# A complete picture of the four-party linear inequalities in terms of the
$0$-entropy

Zhiwei Song<EMAIL_ADDRESS>LMIB(Beihang University), Ministry of
Education, and School of Mathematical Sciences, Beihang University, Beijing
100191, China  Lin Chen<EMAIL_ADDRESS>(corresponding author)
LMIB(Beihang University), Ministry of Education, and School of Mathematical
Sciences, Beihang University, Beijing 100191, China  International Research
Institute for Multidisciplinary Science, Beihang University, Beijing 100191,
China  Yize Sun  sunyize@buaa.edu.cn(corresponding author) LMIB(Beihang
University), Ministry of Education, and School of Mathematical Sciences,
Beihang University, Beijing 100191, China  Mengyao Hu
mengyaohu@buaa.edu.cn(corresponding author) LMIB(Beihang University), Ministry
of Education, and School of Mathematical Sciences, Beihang University, Beijing
100191, China

###### Abstract

Multipartite quantum system is complex. Characterizing the relations among the
three bipartite reduced density operators $\rho_{AB}$, $\rho_{AC}$ and
$\rho_{BC}$ of a tripartite state $\rho_{ABC}$ has been an open problem in
quantum information. One of such relations has been reduced by [Cadney et al,
LAA. 452, 153, 2014] to a conjectured inequality in terms of matrix rank,
namely $r(\rho_{AB})\cdot r(\rho_{AC})\geq r(\rho_{BC})$ for any $\rho_{ABC}$.
It is denoted as open problem $41$ in the website ”Open quantum problems-IQOQI
Vienna”. We prove the inequality, and thus establish a complete picture of the
four-party linear inequalities in terms of the $0$-entropy. Our proof is based
on the construction of a novel canonical form of bipartite matrices under
local equivalence. We apply our result to the marginal problem and the
extension of inequalities in the multipartite systems, as well as the
condition when the inequality is saturated.

###### pacs:

03.65.Ud, 03.67.Mn

##  I Introduction

Multipartite systems play a key role in quantum-information processing. For
example, the inequality for the von Neumann entropy of reduced density
operators of a multipartite has been proposed in 2005A ; 2012Infinitely , and
the multipartite state conversion under many-copy cases has been shown under
stochastic local operations and classical communications ccd2010 . Further,
the relation between the distillability of entanglement of three bipartite
reduced density matrices from a tripartite pure state has been studied
Chen2011Multicopy ; Chen2012NONDISTILLABLE ; HayashiWeaker . However, it is
not easy to extend the relation to the tripartite mixed state, even we merely
consider the rank of reduced density operators. It has been conjectured in
chl14  that the following inequality in terms of matrix rank may hold for any
tripartite mixed state $\rho_{ABC}$,

$\displaystyle r(\rho_{AB})\cdot r(\rho_{AC})\geq r(\rho_{BC}),$ (1)

where $r(M)$ denotes the rank of matrix $M$, see Figure 1. The conjectured
inequality in (1) has been listed as the open problem in
111https://oqp.iqoqi.univie.ac.at/all-rank-inequalities-for-reduced-states-of-
quadripartite-quantum-states/. It has been proven true when $r(\rho_{AB})$ is
at most two and three in chl14  and 2020The , respectively. In this paper we
prove the inequality (1) for any $\rho_{ABC}$ in Theorem 9. The inequality
together with the inequalities constructed in chl14 , establish basic
inequalities for the tradeoff among the ranks of three bipartite reduced
density operators, see (89)-(91). This is another point of view in contrast to
the monogamy trade-off by Bell inequalities Clauser1971Proposed ;
Kurzy2011Correlation ; Pawlowski2009Monogamy ; REPRESENTATIONS ;
Butterley2006Compatibility ; Coffman1999Distributed ; HiguchiOne . We thus
manage to extend the results in Chen2011Multicopy ; Chen2012NONDISTILLABLE ;
HayashiWeaker  from tripartite pure states to mixed states. Next, our results
present a novel necessary condition for the marginal problem, i.e., three
bipartite reduced density operators from the same tripartite state satisfy
(1). Further we extend (89)-(91) to multipartite systems in Lemma 10. We also
discuss the condition when the inequality (1) is saturated in Lemma 11.

Figure 1: The tripartite state $\rho_{ABC}$ is conjectured to satisfy the
inequality $r(\rho_{AB})\cdot r(\rho_{BC})\geq r(\rho_{AC})$ in terms of
bipartite reduced density operators $\rho_{AB}$, $\rho_{BC}$, and $\rho_{AC}$,
and $r(M)$ denotes the rank of matrix $M$. The inequality is known to be
equivalent to the $0$-entropy inequality $S_{0}(AB)+S_{0}(BC)\geq S_{0}(AC)$.
We prove the inequality in this paper.

We review the meaning of (1) in terms of the $0$-entropy. Let
$\alpha\in(0,1)\cup(1,\infty)$ and the logarithm have base two. The known
$\alpha$-Rényi entropy $S_{\alpha}$ of a quantum state $\rho$ is defined as
$S_{\alpha}(\rho):=\frac{1}{1-\alpha}\log\mathop{\rm Tr}\rho^{\alpha}.$ One
can verify that ${\lim_{\alpha\rightarrow 1}}S_{\alpha}(\rho)$ is exactly the
von Neumann entropy $S(\rho):=-\mathop{\rm Tr}(\rho\log\rho)$. In contrast,
authors in Ref. chl14  has defined the $0$-entropy
$S_{0}(A):=\lim_{\alpha\rightarrow 0}S_{\alpha}(\rho_{A})=\log r(\rho_{A})$,
the $0$-entropy vector as the $7$-dimensional vector
$(r(\rho_{A}),r(\rho_{B}),r(\rho_{C}),r(\rho_{D}),r(\rho_{AB}),r(\rho_{AC}),r(\rho_{AD}))$
of a $4$-partite pure state of system $A,B,C,D$, as well as the subadditivity
$S_{0}(A)+S_{0}(B)\geq S_{0}(AB)$, $S_{0}(AB)+S_{0}(AC)\geq S_{0}(A)$ as well
as more inequalities, just like the counterpart inequalities of von Neumann
entropy. Further, the $0$-entropy inequalities determine a cone with extremal
rays characterized by eight 0-entropy vectors. In chl14 , focusing on the
four-party case, authors have found six 0-entropy vectors corresponding to
extremal rays, as well as the set of inequalities they correspond to. The set
turns out to be the known 0-entropy inequalities and that in (1). Hence, our
proof on (1) helps complete the forementioned characterization and construct a
complete picture of the four-party linear inequalities for the $0$-entropy.

Technically speaking, our strategy of proving (1) in Theorem 9 is to prove an
equivalent form in terms of the partial transpose and Schmidt rank, see
Conjecture 1. The equivalence is proven in chl14 , and the proof of this
conjecture is supported via a few basic facts from linear algebra in Lemmas 3
\- 7, as well as the construction of a novel canonical form of bipartite
matrices under local equivalence in Theorem 8.

The partial transpose is a positive map of extensive applications in quantum
information. Firstly it is known that a separable state is positive partial
transpose (PPT), and it is the most efficient method of detecting entanglement
so far hhh96 . Next, the two-qutrit PPT entangled states were constructed in
1997 horodecki1997 , and such states of rank four have been characterized
Chen2012Equivalence ; 2011Three . The PPT entanglement represents quantum
resources which cannot be distillable into pure entangled states under local
operations and classical communications (LOCC). What’s more, bipartite non-PPT
states of rank at most four turn out to be distillable Chen2008Rank ;
Chen2012Distillability ; Lin2016Non , and some non-PPT states are conjectured
to be non distillable hh1999 ; Divincenzo2000Evidence ; 5508622 ; QIAN2021139
; 2020Five . On the other hand, the Schmidt rank is a basic parameter of
characterizing bipartite pure states, and has been extended to multipartite
pure states as an entanglement monotone Eisert2000The ; ccd2010 . Our result
of proving (1) shows novel understanding of the forementioned quantum-
information applications in terms of partial transpose and Schmidt rank.

The rest of this paper is organized as follows. In Sec. II we introduce the
preliminary facts in Lemmas 3 \- 7 for the proof of inequality (1). We show
the proof in Theorem 9 supported by Theorem 8 in Sec. III. Then we apply our
result in Lemmas 10 and 11 in Sec. IV. Finally we conclude in Sec. V.

##  II Preliminaries

In this section we introduce the preliminary knowledge and facts of this
paper. Let $\mathbb{M}_{m,n}$ be the set of $m\times n$ complex matrices, and
$\mathbb{M}_{n}:=\mathbb{M}_{n,n}$. Let $I_{n}$ be the order-$n$ identity
matrix. We denote $M^{T}$ as the transpose of matrix $M$. We denote the
bipartite operator
$M\in\mathbb{M}_{m_{1},n_{1}}\otimes\mathbb{M}_{m_{2},n_{2}}$ as
$M=\sum^{m_{1}}_{i=1}\sum^{n_{1}}_{j=1}|i\rangle\\!\langle j|\otimes M_{i,j}$
with $M_{i,j}\in\mathbb{M}_{m_{2},n_{2}}.$ We denote the partial transpose of
$M$ w.r.t. system $A$ and $B$ as
$M^{\Gamma_{A}}=\sum^{m_{1}}_{i=1}\sum^{n_{1}}_{j=1}|j\rangle\\!\langle
i|\otimes M_{i,j}$, and
$M^{\Gamma_{B}}=\sum^{m_{1}}_{i=1}\sum^{n_{1}}_{j=1}|i\rangle\\!\langle
j|\otimes M_{i,j}^{T}$, respectively. Next, the number of linearly independent
blocks $M_{i,j}$’s is referred to the Schmidt rank $Sr:=Sr(M)$ of $M$. One can
derive that $Sr(M)=Sr(M^{\Gamma_{A}})=Sr(M^{\Gamma_{B}})=Sr(M^{T}).$ It has
been proven by chl14  that the conjectured inequality in (1) is equivalent to
the following conjecture.

###### Conjecture 1

Let $M\in\mathbb{M}_{m_{1},n_{1}}\otimes\mathbb{M}_{m_{2},n_{2}}$. Then

$\displaystyle r(M^{\Gamma_{B}})\leq Sr(M)\cdot r(M).$ (2)

The conjecture is equivalent to $r(M^{\Gamma_{A}})\leq Sr(M)\cdot r(M)$. If we
write $M=\sum^{K}_{i=1}R_{i}\otimes S_{i}^{T}$, where
$R_{1},\dots,R_{K}\in\mathbb{M}_{m_{1},n_{1}}$ are linearly independent, and
$S_{1},\dots,S_{K}\in\mathbb{M}_{m_{2},n_{2}}$ are linearly independent, then
the inequality is also equivalent to $r\bigg{(}\sum^{K}_{i=1}R_{i}\otimes
S_{i}\bigg{)}\leq K\cdot r\bigg{(}\sum^{K}_{i=1}R_{i}\otimes
S_{i}^{T}\bigg{)}$.  $\sqcap$$\sqcup$

Next it follows from Theorem 5 of chl14  that Conjecture 1 holds for
$Sr(M)\leq 2$. It has been proved in Theorem 2 of 2020The  that Conjecture 1
holds for $Sr(M)=3$. Nevertheless, these proven cases do not contribute to our
proof in the next section. To explain our proof, we present the following
definition.

###### Definition 2

We denote locally equivalent $M,N$ as $M\sim N$ if there exist invertible
product matrices $U\otimes V$ and $W\otimes X$ such that $(U\otimes
V)M(W\otimes X)=N$.  $\sqcap$$\sqcup$

So $M$ and $N$ have the same rank and Schmidt rank, one can also show that
$M^{\Gamma}$ and $N^{\Gamma}$ have the same rank and Schmidt rank. Hence
proving $M$ satisfies Conjecture 1 is equivalent to prove $N$ satisfies
Conjecture 1, and we shall frequently use this fact in the next section. In
the rest of this section, we present five preliminary lemmas used for the
proof of next section. The following two lemmas can be straightforwardly
proven using the basic matrix theory.

###### Lemma 3

The following inequalities hold for any block matrix

$\displaystyle r(A_{1})\leq
r(\begin{bmatrix}A_{1}&A_{2}&\cdots&A_{n}\end{bmatrix})\leq
r(A_{1})+\cdots+r(A_{n}),$ (3) $\displaystyle r(A)+r(C)\leq
r(\begin{bmatrix}A&0\\\ B&C\end{bmatrix}).$ (4)

$\sqcap$$\sqcup$

###### Lemma 4

(i) Suppose $A_{1},\cdots,A_{n}\in\mathbb{M}_{m_{2},n_{2}}$ are linearly
independent, $R\in\mathbb{M}_{n_{2}}$ is invertible. Then $A_{1}\cdot
R,\cdots,A_{n}\cdot R$ are linearly independent.

(ii) Suppose $A_{1},\cdots,A_{n},A_{n+1}\in\mathbb{M}_{m_{2},n_{2}}$, and
$R\in\mathbb{M}_{n_{2}}$ is invertible. If $A_{n+1}\in\mathop{\rm
span}\\{A_{1},\cdots,A_{n}\\}$, then $A_{n+1}\cdot R\in\mathop{\rm
span}\\{A_{1}\cdot R,\cdots,A_{n}\cdot R\\}$.  $\sqcap$$\sqcup$

The following lemma is assertion (a) of Theorem 6 in 2020The .

###### Lemma 5

Suppose one block-row or block-column of the Schmidt-rank-K block matrix $M$
has $K$ linearly independent blocks. Then $M$ satisfies Conjecture 1.
$\sqcap$$\sqcup$

We next present another lemma.

###### Lemma 6

(i) Suppose

$\displaystyle M=\begin{bmatrix}M_{1,1}&M_{1,2}&M_{1,3}&\cdots&M_{1,n_{1}}\\\
\vdots&\vdots&\vdots&\ddots&\vdots\\\
M_{t,1}&M_{t,2}&M_{t,3}&\cdots&M_{t,n_{1}}\\\
0&M_{t+1,2}&M_{t+1,3}&\cdots&M_{t+1,n_{1}}\\\
\vdots&\vdots&\vdots&\ddots&\vdots\\\
0&M_{m_{1},2}&M_{m_{1},3}&\cdots&M_{m_{1},n_{1}}\end{bmatrix},$

where $M_{1,1},\cdots,M_{t,1}$ are linearly independent and there exists a
block $M_{i,j}$ that is linearly independent with the $t$ blocks for $t+1\leq
i\leq m_{1},2\leq j\leq n_{1}$. Then by multiplying an appropriate constant to
the $j$-th block-column of $M$ and adding to the first block-column, the new
first block-column has at least $t+1$ linearly independent blocks.

