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Bounds for permanents anddeterminantsHenryk MincPublished online: 30 May 2007.

To cite this article: Henryk Minc (1980) Bounds for permanents and determinants,Linear and Multilinear Algebra, 9:1, 5-16

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Linear and Multilinear Algebra, 1980, Vol. 9, pp. 5-16 0308-1087/80/0901-005 $04.5010 sc, 1980 Gordon and Breach Science Publishers, Inc. Printed in Great Britain

Bounds for Permanents and Determinants

H E N R Y K MlNCt Department of Mathematics and Institute for Interdisciplinary Applications of Algebra and Combinatorics, University of California, Santa Barbara

(Receiued September 10,1979)

An inequality of Johnson and Newman for determinants of real matrices is extended to complex matrices. A related inequality for permanents of real matrices is improved by means of a new rearrangement theorem.

1. INTRODUCTION

If A = (a,,) is an n-square matrix with complex entries, then evidently

and also

These bounds are rather crude, particularly in the case of determinants. Inequality (2) has been substantially improved for the permanents of (0-1)- matrices [2, p. 1081, nonnegative matrices [2, p. 1101, and complex matrices [2, p. 1131. Recently Schinzel[3] obtained the following remarkable inequality for the determinants of real matrices :

If the matrix A contains both positive and negative entries, inequality (3) yields a considerably stronger bound than in (1). For nonnegative matrices the two inequalities produce the same bound.

t This work was supported by AFOSR Contract AF-F4962078-C-0030.

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6 H. MINC

Schinzel's formula was further improved by Johnson and Newman [I] : n

Idet(A)I < n max p = 1

a p q > O o p q < O

In this paper I use inequality (4) to obtain bounds for the absolute value of the determinants of complex matrices.

Johnson and Newman [ibid.] also applied their method to permanents of real matrices and obtained the following result: If A is an n x n matrix such that the sum of the positive entries and the absolute value of the sum of the negative entries in each row do not exceed 1, then

(per(A)( < 2'"'. ( 5 )

In Section 2, I obtain a rearrangement inequality (Theorem 1) and use it in Section 3, together with a lemma, analogous to Lemma 2 in [I], to strengthen the bound in (5).

2. REARRANGEMENT INEQUALITIES

Our main result in this section is the following inequality which appears to be new,

THEOREM 1 Let (ail, .. . , a,,), i = 1, .. . , k, be sequences of length ti 2 2, i = 1, ..., k, of nonnegative numbers aij < 1. Let a f , a;, ..., a,*, n = x:=, ti, be the a i j , j = 1, ..., t i , i = 1, ..., k, arranged in non-increasing order: af 2 a: 9 B a,*. Then

k n (1 + .Ti - , a&) i f n is even,

i = 1 n (1 +ai1 . . . cciti) < i = 1

(n- 3 ~ 2 ( 6 ) n (1 + a$ - la:i)(l + a$ , a,*- ,a,*) i f n is odd. r i = l

First we prove a sequence of lemmas. We use the notation in the statement of Theorem 1. The numbers aij are assumed to be nonnegative, but not necessarily bounded above by 1, unless it is so stated.

LEMMA 1 al la12+a21a22 < u:a;+a:af. (7)

Proof We can assume without a loss of generality that a l l = a: and a,, 2 a,,. If a,, = a: there is nothing to prove. If a , , # a f , then dj = a,,

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PERMANENTS AND DETERMINANTS

and u:a:+a:af = a , , ~ , , +al2a,,.

But

( ~ l l ~ 2 1 + ~ 1 2 a 2 2 ) - ( ~ 1 1 ~ 1 , +a,,a,,) = (a11 -a,,)(a21 -a,,) 2 0, and the lemma is proved.

Proof For k = 2 Lemma 2 is equivalent to Lemma 1. In other words, if P I , p z , B,, P,, are any nonegative numbers, then

(1 +BlP2)(1+83P4) G (1 + K P : ) ( l + P : P f ) , where PT, /I:, p:, Pf are the numbers P I , f l , P , , P, arranged in non-increasing order. But the product on the left-hand side of (8) can be transformed into the product on the right-hand side of (8) by a finite sequence of such rearrange- ments. The lemma follows. The inequality can be proved more formally by induction on k.

