger$ and $\rho^\Gamma$ have the same inertia when $A$ is an elementary matrix. We obtain that $S=B\rho^\Gamma B^\dagger=\begin{bmatrix}
	1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
	0 & \left| b \right|^2-\left| ab^*-a^* \right|^2-1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
	0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\
	0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\
	0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\
	0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\
	0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\
	0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\
	0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\
	\end{bmatrix}$ has the same inertia as $\rho^\Gamma$, where $B$ is a product of some elementary matrices.  Then we get that the six eigenvalues of $\rho^\Gamma$ are $-1,-1,1,1,1,1,1,1, \left| b \right|^2-\left| ab^*-a^* \right|^2-1$. $\left| b \right|^2-\left| ab^*-a^* \right|^2-1$ can be a positive or zero or negative number. Then $\ln \rho^\Gamma=(3,0,6) $ when $\left| b \right|^2-\left| ab^*-a^* \right|^2-1<0$, $\ln \rho^\Gamma=(2,1,6) $ when $\left| b \right|^2-\left| ab^*-a^* \right|^2-1=0$, $\ln \rho^\Gamma=(2,0,7) $ when $\left| b \right|^2-\left| ab^*-a^* \right|^2-1>0$.

	(iv.b) Suppose $\ket{\beta}=\ket{0}(a\ket{0}+c\ket{2})+\ket{1,0}$ in Eq\eqref{eq:NPT}. If $c=0$, then SR$(\ket{\beta})=1$ and we reduce to (iii). Then we have $c\neq 0$ and $\begin{bmatrix}
	1 & 0 & 0 \\
	0 & 1 & 0 \\
	0 & 0 & c \\
	\end{bmatrix} \otimes \begin{bmatrix}
	1 & 0 & 0 \\
	0 & 1 & 0 \\
	0 & 0 & c^{-1} \\
	\end{bmatrix} \ket{\beta}=\ket{0}(a\ket{0}+\ket{2})+\ket{1,0}$.  So we have $\ket{\beta}$ is equivalent to $\ket{0}(a\ket{0}+\ket{2})+\ket{1,0}$. Then we have $\rho^\Gamma=\begin{bmatrix}
	1+\left| a \right|^2 & 0 & a & a^* & 0 & 1 & 0 & 0 & 0 \\
	0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\
	a^* & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 \\
	a & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\
	0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\
	1 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\
	0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\
	0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\
	0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\
	\end{bmatrix}$. We have known that $A\rho^\Gamma A^\dagger$ and $\rho^\Gamma$ have the same inertia when $A$ is an elementary matrix. We obtain that $S=B\rho^\Gamma B^\dagger=\begin{bmatrix}
	1+\left| a \right|^2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
	0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\
	0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 \\
	0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
	0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\
	0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\
	0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\
	0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\
	0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\
	\end{bmatrix}$ has the same inertia as $\rho^\Gamma$, where $B$ is a product of some elementary matrices. Then we get that nine enigenvalues are $-1,-1,-1,1,1,1,1,1,1+\left| a \right|^2$. Then $\ln \rho^\Gamma=\ln S =(3,0,6)$.

	(iv.c) Suppose $\ket{\beta}=\ket{0}(a\ket{0}+b\ket{1})+e\ket{1,1}$ in Eq\eqref{eq:NPT}. Then we have $\rho^\Gamma=\begin{bmatrix}
	1+\left| a \right|^2 & ab^* & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
	ba^* & \left| b \right|^2 & 0 & 1+ea^* & eb^* & 0 & 0 & 0 & 0 \\
	0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\
	0 & 1+ae^* & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
	0 & be^* & 0 & 0 & 1+\left| e \right|^2 & 0 & 0 & 0 & 0 \\
	0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\
	0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\
	0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\
	0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\
	\end{bmatrix}$. Suppose $1+ae^*\neq 0$.  Similar to (iv.a), we get that  $S=B\rho^\Gamma B^\dagger=\begin{bmatrix}
	1+\left| a \right|^2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
	0 & 0 & 0 & 1+ea^* & 0 & 0 & 0 & 0 & 0 \\
	0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\
	0 & 1+ae^* & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
	0 & 0 & 0 & 0 & 1+\left| e \right|^2 & 0 & 0 & 0 & 0 \\
	0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\
	0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\
	0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\
	0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\
	\end{bmatrix}$ has the same inertia as $\rho^\Gamma$, where $B$ is a product of some elementary matrices.  Then we get that nine enigenvalues are $-1,-1,-\sqrt{1+ae^*}\sqrt{1+ea^*},1,1,1,1+a^2,1+e^2,\sqrt{1+ae^*}\sqrt{1+ea^*}$. Then $\ln (\rho^\Gamma)=(3,0,6)$.
	If $1+ae^*=0$, we get that  $S'=B\rho^\Gamma B^\dagger=\begin{bmatrix}
	1+\left| a \right|^2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
	0 & \left| b \right|^2-\frac{\left| be \right|^2}{1+\left| e \right|^2}-\frac{\left| ab \right|^2}{1+\left| a \right|^2} & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
	0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\
	0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
	0 & 0 & 0 & 0 & 1+\left| e \right|^2 & 0 & 0 & 0 & 0 \\
	0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\
	0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\
	0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\
	0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\
	\end{bmatrix}$ has the same inertia as $\rho^\Gamma$, where $B$ is a product of some elementary matrices. Then we get that nine enigenvalues are $-1,-1,0,1,1,1,1+a^2,1+e^2,\left| b \right|^2-\frac{\left| be \right|^2}{1+\left| e \right|^2}-\frac{\left| ab \right|^2}{1+\left| a \right|^2}$. $\left| b \right|^2-\frac{\left| be \right|^2}{1+\left| e \right|^2}-\frac{\left| ab \right|^2}{1+\left| a \right|^2}$ can be a positive or zero or negative number. Then $\ln (\rho^\Gamma)=(3,1,5),(2,2,5),(2,1,6)$.
	In conclusion, we obtain that $\ln (\rho^\Gamma)=(3,0,6),(3,1,5),(2,2,5),(2,1,6)$.