(ii) Suppose

$\displaystyle M=\begin{bmatrix}M_{1,1}&M_{1,2}&M_{1,3}&\cdots&M_{1,n_{1}}\\\
\vdots&\vdots&\vdots&\ddots&\vdots\\\
M_{t,1}&M_{t,2}&M_{t,3}&\cdots&M_{t,n_{1}}\\\
M_{t+1,1}&M_{t+1,2}&M_{t+1,3}&\cdots&M_{t+1,n_{1}}\\\
\vdots&\vdots&\vdots&\ddots&\vdots\\\
M_{s,1}&M_{s,2}&M_{s,3}&\cdots&M_{s,n_{1}}\end{bmatrix},$

where $s>t$, and the $s+t$ matrices
$M_{1,1},\cdots,M_{s,1},M_{1,2},\cdots,M_{t,2}$ are linearly independent. Any
$M_{i,2}$ is spanned by $M_{1,1},\cdots,M_{s,1}$, where $t+1\leq i\leq s$. At
the same time, there exists a block $M_{i,j}$ that is linearly independent
with the $s+t$ matrices for $t+1\leq i\leq s,3\leq j\leq n_{1}$. Then by
multiplying an appropriate constant to the $j$-th block column of $M$ and
adding to the second block-column, the new second block-column has at least
$t+1$ linearly independent blocks and they are linearly independent with
$M_{1,1},\cdots,M_{s,1}$.

(iii) Suppose

$\displaystyle
M=\begin{bmatrix}M_{1,1}&M_{1,2}&\cdots&M_{1,n}&\cdots&M_{1,n_{1}}\\\
\vdots&\vdots&\ddots&\vdots&\ddots&\vdots\\\
M_{t_{n},1}&M_{t_{n},2}&\cdots&M_{t_{n},n}&\cdots&M_{t_{n},n_{1}}\\\
M_{t_{n}+1,1}&M_{t_{n}+1,2}&\cdots&M_{t_{n}+1,n}&\cdots&M_{t_{n}+1,n_{1}}\\\
\vdots&\vdots&\ddots&\vdots&\ddots&\vdots\\\
M_{t_{2},1}&M_{t_{2},2}&\cdots&M_{t_{2},n}&\cdots&M_{t_{2},n_{1}}\\\
\vdots&\vdots&\ddots&\vdots&\ddots&\vdots\\\
M_{t_{1},1}&M_{t_{1},2}&\cdots&M_{t_{1},n}&\cdots&M_{t_{1},n_{1}}\end{bmatrix},$

where $t_{1}\geq t_{2}\geq\cdots\geq t_{n-1}>t_{n}$, and the
$t_{1}+t_{2}+\cdots+t_{n}$ matrices $M_{1,1},\cdots,M_{t_{1},1}$,
$M_{1,2},\cdots,M_{t_{2},2},\cdots$, $M_{1,n},\cdots,M_{t_{n},n}$ are linearly
independent. Any $M_{i,n}$ is spanned by $M_{1,1},\cdots,M_{t_{1},1}$,
$M_{1,2},\cdots,M_{t_{2},2},\cdots$, $M_{1,n-1},\cdots,M_{t_{n-1},n-1}$, where
$t_{n}+1\leq i\leq t_{1}$. At the same time, there exists a block $M_{i,j}$
that is linearly independent with the $t_{1}+t_{2}+\cdots+t_{n}$ matrices for
$t_{n}+1\leq i\leq t_{1},n+1\leq j\leq n_{1}$. Then by multiplying an
appropriate constant to the $j$-th block column of $M$ and adding to the
$n$-th block-column of $M$, the new $n$-th block-column has at least $t_{n}+1$
linearly independent blocks and they are linearly independent with
$M_{1,1},\cdots,M_{t_{1},1},M_{1,2},\cdots,M_{t_{2},2}$,
$\cdots,M_{1,n-1},\cdots,M_{t_{n-1},n-1}$.

###### Proof.

(i) Without loss of generality, we assume that $M_{t+1,2}$ is linearly
independent with $M_{1,1},\cdots,M_{t,1}$. We have two cases, namely (A) and
(B), as follows.

(A). Suppose any $M_{i,2}$ is spanned by $M_{1,1},\cdots,M_{t,1},M_{t+1,2}$,
where $1\leq i\leq t$. We first apply block-row operations on $M$ and still
use $M_{i,j}$ to denote the new blocks, such that $M_{1,1},\cdots,M_{t,1}$ are
linearly independent, and any $M_{i,2}$, where $1\leq i\leq t$, is the
combinations of the $t$ matrices. Thus we set

$\displaystyle M_{i,2}=a_{1i}M_{1,1}+\cdots+a_{ti}M_{t,1},$ (5)

where $a_{1i},\cdots,a_{ti}$ are coefficients, $1\leq i\leq t$. Next, we
multiply a nonzero constant $k$ to the second block-column of $M$ and add to
the first block-column. Denote
$M_{1,1}^{\prime},\cdots,M_{t,1}^{\prime},M_{t+1,1}^{\prime}$ as the new
blocks in the first block-column, we have

$\displaystyle M_{i,1}^{\prime}=M_{i,1}+kM_{i,2}$ (6)

holds for $1\leq i\leq t+1$. Assume that

$\displaystyle\eta_{1}M_{1,1}^{\prime}+\eta_{2}M_{2,1}^{\prime}\cdots+\eta_{t}M_{t,1}^{\prime}+\eta_{t+1}M_{t+1,1}^{\prime}=0,$
(7)

where $\eta_{1},\eta_{2},\cdots,\eta_{t},\eta_{t+1}$ are complex numbers.
Using (5) and (6), we have

$\displaystyle(\eta_{1}(ka_{11}+1)+\eta_{2}ka_{12}+\cdots+\eta_{t}ka_{1t})M_{1,1}$
$\displaystyle+$
$\displaystyle(\eta_{1}ka_{21}+\eta_{2}(ka_{22}+1)+\cdots+\eta_{t}ka_{2t})M_{2,1}$
$\displaystyle+$ $\displaystyle\cdots$ $\displaystyle+$
$\displaystyle(\eta_{1}ka_{t1}+\eta_{2}ka_{22}+\cdots+\eta_{t}(ka_{tt}+1))M_{t,1}$
$\displaystyle+$ $\displaystyle\eta_{t+1}kM_{t+1,2}=0.$ (8)

Since $M_{1,1},\cdots,M_{t,1},M_{t+1,2}$ are linearly independent, from (II)
we have

$\displaystyle\begin{bmatrix}\begin{array}[]{ccccc}ka_{11}+1&ka_{12}&\cdots&ka_{1t}&0\\\
ka_{21}&ka_{22}+1&\cdots&ka_{2t}&0\\\ \vdots&\vdots&\ddots&\vdots&\vdots\\\
ka_{t1}&ka_{t2}&\cdots&ka_{tt}+1&0\\\
0&0&\cdots&0&k\end{array}\end{bmatrix}\cdot\begin{bmatrix}\eta_{1}\\\
\eta_{2}\\\ \vdots\\\ \eta_{t}\\\ \eta_{t+1}\end{bmatrix}=\begin{bmatrix}0\\\
0\\\ \vdots\\\ 0\\\ 0\end{bmatrix}.$ (9)

For the linear equations (9), we can find an appropriate $k^{\prime}$ such
that

$\displaystyle\det\left|\begin{array}[]{ccccc}k^{\prime}a_{11}+1&k^{\prime}a_{12}&\cdots&k^{\prime}a_{1t}&0\\\
k^{\prime}a_{21}&k^{\prime}a_{22}+1&\cdots&k^{\prime}a_{2t}&0\\\
\vdots&\vdots&\ddots&\vdots&\vdots\\\
k^{\prime}a_{t1}&k^{\prime}a_{t2}&\cdots&k^{\prime}a_{tt}+1&0\\\
0&0&\cdots&0&k^{\prime}\end{array}\right|\neq 0,$ (15)

and hence $\eta_{1}=\eta_{2}=\cdots=\eta_{t}=\eta_{t+1}=0$. Let
$k=k^{\prime}$, then
$M_{1,1}^{\prime},\cdots,M_{t,1}^{\prime},M_{t+1,1}^{\prime}$ are linearly
independent in terms of (7). We have finished the proof.

(B). Suppose there exists at least one $M_{i,2}$ that is linearly independent
with $M_{1,1},\cdots,M_{t,1},M_{t+1,2}$, where $1\leq i\leq t$. Without loss
of generality, assume that
$M_{1,1},\cdots,M_{t,1},M_{t+1,2},M_{1,2},\cdots,M_{v,2}$ are linearly
independent, where $1\leq v\leq t$. At the same time, any $M_{i,2}$ is spanned
by the $t+1+v$ matrices for $v+1\leq i\leq t$. Next, apply block-row
operations on $M$ such that $M_{i,2}$, where $v+1\leq i\leq t$, is the
combination of $M_{1,1},\cdots,M_{t,1}$, and the coefficients are still
denoted by those in (5). By multiplying $k^{\prime}$ in case (A) to the second
block-column of $M$ and adding to the first block-column, we obtain that
$M_{v+1,1}^{\prime},\cdots,M_{t+1,1}^{\prime}$ are linearly independent. Since
$k^{\prime}$ is nonzero, we also have
$M_{1,1}^{\prime},\cdots,M_{v,1}^{\prime}$ are linearly independent, and they
are linearly independent with $M_{v+1,1}^{\prime},\cdots,M_{t,1}^{\prime}$. We
have finished the proof.

(ii) Without loss of generality, assume that $M_{t+1,3}$ is linearly
independent with the $s+t$ matrices. We have two cases, namely (C) and (D), as
follows.

(C). Suppose any $M_{i,3}$ is spanned by
$M_{1,1},\cdots,M_{s,1},M_{1,2},\cdots,M_{t,2},M_{t+1,3}$, where $1\leq i\leq
t$. We first apply block-row operations on $M$, such that any $M_{i,3}$ is
spanned by $M_{1,1},\cdots,M_{s,1},M_{1,2},\cdots,M_{t,2}$, where $1\leq i\leq
t$. At the same time, $M_{t+1,2}$ is still spanned by
$M_{1,1},\cdots,M_{s,1}$. Thus we set

$\displaystyle
M_{i,3}=b_{1i}M_{1,1}+\cdots+b_{si}M_{s,1}+c_{1i}M_{1,2}+\cdots+c_{ti}M_{t,2},$
(16)

where $b_{1i},\cdots,b_{si},c_{1i},\cdots,c_{ti}$ are coefficients, $1\leq
i\leq t$. Set

$\displaystyle M_{t+1,2}=d_{1}M_{1,1}+\cdots+d_{s}M_{s,1},$ (17)

where $d_{1},\cdots,d_{s}$ are coefficients. Next, we multiply a nonzero
constant $k$ to the third block-column of $M$ and add to the second block-
column. Denote
$M_{1,2}^{\prime\prime},\cdots,M_{t,2}^{\prime\prime},M_{t+1,2}^{\prime\prime}$
as the new blocks in the second block-column, we have

$\displaystyle M_{i,2}^{\prime\prime}=M_{i,2}+kM_{i,3}$ (18)

holds for $1\leq i\leq t+1$. Assume that

$\displaystyle\zeta_{1}M_{1,1}+\cdots+\zeta_{s}M_{s,1}+\gamma_{1}M_{1,2}^{\prime\prime}+\cdots+\gamma_{t}M_{t,2}^{\prime\prime}+\gamma_{t+1}M_{t+1,2}^{\prime\prime}=0,$
(19)

where $\zeta_{1},\cdots,\zeta_{s},\gamma_{1},\cdots,\gamma_{t},\gamma_{t+1}$
are complex numbers. Using (16), (17) and (18), we have

$\displaystyle(\zeta_{1}+\gamma_{1}kb_{11}+\cdots+\gamma_{t}kb_{1t}+\gamma_{t+1}d_{1})M_{1,1}$
$\displaystyle+$ $\displaystyle\cdots$ $\displaystyle+$
$\displaystyle(\zeta_{s}+\gamma_{1}kb_{s1}+\cdots+\gamma_{t}kb_{st}+\gamma_{t+1}d_{s})M_{s,1}$
$\displaystyle+$
$\displaystyle(\gamma_{1}(kc_{11}+1)+\cdots+\gamma_{t}kc_{1t})M_{1,2}$
$\displaystyle+$ $\displaystyle\cdots$ $\displaystyle+$
$\displaystyle(\gamma_{1}kc_{t1}+\cdots+\gamma_{t}(kc_{tt}+1))M_{t,2}$
$\displaystyle+$ $\displaystyle k\eta_{t+1}M_{t+1,3}=0.$ (20)

Since $M_{1,1},\cdots,M_{s,1}$, $M_{1,2},\cdots,M_{t,2},M_{t+1,3}$ are
linearly independent, from (II) we have

$\displaystyle\begin{bmatrix}\begin{array}[]{ccccccccc}\lx@intercol\hfil\hbox{\multirowsetup\Large${I_{s}}$}\hfil\lx@intercol&kb_{11}&\cdots&kb_{1t}&d_{1}\\\
&&&\vdots&\ddots&\vdots&\vdots\\\ &&&kb_{s1}&\cdots&kb_{st}&d_{s}\\\
0&\cdots&0&kc_{11}+1&\cdots&kc_{1t}&0\\\
\vdots&\ddots&\vdots&\vdots&\ddots&\vdots&\vdots\\\
0&\cdots&0&kc_{t1}&\cdots&kc_{tt}+1&0\\\
0&\cdots&0&0&\cdots&0&k\end{array}\end{bmatrix}\cdot\begin{bmatrix}\zeta_{1}\\\
\vdots\\\ \zeta_{s}\\\ \gamma_{1}\\\ \vdots\\\ \gamma_{t}\\\
\gamma_{t+1}\end{bmatrix}=\begin{bmatrix}0\\\ \vdots\\\ 0\\\ 0\\\ \vdots\\\
0\\\ 0\end{bmatrix}.$ (21)

For the linear equations (21), we can find an appropriate $k^{\prime\prime}$
such that

$\displaystyle\det\left|\begin{array}[]{ccccc}k^{\prime\prime}c_{11}+1&k^{\prime\prime}c_{12}&\cdots&k^{\prime\prime}c_{1t}&0\\\
k^{\prime\prime}c_{21}&k^{\prime\prime}c_{22}+1&\cdots&k^{\prime\prime}c_{2t}&0\\\
\vdots&\vdots&\ddots&\vdots&\vdots\\\
k^{\prime\prime}c_{t1}&k^{\prime\prime}c_{t2}&\cdots&k^{\prime\prime}c_{tt}+1&0\\\
0&0&\cdots&0&k^{\prime\prime}\end{array}\right|\neq 0,$ (27)

and hence
$\zeta_{1}=\cdots=\zeta_{s}=\gamma_{1}=\cdots=\gamma_{t}=\gamma_{t+1}=0$. Let
$k=k^{\prime\prime}$, then
$M_{1,1},\cdots,M_{s,1},M_{1,2}^{\prime\prime},\cdots,M_{t,2}^{\prime\prime},M_{t+1,2}^{\prime\prime}$
are linearly independent in terms of (19). We have finished the proof.