LEMMA 3 If a l , a,, a,, a,, a , are nonnegative numbers not exceeding 1 then

(1 +a1a2)( l +a,a,a,) G (1 +a:a;)(l +a:ata:). ( 9 )

Proof We can assume that a , 2 a , and a , 2 a, 2 a,. If a , 2 a,, then the two sides in (9) are actually equal. Suppose that a , > a,. With this assump- tion, inequality (9) is implied by

a1a2+a3a4a5 G a la3+a2a4a5 , i.e.,

( ~ ( 3 - ~ 2 ) ~ 4 ~ 5 < ( a 3 - a 2 b 1 . ( l o ) Now, a , - a, > 0 and a, < 1. Hence inequality (10) (and thus (9)) holds if a , B a,. If a , > a,, then a , 2 a, 2 a , > a , 2 a,, and therefore

(1 +a la2 ) ( l +u,a,a,) = (1 +afa: ) ( l +a:a:a:). (11) But

This together with (1 1) completes the proof.

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8 H. MINC

LEMMA 4 If al , a z , a3 , a4, a 5 , a6 are nonnegative numbers not exceeding 1 (or alternatively, if ai 2 1, i = 1 , . . ., 6), then

(1 + ~ 1 ~ 2 a 3 ) ( 1 + a 4 a ~ a 6 ) 6 (1 +a1~2) (1 +a3a4)U + U S % ) (12) The inequality is strict unless at least two of the ai are zero.

Proof If ai ,< 1 for all i, then

where the second inequality is strict unless at least two of the ai are zero.

If ai 2 1 for all i, then

( 1 +alu2a3)(1 +a4a5a6) < 1 +a la2a3a4+a3a4a5a6+a la2a3~4a5a6

< (1 +ala2)(l +a3a4)(1 +asas) .

Proof of Theorem 1 For any nonnegative numbers xl , . . . , x , we clearly have

k I2 n (1 + x z i - l x 2 i ) if k is even, i = 1

1 +xl X k (k- 3,/2 i n (1 +xz i - 1 ~ 2 i ) ( l +xk-Zxk- l ~ k ) if k is odd. i = 1

Hence each factor 1 +ai1 . . . ai,, on the right-hand side of (6) is bounded about by a product of factors of the form 1 possibly multiplied, if ti happens to be odd, by the factor 1 - 2 C L ~ , ~ ; - lai,tl. Thus the product

k n (1 +ail . . . atti) i = 1

is bounded above by a product of factors of the form 1 +ailjlai2j2 and factors of the form 1 + ailjlai2j,ai,j,; each of the numbers aij appearing in exactly one of these factors. By a repeated application of Lemma 4 we construct an upper bound for this product: it is a product of n/2 factors of the form 1 +ailjlai,j2 if n is even, or a product of (n - 3)/2 of such factors multiplied by a factor of the form 1 +ailjlai,j,ai,j, f n is odd. We now apply Lemmas 2 and 3 to complete the proof of inequality (6).

3. BOUNDS FOR PERMANENTS

Let A = (a i j ) be a real n x n matrix. Define

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PERMANENTS AND DETERMINANTS 9

i = 1, ..., n. We shall prove an upper bound for Iper(A)I in terms of the numbers R:, R; , i = 1, . . ., n. We can assume without loss of generality that R: a R;, i = 1, ..., n, and that A has no zero rows.

THEOREM 2 Let A be an n x n real matrix without zero rows, and satisfying R: a R;, i = 1, ..., n. Then

+ R,, - 2&, - , ) R , , ) ) if n is odd7 where cr is a permutation that arranges the numbers R;/R: , i = 1, ..., n, in non-increasing order.

COROLLARY If A is a real n x n matrix and i f the sum of the positive entries and the absolute value of the sum of the negative entries in each row of A do not exceed 1, then

Iper(A)( < 2[""].

The corollary is a slight improvement of the bound in (5) obtained by Johnson and Newman.

We shall require the following lemma.

LEMMA 5 Let A be a real n x n matrix. There exists an n x n matrix K satisfying

Iper(A)J 6 Iper(K)I, and whose entries in the i-th row are R', - R; and n -2 zeros, in some order, i = 1, ..., n.