	(iv.d) Suppose $\ket{\beta}=\ket{0}(b\ket{1}+c\ket{2})+e\ket{1,1}$ in Eq\eqref{eq:NPT}. If $c=0$, then SR$(\ket{\beta})=1$ and we reduce to (iii).  Then we have $c\neq 0$ and $\begin{bmatrix}
	1 & 0 & 0 \\
	0 & 1 & 0 \\
	0 & 0 & c \\
	\end{bmatrix} \otimes \begin{bmatrix}
	1 & 0 & 0 \\
	0 & 1 & 0 \\
	0 & 0 & c^{-1} \\
	\end{bmatrix} \ket{\beta}=\ket{0}(a\ket{0}+\ket{2})+\ket{1,0}$. So $\ket{\beta}$ is equivalent to $\ket{0}(b\ket{1}+\ket{2})+e\ket{1,1}$. Then we have $\rho^\Gamma=\begin{bmatrix}
	1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
	0 & \left| b \right|^2 & b & 1 & eb^* & e & 0 & 0 & 0 \\
	0 & b^* & 1 & 0 & 0 & 0 & 1 & 0 & 0 \\
	0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
	0 & be^* & 0 & 0 & 1+\left| e \right|^2 & 0 & 0 & 0 & 0 \\
	0 & e^* & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\
	0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\
	0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\
	0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\
	\end{bmatrix}$. Similar to (iv.b), we get that  $S=B\rho^\Gamma B^\dagger=\begin{bmatrix}
	1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
	0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\
	0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\
	0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
	0 & 0 & 0 & 0 & 1+\left| e \right|^2 & 0 & 0 & 0 & 0 \\
	0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\
	0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\
	0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\
	0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\
	\end{bmatrix}$ has the same inertia as $\rho^\Gamma$, where $B$ is a product of some elementary matrices.  Then we get that nine enigenvalues are $-1,-1,-1,1,1,1,1,1,1+\left| e \right|^2$. Then $\ln \rho^\Gamma=\ln S =(3,0,6)$.

	In conlusion, if SR$(\ket{\alpha})=3$ and SR$(\ket{\beta})=2$, then we have $\ln \rho^\Gamma=(2,2,5),(2,1,6),(2,0,7),(3,0,6),(3,1,5)$.

\end{proof}
\subsection{Main conclusion}
\label{subsec:4_2}
We have proposed some observations about two-qutrit bipartite NPT states  in Subsec. \ref{subsec:4_1}. Then we investigate the set $\mathcal{N}_{3,3}$. We have the following observations.
\begin{theorem}
	\label{th:N_33}
	(i) $\mathcal{N}_{3,3}$ has the following $13$ arrays,
	$(1,0,8)$, $(1,1,7)$, $(1,2,6)$, $(1,3,5)$, $(1,4,4)$, $(1,5,3)$, $(2,0,7)$, $(2,1,6)$, $(2,2,5)$, $(2,3,4)$, $(3,0,6)$, $(3,1,5)$, $(4,0,5)$.

	(ii) $\mathcal{N}_{3,3}$  does not have the following three arrays,
	$(2,4,3)$, $(3,3,3)$, $(4,2,3)$.

	(iii) $\mathcal{N}_{3,3}$ might have the following two arrays,
	$(3,2,4)$, $(4,1,4)$.

	(iv) $13\leq |\mathcal{N}_{3,3}|\leq 15$,  $\mathcal{N}_{3,3}$ is a subset of above (i) and (iii)'s arrays.

\end{theorem}
\begin{proof}

	(i) We obtain that $(1,2,3), (2,0,4)\in \mathcal{N}_{2,3}$ by \cite[Corollary $11$]{2020Inertias}. Using Lemma \ref{le:rel} we obtain that,
	$(1,0,8), (1,1,7), (1,2,6), (1,3,5), (1,4,4), (1,5,3), (2,0,7), (2,1,6),\\ (2,2,5), (2,3,4)\in \mathcal{N}_{3,3}$. We obtain that $(3,0,6)\in \mathcal{N}_{3,3}$ by Lemma \ref{le:pre}. We obtain that $(4,0,5)\in \mathcal{N}_{3,3}$ by \cite[Example $2$]{2013Negative}. We obtain that $(3,1,5)\in \mathcal{N}_{3,3}$ by Theorem \ref{th:NPT2}.

	(ii) We have known that $v_-\leq 4$ by \cite[Lemma $5$]{2020Inertias} and $v_+\geq 3$ by Lemma \ref{th:N_33}. We have that $v_-+v_0+v_+=9$. Then  $\mathcal{N}_{3,3}$ migtht have the following five arrays,
	$(2,4,3)$, $(3,2,4)$, $(3,3,3)$, $(4,1,4)$, $(4,2,3)$. Then we have that $(2,4,3), (3,3,3), (4,2,3)\notin \mathcal{N}_{3,3}$ by Lemma \ref{le:12} and \ref{le:13}.

	(iii) From (ii), we get that $\mathcal{N}_{3,3}$ might have the following two arrays,
	$(3,2,4)$, $(4,1,4)$.

