Hindawi Publishing Corporation
Journal of Applied Mathematics
Volume 2013, Article ID 271978, 5 pages
http://dx.doi.org/10.1155/2013/271978
Research Article
Completing a 2 × 2 Block Matrix of Real Quaternions with
a Partial Specified Inverse
Yong Lin1,2 and Qing-Wen Wang1
1
2
Department of Mathematics, Shanghai University, Shanghai 200444, China
School of Mathematics and Statistics, Suzhou University, Suzhou 234000, China
Correspondence should be addressed to Qing-Wen Wang; wqw858@yahoo.com.cn
Received 4 December 2012; Revised 23 February 2013; Accepted 20 March 2013
Academic Editor: K. Sivakumar
Copyright © 2013 Y. Lin and Q.-W. Wang. This is an open access article distributed under the Creative Commons Attribution
License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly
cited.
This paper considers a completion problem of a nonsingular 2 × 2 block matrix over the real quaternion algebra H: Let
𝑚1 , 𝑚2 , 𝑛1 , 𝑛2 be nonnegative integers, 𝑚1 + 𝑚2 = 𝑛1 + 𝑛2 = 𝑛 > 0, and 𝐴 12 ∈ H𝑚1 ×𝑛2 , 𝐴 21 ∈ H𝑚2 ×𝑛1 , 𝐴 22 ∈ H𝑚2 ×𝑛2 , 𝐵11 ∈ H𝑛1 ×𝑚1
be given. We determine necessary and sufficient conditions so that there exists a variant block entry matrix 𝐴 11 ∈ H𝑚1 ×𝑛1 such that
𝐴 12
𝐴 = ( 𝐴𝐴 11
) ∈ H𝑛×𝑛 is nonsingular, and 𝐵11 is the upper left block of a partitioning of 𝐴−1 . The general expression for 𝐴 11 is also
21 𝐴 22
obtained. Finally, a numerical example is presented to verify the theoretical findings.
1. Introduction
The problem of completing a block-partitioned matrix of a
specified type with some of its blocks given has been studied
by many authors. Fiedler and Markham [1] considered the
following completion problem over the real number field R.
Suppose 𝑚1 , 𝑚2 , 𝑛1 , 𝑛2 are nonnegative integers, 𝑚1 + 𝑚2 =
𝑛1 +𝑛2 = 𝑛 > 0, 𝐴 11 ∈ R𝑚1 ×𝑛1 , 𝐴 12 ∈ R𝑚1 ×𝑛2 , 𝐴 21 ∈ R𝑚2 ×𝑛1 ,
and 𝐵22 ∈ R𝑛2 ×𝑚2 . Determine a matrix 𝐴 22 ∈ R𝑚2 ×𝑛2 such
that
𝐴 𝐴
𝐴 = ( 11 12 )
𝐴 21 𝐴 22
(1)
is nonsingular and 𝐵22 is the lower right block of a partitioning of 𝐴−1 . This problem has the form of
−1
𝐴 𝐴
( 11 12 )
𝐴 21 ?
=(
? ?
),
? 𝐵22
(2)
and the solution and the expression for 𝐴 22 were obtained in
[1]. Dai [2] considered this form of completion problems with
symmetric and symmetric positive definite matrices over R.
Some other particular forms for 2 × 2 block matrices over
R have also been examined (see, e.g., [3]), such as
−1
𝐵 ?
= ( 11 ) ,
? ?
−1
? ?
= (
),
? 𝐵22
𝐴 𝐴
( 11 12 )
𝐴 21 ?
𝐴 ?
( 11 )
? ?
−1
?
𝐴
( 11
)
? 𝐴 22
(3)
? 𝐵12
= (
).
𝐵21 ?
The real quaternion matrices play a role in computer
science, quantum physics, and so on (e.g., [4–6]). Quaternion
matrices are receiving much attention as witnessed recently
(e.g., [7–9]). Motivated by the work of [1, 10] and keeping
such applications of quaternion matrices in view, in this paper
we consider the following completion problem over the real
quaternion algebra:
H = {𝑎0 + 𝑎1 𝑖 + 𝑎2 𝑗 + 𝑎3 𝑘 |
𝑖2 = 𝑗2 = 𝑘2 = 𝑖𝑗𝑘 = −1 and 𝑎0 , 𝑎1 , 𝑎2 , 𝑎3 ∈ R} .