Proof First we show, by a method analogous to the one used in [I], how to find a matrix K satisfying the conditions in the statement of the lemma in the case per(A) 3 0. Let

per(A(lls)) = max per( 411 j), I

and per(A(1 It) = min pe r (~ ( l1 j)),

1

s # t . Then

< R: per(~(1ls)) - R ; per(A(1 It)) (see [I, Lemma 11). Thus the permanent of A does not exceed the permanent

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10 H. MINC

of the matrix in which the first row of A has been replaced by a vector which has R: in the sth position, - R; in the tth position and zeros elsewhere. Performing similar operations on rows 2,3, . . ., n in order, we obtain K.

Ifper(A) < 0, we cannot use exactly the same method, since per(A) < per(K) does not imply Iper(A)I d Iper(~)I, in general. Let D = diag(- 1, 1, . . ., 1). By the method outlined above we can find a matrix H satisfying

and whose first row contains - R : , R, and n-2 zeros, and the ith row, i = 2, . . . , n, contains R', - R; and n - 2 zeros as entries. Let K = DH. Then

and K satisfies the condition in the statement of the lemma.

Proof of Theorem 2 We follow mutatis mutandis the method used by Johnson and Newman [I] for their proof of inequality (4). However, the evaluation of the bound for permanents is more involved. Inequality (13) is proved by means of Theorem 1.

By Lemma 5 there exists a matrix K such that

and the ith row of K contains R,?, - R ; and n-2 zeros as its entries, i = 1, . . . , n. We can assume that A, and therefore K, has no zero rows. Let R = diag (R:, ..., R i ) and let S = R-'K. Then S satisfies

and the ith row of S contains 1, - R;/R,! and n-2 zeros as its entries, i = 1, ..., n. The matrix S is permutation equivalent to the block triangular matrix.

where each block S, is square, and either is a zero block or is irreducible. If Si = 0 for some i, then per(S) = 0, and therefore per(K) = 0 and thus

per(A) = 0. Otherwise, either Si = [I] or Si is permutation equivalent to a

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PERMANENTS AND DETERMINANTS

bidiagonal ti x ti matrix of the form

where ti 2 2 and the nonzero entries in the sth row are either us = 1 and v, = -R,,,/R; ,,,, (i,s) = (zij;;tj)+s, or us = -R;,,,/R~,,) and v, = 1. Denote R,, , /R~,, , by a,, or simply by ai if Si happens to be [I]. Then cci, < 1, and we have

Iper(Si)/ = lulu2 ... ut,+v1v2 ... v t i ( < 1 +ailai2 ... aif,. (16) Let a:, . . ., a,* be the numbers R,:/R', i = 1, . . ., n, arranged in non-increasing order. Suppose that Sil = ... = Sir = [I], and ti 2 2 for i # {i,, ..., i,). Let W = {il, ..., i,), and let or*,, ...,or*,-, be the numbers R,:/RT, i = 1, ..., n, i # arranged in non-increasing order. If r and n are both even, then by (16) and Theorem 1,

k

Jper(S)I = n /per(S,)I i= 1

k

< (1 +ai l . . . aiti) i = 1 i+!W

(n - W2

< n (1 +4 - 10iZi) i = l

If r is even and n is odd then by (16) and Theorem 1, k

Jper(S)J = n Iper(S,)J i = l

k

< n (1 + ccil . . . aiti) i = 1 iQW

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12 H. MINC

Similarly, if r is odd and n is even,

Lastly, if n and r are both odd, a similar computation yields the inequality

Hence in all cases

n (1 + ali - la,*i) if n is even, i = l

l ~ ~ ~ ( ~ ) l (n- 3)/2 1 n (1 + a:i - lafi)(l +a,*-, a,*- ,a,*) if n is odd. i = 1

The result follows from (14), (15) and (17).

4. BOUNDS FOR DETERMINANTS

Let C = (c,,) be a complex n x n matrix and let C = A+iB ,

where A = (a,,) and B = (b,,) are real matrices, that is, a,, = Re(c,,) and b,, = Im(c,,), p, q = 1, . . ., n. For any real n x n matrix X = (x,,) we introduce

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PERMANENTS AND DETERMINANTS

the following notation

THEOREM 3 Let C be a complex n x n matrix, and let C = A + iB, where A and B are real matrices. Then