	(iii) We have $13\leq |\mathcal{N}_{3,3}|\leq 15$ by the results in (i),(ii),(iii) and $ \mathcal{N}_{3,3}$ is a subset of above (i) and (iii)'s $15$ arrays.
\end{proof}
Then we propose the following example to illustrate the Theorem \ref{th:N_33}(i).
\begin{example}
	\label{ex:13arrays}

Suppose $\rho$ is a two-qutrit bipartite NPT state.
	(i) Suppose $\rho=(\ket{0,0}+\ket{1,1})(\bra{0,0}+\bra{1,1})+\frac{1}{10}I_9$. Then $\ln \rho^\Gamma=(1,0,8)$.

	(ii) Suppose $\rho=(\ket{0,0}+\ket{1,1})(\bra{0,0}+\bra{1,1})+\frac{1}{10}I_9-\frac{1}{10}\ket{2,2}\bra{2,2}$. Then $\ln \rho^\Gamma=(1,1,7)$.

	(iii) Suppose $\rho=(\ket{0,0}+\ket{1,1})(\bra{0,0}+\bra{1,1})+\frac{1}{10}I_9-\frac{1}{10}(\ket{2,2}\bra{2,2}+\ket{2,1}\bra{2,1})$. Then $\ln \rho^\Gamma=(1,2,6)$.

	(iv) Suppose $\rho=(\ket{0,0}+\ket{1,1})(\bra{0,0}+\bra{1,1})+\frac{1}{10}I_9-\frac{1}{10}(\ket{2,2}\bra{2,2}+\ket{2,1}\bra{2,1}-\ket{2,0}\bra{2,0})$. Then $\ln \rho^\Gamma=(1,3,5)$.

	(v) Suppose $\rho=(\ket{0,0}+\ket{1,1})(\bra{0,0}+\bra{1,1})+(\ket{2,2})(\bra{2,2})$. Then $\ln \rho^\Gamma=(1,4,,4)$.

	(vi) Suppose $\rho=(\ket{0,0}+\ket{1,1})(\bra{0,0}+\bra{1,1})$. Then $\ln \rho^\Gamma=(1,5,3)$.

	(vii) Suppose $\rho=(\ket{0,0}+\ket{1,1}+\ket{2,2})(\bra{0,0}+\bra{1,1}+\bra{2,2})+(\ket{0,1}+\ket{1,0})(\bra{0,1}+\bra{1,0})$. Then $\ln \rho^\Gamma=(2,0,7)$.

	(viii) Suppose $\rho=(\ket{0,0}+\ket{1,1}+\ket{2,2})(\bra{0,0}+\bra{1,1}+\bra{2,2})+(\ket{0,0}+\ket{0,1}+\ket{1,1})(\bra{0,0}+\bra{0,1}+\bra{1,1})$. Then $\ln \rho^\Gamma=(2,1,6)$.

	(ix) Suppose $\rho=(\ket{0,0}+\ket{1,1})(\bra{0,0}+\bra{1,1})+(\ket{0,0}+\ket{0,1}+\ket{2,2})(\bra{0,0}+\bra{0,1}+\bra{2,2})$. Then $\ln \rho^\Gamma=(2,2,5)$.

	(x) Suppose $\rho=(\ket{0,0}+\ket{1,1})(\bra{0,0}+\bra{1,1})+(\ket{0,1}+\ket{0,1})(\bra{1,2}+\bra{1,2})$. Then $\ln \rho^\Gamma=(2,3,4)$.

	(xi) Suppose $\rho=(\ket{0,0}+\ket{1,1}+\ket{2,2})(\bra{0,0}+\bra{1,1}+\bra{2,2})+\ket{1,1}\bra{1,1}$. Then $\ln \rho^\Gamma=(3,0,6)$.

	(xii) Suppose $\rho=(\ket{0,0}+\ket{1,1}+\ket{2,2})(\bra{0,0}+\bra{1,1}+\bra{2,2})+(\ket{0,0}+2\ket{0,1}+2\ket{1,1})(\bra{0,0}+2\bra{0,1}+2\bra{1,1})$. Then $\ln \rho^\Gamma=(3,1,5)$.

	(xiii) Suppose $\rho=(\ket{0,0}+\frac{1}{4}\ket{1,1}+\ket{2,2})(\bra{0,0}+\frac{1}{4}\bra{1,1}+\bra{2,2})+(\ket{0,1}+\frac{1}{3}\ket{1,2}+\frac{1}{3}\ket{2,0})(\bra{0,1}+\frac{1}{3}\bra{1,2}+\frac{1}{3}\bra{2,0})+(\ket{0,2}+\frac{1}{2}\ket{1,0}+\ket{2,1})(\bra{0,2}+\frac{1}{2}\bra{1,0}+\bra{2,1})$. Then $\ln \rho^\Gamma=(4,0,5)$.
\end{example}

\section{application}
\label{sec:app}
In this section, we study the relationship between the inertias of $2\times 3$ states and those of $3\times 3$ states firstly. Next we extend some conclusions on the inertias  from  $2\times n$ states to
 $3\times n$ states. We also partially test the existence of two unverified inertias using python program.

 We have obtained that $\mathcal{N}_{2,3}=\{(1,2,3),(1,1,4),(1,0,5),(2,0,4)\}$ \cite{2020Inertias}. Then we have obtained that $(1,0,8)$, $(1,1,7)$, $(1,2,6)$, $(1,3,5)$, $(1,4,4)$, $(1,5,3)$, $(2,0,7)$, $(2,1,6)$, $(2,2,5)$, $(2,3,4)$, $(3,0,6)$, $(3,1,5)$, $(4,0,5)$ $\in \mathcal{N}_{3,3}$ in Theorem \ref{th:N_33}. By Theorem \ref{th:N_33} (i), we have also obtained that $(1,0,8)$, $(1,1,7)$, $(1,2,6)$, $(1,3,5)$, $(1,4,4)$, $(1,5,3)$ are from $(1,2,3)$, and $(2,0,7)$, $(2,1,6)$, $(2,2,5)$, $(2,3,4)$ are from $(2,0,4)$. Next we consider the remaining three inertias,  $(3,0,6)$, $(3,1,5)$ and $(4,0,5)$. From Example \ref{ex:13arrays} (xi), (xii), (xiii), we propose the following example.
\begin{example}
	\label{ex:3arrays}
	Suppose $\rho$ is a $2 \times 3$ bipartite NPT state.