(4)
Problem 1. Suppose 𝑚1 , 𝑚2 , 𝑛1 , 𝑛2 are nonnegative integers, 𝑚1 + 𝑚2 = 𝑛1 + 𝑛2 = 𝑛 > 0, and 𝐴 12 ∈ H𝑚1 ×𝑛2 ,
2
Journal of Applied Mathematics
𝐴 21 ∈ H𝑚2 ×𝑛1 , 𝐴 22 ∈ 𝑅𝑚2 ×𝑛2 , 𝐵11 ∈ H𝑛1 ×𝑚1 . Find a matrix
𝐴 11 ∈ H𝑚1 ×𝑛1 such that
𝐴 𝐴
𝐴 = ( 11 12 ) ∈ H𝑛×𝑛
𝐴 21 𝐴 22
−1
𝐵 ?
= ( 11 ) ,
? ?
𝐵 𝐵
( 22 21 ) ,
𝐵12 𝐵11
(5)
is nonsingular, and 𝐵11 is the upper left block of a partitioning
of 𝐴−1 . That is
? 𝐴 12
)
(
𝐴 21 𝐴 22
Proof. It is readily seen that
(6)
where H𝑚×𝑛 denotes the set of all 𝑚 × 𝑛 matrices over H and
𝐴−1 denotes the inverse matrix of 𝐴.
Throughout, over the real quaternion algebra H, we
denote the identity matrix with the appropriate size by 𝐼, the
transpose of 𝐴 by 𝐴𝑇 , the rank of 𝐴 by 𝑟(𝐴), the conjugate
transpose of 𝐴 by 𝐴∗ = (𝐴)𝑇 , a reflexive inverse of a matrix 𝐴
over H by 𝐴+ which satisfies simultaneously 𝐴𝐴+ 𝐴 = 𝐴 and
𝐴+ 𝐴𝐴+ = 𝐴+ . Moreover, 𝐿 𝐴 = 𝐼−𝐴+ 𝐴, 𝑅𝐴 = 𝐼−𝐴𝐴+ , where
𝐴+ is an arbitrary but fixed reflexive inverse of 𝐴. Clearly, 𝐿 𝐴
and 𝑅𝐴 are idempotent, and each is a reflexive inverse of itself.
R(𝐴) denotes the right column space of the matrix 𝐴.
The rest of this paper is organized as follows. In Section 2,
we establish some necessary and sufficient conditions to solve
Problem 1 over H, and the general expression for 𝐴 11 is also
obtained. In Section 3, we present a numerical example to
illustrate the developed theory.
𝐴 𝐴
( 22 21 )
𝐴 12 𝐴 11
are inverse to each other, so we may suppose that 𝑛(𝐴 11 ) <
𝑛(𝐵22 ).
If 𝑛(𝐵22 ) = 0, necessarily 𝑛(𝐴 11 ) = 0 and we are finished.
Let 𝑛(𝐵22 ) = 𝑐 > 0, then there exists a matrix 𝐹 with 𝑐 right
linearly independent columns, such that 𝐵22 𝐹 = 0. Then,
using
Lemma 1 (singular-value decomposition [9]). Let 𝐴 ∈ H𝑚×𝑛
be of rank 𝑟. Then there exist unitary quaternion matrices 𝑈 ∈
H𝑚×𝑚 and 𝑉 ∈ H𝑛×𝑛 such that
(7)
where 𝐷𝑟 = diag(𝑑1 , . . . , 𝑑𝑟 ) and the 𝑑𝑗 ’s are the positive
singular values of 𝐴.
Let H𝑛𝑐 denote the collection of column vectors with 𝑛
components of quaternions and 𝐴 be an 𝑚 × 𝑛 quaternion
matrix. Then the solutions of 𝐴𝑥 = 0 form a subspace of H𝑛𝑐
of dimension 𝑛(𝐴). We have the following lemma.
Lemma 2. Let
(8)
be a partitioning of a nonsingular matrix 𝐴 ∈ H𝑛×𝑛 , and let
𝐵 𝐵
( 11 12 )
𝐵21 𝐵22
(11)
𝐴 11 𝐵12 𝐹 = 0.