(det(C)I2 6 n (max(Rl (A), R; (A)) + max(R; (B), R, (B)))' p = 1

- n (min(R; (A), R; (A)) + min(R; (B), R, (B)))'. (1 8) p = l

Proof We may assume without loss of generality that

R; (A) 2 R, (A), R; (B) 2 R, (B), (19) p = 1, . . ., n. For, both sides of inequality (18) are invariant under multiplica- tion by - 1 or f i. Thus if it so happens that, for some p, Rl(A) < R;(A) but Ri(B) 3 R;(B), we can multiply thepth row of C by - i ; ifRl(A) 3 R,(A) and R; (B) < R; (B) we can multiply it by i ; and if Rl(A) < R; ( A ) and R;(B) < R,(B) we can multiply the row by - 1. In each case the resulting row satisfies inequalities (19). Henceforth it is assumed that conditions (19) hold. We have to prove that

n n

Jdet(C)12 < n (R,' (A) + R,'(B))' - n (R; (A) + R; (B))'. (20) p= 1 p = l

Let G be the 2n x 2n real matrix

It is known that

Idet(c)l2 = det(G).

For, if U is the unitary matrix

then

It follows from (21) and (4) that

p = 1 p= 1

Now, for n + 1 < p 6 2% max(R; (G), R; (G)) = Ri-,(A) + R,'-,(B) and

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14 H. MINC

min(Rl(G), R,(G)) = R; ,(A) + R;,(B), by (19). On the other hand, if 1 < p 6 n, Rd(G) = Rl(A) + R; (B) which may or may not exceed R;(G) = R;(A)+ R;(B). However, in either case, conditions (19) imply that

R; (A) + R; (B) < min(R; (G), R; (G)) < max(R; (G), R; (G))

G R:(A)+Rl(B) (23) for p = 1, ..., n. But, we also have

R; (4 +R; ( B ) = min(RT+,(G), R,Lp(G)) d max(RAp(G), RLp(G)) = Rl(A)+R,f(B) (24)

p = 1 , . . ., n. Inequality (20) now follows from (21), (22), (23) and (24).

Theorem 3 can be improved by obtaining a stronger upper bound for max(R:(G), R;(G)), and, for some matrices, a sharper lower bound for min(Ri(G), R;(G)), than those given in (23) and (24).

THEOREM 4 Let C be an n x n complex matrix, and let C = A + iB, where A and B are real matrices. Then

1 " ldet(c)i2 G n (R, ( A ) + R~ (B) + m a x p ; (A) - R; (A)I, IR: (B) - R; ( B ) I ) ) ~

p = 1 n

- n (min(R; (A), R; (A) ) + min(R; (B), R; (B)))2 ; p = 1

(25)

and also

Proof We can assume, as we did in the proof of Theorem 3, that R ~ ( A ) 2 R;(A) and R;(B) 2 R, (B). Inequalities (25) and (26) then become

n

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PERMANENTS AND DETERMINANTS

and

We have as before,

For each p, either R; (A) + R; (B) 2 R; (A) + R,+ (B) or R,+ ( A ) + R; (B) d R;(A) + R i (B). If the first alternative holds, then

max(R,i (G), R; (G)) = R,+ (A) + R; (B), and since

we have

max(Ri (GI. R; (GI) . max(RL R, p(G)) = (R i (A)) + R; (B))(R,+ ( A )

+ R,+ (B)) 6 (R,+(A) +eRp(B))2.

If R,+ (A) + R; (B) d R; (A) + R; (B), then

max(R; (G), R; (GI) . max(RL ,(GI, R,L ,(GI) = (R; (A) + R,f (B))(R; (4 + R,+ (B))

G (R;(B) ++R,(A))~. This establishes the first product 011 the right-hand side of (27). To complete the proof of (27) we prove exactly as in the proof of Theorem 3 that

2 n n

n (G), R; (G)) 2 n (R; (A) + R; (B))2. p = l i = 1

Since the first product on the right-hand of (28) is the same as the first product on the right side of (271, in order to prove inequality (28) it remains to show that

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References

[l] C. R. Johnson and M. Newman, A surprising determinantal inequality for real matrices (to appear).

[2] H. Minc, Permunents, Encyclopedia of Mathematics and its Applications, Vol. 6, Addison- Wesley, Reading, 1978.

[3] A. Schinzel, An inequality for determinants with real entries, Coll. Math. 38 (1978), 319-321.

AMS-MOS numbers: 15A15,26A87

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