(D). Suppose there exists at least one $M_{i,3}$ that is linearly independent
with $M_{1,1},\cdots,M_{s,1},M_{1,2},\cdots,M_{t,1},M_{t+1,3}$, where $1\leq
i\leq t$. Without loss of generality, assume that
$M_{1,1},\cdots,M_{s,1},M_{1,2},\cdots,M_{t,2},M_{t+1,3},M_{1,3},\cdots,M_{v,3}$
are linearly independent, where $1\leq v\leq t$. At the same time, any
$M_{i,3}$ is spanned by the $s+t+1+v$ matrices for $v+1\leq i\leq t$. Next,
apply block-row operations on $M$ such that $M_{i,3}$, where $v+1\leq i\leq
t$, is the combination of $M_{1,1},\cdots,M_{s,1},M_{1,2},\cdots,M_{t,2}$, and
the coefficients are still denoted by those in (16). By multiplying
$k^{\prime\prime}$ in case (C) to the third block-column of $M$ and adding to
the second block-column, we obtain that
$M_{v+1,2}^{\prime\prime},\cdots,M_{t+1,2}^{\prime\prime}$ are linearly
independent and they are linearly independent with $M_{1,1},\cdots,M_{s,1}$.
Since $k^{\prime\prime}$ is nonzero, we also have
$M_{1,2}^{\prime\prime},\cdots,M_{v,2}^{\prime\prime}$ are linearly
independent and they are linearly independent with
$M_{v+1,2}^{\prime\prime},\cdots,M_{t+1,2}^{\prime\prime}$ and the $s$ blocks
in the first block-column. We have finished the proof.

(iii) The proof is similar to that of (ii).  $\sqcap$$\sqcup$

Finally we present another observation.

###### Lemma 7

Suppose the $m\times 1$ block matrix $\begin{bmatrix}P_{1}\\\ P_{2}\\\
\vdots\\\ P_{m}\end{bmatrix}$ has $n$ columns and $k$ linearly independent
columns, $k\leq n$. Then for any matrix $Q\in\mathop{\rm
span}\\{P_{1},\cdots,P_{m}\\}$, Q has $k$ linearly independent columns at
most. Through elementary column operations on $Q$, $Q$ has $k$ nonzero columns
at most.

###### Proof.

Without loss of generality, assume that the first $k$ columns of
$\begin{bmatrix}P_{1}\\\ P_{2}\\\ \vdots\\\ P_{m}\end{bmatrix}$ are linearly
independent and the remaining columns are in the span of them. Define
$\beta_{j}$ as the $j$-th column of $\begin{bmatrix}P_{1}\\\ P_{2}\\\
\vdots\\\ P_{m}\end{bmatrix}$, where $1\leq j\leq n$. We set

$\displaystyle\beta_{k+1}$ $\displaystyle=$ $\displaystyle
c_{1,k+1}\beta_{1}+c_{2,k+1}\beta_{2}+\cdots+c_{k,k+1}\beta_{k},$
$\displaystyle\beta_{k+2}$ $\displaystyle=$ $\displaystyle
c_{1,k+2}\beta_{1}+c_{2,k+2}\beta_{2}+\cdots+c_{k,k+2}\beta_{k},$
$\displaystyle\vdots$ $\displaystyle\beta_{n}$ $\displaystyle=$ $\displaystyle
c_{1,n}\beta_{1}+c_{2,n}\beta_{2}+\cdots+c_{k,n}\beta_{k},$ (28)

where $c_{a,b}$ is the combination cofficient, $1\leq a\leq k$, and $k+1\leq
b\leq n$. Define $\beta_{j,i}$ as the $j$-th column of $P_{i}$, where $1\leq
j\leq n$. From (II) we obtain that

$\displaystyle\beta_{k+1,i}$ $\displaystyle=$ $\displaystyle
c_{1,k+1}\beta_{1,i}+c_{2,k+1}\beta_{2,i}+\cdots+c_{k,k+1}\beta_{k,i},$
$\displaystyle\beta_{k+2,i}$ $\displaystyle=$ $\displaystyle
c_{1,k+2}\beta_{1,i}+c_{2,k+2}\beta_{2,i}+\cdots+c_{k,k+2}\beta_{k,i},$
$\displaystyle\vdots$ $\displaystyle\beta_{n,i}$ $\displaystyle=$
$\displaystyle
c_{1,n}\beta_{1,i}+c_{2,n}\beta_{2,i}+\cdots+c_{k,n}\beta_{k,i}$ (29)

hold for any $1\leq i\leq m$. Since $Q$ is in the span of
$P_{1},\cdots,P_{m}$, assume

$\displaystyle Q=p_{1}P_{1}+\cdots+p_{m}P_{m},$ (30)

with the combination cofficients $p_{1},\cdots,p_{m}$. Define $\alpha_{j}$ as
the $j$-th column of $Q$, where $1\leq j\leq n$. From (30) we have

$\displaystyle\alpha_{k+1}$
$\displaystyle=p_{1}\beta_{k+1,1}+\cdots+p_{m}\beta_{k+1,m}$ (31)
$\displaystyle=p_{1}(c_{1,k+1}\beta_{1,1}+c_{2,k+1}\beta_{2,1}+\cdots+c_{k,k+1}\beta_{k,1})$
$\displaystyle+\cdots$
$\displaystyle+p_{m}(c_{1,k+1}\beta_{1,m}+c_{2,k+1}\beta_{2,m}+\cdots+c_{k,k+1}\beta_{k,m}).$

From (II) and (31), we have

$\displaystyle\alpha_{k+1}$
$\displaystyle=c_{1,k+1}(p_{1}\beta_{1,1}+\cdots+p_{m}\beta_{1,m})$ (32)
$\displaystyle+\cdots$
$\displaystyle+c_{k,k+1}(p_{1}\beta_{k,1}+\cdots+p_{m}\beta_{k,m})$
$\displaystyle=c_{1,k+1}\alpha_{1}+\cdots+c_{k,k+1}\alpha_{k}.$

We can use the same way of obtaining (32) to prove that

$\displaystyle\alpha_{k+2}$ $\displaystyle=$ $\displaystyle
c_{1,k+2}\alpha_{1}+\cdots+c_{k,k+2}\alpha_{k},$ $\displaystyle\vdots$
$\displaystyle\alpha_{n}$ $\displaystyle=$ $\displaystyle
c_{1,n}\alpha_{1}+\cdots+c_{k,n}\alpha_{k}.$ (33)

Hence any $\alpha_{j}$ is spanned by $\alpha_{1},\cdots,\alpha_{k}$, where
$k+1\leq j\leq n$. By elementary column operations on $Q$, we obtain that
$\alpha_{j}$ becomes zero column for any $k+1\leq j\leq n$. And it has $k$
nonzero columns at most. We finished the proof.  $\sqcap$$\sqcup$

Using preceding lemmas, we present the proof of Conjecture 1 in the next
section. This is equivalent to proving (1) in terms of chl14 .

##  III Proof

For any block matrix
$M\in\mathbb{M}_{m_{1},n_{1}}\otimes\mathbb{M}_{m_{2},n_{2}}$ with Schmidt
rank $K\leq m_{1}\cdot n_{1}$, we write

$\displaystyle M=\begin{bmatrix}M_{1,1}&M_{1,2}&M_{1,3}&\cdots&M_{1,n_{1}}\\\
M_{2,1}&M_{2,2}&M_{2,3}&\cdots&M_{2,n_{1}}\\\
\vdots&\vdots&\vdots&\ddots&\vdots\\\
M_{m_{1},1}&M_{m_{1},2}&M_{m_{1},3}&\cdots&M_{m_{1},n_{1}}\end{bmatrix},$ (34)

where $M$ has $K$ linearly independent blocks.

Our proof of Conjecture 1 is divided into three subsections. In subsection
III.1, we proof Theorem 8, transforming $M$ to a canonical form $N$ in (35) up
to local equivalence. In subsection III.2, we present an equivalent form of
$N$ in (III.2). In subsection III.3, we prove the conjecture by induction.

###  III.1 A canonical form of $M$

We first present a set
$\mathbb{M}_{canonical}\subseteq\mathbb{M}_{m_{1},n_{1}}\otimes\mathbb{M}_{m_{2},n_{2}}$.
Any block matrix $N\in\mathbb{M}_{canonical}$ can be written as

$\displaystyle
N=\begin{bmatrix}\begin{array}[]{c|c|ccc|cc}N_{1,1}&N_{1,2}&\lx@intercol\hfil
N_{1,3}\hfil\lx@intercol\vrule\lx@intercol&\lx@intercol\hfil\cdots\hfil\lx@intercol\vrule\lx@intercol&N_{1,p}&\lx@intercol\hfil\hbox{\multirowsetup$A_{p}$}\hfil\lx@intercol\\\
\vdots&\vdots&\lx@intercol\hfil\vdots\hfil\lx@intercol\vrule\lx@intercol&\lx@intercol\hfil\ddots\hfil\lx@intercol\vrule\lx@intercol&\vdots\\\
N_{{k_{p}},1}&N_{{k_{p}},2}&\lx@intercol\hfil
N_{{k_{p}},3}\hfil\lx@intercol\vrule\lx@intercol&\lx@intercol\hfil\cdots\hfil\lx@intercol\vrule\lx@intercol&N_{{k_{p}},p}\\\
\cline{5-5}\cr\vdots&\vdots&\lx@intercol\hfil\vdots\hfil\lx@intercol\vrule\lx@intercol&\lx@intercol\hfil\vdots\hfil\lx@intercol\vrule\lx@intercol&\lx@intercol\hfil\hbox{\multirowsetup$A_{p-1}$}\hfil\lx@intercol\vrule\lx@intercol\\\
\cline{4-4}\cr\cline{6-7}\cr\vdots&\vdots&\lx@intercol\hfil\vdots\hfil\lx@intercol\vrule\lx@intercol&\lx@intercol\hfil\hfil\lx@intercol\vrule\lx@intercol&\lx@intercol\hfil\hfil\lx@intercol\\\
\cline{5-7}\cr N_{{k_{3}},1}&N_{{k_{3}},2}&\lx@intercol\hfil
N_{{k_{3}},3}\hfil\lx@intercol\vrule\lx@intercol&\lx@intercol\hfil\hbox{\multirowsetup$\cdots$}\hfil\lx@intercol\\\
\cline{3-3}\cr\vdots&\vdots&\lx@intercol\hfil\hfil\lx@intercol\vrule\lx@intercol\\\
N_{{k_{2}},1}&N_{{k_{2}},2}&\lx@intercol\hfil\hfil\lx@intercol\vrule\lx@intercol&\\\
\cline{2-2}\cr\cline{4-7}\cr\vdots&&\lx@intercol\hfil\hbox{\multirowsetup$A_{2}$}\hfil\lx@intercol\\\
N_{k_{1},1}&\\\ \cline{1-1}\cr\cline{3-7}\cr
0&\lx@intercol\hfil\hbox{\multirowsetup$A_{1}$}\hfil\lx@intercol\\\ \vdots\\\
0\end{array}\end{bmatrix},$ (35)

where
$N_{1,1},\cdots,N_{k_{1},1},N_{1,2},\cdots,N_{k_{2},2},\cdots,N_{1,p},\cdots,N_{k_{p},p}$
are linearly independent, and other blocks are in the span of them, $1\leq
p\leq n_{1}$, and

$\displaystyle 1\leq k_{p}\leq\cdots\leq k_{2}\leq k_{1}\leq m_{1},$ (36)
$\displaystyle k_{1}+k_{2}+\cdots+k_{p}=K.$ (37)

At the same time, every block in $A_{i}$ is spanned by
$N_{1,1},\cdots,N_{k_{1},1}$,
$N_{1,2},\cdots,N_{k_{2},2},\cdots,N_{1,i},\cdots,N_{k_{i},i}$, namely

$\displaystyle A_{i}\tilde{\in}\mathop{\rm
span}\\{N_{1,1},\cdots,N_{k_{1},1},N_{1,2},\cdots,N_{k_{2},2},N_{1,i},\cdots,N_{k_{i},i}\\}$
(38)

holds for any $1\leq i\leq p$.

Note that if $m_{1}=k_{1}$, then the zero blocks below $N_{k_{1},1}$ disappear
and $A_{1}$ becomes a $(k_{1}-k_{2})\times 1$ rectangular block matrix. If
$k_{1}=k_{2}$, then $A_{1}$ becomes a $(m_{1}-k_{1})\times(n_{1}-1)$
rectangular block matrix and $A_{2}$ becomes a $(k_{2}-k_{3})\times 1$
rectangular block matrix. If there exists $i$ ($2\leq i<p-1$) such that
$k_{i}=k_{i+1}$, then $A_{i}$ becomes a $(k_{i-1}-k_{i})\times(n_{1}-i)$
rectangular block matrix, and $A_{i+1}$ becomes a $(k_{i+1}-k_{i+2})\times 1$
rectangular block matrix. If $k_{p-1}=k_{p}$, then $A_{p-1}$ becomes a
$(k_{p-2}-k_{p-1})\times(n_{1}-p+1)$ rectangular block matrix, and $A_{p}$
becomes a $k_{p}\times(n_{1}-p)$ rectangular block matrix.

We present the main result of this subsection.

###### Theorem 8

For any bipartite matrix $M$ in (34), there exists a block matrix
$N\in\mathbb{M}_{canonical}$ in (35), such that $M$ is locally equivalent to
$N$.

###### Proof.

Our first aim is to obtain that

$\displaystyle M\sim M^{\prime}=\begin{bmatrix}\begin{array}[]{c|
cccc}M_{1,1}&M_{1,2}&M_{1,3}&\cdots&M_{1,n_{1}}\\\
M_{2,1}&M_{2,2}&M_{2,3}&\cdots&M_{2,n_{1}}\\\
\vdots&\vdots&\vdots&\ddots&\vdots\\\
M_{k_{1},1}&M_{k_{1},2}&M_{k_{1},3}&\cdots&M_{k_{1},n_{1}}\\\ \hline\cr
0&\vrule\lx@intercol\hfil\hbox{\multirowsetup$A_{1}$}\hfil\lx@intercol\\\
\vdots&\\\ 0\end{array}\end{bmatrix},$ (39)

where $k_{1}\leq m_{1}$, and $M_{1,1},\cdots,M_{k_{1},1}$ are linearly
independent, and all the blocks in $A_{1}$ are spanned by
$M_{1,1},\cdots,M_{k_{1},1}$, namely

$\displaystyle A_{1}\tilde{\in}\mathop{\rm
span}\\{M_{1,1},\cdots,M_{k_{1},1}\\}.$ (40)

Note that the blocks $M_{i,j}$ in (39) may be different from those in (34),
and we use the same symbols $M_{i,j}$, when there is no confusion, throughtout
the proof. We apply the following three steps, namely Steps 1-3, to achieve
this aim.