	(xi) Suppose $\rho=(\ket{0,0}+\ket{1,1}+\ket{1,2})(\bra{0,0}+\bra{1,1}+\bra{1,2})+\ket{1,1}\bra{1,1}$. Then $\ln \rho^\Gamma=(1,1,4)$.

	(xii) Suppose $\rho=(\ket{0,0}+\ket{1,1}+\ket{1,2})(\bra{0,0}+\bra{1,1}+\bra{2,2})+(\ket{0,0}+2\ket{0,1}+2\ket{1,1})(\bra{0,0}+2\bra{0,1}+2\bra{1,1})$. Then $\ln \rho^\Gamma=(2,0,4)$.

	(xiii) Suppose $\rho=(\ket{0,0}+\frac{1}{4}\ket{1,1}+\ket{1,2})(\bra{0,0}+\frac{1}{4}\bra{1,1}+\bra{1,2})+(\ket{0,1}+\frac{1}{3}\ket{1,2}+\frac{1}{3}\ket{1,0})(\bra{0,1}+\frac{1}{3}\bra{1,2}+\frac{1}{3}\bra{1,0})+(\ket{0,2}+\frac{1}{2}\ket{1,0}+\ket{1,1})(\bra{0,2}+\frac{1}{2}\bra{1,0}+\bra{1,1})$. Then $\ln \rho^\Gamma=(2,0,4)$.
\end{example}

   From this example, we get that $(3,0,6)$ is from $(1,1,4)$, and $(3,1,5)$, $(4,0,5)$ are from $(2,0,4)$ by Example \ref{ex:3arrays}. We conclude the above findings in the following table.
   \begin{center}
   	\label{ta:1}
   \captionof{table}{the relationship between $2\times 3$ states and $3\times 3$ states}
   \begin{tabular}{|l|c|}
   	\hline
   	$\mathcal{N}_{2,3}$ & $\mathcal{N}_{3,3}$ \\
   	\hline
   	$(1,2,3)$ & $(1,5,3)$ \\
   	& $(1,4,4)$ \\
   	& $(1,3,5)$ \\
   	& $(1,2,6)$ \\
   	& $(1,1,7)$ \\
   	& $(1,0,8)$ \\
   	\hline
   	$(1,1,4)$ & $(3,0,6)$ \\
   	\hline
   	$(2,0,4)$ & $(2,3,4)$ \\
   	 & $(2,2,5)$ \\
   	 & $(2,1,6)$ \\
   	 & $(2,0,7)$ \\
   	 & $(3,1,5)$ \\
   	 & $(4,0,5)$ \\
   	 \hline
   \end{tabular}
	\end{center}

	 So far we have investigated the $3\times 3$ states. Next we will extend the conclusions from low dimensional states to high dimensional states. For extending the conclusions on the inertias  from $2\times n$ states to $3\times n$ states, we consider the inertias of $\mathcal{N}_{3,n}$ in the following.
	\begin{lemma}
		\label{le:3n}
		There are at least $(n-1)(2n-1)$ inertias in $\mathcal{N}_{3,n}$, i.e.,

		$|\mathcal{N}_{3,n}|\geq (n-1)(2n-1) $, for $\forall n\geq 2$.

		Furthermore, these  $(n-1)(2n-1)$ inertias in $\mathcal{N}_{3,n}$ are as follows.

		$(1,3n-4-j,j+3)$, $\forall 0\leq j\leq 3n-4$,

		$(2,3n-6-j,j+4)$, $\forall 0\leq j\leq 3n-6$,

		...

		$(n-1,n-j,n+1+j)$, $\forall 0\leq j\leq n$.
	\end{lemma}
\begin{proof}
	 	We have obtained that $(1,2n-4,3)\in \mathcal{N}_{2,n}$ \cite{2020Inertias}.  Using Lemma \ref{le:rel} we have that $(1,3n-4-j,j+3)\in  \mathcal{N}_{3,n}$, $\forall 0\leq j\leq 3n-4$.

	 Similarly, we also have that

	 $(2,3n-6-j,j+4)\in  \mathcal{N}_{3,n}$, $\forall 0\leq j\leq 3n-6$,

	 ...

	 $(n-1,n-j,n+1+j)$, $\forall 0\leq j\leq n$.

	 So we obtain that $\mathcal{N}_{3,n}$ obtains at least $(3n-3)+(3n-5)+...+(n+1)$, namely $(n-1)(2n-1)$ inertias.
\end{proof}

Finally, we partially investigate the existence of inertia $(4,1,4)$ and $(3,2,4)$ in terms of numerical test by using a python program. We utilize a program package named numpy \footnote{Oliphant,T.E. A guide to NumPy (Vol.1, p.85).USA: Trelgol Publishing.(2006).} to generate a $9\times3$ real random matrix $R$ \footnote{https://machinelearningmastery.com/how-to-generate-random-numbers-in-python/(2018).}. Then we multiply $R$ by its transpose $R^T$ to obtain a random positive semi-definite matrix $M=RR^T$ \footnote{https://en.m.wikipedia.org/wiki/Definite matrixNegative-definite.2C semidefinite and indefinite matrices(2002).}. After transforming $M$ into $M^\Gamma$, the program only need to calculate the feature values of $\rm M^\Gamma$ and judge its inertia index, which turns out to be neither of the two inertias $(4,1,4)$ or $(3,2,4)$. Such a numerical test supports the conjecture that neither of the two inertias exists.