(12)
𝐴 21 𝐵12 + 𝐴 22 𝐵22 = 𝐼,
(13)
𝐴 21 𝐵12 𝐹 = 𝐹.
(14)
From
we have
It follows that the rank 𝑟(𝐵12 𝐹) ≥ 𝑐. In view of (12), this
implies
𝑛 (𝐴 11 ) ≥ 𝑟 (𝐵12 𝐹) ≥ 𝑐 = 𝑛 (𝐵22 ) .
(15)
𝑛 (𝐴 11 ) = 𝑛 (𝐵22 ) .
(16)
Thus
In this section, we begin with the following lemmas.
𝐴 𝐴
( 11 12 )
𝐴 21 𝐴 22
𝐴 11 𝐵12 + 𝐴 12 𝐵22 = 0,
we have
2. Main Results
𝐷 0
𝑈𝐴𝑉 = ( 𝑟 ) ,
0 0
(10)
(9)
be the corresponding (i.e., transpose) partitioning of 𝐴−1 . Then
𝑛(𝐴 11 ) = 𝑛(𝐵22 ).
Lemma 3 (see [10]). Let 𝐴 ∈ H𝑚×𝑛 , 𝐵 ∈ H𝑝×𝑞 , 𝐷 ∈ H𝑚×𝑞 be
known and 𝑋 ∈ H𝑛×𝑝 unknown. Then the matrix equation
𝐴𝑋𝐵 = 𝐷
(17)
𝐴𝐴+ 𝐷𝐵+ 𝐵 = 𝐷.
(18)
is consistent if and only if
In that case, the general solution is
𝑋 = 𝐴+ 𝐷𝐵+ + 𝐿 𝐴𝑌1 + 𝑌2 𝑅𝐵 ,
(19)
where 𝑌1 , 𝑌2 are any matrices with compatible dimensions over
H.
By Lemma 1, let the singular value decomposition of the
matrix 𝐴 22 and 𝐵11 in Problem 1 be
𝐴 22 = 𝑄 (
Λ 0 ∗
)𝑅 ,
0 0
(20)
Σ 0 ∗
𝐵11 = 𝑈 (
)𝑉 ,
0 0
(21)
Journal of Applied Mathematics
3
where Λ = diag(𝜆 1 , 𝜆 2 , . . . , 𝜆 𝑠 ) is a positive diagonal matrix,
𝜆 𝑖 ≠ 0 (𝑖 = 1, . . . , 𝑠) are the singular values of 𝐴 22 , 𝑠 =
𝑟(𝐴 22 ), Σ = diag(𝜎1 , 𝜎2 , . . . , 𝜎𝑟 ) is a positive diagonal matrix,
𝜎𝑖 ≠ 0 (𝑖 = 1, . . . , 𝑟) are the singular values of 𝐵11 and 𝑟 =
𝑟(𝐵11 ).
𝑄 = (𝑄1 𝑄2 ) ∈ H𝑚2 ×𝑚2 , 𝑅 = (𝑅1 𝑅2 ) ∈ H𝑛2 ×𝑛2 ,
𝑈 = (𝑈1 𝑈2 ) ∈ H𝑛1 ×𝑛1 , 𝑉 = (𝑉1 𝑉2 ) ∈ H𝑚1 ×𝑚1 are unitary
quaternion matrices, where 𝑄1 ∈ H𝑚2 ×𝑠 , 𝑅1 ∈ H𝑛2 ×𝑠 , 𝑈1 ∈
H𝑛1 ×𝑟 , and 𝑉1 ∈ H𝑚1 ×𝑟 .
It follows that
Σ 0 ∗
Λ 0 ∗
) 𝑉 𝑉 = 𝑄∗ 𝑄 (
) 𝑅 𝐾𝑉.
𝑄∗ 𝐴 21 𝑈 (
0 0
0 0
This implies that
(
∗
(d) R(𝐴∗12 𝐵11
) ⊂ R(𝐴∗22 ).
×
(30)
̂=(
𝐾
Λ−1 𝑄1∗ 𝐴 21 𝑈1 Σ
0
+ 𝐴 12 𝑅 (
−1 )
𝐻
−(𝑉2∗ 𝐴 12 𝑅2 )
(22)
+𝑌−
+
𝑌𝐵11 𝐵11
,
Λ−1 𝑄1∗ 𝐴 21 𝑈1 Σ
0
𝐻
𝐾22
),
(32)
𝐻 ∈ H(𝑛2 −𝑠)×𝑟 , 𝐾22 ∈ H(𝑛2 −𝑠)×(𝑚1 −𝑟) .