Step 1 Consider the first block-column of $M$ in (34). Assume that it has
$s_{1}$ linearly independent blocks, $s_{1}\leq m_{1}$. Then by block-row
operations on $M$, we obtain that

$\displaystyle M\sim
M_{1}=\begin{bmatrix}M_{1,1}&M_{1,2}&M_{1,3}&\cdots&M_{1,n_{1}}\\\
M_{2,1}&M_{2,2}&M_{2,3}&\cdots&M_{2,n_{1}}\\\
\vdots&\vdots&\vdots&\ddots&\vdots\\\
M_{s_{1},1}&M_{s_{1},2}&M_{s_{1},3}&\cdots&M_{s_{1},n_{1}}\\\
0&M_{s_{1}+1,2}&M_{s_{1}+1,3}&\cdots&M_{s_{1}+1,n_{1}}\\\
\vdots&\vdots&\vdots&\ddots&\vdots\\\
0&M_{m_{1},2}&M_{m_{1},3}&\cdots&M_{m_{1},n_{1}}\end{bmatrix},$ (41)

where $M_{1,1},\cdots,M_{s_{1},1}$ are linearly independent. From (41), it is
obvious that $M$ satisfies (39) if $s_{1}=m_{1}$ or $M_{i,j}$ is linearly
dependent with $M_{1,1},\cdots,M_{s_{1},1}$ for any $s_{1}+1\leq i\leq m_{1}$
and $2\leq j\leq n_{1}$.

Step 2 Suppose $s_{1}<m_{1}$ and there exists a block $M_{i,j}$ that is
linearly independent with $M_{1,1},\cdots,M_{s_{1},1}$ for $s_{1}+1\leq i\leq
m_{1}$ and $2\leq j\leq n_{1}$. Using Lemma 6, we multiply an appropriate
constant to the $j$-th block-column of $M_{1}$ and add to the first block-
column, such that the new first block-column has $s_{1}+a$ linearly
independent blocks, $1\leq a\leq m_{1}-s_{1}$. Next, by repeating Step 1, we
have

$\displaystyle M_{1}\sim
M_{2}=\begin{bmatrix}M_{1,1}&M_{1,2}&M_{1,3}&\cdots&M_{1,n_{1}}\\\
M_{2,1}&M_{2,2}&M_{2,3}&\cdots&M_{2,n_{1}}\\\
\vdots&\vdots&\vdots&\ddots&\vdots\\\
M_{s_{1}+a,1}&M_{s_{1}+a,2}&M_{s_{1}+a,3}&\cdots&M_{s_{1}+a,n_{1}}\\\
0&M_{s_{1}+a+1,2}&M_{s_{1}+a+1,3}&\cdots&M_{s_{1}+a+1,n_{1}}\\\
\vdots&\vdots&\vdots&\ddots&\vdots\\\
0&M_{m_{1},2}&M_{m_{1},3}&\cdots&M_{m_{1},n_{1}}\end{bmatrix},$ (42)

where $M_{1,1},\cdots,M_{s_{1}+a,1}$ are linearly independent.

Step 3 For $M_{2}$, if $s_{1}+a<m_{1}$ and there exists a block $M_{i,j}$ that
is linearly independent with $M_{1,1},\cdots,M_{s_{1}+a,1}$ for $s_{1}+a+1\leq
i\leq m_{1}$ and $2\leq j\leq n_{1}$, then repeat Steps 2 and 1. Eventually we
obtain that $M\sim M^{\prime}$ in (39).

Recall that $M\in\mathbb{M}_{m_{1},n_{1}}\otimes\mathbb{M}_{m_{2},n_{2}}$ in
(34) has Schmidt rank $K$, and hence $M^{\prime}$ has Schmidt rank $K$. If
$k_{1}=K$ in $M^{\prime}$ in (35), then $M^{\prime}\sim N$ with $p=1$ in terms
of (39). Hence we suppose

$\displaystyle k_{1}<K$ (43)

in $M^{\prime}$. Then our next aim is to obtain that

$\displaystyle M^{\prime}\sim
M^{\prime\prime}=\begin{bmatrix}\begin{array}[]{c|c|ccc}M_{1,1}&M_{1,2}&M_{1,3}&\cdots&M_{1,n_{1}}\\\
M_{2,1}&M_{2,2}&M_{2,3}&\cdots&M_{2,n_{1}}\\\
\vdots&\vdots&\vdots&\ddots&\vdots\\\
M_{{k_{2}},1}&M_{{k_{2}},2}&M_{{k_{2}},3}&\cdots&M_{{k_{2}},n_{1}}\\\
\cline{2-5}\cr
M_{{k_{2}}+1,1}&&\lx@intercol\hfil\hbox{\multirowsetup$A_{2}$}\hfil\lx@intercol\\\
\vdots&&&&\\\ M_{k_{1},1}&&&&\\\ \cline{1-1}\cr\cline{3-5}\cr
0&\lx@intercol\hfil\hbox{\multirowsetup$A_{1}$}\hfil\lx@intercol\\\ \vdots\\\
0\end{array}\end{bmatrix},$ (44)

where $1\leq k_{2}\leq k_{1}\leq m_{1}$, and
$M_{1,1},\cdots,M_{k_{1},1},M_{1,2},\cdots,M_{k_{2},2}$ are linearly
independent. At the same time, (40) still holds and

$\displaystyle A_{2}\tilde{\in}\mathop{\rm
span}\\{M_{1,1},\cdots,M_{k_{1},1},M_{1,2},\cdots,M_{k_{2},2}\\}.$ (45)

We next apply the following four steps, namely Steps 4-7, to achieve this aim.

Step 4 Consider the matrix $M^{\prime}$ in (39). Because of (40) and (43), we
obtain that there exists a block $M_{i,j}$ that is linearly independent with
$M_{1,1},\cdots,M_{k_{1},1}$ for $1\leq i\leq k_{1}$ and $2\leq j\leq n_{1}$.
By block-row and block-column switches on $M^{\prime}$ in (39), $M_{i,j}$
becomes $M_{1,2}$, and $M_{1,1},\cdots,M_{k_{1},1},M_{1,2}$ are linearly
independent.

Step 5 For the new $M^{\prime}$ in (39), if there exists a block $M_{i,2}$
that is linearly independent with $M_{1,1},\cdots,M_{k_{1},1},M_{1,2}$ for
$2\leq i\leq k_{1}$, then by block-row switch, this block becomes $M_{2,2}$.
We repeat from the beginning of Step 5, until we obtain that

$\displaystyle M^{\prime}\sim M_{3}=\begin{bmatrix}\begin{array}[]{c|
cccc}M_{1,1}&M_{1,2}&M_{1,3}&\cdots&M_{1,n_{1}}\\\
M_{2,1}&M_{2,2}&M_{2,3}&\cdots&M_{2,n_{1}}\\\
\vdots&\vdots&\vdots&\ddots&\vdots\\\
M_{{s_{2}},1}&M_{{s_{2}},2}&M_{{s_{2}},3}&\cdots&M_{{s_{2}},n_{1}}\\\
M_{{s_{2}}+1,1}&M_{{s_{2}}+1,2}&M_{{s_{2}}+1,3}&\cdots&M_{{s_{2}}+1,n_{1}}\\\
\vdots&\vdots&\vdots&\ddots&\vdots\\\
M_{k_{1},1}&M_{k_{1},2}&M_{k_{1},3}&\cdots&M_{k_{1},n_{1}}\\\ \hline\cr
0&\vrule\lx@intercol\hfil\hbox{\multirowsetup$A_{1}$}\hfil\lx@intercol\\\
\vdots&\\\ 0\end{array}\end{bmatrix},$ (46)

where $1\leq s_{2}\leq k_{1}$, and
$M_{1,1},\cdots,M_{k_{1},1},M_{1,2},\cdots,M_{s_{2},2}$ are linearly
independent. At the same time,

$\displaystyle M_{i,2}\in\mathop{\rm
span}\\{M_{1,1},\cdots,M_{k_{1},1},M_{1,2},\cdots,M_{s_{2},2}\\}$ (47)

holds for any $s_{2}+1\leq i\leq k_{1}$.

Step 6 By (47), we apply block-row operations on $M_{3}$ in (46) such that

$\displaystyle M_{i,2}\in\mathop{\rm span}\\{M_{1,1},\cdots,M_{k_{1},1}\\}$
(48)

holds for any $s_{2}+1\leq i\leq k_{1}$.

Step 7 From (40) and (48), one can obtain that $M_{3}\sim M^{\prime\prime}$ in
(44) if $s_{2}=k_{1}$ or $M_{i,j}$ is linearly dependent with
$M_{1,1},\cdots,M_{k_{1},1},M_{1,2},\cdots,M_{s_{2},2}$ for any $s_{2}+1\leq
i\leq k_{1}$ and $3\leq j\leq n_{1}$.

Suppose $s_{2}<k_{1}$ and there exists a block $M_{ij}$ ($s_{2}+1\leq i\leq
k_{1}$, $3\leq j\leq n_{1}$) that is linearly independent with
$M_{1,1},\cdots,M_{k_{1},1}$, $M_{1,2},\cdots,M_{s_{2},2}$ in (46). Using
Lemma 6, we multiply an appropriate constant to the $j$-th block-column of
$M_{3}$ and add to the second block-column, such that the new second block-
column has $s_{2}+b$ linearly independent blocks, $1\leq b\leq k_{1}-s_{2}$,
and they are linearly independent with $M_{1,1},\cdots,M_{k_{1},1}$. For the
new $M_{3}$, we next repeat Steps 5, 6 and 7. Finally we obtain that
$M^{\prime}\sim M^{\prime\prime}$ in (44), and $M^{\prime\prime}$ has Schmidt
rank $K$.

If $k_{1}+k_{2}=K$, then $M^{\prime\prime}\sim N$ with $p=2$ in (35). On the
other hand if $k_{1}+k_{2}<K$, then we consider the third block-column of
$M^{\prime\prime}$ using the same way from Step 4 to 7. Continue this process
until we obtain that $M^{\prime\prime}\sim N$.

By achieving (39) and (44), we have shown that $M\sim M^{\prime}\sim
M^{\prime\prime}\sim N$ in (35). Hence we have finished the proof.
$\sqcap$$\sqcup$

###  III.2 Equivalent form of $M$

In subsection III.1, we have shown that any bipartite matrix $M$ has a
canonical form $N\in\mathbb{M}_{canonical}$ in (35) up to local equivalence.
In this subsection, we continue to apply three steps on $N$ , namely Steps
8-10, to obtain another form of $M$ in (III.2) up to local equivalence.

Step 8 Denote $*_{i}$ as a matrix that contains exactly $i$ columns,
$\lambda_{i}$ as a matrix that contains exactly $i$ column vectors and they
are linearly independent, $0_{i}$ as a matrix that contains exactly $i$
columns and they are all zero column vectors. Denote $*_{i}^{T}$ as a matrix
that contains exactly $i$ rows, $0_{i}^{T}$ as a matrix that contains exactly
$i$ zero rows. Denote $0_{i}^{\Gamma_{B}}$ as the partial transpose of system
$B$ of a block matrix $0_{i}$.

Recall that $N\in\mathbb{M}_{m_{1},n_{1}}\otimes\mathbb{M}_{m_{2},n_{2}}$ in
(35). Consider the $k_{1}$ linearly independent blocks in the first block-
column of $N$, and they form a matrix of $n_{2}$ columns. If the matrix has
$r_{1}$ linearly independent column vectors, then

$\displaystyle r_{1}\leq\mathop{\rm min}\\{n_{2},r(N)\\}.$ (49)

Next, we can find an order-$n_{2}$ invertible matrix $R_{1}$ such that

$\displaystyle\begin{bmatrix}N_{1,1}\\\ \vdots\\\
N_{k_{1},1}\end{bmatrix}\cdot
R_{1}=\begin{bmatrix}\begin{array}[]{c}\hbox{\multirowsetup$\lambda_{r_{1}}$
$0_{n_{2}-r_{1}}$}\\\ \\\ \\\ \end{array}\end{bmatrix},$ (50)

where the leftmost $r_{1}$ column vectors are linearly independent, and the
rightmost $n_{2}-r_{1}$ column vectors are zero vectors. Denote
$N_{i,j}^{(A_{k})}$ as a block $N_{i,j}$ in $A_{k}$, where $1\leq k\leq p$ in
(35). Recall that $A_{1}\tilde{\in}\mathop{\rm
span}\\{N_{1,1},\cdots,N_{k_{1},1}\\}$ from (38). Using Lemma 7 and (50), we
have

$\displaystyle N_{i,j}^{(A_{1})}\cdot
R_{1}=\begin{bmatrix}\begin{array}[]{c}\hbox{\multirowsetup$*_{r_{1}}$
$0_{n_{2}-r_{1}}$}\\\ \\\ \end{array}\end{bmatrix},$ (51)

where the rightmost $n_{2}-r_{1}$ columns are zero column vectors. Let

$\displaystyle N_{1}=N\cdot(I_{n_{1}}\otimes R_{1}),$ (52)

From (35), (50) and (51) we have

$\displaystyle N\sim
N_{1}=\begin{bmatrix}\begin{array}[]{c|c|ccc|cc}\hbox{\multirowsetup$\lambda_{r_{1}}$
$0_{n_{2}-r_{1}}$}&N_{1,2}&\lx@intercol\hfil
N_{1,3}\hfil\lx@intercol\vrule\lx@intercol&\lx@intercol\hfil\cdots\hfil\lx@intercol\vrule\lx@intercol&N_{1,p}&\lx@intercol\hfil\hbox{\multirowsetup$A_{p}$}\hfil\lx@intercol\\\
&\vdots&\lx@intercol\hfil\vdots\hfil\lx@intercol\vrule\lx@intercol&\lx@intercol\hfil\ddots\hfil\lx@intercol\vrule\lx@intercol&\vdots\\\
&N_{{k_{p}},2}&\lx@intercol\hfil
N_{{k_{p}},3}\hfil\lx@intercol\vrule\lx@intercol&\lx@intercol\hfil\cdots\hfil\lx@intercol\vrule\lx@intercol&N_{{k_{p}},p}\\\
\cline{5-5}\cr&\vdots&\lx@intercol\hfil\vdots\hfil\lx@intercol\vrule\lx@intercol&\lx@intercol\hfil\vdots\hfil\lx@intercol\vrule\lx@intercol&\lx@intercol\hfil\hbox{\multirowsetup$A_{p-1}$}\hfil\lx@intercol\vrule\lx@intercol\\\
\cline{4-4}\cr\cline{6-7}\cr&\vdots&\lx@intercol\hfil\vdots\hfil\lx@intercol\vrule\lx@intercol&\lx@intercol\hfil\hfil\lx@intercol\vrule\lx@intercol&\lx@intercol\hfil\hfil\lx@intercol\\\
\cline{5-7}\cr&N_{{k_{3}},2}&\lx@intercol\hfil
N_{{k_{3}},3}\hfil\lx@intercol\vrule\lx@intercol&\lx@intercol\hfil\hbox{\multirowsetup$\cdots$}\hfil\lx@intercol\\\
\cline{3-3}\cr&\vdots&\lx@intercol\hfil\hfil\lx@intercol\vrule\lx@intercol\\\
&N_{{k_{2}},2}&\lx@intercol\hfil\hfil\lx@intercol\vrule\lx@intercol&\\\
\cline{2-2}\cr\cline{4-7}\cr&\hbox{\multirowsetup$*_{r_{1}}$
$0_{n_{2}-r_{1}}$}&\lx@intercol\hfil\hbox{\multirowsetup$A_{2}$}\hfil\lx@intercol\\\
&\\\
\cline{1-1}\cr\cline{3-7}\cr\hbox{\multirowsetup$0_{n_{2}}$}&\lx@intercol\hfil\hbox{\multirowsetup$*_{r_{1}}$
$0_{n_{2}-r_{1}}$}\hfil\lx@intercol&\lx@intercol\hfil\hbox{\multirowsetup$*_{r_{1}}$
$0_{n_{2}-r_{1}}$}\hfil\lx@intercol&\hbox{\multirowsetup$\cdots$}&\lx@intercol\hfil\hbox{\multirowsetup$*_{r_{1}}$
$0_{n_{2}-r_{1}}$}\hfil\lx@intercol&\lx@intercol\hfil\hbox{\multirowsetup$\cdots$}\hfil\lx@intercol\\\
&\lx@intercol\hfil\hfil\lx@intercol\\\
&\lx@intercol\hfil\hfil\lx@intercol\end{array}\end{bmatrix},$ (53)

where (36), (37) and (38) still hold by Lemma 4. Note that $N_{i,j}\cdot
R_{1}$ has been denoted by $N_{i,j}$ still, when there is no confusion. The
same denotion applies also to $A_{2},\cdots,A_{p}$, as well as the end of
Steps 9 and 10.