\section{Conclusion}
\label{sec:con}
 We have investigated two-qutrit EWs constructed by the partial transpose of NPT states.  Furthermore, we investigated the inertias of the two-qutrit bipartite NPT states.  One open problem is to deterimine whether $(3,2,4), (4,1,4)$ are in $\mathcal{N}_{3,3}$. In the future, we need to extend more conclusions on the inertias from low dimensional states to high dimensional states.

\bibliographystyle{unsrt}

redpajama_set_name: RedPajamaArXiv


According to IBIS World research, 24,440 restaurants were opened in Australia from 2013 – 2015.
This shows how competitive the food business really is. Either way, don't let that number scare you. Competition is not necessarily a bad thing.
Competition is often pushing other businesses to be better, smarter and more efficient. The cleverest way to observe your competition is to use it as motivation for your business growth and, why not – an inspiration!
This doesn't mean you should copy other people's ideas. Every business owner needs to find their place on the market in order to prosper. Don't be impatient if you are not getting your Michelin stars after your first week of business.
Nobody is having too much fun if their business investment is not providing sufficient income. To avoid this from happening, you need a good plan. Therefore, the best way to start is to create a well-functioning, sustainable and effective business plan. Your business plan needs to be tailored to your own idea, resources and abilities.
Nevertheless, there are some common rules that are applicable to almost all restaurant business plans. Let's talk about these 5 brilliant tips that will work on your plan, as well.
People with consistent employment history are more likely to be loyal.
Potentially loyal employees don't bad mouth their previous employers and are honest about the reasons for leaving their previous workplace.
Excellent variety of colours and styles of aprons for your wait staff. Sturdy, yet classy design.
Simple black pants will match any waiter uniform. They are versatile and comfortable to wear.
Nice mandarin collar on a spotless waiter jacket is a perfect touch and comes with different styles.
How do you even know if you are making money with your business? It is obvious from your reports, isn't it? So why don't you make more reports? You can introduce different types of reports to make sure every dollar you invest is well spent and accounted for.
You can introduce a menu item sales report that shows which ingredients are the best sellers. In that way, you are able to plan your meals more efficiently and offer people the dishes they really want to order.
For example, you may notice that a significant number of your customers order chicken-based meals. This information can help you plan your next menu item and you can make sure to include chicken in it. You can use this calculator to find out how much that item will cost you.
Groceries are not the only items that cost you money. You also need to pay your employees. To have decent workers, you need to pay them decent wages. Consequently, you need to make sure you are getting your money's worth.
Introducing an hourly labour report doesn't mean you distrust your employees. It will help you see what needs the most work and how many members of the staff you really need. This will let you see if your team is working 100% efficiently.
The best possible marketing formula for your restaurant needs to involve both offline and online efforts.
Both of these need to be elaborated in your restaurant business plan. Build good relations with your returning customers by creating loyalty programs or offering them birthday specials. Organize and promote wine tastings and national cuisine evenings. In other words, give people a reason to keep coming back to your restaurant.
You already know that the future is here and that your marketing game needs to be strong both offline and online. You need a good website, created by a professional and active and well managed social media profiles.
The video is interesting, the music is dynamic and engaging. The viewers will want to check out this restaurant in person, not that they have seen how a day in Sepia looks like. They have experienced a good story and now they want to be a part of it.
People are more than able to cook their own food or go to a friend's house for a dinner. However, when there is a special date in their lives, they choose to go to a restaurant and make that even more special. Your customers will come for the food, but they will bring their friends and family because your restaurant is a pleasant and the ambiance is lovely.
The first thing you need to think about is your restaurant location. Can you get a venue on a corner? Is there a parking lot available? What is the neighbourhood like?
Don't be desperate if your perfectly positioned restaurant doesn't have enough space. You can make up for the lack of space with better organisation. Make sure your kitchen is perfectly designed so that it makes working there very easy.
For example, the completed orders should be on the very entrance of the kitchen, so that the waiters don't need to enter it and make it even more crowded.
Also, if you don't have enough space, work on your takeaway menu and promotion. Allow people to book a table in advance using your website and prevent the crowding up of your place.
However, when it comes to designing your restaurant, we strongly recommend talking to a professional. Everything from your wall art to the choice of your crispy clean tea towels has to be perfect. Even your washroom needs to be presentable. It goes without saying that your staff needs lovely and spotless uniforms, as well.
Choose among the 12 different styles and two sizes of high-quality tea towels.
Spotless and well-stocked washroom is important for your business image.
Leave an impression of class and professionalism. Rent amazing wait staff uniforms.
Your marketing and social media efforts will bloom if you get your customers to take photos, share them, brag a bit about their experience, tag their friends, check in and recommend your restaurant.
For this to happen, you need to have something Instagram or Pinterest worthy. It can be the interior itself, the way you serve your food, the way you fold your napkins or something entirely different. Take a look at these ideas from already existing restaurants.
There is a Japanese restaurant in Victoria that features interesting chairs. Photo ideas are popping up as soon as you land your eyes on these interestingly designed chairs.
This eye-catching ceiling at Nobu Restaurant, Perth, Australia provides excellent mood lighting for the restaurant and a lovely photo opportunity that the customers will surely take.
Go to Facebook or Instagram and search it for hashtags #nobu and #perth. Almost every fifth photo is the photo of this mesmerising ceiling.
Your idea doesn't have to be extremely expensive in order to be effective. How about some well-positioned ropes all over your restaurant? That's what The Grand Old Lady of Melbourne's Abbotsford did. These are not for free, but they might be quite cheaper than installing the designer chandeliers.
On the other hand, you can place your efforts into serving your food. This truly has to be unique for your restaurant and you need to find your own style. For example, this is taken from the Attica Restaurant in Melbourne.
You can come up with your own ideas, but it is highly advisable you hire a good interior designer to help you with the entire concept and the individual technicalities.
Do you want more ideas for your restaurant?
You can look at these 5 tips as 5 main points in your restaurant business plan. However, each and every one of them can be elaborated much further and broken down to many details. Take it from the experts and learn from the experience. Get 126 expert ideas for your restaurant business and work out how you can create a restaurant you always dreamed about and make it a success.