In the same way, from (d), we can obtain
where 𝐻 is an arbitrary matrix in H
arbitrary matrix in H𝑚1 ×𝑛1 .
(𝑛2 −𝑠)×𝑟
and 𝑌 is an
Proof. If there exists an 𝑚1 × 𝑛1 matrix 𝐴 11 such that 𝐴 is
nonsingular and 𝐵11 is the corresponding block of 𝐴−1 , then
(a) is satisfied. From 𝐴𝐵 = 𝐵𝐴 = 𝐼, we have that
𝐴 21 𝐵11 + 𝐴 22 𝐵21 = 0,
𝐵11 𝐴 12 + 𝐵12 𝐴 22 = 0,
𝑉1∗ 𝐴 12 𝑅2 = 0.
(33)
12
Notice that ( 𝐴
𝐴 22 ) in (a) is a full column rank matrix. By (20),
(21), and (33), we have
Λ
0
0
0
0 𝑄
𝐴
( ∗
) ( 12 ) 𝑅 = (𝑉∗ 𝐴 𝑅 𝑉∗ 𝐴 𝑅 ) ,
𝑉 0
𝐴 22
12
1
12 2
1
1
∗
∗
𝑉2 𝐴 12 𝑅1 𝑉2 𝐴 12 𝑅2
∗
(23)
so that (c) and (d) are satisfied.
By (11), we have
𝑟 (𝐴 22 ) + 𝑛 (𝐴 22 ) = 𝑛2 ,
(31)
̂ From (29), (30), we have
Let 𝑅∗ 𝐾𝑉 = 𝐾.
In that case, the general solution has the form of
+
𝑉∗ 𝐵11
Σ 0
)(
)
0 0
𝑄2∗ 𝐴 21 𝑈1 = 0.
(c) R(𝐴 21 𝐵11 ) ⊂ R(𝐴 22 ),
𝐴 11 =
𝑄2∗ 𝐴 21 𝑈1 𝑄2∗ 𝐴 21 𝑈2
Comparing corresponding blocks in (30), we obtain
(b) 𝑛2 − 𝑟(𝐴 22 ) = 𝑚1 − 𝑟(𝐵11 ), that is 𝑛2 − 𝑠 = 𝑚1 − 𝑟,
+
𝐵11
𝑄1∗ 𝐴 21 𝑈1 𝑄1∗ 𝐴 21 𝑈2
∗
∗
Λ 0 𝑅1 𝐾𝑉1 𝑅1 𝐾𝑉2
).
=(
)( ∗
0 0 𝑅 𝐾𝑉1 𝑅∗ 𝐾𝑉2
2
2
Theorem 4. Problem 1 has a solution if and only if the
following conditions are satisfied:
12
(a) 𝑟 ( 𝐴
𝐴 22 ) = 𝑛2 ,
(29)
(34)
so that
𝐴
0 𝑄∗ 𝐴 12
) 𝑅)
𝑛2 = 𝑟 ( 12 ) = 𝑟 (( ∗
)(
𝐴 22
𝑉 0
𝐴 22
𝑟 (𝐵11 ) + 𝑛 (𝐵11 ) = 𝑚1 . (24)
11
From Lemma 2, Notice that ( 𝐴
𝐴 21
−1
partitioning of 𝐵 , we have
𝐴 12
𝐴 22
) is the corresponding
𝑛 (𝐵11 ) = 𝑛 (𝐴 22 ) ,
Λ
0
0
0
= 𝑟 (𝑉 ∗ 𝐴 𝑅 𝑉 ∗ 𝐴 𝑅 )
12 1
12 2
1
1
(25)
𝑉2∗ 𝐴 12 𝑅1
implying that (b) is satisfied.
Conversely, from (c), we know that there exists a matrix
𝐾 ∈ H𝑛2 ×𝑚1 such that
𝐴 21 𝐵11 = 𝐴 22 𝐾.
(26)
𝐵21 = −𝐾.
(27)
Let
From (20), (21), and (26), we have
Σ 0 ∗
Λ 0 ∗
𝐴 21 𝑈 (
)𝑉 = 𝑄(
) 𝑅 𝐾.
