S. Ü. Fen-Edebiyat Fakültesi Fen Dergisi Sayı 17[2000]51-54,KONYA
Some Inequalities on The Permanents
Dursun TAŞCI 1
Abstract: In this paper we obtained some inequalities about permanents of Hadamard product of matrices
and permanents of sum of matrices.
Key Words: Permanent, Hadamard product, Positive Semidefinite Hermitian matrices
Permanentler Üzerine Bazı Eşitsizlikler
Özet: Bu çalışmada matrislerin Hadamard çarpımının Permanentleri ve matrislerin toplamının Permanentleri
ile ilgili bazı eşitsizlikler elde edildi.
Anahtar Kelimeler: Permanent, Hadamard Çarpımı, Pozitif Yarı tanımlı Hermityen Matrisler.
Introduction and the Main Results
Definition 1.[1] The Permanent of real n×n matrix A=(aij)∈Mn is defined by
∑ ∏
∈ σ = σ = n S n i ) i ( i a ) A ( per 1 , where Sn is the symmetric group of order n.The permanent can thus be thought of as a function whose domain is the set of n×n real matrices and whose range is the set of real numbers.
Definition 2.[2] If A=(aij) and B=(bij) are n×n matrices then their Hadamard product is the n×n matrix C=AοBwhose (i,j) entry is aijbij.
Lemma 1.[2] If A and B positive semidefinite Hermitian matrices then so is AοB. Theorem 1.[1] If A=(aij) is an n×n matrix then for any i, 1≤i≤n,
∑
= = n j ij ijper(A ) a ) A ( per 1 ,where Aij denotes the submatrix obtained from A by deleting rows i and colums j.
1Selcuk University, Department of Mathematics, [42031]Campus/Konya/TURKEY
D. TAŞCI
Theorem 2. Let A∈Mn be positive semidefinite Hermitian matrix and define
⎪ ⎩ ⎪ ⎨ ⎧ = μ otherwise , Hermitian te semidefini positive is A if , ) A ( per ) A ( per ) A ( 11 0 11
where is the (n-1)×(n-1) principal submatrix of A that results from deleting the first row and column of A and denotes n×n matrices. Then
11 A n M ) B ( ) A ( a ) B ( b ) A ( ) B A ( ο ≥μ +μ −μ μ μ 11 11 (1)
Proof. It suffices to prove that
0 11 11 11 11 11 11 11 11 ≥ + − − ο ο ) B ( per ) B ( per ) A ( per ) A ( per a ) B ( per ) B ( per b ) A ( per ) A ( per ) B A ( per ) B A ( per . (2) We have ) ( . b ) B ( per ) B ( per a ) A ( per ) A ( per b a ) B A ( per ) B A ( per ) B ( per ) B ( per ) A ( per ) A ( per a ) B ( per ) B ( per b ) A ( per ) A ( per ) B A ( per ) B A ( per 3 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − + ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − ο ο = + − − ο ο
Now we must show that
0 11 11 11 11 ≥ − ο ο b a ) B A ( per ) B A ( per , (4) 0 11 11 ≥ − a ) A ( per ) A ( per , (5) and 0 11 11 ≥ − b ) B ( per ) B ( per , (6) respectively. Considering Theorem 1 we have
) B A ( per b a ) B A ( per b a ) B A ( per b a ) B A ( per ο = 11 11 11ο 11 + 12 12 12ο 12 +L+ 1n 1n 1nο 1n , (7) where A1j and B1j ,
1
≤
j
≤
n
, denote the submatrices obtained from A and B by deleting row 1 and columns j respectively. Now from (7 ) we write) B A ( per b a ) B A ( per ο ≥ 11 11 11ο 11 or 0 11 11 11 11 ≥ − ο ο b a ) B A ( per ) B A ( per .
Similarly the inequalities (5) and (6) are satisfied . From (4), (5), and (6), the inequality (1) holds and thus the proof is complete.