Step 9 Consider the $k_{2}$ linearly independent blocks in the second block-
column of $N_{1}$ in (53). We write

$\displaystyle\begin{bmatrix}N_{1,2}\\\ \vdots\\\
N_{k_{2},2}\end{bmatrix}=\begin{bmatrix}\begin{array}[]{c}\hbox{\multirowsetup$*_{r_{1}}$
$*_{n_{2}-r_{1}}$}\\\ \\\ \\\ \end{array}\end{bmatrix},$ (54)

and next consider the rightmost $n_{2}-r_{1}$ columns in (54). Suppose they
form a matrix of $r_{2}$ linearly independent column vectors, then

$\displaystyle r_{2}\leq\mathop{\rm min}\\{n_{2}-r_{1},r(N_{1})\\}.$ (55)

Thus we can find an order-($n_{2}-r_{1}$) invertible matrix $R_{2}$ such that

$\displaystyle\begin{bmatrix}N_{1,2}\\\ \vdots\\\
N_{k_{2},2}\end{bmatrix}\cdot\begin{bmatrix}I_{r_{1}}&0\\\
0&R_{2}\end{bmatrix}=\begin{bmatrix}\begin{array}[]{c}\hbox{\multirowsetup$*_{r_{1}}$
$\lambda_{r_{2}}$ $0_{n_{2}-r_{1}-r_{2}}$}\\\ \\\ \\\
\end{array}\end{bmatrix},$ (56)

where the middle $r_{2}$ column vectors are linearly independent and the
rightmost $n_{2}-r_{1}-r_{2}$ column vectors are zero vectors. Further, for
$N_{1}$ in (53), we have

$\displaystyle N_{i,j}^{(A_{1})}\cdot\begin{bmatrix}I_{r_{1}}&0\\\
0&R_{2}\end{bmatrix}=\begin{bmatrix}\begin{array}[]{c}\hbox{\multirowsetup$*_{r_{1}}$
$0_{n_{2}-r_{1}}$}\\\ \\\
\end{array}\end{bmatrix}\cdot\begin{bmatrix}I_{r_{1}}&0\\\
0&R_{2}\end{bmatrix}=N_{i,j}^{(A_{1})}.$ (57)

Recall that $A_{2}\tilde{\in}\mathop{\rm
span}\\{N_{1,1},\cdots,N_{k_{1},1},N_{1,2},\cdots,N_{k_{2},2}\\}$ from (38).
Using Lemma 7 and (56), we have

$\displaystyle N_{i,j}^{(A_{2})}\cdot\begin{bmatrix}I_{r_{1}}&0\\\
0&R_{2}\end{bmatrix}=\begin{bmatrix}\begin{array}[]{c}\hbox{\multirowsetup$*_{r_{1}+r_{2}}$
$0_{n_{2}-r_{1}-r_{2}}$}\\\ \\\ \end{array}\end{bmatrix}.$ (58)

Let

$\displaystyle N_{2}=N_{1}\cdot(I_{n_{1}}\otimes\begin{bmatrix}I_{r_{1}}&0\\\
0&R_{2}\end{bmatrix}).$ (59)

From (53), (56), (57) and (58), we have

$\displaystyle N_{1}\sim N_{2}=$
$\displaystyle\begin{bmatrix}\begin{array}[]{c|c|c|ccc|cc}\hbox{\multirowsetup$\lambda_{r_{1}}$
$0_{n_{2}-r_{1}}$}&\hbox{\multirowsetup$*_{r_{1}}$ $\lambda_{r_{2}}$
$0_{n_{2}-r_{1}-r_{2}}$}&N_{1,3}&\lx@intercol\hfil
N_{1,4}\hfil\lx@intercol\vrule\lx@intercol&\lx@intercol\hfil\cdots\hfil\lx@intercol\vrule\lx@intercol&N_{1,p}&\lx@intercol\hfil\hbox{\multirowsetup$A_{p}$}\hfil\lx@intercol\\\
&&\vdots&\lx@intercol\hfil\vdots\hfil\lx@intercol\vrule\lx@intercol&\lx@intercol\hfil\ddots\hfil\lx@intercol\vrule\lx@intercol&\vdots\\\
&&N_{{k_{p}},3}&\lx@intercol\hfil
N_{{k_{p}},4}\hfil\lx@intercol\vrule\lx@intercol&\lx@intercol\hfil\cdots\hfil\lx@intercol\vrule\lx@intercol&N_{{k_{p}},p}\\\
\cline{6-6}\cr&&\vdots&\lx@intercol\hfil\vdots\hfil\lx@intercol\vrule\lx@intercol&\lx@intercol\hfil\vdots\hfil\lx@intercol\vrule\lx@intercol&\hbox{\multirowsetup$A_{p-1}$}&&\\\
\cline{5-5}\cr\cline{7-8}\cr&&\vdots&\lx@intercol\hfil\vdots\hfil\lx@intercol\vrule\lx@intercol&\lx@intercol\hfil\hfil\lx@intercol\vrule\lx@intercol&\lx@intercol\hfil\hfil\lx@intercol&\\\
\cline{6-8}\cr&&N_{k_{4},3}&\lx@intercol\hfil
N_{k_{4},4}\hfil\lx@intercol\vrule\lx@intercol&\lx@intercol\hfil\hbox{\multirowsetup$\cdots$}\hfil\lx@intercol\\\
\cline{4-4}\cr&&\vdots&\lx@intercol\hfil\hfil\lx@intercol\vrule\lx@intercol&\lx@intercol\hfil\hfil\lx@intercol&\lx@intercol\hfil\hfil\lx@intercol&\\\
&&N_{k_{3},3}&\lx@intercol\hfil\hfil\lx@intercol\vrule\lx@intercol&\lx@intercol\hfil\hfil\lx@intercol\\\
\cline{3-3}\cr\cline{5-8}\cr&&\hbox{\multirowsetup$*_{r_{1}+r_{2}}$
$0_{n_{2}-r_{1}-r_{2}}$}&\lx@intercol\hfil\hbox{\multirowsetup$A_{3}$}\hfil\lx@intercol\\\
&&\lx@intercol\hfil\hfil\lx@intercol\vrule\lx@intercol&\lx@intercol\hfil\hfil\lx@intercol\\\
\cline{2-2}\cr\cline{4-8}\cr&\hbox{\multirowsetup$*_{r_{1}}$
$0_{n_{2}-r_{1}}$}&\lx@intercol\hfil\hbox{\multirowsetup$*_{r_{1}+r_{2}}$
$0_{n_{2}-r_{1}-r_{2}}$}\hfil\lx@intercol&\hbox{\multirowsetup$\cdots$}&\lx@intercol\hfil\hbox{\multirowsetup$\cdots$}\hfil\lx@intercol&\lx@intercol\hfil\hbox{\multirowsetup$*_{r_{1}+r_{2}}$
$0_{n_{2}-r_{1}-r_{2}}$}\hfil\lx@intercol&\lx@intercol\hfil\hbox{\multirowsetup$\cdots$}\hfil\lx@intercol\\\
&&\lx@intercol\hfil\hfil\lx@intercol&&\lx@intercol\hfil\hfil\lx@intercol&\lx@intercol\hfil\hfil\lx@intercol&\\\
\cline{1-1}\cr\cline{3-8}\cr\hbox{\multirowsetup$0_{n_{2}}$}&\lx@intercol\hfil\hbox{\multirowsetup$*_{r_{1}}$
$0_{n_{2}-r_{1}}$}\hfil\lx@intercol&\lx@intercol\hfil\hbox{\multirowsetup$*_{r_{1}}$
$0_{n_{2}-r_{1}}$}\hfil\lx@intercol&\hbox{\multirowsetup$\cdots$}&\lx@intercol\hfil\hbox{\multirowsetup$\cdots$}\hfil\lx@intercol&\lx@intercol\hfil\hbox{\multirowsetup$*_{r_{1}}$
$0_{n_{2}-r_{1}}$}\hfil\lx@intercol&\lx@intercol\hfil\hbox{\multirowsetup$\cdots$}\hfil\lx@intercol\\\
&\lx@intercol\hfil\hfil\lx@intercol&\lx@intercol\hfil\hfil\lx@intercol&&\lx@intercol\hfil\hfil\lx@intercol&\lx@intercol\hfil\hfil\lx@intercol\\\
&\lx@intercol\hfil\hfil\lx@intercol&\lx@intercol\hfil\hfil\lx@intercol&&\lx@intercol\hfil\hfil\lx@intercol&\lx@intercol\hfil\hfil\lx@intercol\\\
\end{array}\end{bmatrix},$

where (36), (37) and (38) still hold by Lemma 4. Note that if $r_{1}=n_{2}$,
then we have $N_{1}=N_{2}$, i.e., the rightmost $n_{2}-r_{1}$ columns of each
block in $N_{2}$ disappear.

Step 10 Consider the $k_{3}$ linearly independent blocks in the third block-
column of $N_{2}$ using the same way in Steps 8 and 9. Suppose the rightmost
$n_{2}-r_{1}-r_{2}$ columns of the block matrix have $r_{3}$ linearly
independent column vectors. Then

$\displaystyle r_{3}\leq\mathop{\rm min}\\{n_{2}-r_{1}-r_{2},r(N_{2})\\}.$
(61)

We can find an order-($n_{2}-r_{1}-r_{2}$) invertible matrix $R_{3}$ such that

$\displaystyle\begin{bmatrix}N_{1,3}\\\ \vdots\\\
N_{k_{3},3}\end{bmatrix}\cdot\begin{bmatrix}I_{r_{1}+r_{2}}&0\\\
0&R_{3}\end{bmatrix}=\begin{bmatrix}\begin{array}[]{c}\hbox{\multirowsetup$*_{r_{1}+r_{2}}$
$\lambda_{r_{3}}$ $0_{n_{2}-r_{1}-r_{2}-r_{3}}$}\\\ \\\ \\\
\end{array}\end{bmatrix}.$ (62)

Let

$\displaystyle
N_{3}=N_{2}\cdot(I_{n_{1}}\otimes\begin{bmatrix}I_{r_{1}+r_{2}}&0\\\
0&R_{3}\end{bmatrix}).$ (63)

Hence $N_{2}\sim N_{3}$, and we next consider the fourth block-column of
$N_{3}$. Continuing this process and using (53), (III.2), finally we obtain
that

$\displaystyle N\sim N_{1}\sim\cdots\sim N_{p}=$
$\displaystyle\begin{bmatrix}\begin{array}[]{c|c|c|c|c|c|c}\hbox{\multirowsetup$\lambda_{r_{1}}$
$0_{n_{2}-r_{1}}$}&\hbox{\multirowsetup$*_{r_{1}}$ $\lambda_{r_{2}}$
$0_{n_{2}-r_{1}-r_{2}}$}&\hbox{\multirowsetup$*_{r_{1}+r_{2}}$
$\lambda_{r_{3}}$
$0_{n_{2}-r_{1}-r_{2}-r_{3}}$}&\cdots&\hbox{\multirowsetup$*_{r_{1}+\cdots+r_{p-1}}\lambda_{r_{p}}0_{n_{2}-r_{1}-\cdots-
r_{p}}$}&\lx@intercol\hfil\hbox{\multirowsetup{$A_{p}$}}\hfil\lx@intercol\\\
&&&\ddots&&\lx@intercol\hfil\hfil\lx@intercol\\\
\cline{5-5}\cr&\hbox{\multirowsetup}\hfil&\hbox{\multirowsetup}\hfil&\cdots&\hbox{\multirowsetup$*_{r_{1}+\cdots+r_{p-1}}0_{n_{2}-r_{1}-\cdots
r_{p-1}}$}&\lx@intercol\hbox{\multirowsetup}\hfil\hfil\lx@intercol\\\
&\hbox{\multirowsetup}\hfil&\hbox{\multirowsetup}\hfil&\vdots&&\lx@intercol\hfil\hfil\lx@intercol\\\
\cline{4-4}\cr\cline{6-7}\cr&&&&\lx@intercol\hfil\hbox{\multirowsetup$*_{r_{1}+\cdots+r_{p-1}}0_{n_{2}-r_{1}-\cdots
r_{p-1}}$}\hfil\lx@intercol&\lx@intercol\hfil\hbox{\multirowsetup$\cdots$}\hfil\lx@intercol\\\
&&&\lx@intercol\hfil\hfil\lx@intercol\vrule\lx@intercol&\lx@intercol\hfil\hfil\lx@intercol&\lx@intercol\hfil\hfil\lx@intercol\\\
\cline{3-3}\cr\cline{5-7}\cr&\hbox{\multirowsetup}\hfil&\hbox{\multirowsetup$*_{r_{1}+r_{2}}$
$0_{n_{2}-r_{1}-r_{2}}$}&\lx@intercol\hfil\hbox{\multirowsetup$\cdots$}\hfil\lx@intercol\\\
&\hbox{\multirowsetup}\hfil&&\lx@intercol\hfil\hfil\lx@intercol&\lx@intercol\hfil\hfil\lx@intercol&\lx@intercol\hfil\hfil\lx@intercol\\\
\cline{2-2}\cr\cline{4-7}\cr&\hbox{\multirowsetup$*_{r_{1}}$
$0_{n_{2}-r_{1}}$}&\lx@intercol\hfil\hbox{\multirowsetup$*_{r_{1}+r_{2}}$
$0_{n_{2}-r_{1}-r_{2}}$}\hfil\lx@intercol&\lx@intercol\hfil\hbox{\multirowsetup$\cdots$}\hfil\lx@intercol&\lx@intercol\hfil\hbox{\multirowsetup$*_{r_{1}+r_{2}}$
$0_{n_{2}-r_{1}-r_{2}}$}\hfil\lx@intercol&\lx@intercol\hfil\hbox{\multirowsetup$\cdots$}\hfil\lx@intercol\\\
&&\lx@intercol\hfil\hfil\lx@intercol&\lx@intercol\hfil\hfil\lx@intercol&\lx@intercol\hfil\hfil\lx@intercol&\lx@intercol\hfil\hfil\lx@intercol\\\
\cline{1-1}\cr\cline{3-7}\cr\hbox{\multirowsetup$0_{n_{2}}$}&\lx@intercol\hfil\hbox{\multirowsetup$*_{r_{1}}$
$0_{n_{2}-r_{1}}$}\hfil\lx@intercol&\lx@intercol\hfil\hbox{\multirowsetup$*_{r_{1}}$
$0_{n_{2}-r_{1}}$}\hfil\lx@intercol&\lx@intercol\hfil\hbox{\multirowsetup$\cdots$}\hfil\lx@intercol&\lx@intercol\hfil\hbox{\multirowsetup$*_{r_{1}}$
$0_{n_{2}-r_{1}}$}\hfil\lx@intercol&\lx@intercol\hfil\hbox{\multirowsetup$\cdots$}\hfil\lx@intercol\\\
&\lx@intercol\hfil\hfil\lx@intercol&\lx@intercol\hfil\hfil\lx@intercol&\lx@intercol\hfil\hfil\lx@intercol&\lx@intercol\hfil\hfil\lx@intercol&\lx@intercol\hfil\hfil\lx@intercol\\\
&\lx@intercol\hfil\hfil\lx@intercol&\lx@intercol\hfil\hfil\lx@intercol&\lx@intercol\hfil\hfil\lx@intercol&\lx@intercol\hfil\hfil\lx@intercol&\lx@intercol\hfil\hfil\lx@intercol\end{array}\end{bmatrix},$