text:
puretaboo.com
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redpajama_set_name: RedPajamaCommonCrawl


text:
Gabrielle Union defends Ayesha Curry over viral comments about Steph Curry marriage
By Herb Scribner
Actress Gabrielle Union defended Ayesha Curry on Wednesday after Curry's comments on her marriage to Golden State Warriors star Steph Curry went viral.
"She was honest and she told her truth, and no matter what your truth is, there are going to be people who do not like it," Union said on the BuzzFeed News Twitter morning show, "AM to DM."
Curry appeared on an episode of the Facebook Watch series "Red Table Talk," which is hosted by Jada Pinkett Smith. During the episode, Curry talked about how she sometimes feels self-doubt when women give her husband attention.
"Something that really bothers me, and honestly has given me a sense of a little bit of an insecurity, is the fact that yeah, there are all these women, like, throwing themselves (at him), but me, like, the past 10 years, I don't have any of that," she said on "Red Table Talk." "I have zero — this sounds weird — but, like, male attention, and so then I begin to internalize it, and I'm like, 'Is something wrong with me?'"
The comments sparked a heavy backlash on social media. Ayesha Curry defended her comments in an Instagram post the day after the backlash began. More recently, Curry said she knew what she said and she doesn't regret it.
Union told BuzzFeed that Curry's comments were making the rounds in her own social circles. She said "there was not one mother who did not understand what she was talking about."
She also said critics of Curry didn't "care to try to understand her perspective."
She told BuzzFeed that one mother asked her why Curry made the comments publicly to which Union responded: "I said, 'But if she doesn't share her truth publicly there's a lot of other people that are out there feeling like they're alone on an island when really that island is over-crowded.'"
Union isn't the only one to defend Curry's comments. Celebrity Wendy Williams defended Curry. And Steph Curry himself also praised his wife for speaking her truth.
bonneville	2019-05-17T16:01:09+00:00


text:
#ifndef CompareUtilsH
#define CompareUtilsH

enum EComparisonType {
  EQ,
  NOT_EQ,
  LESS,
  GREATER,
  LESS_OR_EQ,
  GREATER_OR_EQ
};

template <const EComparisonType>
// Functor
struct Compare {};

template<>
struct Compare<EComparisonType::EQ> {
  template <typename T, class U>
  bool operator()(const T& first, const U& second) throw() {
    return first == second;
  }
};

template<>
struct Compare<EComparisonType::NOT_EQ> {
  template <typename T, class U>
  bool operator()(const T& first, const U& second) throw() {
    return first != second;
  }
};

template<>
struct Compare<EComparisonType::LESS> {
  template <typename T, class U>
  bool operator()(const T& first, const U& second) throw() {
    return first < second;
  }
};

template<>
struct Compare<EComparisonType::LESS_OR_EQ> {
  template <typename T, class U>
  bool operator()(const T& first, const U& second) throw() {
    return first <= second;
  }
};

template<>
struct Compare<EComparisonType::GREATER> {
  template <typename T, class U>
  bool operator()(const T& first, const U& second) throw() {
    return first > second;
  }
};

template<>
struct Compare<EComparisonType::GREATER_OR_EQ> {
  template <typename T, class U>
  bool operator()(const T& first, const U& second) throw() {
    return first >= second;
  }
};

#endif // CompareUtilsH

redpajama_set_name: RedPajamaGithub


text:
Q: IndexedDB - new version of database My PWA uses IndexedDB to store Data.
Now im releasing Version 2.0 for it, which requires the storing of data in a third ObjectStore.
When the PWA was first released, the IndexedDB created had the version "1" like this.
var db;
var request = window.indexedDB.open("database-db", 1);

// Creating ObjectStores
request.onupgradeneeded = function(event) {
    var db = event.target.result;

    var driverTable = db.createObjectStore("driver");
    var claimTable = db.createObjectStore("claim", {
        autoIncrement: "true"
    });
};

For the Release of 2.0 i now went ahead and created another ObjectStore (language) in my "onupgradeneeded" handler and changed the version like this, hoping the IndexedDB would automatically reload itself with the new version in the background.
var db;
var request = window.indexedDB.open("database-db", 2);

// Creating ObjectStores
request.onupgradeneeded = function(event) {
    var db = event.target.result;

    var languageTable = db.createObjectStore("language");
    var driverTable = db.createObjectStore("driver");
    var claimTable = db.createObjectStore("claim", {
        autoIncrement: "true"
    });
};

Unfortunately this only works for 2 scenarios.

*

*I open the PWA in Chrome on my desktop and delete the IndexedDB in the Application-Tab of DevTools and  reload the page afterwards.

*I never opened the site before and therefore have no IndexedDB installed in my browser.
The problem is, that i already have existing users and they have no option of deleting the IndexedDB on their smartphones, apart from that not being an option in terms of user-experience.
Is there a better way to approach this?
Thanks in advance.

