On the Upper Bounds for Permanents
Ahmet Ali ÖÇAL1Abstract: In this paper, considering and λ operator norms, we obtained some upper bounds for permanents.
2 1
,
λ
λ
∞Key Words: λ1,λ2 and λ∞ operator norms, permanent
Permanentlerin Üst Sınırları Üzerine
Özet: Bu çalışmada, ve operatör normları gözönüne alınarak permanentler için bazı üst sınırlar elde edilmiştir.
2 1,λ
λ λ∞
Anahtar Kelimeler: λ1,λ2 ve λ∞ operatör normları, permanent
Introduction and the Statemens of Results
Definition 1. [1] The permanent of a real n×n matrix A=(aij) is defined by
∑
∏
∈ σ = σ = n S n 1 i ) i ( i a ) A ( per ,where S is the symmetric group of order n. n
Definition 2. ([2]) The λ1 operator norm of an n×n matrix A=(aij)∈Cn×n is defined
{
Ax :x , x 1}
max
A n 1
1
1= ∈C = ,
where T, (T denoting the transpoze) and
n 2 1,x , ,x ) x ( x= Κ
∑
= = n 1 i i 1 x x .Definition 3. ([2]) The λ2 operator norm of an n×n matrix A=(aij)∈Cn×n is defined
{
Ax :x , x 1}
max A n 2 2 2 = ∈C = ,where T and n 2 1,x , ,x ) x ( x= Κ 2 1 n 1 i 2 i 2 x x =
∑
= .Definition 4. ([2]) The λ∞ operator norm of an n×n matrix A=(aij)∈Cn×n is defined
{
Ax :x , x 1}
max A = ∈ n = ∞ ∞ ∞ C , where T and n 2 1,x , ,x ) x ( x= Κ i n i 1 x max x ≤ ≤ ∞ = .Lemma 1. Let a1,a2,Κ,an be the columns of A=(aij)∈Cn×n. Then
( )
1 2 2 2 n 2 n a a a n ) A ( per ≤ Κ , where n j , a a n i ij j ≤ ≤ =∑
= 1 2 1 1 2 2 .Proof. We make use of the inequality (see e.g. [1, p.113])
∏
= ≤ n 1 i i c ) A ( perwhere c1,c2,Κ,cn are column sums of A and A=(aij)n×n is a nonnegative matrix. Since
) A ( per ) A ( per ≤
by the triangle inequality, any such bound can be used to produce an upper bound for the permanents of complex matrices. For example from the inequality (1), we obtain
∏
= ≤ n 1 j j q ) A ( per , (2) where n , , 2 , 1 j , a q n 1 i ij j =∑
= Κ = . By the Cauchy-Schwarz Inequality, we have. n a n a a a q j n i ij n i n i ij n i ij j 2 2 1 1 2 2 1 1 2 1 1 2 1 1 = = ≤ =
∑
∑
∑
∑
= = = =So from inequality (2) we obtain
( )
1 2 2 2 n 2 n a a a n ) A ( per ≤ Κand the proof is complete.
Theorem 1. Let
{
Ax :x , x 1}
max A n 2 2 2 = ∈C =be λ2 operator norm of A∈Cn×n. Then
n 2 2 n A n ) A ( per ≤ .
Proof. Denote the columns of A by a1,a2,Κ,an and let e1,e2,Κ,en be the standart basis of Cn. Then we have
aj=Aej, 1≤j≤n. (3) So considering Lemma 1 we have
2 n 2 2 2 1 2 n a a a n ) A ( per ≤ Κ n n n x n n j n j n n j n j n
A
n
Ax
n
Ae
n
a
n
2 2 2 1 2 2 1 2 2 1 2 2max
max
max
=
≤
=
≤
= ≤ ≤ ≤ ≤and thus the theorem is proved.
∏
= ≤ n 1 j 1 j a ) A ( per , where n , , 2 , 1 j , a a n 1 i ij 1 j =∑
= Κ = .Proof. The proof of Lemma is immediately seen from (2). Theorem 2. Let
{
Ax :x , x 1}
max A n 1 1 1= ∈C =be λ1 operator norm of A∈Cn×n. Then
n 1 A ) A ( per ≤ .
Proof. Considering Lemma 2 and the equality (3), we have
per(A) ≤ a1 1 a2 1Κ an 1 1 1 2 1 1 Ae Aen Ae Κ ≤ . A Ax max e A max n n x n j n j 1 1 1 1 1 1 = ≤ = ≤ ≤ ≤
Thus the theorem is proved.
Lemma 3. Let a1,a2,Κ,an be the columns of A=(aij)∈Cn×n. Then
∞ ≤ j 1 j n a a , where n , , 2 , 1 j , a a n 1 i ij 1 j =
∑
= Κ = ,and n j 1 , a max aj ∞ = 1≤i≤n ij ≤ ≤ . Proof. For all j, 1≤j≤n, we have
n 1j 2j nj 1 i ij 1 j a a a a a =
∑
= + + + = Λ . a n a max n j ij n i ∞ ≤ ≤ = ≤ 1Thus the proof is complete.
Theorem 4. Let a1,a2,Κ,an be the columns of A=(aij)∈Cn×n. Then
∏
= ∞ ≤ n 1 j j n a n ) A ( per .Proof. Considering Lemma 2 and Lemma 3 the proof is easily seen. Theorem 5. Let
{
Ax :x , x 1}
max A = ∈ n = ∞ ∞ ∞ Cbe λ∞ operator norm of A .Then
n n A n ) A ( per ≤ ∞ .
Proof. From Theorem 4 and equality (3), we have
∞ ∞ ∞ ≤nn a1 a2 an ) A ( per Κ
, A n Ax max n Ae max n a max n n n n x n n j n j n n j n j n ∞ ∞ = ∞ ≤ ≤ ≤ ≤ = ≤ = ≤ ∞ ∞ 1 1 1
and thus the theorem is proved.
REFERENCES
1. Minc, H., Permanents, In Encyclopedia of Mathematics and Its Applications Vol. 6, Addison-Wesley (1978). 2. Taşcı, D., On a Conjecture by Goldberg and Newman, Linear Algebra and its Appl., 215: 275-277 (1995).