where $*_{i},\lambda_{i},0_{i}$ are defined in Step 8 and
$r_{1}+r_{2}+\cdots+r_{p}\leq n_{2}$. At the same time, (36), (37) and (38)
still hold by Lemma 4. At the same time, the rightmost ($n_{2}-r_{1}-\cdots-
r_{p}$) column vectors of $N_{i,j}^{(A_{p})}$ in (III.2) are zero vectors by
(38) and Lemma 7. One can also verify that if $r_{1}+\cdots+r_{i}=n_{2}$ for
$1\leq i<p$, then $N\sim N_{i}=N_{i+1}=\cdots=N_{p}$.

We have shown that $N\sim N_{p}$ in (35) and (III.2), hence we have presented
an equivalent form of $M$ in (34).

###  III.3 Proof of Conjecture 1

In this subsection, we prove the conjectue by induction.

###### Theorem 9

Conjecture 1 holds. Equivalently, the inequality (1) holds for any tripartite
mixed state $\rho_{ABC}$.

###### Proof.

First, one can show that Conjecture 1 holds for any $M$ of Schmidt rank one.
Next, suppose Conjecture 1 holds for any matrix of Schmidt rank at most $K-1$,
with $K\geq 2$. We will prove that Conjecture 1 holds for any $M$ of Schmidt
rank $K$.

In subsections III.1 and III.2, we have shown that proving $M$ satisfies
Conjecture 1 is equivalent to proving $N_{p}\sim N$ in (III.2) satisfies
Conjecture 1. Further, it has been proved by Lemma 5 that $N$ in (35)
satisfies Conjecture 1 if $k_{1}=K$ in (37). So from (36) and (37), we assume
that

$\displaystyle 1\leq k_{s}<K$ (65)

holds for any $1\leq s\leq p$ in $N$ in (35) and $N_{p}$ in (III.2).

We next decompose $N_{p}$ in (III.2) into the sum of $p$ block matrices.
Firstly, let
$N_{q_{1}}=[N_{i,j}^{(q_{1})}]\in\mathbb{M}_{m_{1},n_{1}}\otimes\mathbb{M}_{m_{2},n_{2}}$,
where the first $r_{1}$ columns of $N_{i,j}^{(q_{1})}$ in $N_{q_{1}}$ are
exactly the first $r_{1}$ columns of $N_{i,j}$ in $N_{p}$ in (III.2), the
remaining $n_{2}-r_{1}$ column vectors of $N_{i,j}^{(q_{1})}$ in $N_{q_{1}}$
are zero vectors. From (III.2), we have

$\displaystyle N_{q_{1}}=$
$\displaystyle\begin{bmatrix}\begin{array}[]{c|c|c|c}\hbox{\multirowsetup$\lambda_{r_{1}}$
$0_{n_{2}-r_{1}}$}&\hbox{\multirowsetup$*_{r_{1}}$
$0_{n_{2}-r_{1}}$}&\hbox{\multirowsetup$\cdots$}&\hbox{\multirowsetup$*_{r_{1}}$
$0_{n_{2}-r_{1}}$}\\\ &&&\\\ &&&\\\
\hline\cr\hbox{\multirowsetup$0_{n_{2}}$}&\hbox{\multirowsetup$*_{r_{1}}$
$0_{n_{2}-r_{1}}$}&\hbox{\multirowsetup$\cdots$}&\hbox{\multirowsetup$*_{r_{1}}$
$0_{n_{2}-r_{1}}$}\\\ &&&\end{array}\end{bmatrix}$ $\displaystyle:=$
$\displaystyle\begin{bmatrix}\begin{array}[]{c|ccc}N_{1,1}^{(q_{1})}&N_{1,2}^{(q_{1})}&\cdots&N_{1,n_{1}}^{(q_{1})}\\\
\vdots&\vdots&\ddots&\vdots\\\
N_{k_{1},1}^{(q_{1})}&N_{k_{1},2}^{(q_{1})}&\cdots&N_{k_{1},n_{1}}^{(q_{1})}\\\
\hline\cr\hbox{\multirowsetup{\large$0_{n_{2}}$}}&\lx@intercol\hfil\hbox{\multirowsetup{\huge$\omega$}${}_{1}$}\hfil\lx@intercol\\\
&&\end{array}\end{bmatrix},$ (66)

where $\omega_{1}$ is an $(m_{1}-k_{1})\times(n_{1}-1)$ rectangular block
matrix, and $N_{i,1}^{(q_{1})}=N_{i,1}$ in terms of (35) and (III.2),
$i=1,\cdots,k_{1}$. From (38), we obtain that

$\displaystyle\text{\Large$\omega_{1}$}\tilde{\in}\mathop{\rm
span}\\{N_{1,1}^{(q_{1})},\cdots,N_{k_{1},1}^{(q_{1})}\\},$ (67)

holds in (66). Thus by (65), we have

$\displaystyle Sr(\text{\Large$\omega_{1}$})\leq k_{1}<K,$ (68)

i.e., the Schmidt rank of $\omega_{1}$ is less than $K$. At the same time,
from (66), using Lemma 3 and the definition of $\lambda_{i}$, we have

$\displaystyle r(\begin{bmatrix}N_{1,1}^{(q_{1})}\\\ \vdots\\\
N_{k_{1},1}^{(q_{1})}\end{bmatrix})+r(\text{\Large$\omega_{1}$})=r_{1}+r(\text{\Large$\omega_{1}$})\leq
r(N_{q_{1}}).$ (69)

Secondly, let
$N_{q_{2}}=[N_{i,j}^{(q_{2})}]\in\mathbb{M}_{m_{1},n_{1}}\otimes\mathbb{M}_{m_{2},n_{2}}$,
where the $k$-th column of $N_{i,j}^{(q_{2})}$ in $N_{q_{2}}$ is exactly the
$k$-th column of $N_{i,j}$ in $N_{p}$ in (III.2), $r_{1}+1\leq k\leq
r_{1}+r_{2}$. At the same time, the remaining $n_{2}-r_{2}$ column vectors of
$N_{i,j}^{(q_{2})}$ in $N_{q_{2}}$ are zero vectors. From (III.2), we have

$\displaystyle N_{q_{2}}=$
$\displaystyle\begin{bmatrix}\begin{array}[]{c|c|c|c|c}\hbox{\multirowsetup$0_{n_{2}}$}&\hbox{\multirowsetup$0_{r_{1}}\lambda_{r_{2}}0_{n_{2}-r_{1}-r_{2}}$}&\hbox{\multirowsetup$0_{r_{1}}*_{r_{2}}0_{n_{2}-r_{1}-r_{2}}$}&\hbox{\multirowsetup$\cdots$}&\hbox{\multirowsetup$0_{r_{1}}*_{r_{2}}0_{n_{2}-r_{1}-r_{2}}$}\\\
&&&&\\\ &&&&\\\
\cline{2-5}\cr&\hbox{\multirowsetup$0_{n_{2}}$}&\hbox{\multirowsetup$0_{r_{1}}*_{r_{2}}0_{n_{2}-r_{1}-r_{2}}$}&\hbox{\multirowsetup$\cdots$}&\hbox{\multirowsetup$0_{r_{1}}*_{r_{2}}0_{n_{2}-r_{1}-r_{2}}$}\\\
&&&&\\\
\cline{2-5}\cr&\hbox{\multirowsetup$0_{n_{2}}$}&\hbox{\multirowsetup$0_{n_{2}}$}&\hbox{\multirowsetup$\cdots$}&\hbox{\multirowsetup$0_{n_{2}}$}\\\
&&&&\end{array}\end{bmatrix}$ $\displaystyle:=$
$\displaystyle\begin{bmatrix}\begin{array}[]{c|c|ccc}\hbox{\multirowsetup$0_{n_{2}}$}&N_{1,2}^{(q_{2})}&N_{1,3}^{(q_{2})}&\cdots&N_{1,n_{1}}^{(q_{2})}\\\
&\vdots&\vdots&\ddots&\vdots\\\
&N_{k_{2},2}^{(q_{2})}&N_{k_{2},3}^{(q_{2})}&\cdots&N_{k_{2},n_{1}}^{(q_{2})}\\\
\cline{2-5}\cr&\hbox{\multirowsetup{\large$0_{n_{2}}$}}&\lx@intercol\hfil\hbox{\multirowsetup{\huge$\omega$}${}_{2}$}\hfil\lx@intercol\\\
&&&&\\\
\cline{2-5}\cr&\lx@intercol\hfil\hbox{\multirowsetup$0_{(n_{1}-1)n_{2}}$}\hfil\lx@intercol\\\
&\lx@intercol\hfil\hfil\lx@intercol&\end{array}\end{bmatrix},$ (70)

where $\omega_{2}$ is a $(k_{1}-k_{2})\times(n_{1}-2)$ rectangular block
matrix. Recall that (38) holds in $N_{p}$ in (III.2), we have
$\text{\Large$\omega_{2}$}\tilde{\in}\mathop{\rm
span}\\{N_{1,1}^{(q_{2})},\cdots,N_{k_{1},1}^{(q_{2})},N_{1,2}^{(q_{2})},\cdots,N_{k_{2},2}^{(q_{2})}\\}$
in $N_{q_{2}}$. Note that $N_{1,1}^{(q_{2})},\cdots,N_{k_{1},1}^{(q_{2})}$ are
zero matrices, therefore

$\displaystyle\text{\Large$\omega_{2}$}\tilde{\in}\mathop{\rm
span}\\{N_{1,2}^{(q_{2})},\cdots,N_{k_{2},2}^{(q_{2})}\\},$ (71)

and by (65), we have

$\displaystyle Sr(\text{\Large$\omega_{2}$})\leq k_{2}<K.$ (72)

At the same time, using Lemma 3 and (70), we have

$\displaystyle r(\begin{bmatrix}N_{1,2}^{(q_{2})}\\\ \vdots\\\
N_{k_{2},2}^{(q_{2})}\end{bmatrix})+r(\text{\Large$\omega_{2}$})=r_{2}+r(\text{\Large$\omega_{2}$})\leq
r(N_{q_{2}}).$ (73)

Similarly, we continue to find
$N_{q_{s}}=[N_{i,j}^{(q_{s})}]\in\mathbb{M}_{m_{1},n_{1}}\otimes\mathbb{M}_{m_{2},n_{2}}$,
where $3\leq s\leq p$. For each $N_{q_{s}}$, the $k$-th column of
$N_{i,j}^{(q_{s})}$ is exactly the $k$-th column of $N_{i,j}$ in $N_{p}$ in
(III.2), $r_{1}+\cdots+r_{s-1}+1\leq k\leq r_{1}+\cdots+r_{s}$, and the
remaining column vectors of $N_{i,j}^{(q_{s})}$ are zero vectors. Thus by (66)
and (70), we obtain that

$\displaystyle N_{q_{s}}$
$\displaystyle=\begin{bmatrix}\begin{array}[]{c|c|c|c|c}\hbox{\multirowsetup$0_{(s-1)n_{2}}$
}&\hbox{\multirowsetup$0_{\sum_{i=1}^{s-1}r_{i}}\lambda_{r_{s}}0_{n_{2}-\sum_{i=1}^{s}r_{i}}$}&\hbox{\multirowsetup$0_{\sum_{i=1}^{s-1}r_{i}}*_{r_{s}}0_{n_{2}-\sum_{i=1}^{s}r_{i}}$}&\hbox{\multirowsetup$\cdots$}&\hbox{\multirowsetup$0_{\sum_{i=1}^{s-1}r_{i}}*_{r_{s}}0_{n_{2}-\sum_{i=1}^{s}r_{i}}$}\\\
&&&&\\\ &&&&\\\
\cline{2-5}\cr&\hbox{\multirowsetup$0_{n_{2}}$}&\hbox{\multirowsetup$0_{\sum_{i=1}^{s-1}r_{i}}*_{r_{s}}0_{n_{2}-\sum_{i=1}^{s}r_{i}}$}&\hbox{\multirowsetup$\cdots$}&\hbox{\multirowsetup$0_{\sum_{i=1}^{s-1}r_{i}}*_{r_{s}}0_{n_{2}-\sum_{i=1}^{s}r_{i}}$}\\\
&&&&\\\
\cline{2-5}\cr&\hbox{\multirowsetup$0_{n_{2}}$}&\hbox{\multirowsetup$0_{n_{2}}$}&\hbox{\multirowsetup$\cdots$}&\hbox{\multirowsetup$0_{n_{2}}$}\\\
&&&&\end{array}\end{bmatrix}$
$\displaystyle:=\begin{bmatrix}\begin{array}[]{c|c|ccc}\hbox{\multirowsetup$0_{(s-1)n_{2}}$}&N_{1,s}^{(q_{s})}&N_{1,s+1}^{(q_{s})}&\cdots&N_{1,n_{1}}^{(q_{s})}\\\
&\vdots&\vdots&\vdots&\vdots\\\
&N_{k_{s},s}^{(q_{s})}&N_{k_{s},s+1}^{(q_{s})}&\cdots&N_{k_{s},n_{1}}^{(q_{s})}\\\
\cline{2-5}\cr&\hbox{\multirowsetup{\large$0_{n_{2}}$}}&\lx@intercol\hfil\hbox{\multirowsetup{\huge$\omega$}${}_{s}$}\hfil\lx@intercol\\\
&&&&\\\
\cline{2-5}\cr&\lx@intercol\hfil\hbox{\multirowsetup$0_{(n_{1}-s+1)n_{2}}$}\hfil\lx@intercol\\\
&\lx@intercol\hfil\hfil\lx@intercol\\\
&\lx@intercol\hfil\hfil\lx@intercol\end{array}\end{bmatrix},$ (74)

where $\omega_{s}$ is a $(k_{s-1}-k_{s})\times(n_{1}-s)$ rectangular block
matrix, and