0 0
0 0
(35)
𝑉2∗ 𝐴 12 𝑅2
= 𝑟 (Λ) + 𝑟 (𝑉2∗ 𝐴 12 𝑅2 )
= 𝑠 + 𝑟 (𝑉2∗ 𝐴 12 𝑅2 ) .
It follows from (b) and (35) that 𝑉2𝑇 𝐴 12 𝑅2 is a full column
rank matrix, so it is nonsingular.
From 𝐴𝐵 = 𝐼, we have the following matrix equation:
𝐴 11 𝐵11 + 𝐴 12 𝐵21 = 𝐼,
(36)
that is
(28)
𝐴 11 𝐵11 = 𝐼 − 𝐴 12 𝐵21 ,
𝐼 ∈ H𝑚1 ×𝑚1 ,
(37)
4
Journal of Applied Mathematics
where 𝐵11 , 𝐴 12 were given, 𝐵21 = −𝐾 (from (27)). By
Lemma 3, the matrix equation (37) has a solution if and only
if
+
𝐵11 = 𝐼 − 𝐴 12 𝐵21 .
(𝐼 − 𝐴 12 𝐵21 ) 𝐵11
(38)
By (21), (27), (32), and (33), we have that (38) is equivalent to:
3. An Example
In this section, we give a numerical example to illustrate the
theoretical results.
Example 5. Consider Problem 1 with the parameter matrices
as follows:
Σ 0 ∗
Σ−1 0
(𝐼 + 𝐴 12 𝐾) 𝑉 (
) 𝑈∗ 𝑈 (
) 𝑉 = 𝐼 + 𝐴 12 𝐾. (39)
0 0
0 0
2+𝑗
𝐴 12 = (
We simplify the equation above. The left hand side reduces to
(𝐼 + 𝐴 12 𝐾)𝑉1 𝑉1∗ and so we have
𝐴 12 𝐾𝑉1 𝑉1∗ − 𝐴 12 𝐾 = 𝐼 − 𝑉1 𝑉1∗ .
(40)
𝐴 21
So,
∗
̂ ∗ 𝑉1 𝑉∗ − 𝐴 12 𝑅𝐾𝑉
̂ ∗ = (𝑉1 𝑉2 ) (𝑉1∗ ) − 𝑉1 𝑉∗ .
𝐴 12 𝑅𝐾𝑉
1
1
𝑉2
(41)
This implies that
∗
∗
̂ (𝑉1∗ 𝑉1 ) 𝑉∗ − 𝐴 12 𝑅𝐾
̂ (𝑉1∗ ) = 𝑉2 𝑉∗ ,
𝐴 12 𝑅𝐾
1
2
𝑉2 𝑉1
𝑉2
3 1
1
1
+ 𝑖 − 𝑗− 𝑘
2 2
2
2
=(
),
1
1
3 1
𝑗+ 𝑘
+ 𝑖
2
2
2 2
2 𝑖
𝐴 22 = (
),
2𝑗 𝑘
(49)
1 𝑖
𝐵11 = (
).
𝑗 𝑘
It is easy to show that (c), (d) are satisfied, and that
(42)
𝐴
𝑛2 = 𝑟 ( 12 ) = 2,
𝐴 22
so that
∗
̂ (𝐼) 𝑉∗ − 𝐴 12 𝑅𝐾
̂ (𝑉1∗ ) = 𝑉2 𝑉∗ .
𝐴 12 𝑅𝐾
1
2
𝑉2
0
−𝑘
1
𝑘
2
),
1
1+ 𝑗
2
(50)
𝑛2 − 𝑟 (𝐴 22 ) = 𝑚1 − 𝑟 (𝐵11 ) = 0,
(43)
so (a), (b) are satisfied too. Therefore, we have
So,
̂ ( 0∗ ) = 𝑉2 𝑉∗ ,
−𝐴 12 𝑅𝐾
2
𝑉2
(44)
+
𝐵11
and hence,
Λ−1 𝑄1∗ 𝐴 21 𝑈1 Σ 0
0
) ( ∗ ) = 𝑉2 𝑉2∗ .