A: Yes there are better ways to approach this. You can do version number based checks, or just check if things exist or not. Here is some simple example code that checks if a store exists and only creates a store if it does not exist.
function myOnUpgradeNeeded(event) {
  const db = event.target.result;

  if (db.objectStoreNames.contains('mystorename') {
    console.log('mystorename already exists, not recreating');
  } else {
    console.log('mystorename does not exist,  creating');
    db.createObjectStore('mystorename');
  }
}

If you want to remove all objects from a store on upgrade, then you can add logic that says that if the table exists, you can clear it.

redpajama_set_name: RedPajamaStackExchange


United® Cutlery, industry leader in licensed adult collectibles, is proud to announce new partnerships with three custom designers: Mykel Hawke, Wesley Hibben and Mikkel Willumsen. Their world-class custom designs and innovative ideas will allow United® Cutlery to discover the newest mechanisms, modern materials and custom finishing techniques.
Mykel Hawke is a U.S. Army Special Forces Veteran and Co-founder and Director of Specops, an international tourism and training company. Hawke stars in Discovery Channel's hit survival series, MAN, WOMAN, WILD. Hawke also serves as a media personality in TV and film and is the author of two bestselling books, Hawke's Green Beret Survival Manual (Running Press) and The Quick and Dirty Guide to Learning Foreign Languages Fast! (Paladin). Renown for his survival skills and tactics, Mykel Hawke brings his talents and real field experience to his exclusive knife designs. Hawke's designs prove to be the most useful all-purpose utilitarian survival knives on the market today.
Wesley Hibben has been making knives since he was eight years old, having learned the craft of knifemaking from his father, Hall of Fame knifemaker Gil Hibben. Hibben makes a variety of knives including fantasy, hunting and tactical knives. Hibben spends his time knifemaking and teaching Kenpo Karate in Anchorage, Alaska. Renown for his survival skills and tactics, Wesley Hibben brings his talents and real life experience as an avid hunter and fisherman in Alaska to his exclusive knife designs. Hibben's designs are quickly becoming recognized on the market today.
Mikkel Willumsen graduated as a teacher in metal and woodwork and decided that since he was always interested in knives as a child, knifemaking was what he would do. Willumsen has been a custom knifemaker for the past ten years in Copenhagen, Denmark, calling his designs Urban Tactical. Specializing in folders, fixed blades and balisongs, Willumsen's designs are based on functionality and quality, mixing classical designs with modern tactics. Willumsen also makes kitchen knives that have been used on Danish National Television and TV2 "Goodmorning Denmark" for several years.
Next articleA Recommitment to the Good Ol' U.S. of A.


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Worth 1,000 Words: Five Gateway Graphic Novels for Lit Lovers
Linnie Greene
Freelancer & marketer @stanfordpress. Bylines @lareviewofbooks, @nytstyles, @pacificstand, elsewhere. Email linnie.w.greene at gmail.
I spent years of my life without graphic novels, save for the grocery store Betty & Veronicas I snuck into my mother's cart (I'll still splurge on an Archie if I've had a bad week). There's got to be a metaphor for that lack: perhaps a thirsty, shipwrecked sailor who's never thought to drink from a coconut, or a dilettante who's never logged on to Pinterest. A lover of all things printed and bound, I shortchanged myself until I started working in a bookstore, where I'd shelve titles like Persepolis and wonder what I was missing. Real readers don't ignore an entire section of a bookstore; they plunge into it, voraciously. I leapt in headfirst with Alison Bechdel's Fun Home and never looked back. If you, like me, didn't grow up flipping through panels, here are a few gateway graphic novels to usher you into this oft-overlooked genre.
If you like: Cerebral, modern novels that pivot between self-awareness and neuroticism (see: David Foster Wallace) and candid explorations of race (like Chimamanda Ngozi Adichie's Americanah)…
Try: Adrian Tomine's Shortcomings, an elegant and spare graphic novel that follows protagonist Ben Tanaka through the labyrinths of his relationships, friendships, and neuroses. Tomine, the man behind the celebrated comic book series Optic Nerve, has had work in The New Yorker and McSweeney's, among others, and when you read his dart-sharp dialogue alongside his drawings — which radiate Ben's awkwardness and self-doubt — you'll see why.
If you like: No-holds-barred, confessional, feminist memoirs like Mary Karr's The Liars' Club, Daisy Hernandez' A Cup of Water Under My Bed or Lena Dunham's Not That Kind of Girl…
Try: Alison Bechdel's Fun Home, the artist's memoir about growing up in her strange and slightly dysfunctional family's funeral home in rural Pennsylvania. Bechdel's trichromatic color palette and clean, simple lines lend themselves perfectly to her story, which centers on her relationship with her father, her sexuality, and the truths a family chooses to ignore. Fun Home proves that pictures and text can be as moving and evocative as any great traditional novel.
If you like: Coming-of-age stories centered on female relationships like Emily Gould's Friendship, Grace Paley short stories, or brash explorations of young female sexuality a la Pamela Erens' The Virgins…
Try: Jillian and Mariko Tamaki's This One Summer. Written and drawn by a pair of sisters, This One Summer follows two teenage girls as they convene at their summer lake houses. What's usually a blissful, innocent affair has suddenly changed — as the two girls navigate their ascent into adulthood, relationships strain and their typically sunny vacation gets clouded by boys, fights and big questions looming like thunderheads. Like Bechdel's Fun Home, Jillian Tamaki keeps it tastefully simple with minimal colors (black, white, and lavender) and clean, effortless lines.
If you like: Fragmented novels or short stories (like Daniel Handler's Adverbs) or flash-fiction by people like Stuart Dybek…
Try: Jeffrey Brown's Clumsy. Alternately sweet and sad, Clumsy traces Brown's real-life relationship with his girlfriend Theresa through a series of vignettes. Their gravity changes from page to page — a heart-rending fight might follow joke about burping. It's so twee you can almost hear The Moldy Peaches playing in the background, but Brown's soul-baring dialogue and messy, scribbled drawings are undeniably winsome, perfect for those who like their stories delivered in short, pointed dispatches.
If you like: Travelogues that explore landscapes both interior and exterior (like Justin Go's The Steady Running of the Hour) with an emphasis on family dynamics in all their glorious oddity (like Yelena Akhtiorskaya's Panic in a Suitcase)…
Try: Rutu Modan's The Property. Chronicling a grandmother & granddaughter's trip to Poland to reclaim a pre-WWII apartment, The Property is a sweet meditation on family and history that takes on gargantuan themes (the Holocaust, namely) while managing both humor and gravity. Modan's explosive use of color imbues the story with movement and wonder — it's as if the reader, too, is walking through a new landscape, figuring it out as they go.
Sunday Sundries: Literary Links from Around the Web (Nov. 23rd)
Nov 23 - Electric Literature Read
How a Comic Book About Feral Elves Got Me Through Middle School
"ElfQuest" wasn't perfect, but it helped me remember who I was—and figure out who I wanted to be
Jul 3 - Madeline ffitch
10 Exuberantly Queer Graphic Novels
Memoirs and fiction that celebrate the lives of queer women in words and pictures
Jun 19 - Chaya Bhuvaneswar
Vivek Shraya's "Death Threat" Turns Online Hatred Into Art
The graphic novel illustrates a series of real-life threatening messages Shraya received
Jun 12 - Roberto Rodriguez-Estrada