$\displaystyle\text{\Large$\omega_{s}$}\tilde{\in}\mathop{\rm
span}\\{N_{1,s}^{(q_{s})},\cdots,N_{k_{s},s}^{(q_{s})}\\},$ (75)
$\displaystyle r(\begin{bmatrix}N_{1,s}^{(q_{s})}\\\ \vdots\\\
N_{k_{s},s}^{(q_{s})}\end{bmatrix})+r(\text{\Large$\omega_{s}$})=r_{s}+r(\text{\Large$\omega_{s}$})\leq
r(N_{q_{s}})$ (76)

hold for any $1\leq s\leq p$. Further, from (65) and (75), we have

$\displaystyle Sr({\Large\text{\Large$\omega_{s}$}})\leq k_{s}<K$ (77)

holds for any $1\leq s\leq p$. On the other hand, from the construction of
$N_{q_{s}}$, where $1\leq s\leq p$, one can obtain that

$\displaystyle N_{p}=N_{q_{1}}+N_{q_{2}}+\cdots+N_{q_{p}},$ (78)

and by Lemma 3,

$\displaystyle r(N_{p})\geq r(N_{q_{s}})$ (79)

holds for any $1\leq s\leq p$.

By (78), we have decomposed $N_{p}$ in (III.2) into the sum of
$N_{q_{1}},\cdots,N_{q_{s}}$. Note that if $r_{1}+\cdots+r_{i}=n_{2}$ for
$1\leq i<p$ in $N_{p}$, then $N_{q_{i+1}},\cdots,N_{q_{p}}$ disappear. We next
consider the partial transpose of system $B$ of each $N_{q_{s}}$. For $s=1$,
from (66), we have

$\displaystyle N_{q_{1}}^{\Gamma_{B}}=$
$\displaystyle\begin{bmatrix}\begin{array}[]{c|ccc}{N_{1,1}^{(q_{1})}}^{T}&{N_{1,2}^{(q_{1})}}^{T}&\cdots&{N_{1,n_{1}}^{(q_{1})}}^{T}\\\
\vdots&\vdots&\ddots&\vdots\\\
{N_{k_{1},1}^{(q_{1})}}^{T}&{N_{k_{1},2}^{(q_{1})}}^{T}&\cdots&{N_{k_{1},n_{1}}^{(q_{1})}}^{T}\\\
\hline\cr\hbox{\multirowsetup{\large$0_{n_{2}}^{\Gamma_{B}}$}}&\lx@intercol\hfil\hbox{\multirowsetup{\huge$\omega$}${}_{1}$${}^{\Gamma_{B}}$}\hfil\lx@intercol\\\
&&\end{array}\end{bmatrix}=\begin{bmatrix}\begin{array}[]{cccc}*_{r_{1}}^{T}&*_{r_{1}}^{T}&\hbox{\multirowsetup$\cdots$}&*_{r_{1}}^{T}\\\
0_{n_{2}-r_{1}}^{T}&0_{n_{2}-r_{1}}^{T}&&0_{n_{2}-r_{1}}^{T}\\\
\hline\cr\vdots&\vdots&\ddots&\vdots\\\
\hline\cr*_{r_{1}}^{T}&*_{r_{1}}^{T}&\hbox{\multirowsetup$\cdots$}&*_{r_{1}}^{T}\\\
0_{n_{2}-r_{1}}^{T}&0_{n_{2}-r_{1}}^{T}&&0_{n_{2}-r_{1}}^{T}\\\
\hline\cr\lx@intercol\hfil\hbox{\multirowsetup{\large$0_{n_{2}}^{\Gamma_{B}}$}}\hfil\lx@intercol\vrule\lx@intercol&\lx@intercol\hfil\hbox{\multirowsetup{\huge$\omega$}${}_{1}$${}^{\Gamma_{B}}$}\hfil\lx@intercol\\\
\lx@intercol\hfil\hfil\lx<EMAIL_ADDRESS>(80)

Using Lemma 3 and (80), we have

$\displaystyle r(N_{q_{1}}^{\Gamma_{B}})\leq k_{1}\cdot
r_{1}+r(\text{\Large$\omega_{1}$}^{\Gamma_{B}}).$ (81)

Similar to $N_{q_{1}}$, from (74), for any $1\leq s\leq p$, one can show that

$\displaystyle N_{q_{s}}^{\Gamma_{B}}$
$\displaystyle=\begin{bmatrix}\begin{array}[]{c|c|ccc}\hbox{\multirowsetup{\large$0_{(s-1)n_{2}}^{\Gamma_{B}}$}}&{N_{1,s}^{(q_{s})}}^{T}&{N_{1,s+1}^{(q_{s})}}^{T}&\cdots&{N_{1,n_{1}}^{(q_{s})}}^{T}\\\
&\vdots&\vdots&\ddots&\vdots\\\
&{N_{k_{s},s}^{(q_{s})}}^{T}&{N_{k_{s},s+1}^{(q_{s})}}^{T}&\cdots&{N_{k_{s},n_{1}}^{(q_{s})}}^{T}\\\
\cline{2-5}\cr&\hbox{\multirowsetup$0_{n_{2}}^{\Gamma_{B}}$}&\lx@intercol\hfil\hbox{\multirowsetup{\huge$\omega$}${}_{s}$${}^{\Gamma_{B}}$}\hfil\lx@intercol\\\
&&&&\\\
\cline{2-5}\cr&\lx@intercol\hfil\hbox{\multirowsetup$0_{(n_{1}-s_{1}+1)n_{2}}^{\Gamma_{B}}$}\hfil\lx@intercol\\\
&\lx@intercol\hfil\hfil\lx@intercol\\\
&\lx@intercol\hfil\hfil\lx@intercol\end{array}\end{bmatrix}$ (82)
$\displaystyle=\begin{bmatrix}\begin{array}[]{c|cccc}\hbox{\multirowsetup$0_{n_{2}}^{T}$}&0_{\sum_{i=1}^{s-1}r_{i}}^{T}&0_{\sum_{i=1}^{s-1}r_{i}}^{T}&\hbox{\multirowsetup$\cdots$}&0_{\sum_{i=1}^{s-1}r_{i}}^{T}\\\
&\hbox{\multirowsetup$*_{r_{s}}^{T}$}&\hbox{\multirowsetup$*_{r_{s}}^{T}$}&&\hbox{\multirowsetup$*_{r_{s}}^{T}$}\\\
&&&&\\\
&0_{n_{2}-\sum_{i=1}^{s}r_{i}}^{T}&0_{n_{2}-\sum_{i=1}^{s}r_{i}}^{T}&&0_{n_{2}-\sum_{i=1}^{s}r_{i}}^{T}\\\
\hline\cr\vdots&\vdots&\vdots&\ddots&\vdots\\\
\hline\cr\hbox{\multirowsetup$0_{n_{2}}^{T}$}&0_{\sum_{i=1}^{s-1}r_{i}}^{T}&0_{\sum_{i=1}^{s-1}r_{i}}^{T}&\hbox{\multirowsetup$\cdots$}&0_{\sum_{i=1}^{s-1}r_{i}}^{T}\\\
&\hbox{\multirowsetup$*_{r_{s}}^{T}$}&\hbox{\multirowsetup$*_{r_{s}}^{T}$}&&\hbox{\multirowsetup$*_{r_{s}}^{T}$}\\\
&&&&\\\
&0_{n_{2}-\sum_{i=1}^{s}r_{i}}^{T}&0_{n_{2}-\sum_{i=1}^{s}r_{i}}^{T}&&0_{n_{2}-\sum_{i=1}^{s}r_{i}}^{T}\\\
\hline\cr\hbox{\multirowsetup$0_{(s-1)n_{2}}^{\Gamma_{B}}$}&\lx@intercol\hfil\hbox{\multirowsetup$0_{n_{2}}^{\Gamma_{B}}$}\hfil\lx@intercol\vrule\lx@intercol&\lx@intercol\hfil\hbox{\multirowsetup{\huge$\omega$}${}_{s}$${}^{\Gamma_{B}}$}\hfil\lx@intercol\\\
&\lx@intercol\hfil\hfil\lx@intercol\vrule\lx@intercol&&&\\\
\cline{2-5}\cr&\lx@intercol\hfil\hbox{\multirowsetup$0_{(n_{1}-s_{1}+1)n_{2}}^{\Gamma_{B}}$}\hfil\lx@intercol\\\
&\lx@intercol\hfil\hfil\lx@intercol\\\
&\lx@intercol\hfil\hfil\lx@intercol\end{array}\end{bmatrix},$

and hence

$\displaystyle r(N_{q_{s}}^{\Gamma_{B}})\leq k_{s}\cdot
r_{s}+r(\text{\Large$\omega_{s}$}^{\Gamma_{B}}).$ (83)

From (77), we obtain that the Schmidt rank of $\omega_{s}$ is at most $K-1$,
where $\omega_{s}$ is from (74). Recalling the assumption of induction that
Conjecture 1 holds for any matrix of Schmidt rank at most $K-1$, hence for any
$1\leq s\leq p$, we have

$\displaystyle r(\text{\Large$\omega_{s}$}^{\Gamma_{B}})\leq
Sr(\text{\Large$\omega_{s}$})\cdot r(\text{\Large$\omega_{s}$}).$ (84)

Further, from (76), (77), (83) and (84) we obtain that

$\displaystyle r(N_{q_{s}}^{\Gamma_{B}})$ $\displaystyle\leq k_{s}\cdot
r_{s}+Sr(\text{\Large$\omega_{s}$})\cdot r(\text{\Large$\omega_{s}$})$ (85)
$\displaystyle\leq k_{s}\cdot(r_{s}+r(\text{\Large$\omega_{s}$}))$
$\displaystyle\leq k_{s}\cdot r(N_{q_{s}})$

holds for any $1\leq s\leq p$. On the other hand, from (78), we have

$\displaystyle
N_{p}^{\Gamma_{B}}=N_{q_{1}}^{\Gamma_{B}}+N_{q_{2}}^{\Gamma_{B}}+\cdots+N_{q_{p}}^{\Gamma_{B}},$
(86)

and by Lemma 3,

$\displaystyle r(N_{p}^{\Gamma_{B}})\leq
r(N_{q_{1}}^{\Gamma_{B}})+r(N_{q_{2}}^{\Gamma_{B}})+\cdots+r(N_{q_{p}}^{\Gamma_{B}}).$
(87)

From (37), (79), (85) and (87), we have

$\displaystyle r(N_{p}^{\Gamma_{B}})$ $\displaystyle\leq k_{1}\cdot
r(N_{q_{1}})+k_{2}\cdot r(N_{q_{2}})+\cdots+k_{p}\cdot r(N_{q_{p}})$ (88)
$\displaystyle\leq k_{1}\cdot r(N_{p})+k_{2}\cdot r(N_{p})+\cdots+k_{p}\cdot
r(N_{p})$ $\displaystyle=(k_{1}+k_{2}+\cdots+k_{p})\cdot r(N_{p})$
$\displaystyle=K\cdot r(N_{p}).$

This implies Conjecture 1 holds for $N_{p}$ in (III.2), and for $M$ in (34) up
to local equivalence.

We have shown by induction that Conjecture 1 holds. Hence we have finished the
proof.  $\sqcap$$\sqcup$

Example To illustrate our proof, we present a boundary case as an example.
Suppose $M$ in (34) has $m_{1}\cdot n_{1}$ linearly independent blocks. In
this case, we have $p=n_{1}$, $m_{1}=k_{1}=k_{2}=\cdots=k_{p}$ and
$A_{1},\cdots,A_{p}$ disappear in $N$ in (35). Further, $\omega_{s}$
disappears in $N_{q_{s}}$ in (74), $1\leq s\leq p$. Therefore from (76) and
(82) we have $r(N_{q_{s}})^{\Gamma_{B}}\leq k_{s}\cdot r_{s}\leq k_{s}\cdot
r(N_{q_{s}})$ for any $1\leq s\leq p$. This implies inequality (85) holds, and
further (88) holds. Hence $M$ satisfies Conjecture 1 up to local equivalence.

##  IV Application

In this section, we apply Theorem 9. Firstly, we know that there are three
inequalities for a tripartite state $\rho_{ABC}$ in terms of the 0-entropy
$S_{0}(A)$ of state $\rho_{A}$ chl14 . Using Theorem 9 the inequalities become

$\displaystyle S_{0}(A)+S_{0}(B)\geq S_{0}(AB),$ (89) $\displaystyle
S_{0}(AB)+S_{0}(AC)+S_{0}(BC)\geq 2S_{0}(A),$ (90) $\displaystyle
S_{0}(AB)+S_{0}(AC)\geq S_{0}(BC).$ (91)

Note that we have omitted the inequality

$\displaystyle S_{0}(AB)+S_{0}(AC)\geq S_{0}(A),$ (92)

obtained in chl14 , which is a corollary of (90) and (91). So we have
established a complete picture of the four-party linear inequalities in terms
of the $0$-entropy.

Next we also point out that, the inequalities (89)-(91) are necessary
conditions by which three bipartite states $\rho_{AB}$, $\rho_{AC}$ and
$\rho_{BC}$ are from a tripartite state. For example, consider three bipartite
states $\rho_{AB}={1\over d^{2}}I_{d}\otimes I_{d}$,
$\rho_{BC}=\rho_{AC}=|\psi\rangle\\!\langle\psi|$ where
$|\psi\rangle={1\over\sqrt{d}}\sum^{d}_{i=1}|ii\rangle$. One can verify that
they share the same one-party reduced density operators, $r_{AB}=d^{2}$ and
$r_{AC}=r_{BC}=1$, so $r_{AB}>r_{AC}r_{BC}$. Thus, the three bipartite states
are not from any tripartite state by Theorem 9. It shows novel understanding
to the long-standing marginal problem.