− (𝐴 12 𝑅1 𝐴 12 𝑅2 ) (
𝑉2
𝐻
𝐾22
(45)
1
1
− 𝑗
2
2
=(
),
1
1
− 𝑖 − 𝑘
2
2
Λ 0 ∗
)𝑅 ,
0 0
𝐴 22 = 𝑄 (
(51)
Σ 0 ∗
𝐵11 = 𝑈 (
)𝑉 ,
0 0
where
Finally, we obtain
𝐴 12 𝑅2 𝐾22 𝑉2∗ = −𝑉2 𝑉2∗ .
(46)
∗
Multiplying both sides of (46) by 𝑉 from the left, considering (33) and the fact that 𝑉2∗ 𝐴 12 𝑅2 is nonsingular, we have
𝐾22 =
−1
−(𝑉2∗ 𝐴 12 𝑅2 ) .
𝑄=
1 0
𝑅=(
),
0 1
(47)
√2 0
),
Σ=(
0 √2
From Lemma 3, (38), (47), Problem 1 has a solution and the
general solution is
Λ−1 𝑄1∗ 𝐴 21 𝑈1 Σ
0
+
+ 𝐴 12 𝑅 (
)
𝐴 11 = 𝐵11
−1
∗
𝐻
−(𝑉2 𝐴 12 𝑅2 )
(48)
+
+
+ 𝑌 − 𝑌𝐵11 𝐵11
,
× 𝑉∗ 𝐵11
where 𝐻 is an arbitrary matrix in H(𝑛2 −𝑠)×𝑟 and 𝑌 is an
arbitrary matrix in H𝑚1 ×𝑛1 .
1 1 𝑖
(
),
√2 𝑗 𝑘
Λ=(
𝑈=
2√2 0
),
0 √2
1 1 𝑖
(
),
√2 𝑗 𝑘
(52)
1 0
𝑉=(
).
0 1
We also have
1 1 𝑖
(
),
√2 𝑗 𝑘
1 0
𝑅1 = (
),
0 1
1 1 𝑖
(
),
𝑈1 =
√2 𝑗 𝑘
1 0
𝑉1 = (
).
0 1
𝑄1 =
(53)
Journal of Applied Mathematics
5
By Theorem 4, for an arbitrary matrices 𝑌 ∈ H2×2 , we
have
+
+
+
+ 𝐴 12 𝑅 (Λ−1 𝑄1∗ 𝐴 21 𝑈1 Σ) 𝑉∗ 𝐵11
+ 𝑌 − 𝑌𝐵11 𝐵11
𝐴 11 = 𝐵11
3 1
1
3 1
3
+ 𝑗+ 𝑘
+ 𝑖− 𝑗
2 4
4
4 4
2
=(
),
1
1
1 1 3
1
−𝑖+ 𝑗− 𝑘
− 𝑖− 𝑗−𝑘
2
4
4 4 4
2
(54)
it follows that
𝐴=
3 1
1
3 1
3
1
+ 𝑗+ 𝑘
+ 𝑖− 𝑗
2+𝑗
𝑘
2 4
4
4 4
2
2
1
1
1 3
1
1
1
−𝑖+ 𝑗− 𝑘
− 𝑖 − 𝑗 − 𝑘 −𝑘 1 + 𝑗
(2
4
4
4 4
2
2 )
(
),
1
1
3 1
+ 𝑖
− 𝑗− 𝑘
2
𝑖
2 2
2
2
1
1
3 1
𝑗+ 𝑘
+ 𝑖
2𝑗
𝑘
( 2 2
)
2 2
1
𝑗
𝐴−1
𝑖
𝑘
−1
−1
0
−1
1 3
3
− 𝑗
= (−1 0
).
4
2 4
1
1 1
1
−1 −1
−𝑖
− 𝑖− 𝑗−𝑘
2
2 2
2
(55)
The results verify the theoretical findings of Theorem 4.
Acknowledgments
The authors would like to give many thanks to the referees
and Professor K. C. Sivakumar for their valuable suggestions
and comments, which resulted in a great improvement of
the paper. This research was supported by Grants from the
Key Project of Scientific Research Innovation Foundation of
Shanghai Municipal Education Commission (13ZZ080), the
National Natural Science Foundation of China (11171205), the
Natural Science Foundation of Shanghai (11ZR1412500), the
Discipline Project at the corresponding level of Shanghai (A.
13-0101-12-005), and Shanghai Leading Academic Discipline
Project (J50101).
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