I've been having Bright Ideas for a while - or, rather, occasional things that occur to me.
Sometimes the ideas are a bit unexpected.
Or if we couldn't reintroduce (and update!) the idea of At Home - common in Victorian times in the UK, this involved mostly middle class ladies giving out a card (a bit like a business card) to their acquaintances letting them know they could call in to visit (and perhaps 'take tea'!) on certain afternoons on particular days of the week or month.
We could update this by thinking of it as a kind of Lounge Share (or Living Room Swap), whereby people living near each other - of any age - who find they are at home a lot in the daytime could be invited to spend a bit of time in each other's homes on a regular basis.
Why might this be good? Well, in the winter months - for example - it would mean those visiting could turn off their heating for a few hours because they were instead at their At Home/Lounge Share host's home.
So, if you invited a few people round to yours every now and then, it might save some money on heating bills and in turn help the environment. Visitors and hosts would also have a bit of company - and there might be all sorts of other benefits, too. This could be good for some older people, and a whole host of others, such as: people working from home; people who are out of work; people studying; people with young children. Etc.
Essentially - this idea might be of interest and value to anyone who is at home anyway, and who would be very pleased to have others 'hang out' (is there a better phrase?) with them for a few hours from time to time.
As a host, you wouldn't necessarily have to 'entertain' everyone - unless you wanted to - but you might have a few things out for people to do if they wanted (I defy anyone to walk past a table with an unfinished jigsaw puzzle on it and not try to put at least one piece in place), or you might put a DVD on (I have a good selection of Hitchcock, and Powell/Pressburger, films, for example), or you might have juggling balls to try out (in the garden?), or you might suggest people bring something with them they want to do or want to share (appropriate things, of course!).
It's not day care - but it has some similarities with Shared Lives day support. It's not Sunday tea with Contact the Elderly - but it has some similarities with home-based, pop-up tea shops or supper clubs. It's not a one-off communal meal for the neighbourhood - but it has some similarities with the Big Lunch.
All the normal 'rules' of your life would apply - such as, don't invite anyone into your home if you're not sure about them or don't know who they are. Tell people if they can smoke indoors or not. Ask them to take off their shoes if that's what you usually do (and, in which case, invite them to bring their slippers with them). Be clear what time the visit starts, and ends. If you're on a low income and you're going to do this regularly, you might set up some kind of tea and coffee 'kitty' so you're not having to pay out all the costs, or ask people to bring something with them to contribute - like some milk. Or a loo roll.
Be sensible, be clear, be innovative, be generous - and be safe!
And a really good way to start this up, if you would like to do this but lack a bit of confidence? Find a charity you want to support - whether local or national - and invite people round for a coffee morning or an afternoon cup of tea (you get the general idea) to raise a bit of money for that charity. Many charities have a specific coffee morning fundraiser - so you could always join in with that. Or, you could just decide to have your own event, and not wait until the charity day comes round.
Pop a simple written invitation through the doors of neighbours and others living nearby, or - better - knock on the door and speak to them face-to-face. Ask them to let you know if they're coming (for the purposes of counting out the right number of biscuits, or cutting the cake appropriately!) - and take it from there.
When people call in for the first charity event, you could say you had wondered about doing this sort of get together a bit more frequently (perhaps minus the charity angle - or not!) - maybe once a month - and just see how it goes.
You might have to do it for a while until people relax into the idea. Or you might find only one person wants to do it - that would be fine. Or, you could widen the invitation 'network'.
Now, for the people who say "this won't work" - feel free to come up with something that will. Please - seriously. We all need more Bright Ideas.


ST. LOUIS — Equifax Workforce Solutions, a provider of human resource services, data, analytics and verification services, today announced that for the first time ever, the company has processed more than 34 million instant income and employment verifications in one year.
The achievement marks a near 40% increase from last year, highlighting a significant accomplishment for Equifax Workforce Solutions.
"Passing the 34 million mark is a great milestone for job-seekers, loan applicants, consumers needing government assistance, and employers around the country, as well as for Equifax," said Craig Leabig, senior director of product management of Equifax Workforce Solutions.
"It takes time and resources to track down an applicant's employer, contact that employer and wait for the response. We simplify the process for applicants across the country by providing income and employment data instantly through The Work Number service. Additionally, we leverage our extensive relationships with human resource departments at thousands of employers to streamline manual verification delivery when an applicant's information is not available instantly through our database."


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