Third, we extend the inequality in (89) to the multipartite system as follows.

$\displaystyle\sum^{n}_{j=1}S_{0}(A_{j})\geq S_{0}(A_{1}...A_{n}).$ (93)

We omit the proof as it is similar to that of (89). In the following we extend
(90) and (91) as follows.

###### Lemma 10

Let $A_{n+1}\equiv A_{1}$. Given an $n$-partite mixed state of systems
$A_{1},...,A_{n}$, we have

$\displaystyle\sum^{n}_{j=1}S_{0}(A_{j}A_{j+1})\geq 2S_{0}(A_{1}),$ (94)
$\displaystyle\sum^{n-1}_{j=1}S_{0}(A_{j}A_{j+1})\geq S_{0}(A_{1}A_{n}),$ (95)
$\displaystyle
S_{0}(A_{1}A_{2}A_{3})+S_{0}(A_{1}A_{2}A_{4})+S_{0}(A_{1}A_{3}A_{4})\geq
S_{0}(A_{2}A_{3}A_{4}),$ (96) $\displaystyle
S_{0}(A_{1}A_{2}A_{3}A_{4})+S_{0}(A_{1}A_{2}A_{3}A_{5})+S_{0}(A_{1}A_{2}A_{4}A_{5})+S_{0}(A_{1}A_{3}A_{4}A_{5})\geq
S_{0}(A_{2}A_{3}A_{4}A_{5}),$ (97)
$\displaystyle\sum^{n}_{k=1}S_{0}(A_{1}\cdots A_{k-1}A_{k+1}\cdots A_{n})\geq
S_{0}(A_{1}).$ (98)

###### Proof.

Firstly we prove (94). Using Theorem 9 we have
$S_{0}(A_{1}A_{2})+S_{0}(A_{2}A_{3})\geq S_{0}(A_{1}A_{3})$. Using the same
idea one can show that

$\displaystyle\sum^{n}_{j=1}S_{0}(A_{j}A_{j+1})$ $\displaystyle\geq$
$\displaystyle S_{0}(A_{1}A_{n-1})+S_{0}(A_{n-1}A_{n})+S_{0}(A_{n}A_{1})\geq
2S_{0}(A_{1}),$ (99)

where the last inequality follows from (90). We have proven (94).

Next we prove (95). We apply the induction to $n$. If $n=2$ then the assertion
holds. Suppose it holds for $n-1$. We have

$\displaystyle\sum^{n-1}_{j=1}S_{0}(A_{j}A_{j+1})=$
$\displaystyle\sum^{n-2}_{j=1}S_{0}(A_{j}A_{j+1})+S_{0}(A_{n-1}A_{n})$
$\displaystyle\geq$ $\displaystyle S_{0}(A_{1}A_{n-1})+S_{0}(A_{n-1}A_{n})$
$\displaystyle\geq$ $\displaystyle S_{0}(A_{1}A_{n}),$ (100)

where the first inequality follows from the induction hypothesis on $n-1$, and
the last inequality follows from Theorem 9. So the assertion holds for $n$.
The induction implies that (95) holds.

Further we prove (96). We have

$\displaystyle
S_{0}(A_{1}A_{2}A_{3})+S_{0}(A_{1}A_{2}A_{4})+S_{0}(A_{1}A_{3}A_{4})$
$\displaystyle\geq$ $\displaystyle S_{0}(A_{1}A_{2})+S_{0}(A_{1}A_{3}A_{4})$
$\displaystyle\geq$ $\displaystyle S_{0}(A_{2}A_{3}A_{4}).$ (101)

Here the first inequality follows from (92), and the second inequality follows
from Theorem 9.

Similarly we prove (10). We can obtain

$\displaystyle
S_{0}(A_{1}A_{2}A_{3}A_{4})+S_{0}(A_{1}A_{2}A_{3}A_{5})+S_{0}(A_{1}A_{2}A_{4}A_{5})+S_{0}(A_{1}A_{3}A_{4}A_{5})$
$\displaystyle\geq$ $\displaystyle S_{0}(A_{4}A_{5})+S_{0}(A_{2}A_{3})$
$\displaystyle\geq$ $\displaystyle S_{0}(A_{2}A_{3}A_{4}A_{5}),$ (102)

where the first inequality follows from (91) and the second inequality follows
from (89).

Next we prove (98). One can show that

$\displaystyle 2\sum^{n}_{k=1}S_{0}(A_{1}\cdots A_{k-1}A_{k+1}\cdots A_{n})$
$\displaystyle\geq$ $\displaystyle\sum^{n}_{j=1}S_{0}(A_{j}A_{j+1})$
$\displaystyle\geq$ $\displaystyle 2S_{0}(A_{1}).$ (103)

Here the first inequality follows from (91), and the second inequality follows
from (94). So we have $\sum^{n}_{k=1}S_{0}(A_{1}\cdots A_{k-1}A_{k+1}\cdots
A_{n})\geq S_{0}(A_{1})$. This completes the proof.  $\sqcap$$\sqcup$

We have constructed a few novel inequalities every multipartite state satisfy.
It shed new light to the marginal problem for multipartite system, as well as
the understanding of von Neumann entropy.

Fourth, we investigate the condition when the inequality in (1) is saturated.
That is, if $r(\rho_{AB})\cdot r(\rho_{AC})\geq r(\rho_{BC})$ then what is the
condition by which $r(\rho_{AB})\cdot r(\rho_{AC})=r(\rho_{BC})$? We partially
answer the problem as follows.

###### Lemma 11

(i) Suppose $\rho_{ABC}$ is a tripartite pure state. Then the condition is
$r(\rho_{B})\cdot r(\rho_{C})=r(\rho_{A})$.

(ii) Suppose $\rho_{ABC}=\rho_{A}\otimes\rho_{BC}$ is a tripartite mixed
state. Then the condition is $r(\rho_{A})=1$ and
$r(\rho_{BC})=r(\rho_{B})\cdot r(\rho_{C})$.

(iii) Suppose $\rho_{ABC}=\rho_{B}\otimes\rho_{AC}$ is a tripartite mixed
state. Then the condition is $r(\rho_{A})\cdot r(\rho_{AC})=r(\rho_{C})$.

(iv) Suppose $\rho_{ABC}=\rho_{C}\otimes\rho_{AB}$ is a tripartite mixed
state. Then the condition is $r(\rho_{A})\cdot r(\rho_{AB})=r(\rho_{B})$.

(v) Suppose $\rho_{ABC}$ is a tripartite PPT state. Then the condition is
$r(\rho_{AB})=r(\rho_{B})$, $r(\rho_{AC})=r(\rho_{C})$ and $r(\rho_{B})\cdot
r(\rho_{C})=r(\rho_{BC})$.

###### Proof.

(i) The assertion follows from the definition of tripartite pure states.

(ii) We have $r(\rho_{A})^{2}\cdot r(\rho_{B})\cdot
r(\rho_{C})=r(\rho_{AB})\cdot r(\rho_{AC})=r(\rho_{BC})\leq r(\rho_{B})\cdot
r(\rho_{C})$. So assertion (i) holds.

(iii) The assertion can be proven straightforwardly.

(iv) The assertion can be proven using the idea of the proof of (ii).

(v) It is known that the rank of a bipartite PPT state is lower bounded by
that of any one of its reduced density operators. Hence (89) implies

$\displaystyle r(\rho_{AB})\cdot r(\rho_{AC})\geq r(\rho_{B})\cdot
r(\rho_{C})\geq r(\rho_{BC}).$ (104)

If $r(\rho_{AB})\cdot r(\rho_{AC})=r(\rho_{BC})$ then we obtain assertion (v).
$\sqcap$$\sqcup$

We point out that the states $\rho_{ABC}$ satisfying the conditions of this
lemma exist, as we show by the following examples. In (i) we assume
$\rho_{ABC}=|0,0,0\rangle\\!\langle 0,0,0|$. Actually the example applies to
all of the five cases in Lemma 11, and we shall show more non-trivial
examples. In (ii) we assume $\rho_{ABC}=|0\rangle\\!\langle
0|_{A}\otimes|\beta\rangle\\!\langle\beta|_{B}\otimes|\gamma\rangle\\!\langle\gamma|_{C}$
where $\beta$ and $\gamma$ are arbitrary states. In (iii), we assume that
$\rho_{AC}$ is a pure state. In (iv) we assume that $\rho_{AB}$ is a pure
state. In (iv), we assume that $\rho_{ABC}=|0\rangle\\!\langle
0|_{A}\otimes|\beta\rangle\\!\langle\beta|_{B}\otimes|\gamma\rangle\\!\langle\gamma|_{C}$
where $\beta$ and $\gamma$ are arbitrary states.

##  V Conclusions

We have proven the inequality $r(\rho_{AB})\cdot r(\rho_{AC})\geq
r(\rho_{BC})$ for any tripartite state $\rho_{ABC}$ by proving an equivalent
conjecture as well as the construction of a novel canonical form of bipartite
matrices. So we have a complete picture of the four-party linear inequalities
in terms of the $0$-entropy. We also have applied our results to the marginal
problem, extended the inequality to the scenario of multipartite systems, and
discussed the condition when the inequality in (1) is saturated.

We believe that the canonical form in Theorem 8 might be applied to more
quantum-information problems concerning bipartite systems. Besides some open
problems from this paper are as follows.

  1. 1.

Whether the lower bounds in the inequalities (93)-(96) are tight enough is
unknown.

  2. 2.

Can we extend the inequality (1) to multipartite system, apart from (93)-(96)?

  3. 3.

Although we have provided some conditions by which the inequality in (1) is
saturated in Lemma 11, a general condition is still missing.

## Acknowledgments

Authors were supported by the NNSF of China (Grant No. 11871089), and the
Fundamental Research Funds for the Central Universities (Grant Nos.
ZG216S2110).

## References

  * [1] N. Linden and A. Winter.  A new inequality for the von neumann entropy.  Communications in Mathematical Physics, 259(1):129–138, 2005.
  * [2] J. Cadney.  Infinitely many constrained inequalities for the von neumann entropy.  IEEE Transactions on Information Theory, 58(6):3657–3663, 2012\.
  * [3] Lin Chen, Eric Chitambar, Runyao Duan, Zhengfeng Ji, and Andreas Winter.  Tensor rank and stochastic entanglement catalysis for multipartite pure states.  Physical Review Letters, 105:200501, 2010.
  * [4] Lin Chen and Masahito Hayashi.  Multicopy and stochastic transformation of multipartite pure states.  Physical Review A, 83(2):4795–4804, 2011.
  * [5] Lin Chen and Masahito Hayashi.  Nondistillable entanglement guarantees distillable entanglement.  International Journal of Modern Physics B, 26(27n28):1243008–, 2012\.
  * [6] Masahito Hayashi and Lin Chen.  Weaker entanglement between two parties guarantees stronger entanglement with a third party.  Physical Review A, 84(1):012325.
  * [7] Josh Cadney, Marcus Huber, Noah Linden, and Andreas Winter.  Inequalities for the ranks of multipartite quantum states.  Linear Algebra and its Applications, 452(0):153 – 171, 2014.
  * [8] Yize Sun and Lin Chen.  The distillability of entanglement of bipartite reduced density matrices of a tripartite state.  Journal of Physics A: Mathematical and Theoretical, 53(27):275304 (19pp), 2020.
  * [9] John F. Clauser, Michael A. Horne, Abner Shimony, Holt, and Richard A.  Proposed experiment to test local hidden-variable theories.  Physical Review Letters, 23(15):880–884, 1971.
  * [10] P. Kurzyński, T. Paterek, R. Ramanathan, W. Laskowski, and D. Kaszlikowski.  Correlation complementarity yields bell monogamy relations.  Physical Review Letters, 106(18):180402, 2011.
  * [11] Marcin Pawlowski and Caslav Brukner.  Monogamy of bell’s inequality violations in nonsignaling theories.  Physical Review Letters, 102(3):030403, 2009.
  * [12] Representations of a group of matrices of the second order with elements from a locally compact field, and special functions on locally compact fields.  Russian Mathematical Surveys, 18(4):29–100.
  * [13] Paul Butterley, Anthony Sudbery, and Jason Szulc.  Compatibility of subsystem states.  Foundations of Physics, 36(1):83–101, 2006.
  * [14] Valerie Coffman, Joydip Kundu, and William K. Wootters.  Distributed entanglement.  Physical Review A, 61(5):052306, 1999.
  * [15] A. Higuchi, A. Sudbery, and J. Szulc.  One-qubit reduced states of a pure many-qubit state: Polygon inequalities.  Physical Review Letters, 90(10):107902.
  * [16] M. Horodecki, P. Horodecki, and R. Horodecki.  Separability of mixed states: necessary and sufficient conditions.  Physics Letters A, 223:1–8, February 1996.
  * [17] P. Horodecki.  Separability criterion and inseparable mixed states with positive partial transposition.  Phys. Lett. A, 232:333, 1997.
  * [18] Lin Chen and Dragomirz Dokovic.  Equivalence classes and canonical forms for two-qutrit entangled states of rank four having positive partial transpose.  Journal of Mathematical Physics, 53(10):805–813, 2012.
  * [19] Lukasz Skowronek.  Three-by-three bound entanglement with general unextendible product bases.  Journal of Mathematical Physics, 52(12):722–725, 2011.
  * [20] Lin Chen and Yi Xin Chen.  Rank three bipartite entangled states are distillable.  Physical Review A, 78(2):3674–3690, 2008.
  * [21] Lin Chen and Dragomir Z Djokovic.  Distillability and ppt entanglement of low-rank quantum states.  Journal of Physics A Mathematical and Theoretical, 44(28):1213–1219, 2012.
  * [22] Chen Lin and Dragomir Z Djokovic.  Non-positive-partial-transpose quantum states of rank four are distillable.  2016\.
  * [23] M. Horodecki and P. Horodecki.  Reduction criterion of separability and limits for a class of distillation protocols.  Phys. Rev. A, 59:4206, 1999.
  * [24] David P. Divincenzo, Peter W. Shor, John A. Smolin, Barbara M. Terhal, and Ashish V. Thapliyal.  Evidence for bound entangled states with negative partial transpose.  Physical Review A, 61(6):200–200, 2000.
  * [25] L. Pankowski, M. Piani, M. Horodecki, and P. Horodecki.  A few steps more towards npt bound entanglement.  IEEE Transactions on Information Theory, 56(8):4085–4100, 2010\.
  * [26] Lilong Qian, Lin Chen, Delin Chu, and Yi Shen.  A matrix inequality for entanglement distillation problem.  Linear Algebra and its Applications, 616:139–177, 2021.
  * [27] P. Horodecki, U. Rudnicki, and K. Zyczkowski.  Five open problems in quantum information.  2020\.
  * [28] Jens Eisert and Hans J. Briegel.  The schmidt measure as a tool for quantifying multi-particle entanglement.  Physical Review A, 64(2):17–18, 2000.