Theorem 3. If A1,A2,K,An are n×n matrices with nonnegative entries then
SOME INEQUALITIES ON THE PERMANENTS
∑
∑
= = ≥ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ n i i n i i ) A ( per A per 1 1 . (8)Proof. We use induction on n for the proof of Theorem. It is true for n = 2 . Indeed the (i,j) entry of is just , where and are n×n matrices with nonnegative entries. Thus a typical term in the sum defining
2 1 A A + a(ij1)+a(ij2) A1=(a(ij1))
A
2=
(
a
(ij2))
(
A1 A2)
per + is(
)
∏
= σ σ + n i ) ( ) i ( i ) ( ) i ( i a a 1 2 1 . (9) Now if we multiply out the product (9) and throw a way all terms expect∏
= σ n i ) ( ) i ( i a 1 1 and∏
= σ n i ) ( ) i ( i a 1 2 we obtain (remember A1 and A2 have nonnegative entries)(
)
∏
∏
∏
= σ = σ = σ σ + ≥ + n i ) ( ) i ( i n i ) ( ) i ( i n i ) ( ) i ( i ) ( ) i ( i a a a a 1 2 1 1 1 2 1 . (10) If we sum all the inequalities (10) for σ∈Sn we get(
)
∑ ∏
∑ ∏
∑ ∏
∈ σ = σ ∈ σ = σ ∈ σ = σ σ + ≥ + n n n S n i ) ( ) i ( i S n i ) ( ) i ( i S n i ) ( ) i ( i ) ( ) i ( i a a a a 1 2 1 1 1 2 1 , that is,(
A1 A2)
per( )
A1 per( )
A2 per + ≥ + .We assume now that the inequality (8) is true for n-1 and show that assumption implies that (8) holds for n. Now, if
( )
∑
∑
− = − = ≥ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ 1 1 1 1 n i i n i i per A A per then( )
( )
( )
∑
( )
∑
∑
∑
∑
= − = − = − = = = + ≥ + ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ≥ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ + = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ n i i n n i i n n i i n i n i n i i A per A per A per A per A per A A per A per 1 1 1 1 1 1 1 1thus we have proved by induction that the inequality (8) holds for all n. Corollary 1. If A is n×n matrix with nonnegative entries and
2 T A A ) A ( H = + then
(
H(A))
per(A) per ≥ 1−n 2 ,where AT denotes the transpose of A.
D. TAŞCI
Proof. By the Theorem 3 we have
(
)
(
)
( )
(
)
(
)
) A ( per ) A ( per A per ) A ( per A A per A A per ) A ( H per n n T n T n T 1 2 1 2 2 1 2 1 2 1 2 − = = + ≥ + = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + =thus the proof is complete.
Corollary 2. If A and B are n×n matrices with nonnegative entries then
[
per(A+B)]
2≥4per(A)per(B).Proof. Using arithmetic-geometric mean inequality and considering Theorem 3 we have
) B ( per ) A ( per ) B ( per ) A ( per ) B A ( per ≥ ⎥⎦ ⎤ ⎢⎣ ⎡ + ≥ ⎥⎦ ⎤ ⎢⎣ ⎡ + 2 2 2 2
and therefore we write
[
per(A+B)]
2≥4per(A)per(B). We conclude the paper with a theorem.Theorem 4. If A and B are n×n matrices with nonnegative entries and A≥B then ) B A ( per ) B ( per ) A ( per − ≥ − (11) and ) A B ( per ) B A ( per − = − . (12) Proof. By Theorem 3 we write
) B ( per ) B A ( per ) B B A ( per ) A ( per = − + ≥ − +
and it follows that the inequality (11) holds. On the other hand we have
(
(B A))
( ) per(B A) per(B A) per ) B A ( per − = − − = −1n − = −Thus the proof is complete. References
1- Minc, H., Permanents, In Encyclopaedia of Mathematics and Its Applications, Vol. 6, Addison-Wesley, Reading MA (1978)
2- Horn, R.A., Johnson, C:R., Topics in Matrix Analysis, Cambridge University Press. (1991)
