Linear Algebra and its Applications 439 (2013) 3555–3560
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Linear Algebra and its Applications www.elsevier.com/locate/laa
On the S-matrix conjecture Roman Drnovšek Department of Mathematics, Faculty of Mathematics and Physics, University of Ljubljana, Jadranska 19, SI-1000 Ljubljana, Slovenia
a r t i c l e
i n f o
a b s t r a c t
Article history: Received 6 June 2013 Accepted 5 September 2013 Available online 27 September 2013 Submitted by X. Zhan
Motivated with a problem in spectroscopy, Sloane and Harwit conjectured in 1976 what is the minimal Frobenius norm of the inverse of a matrix having all entries from the interval [0, 1]. In 1987, Cheng proved their conjecture in the case of odd dimensions, while for even dimensions he obtained a slightly weaker lower bound for the norm. His proof is based on the Kiefer–Wolfowitz equivalence theorem from the approximate theory of optimal design. In this note we give a short and simple proof of his result. © 2013 Elsevier Inc. All rights reserved.
MSC: 15A45 15A60 Keywords: Matrices Frobenius norm Inequalities
A Hadamard matrix is a square matrix with entries in {−1, 1} whose rows and hence columns are mutually orthogonal. In other words, a Hadamard matrix of order n is a {−1, 1}-matrix A satisfying A A T = nI , i.e., √1 A is a unitary matrix. n
An S-matrix of order n is a {0, 1}-matrix formed by taking a Hadamard matrix of order n + 1 in which the entries in the first row and column are 1, changing 1’s to 0’s and −1’s to 1’s, and deleting the first row and column. The Frobenius norm of a real matrix A = [ai , j ]ni, j =1 is defined as
AF =
n n
1/2 a2i , j
.
i =1 j =1
It is associated to the inner product defined by
E-mail address:
[email protected]. 0024-3795/$ – see front matter © 2013 Elsevier Inc. All rights reserved. http://dx.doi.org/10.1016/j.laa.2013.09.012
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R. Drnovšek / Linear Algebra and its Applications 439 (2013) 3555–3560
n n A , B = tr A B T = ai , j b i , j . i =1 j =1
Let Dn denote the set of all matrices A = [ai , j ]ni, j =1 whose entries are in the interval [0, 1]. In 1976, Sloane and Harwit [6] posed the following conjecture. See also [2, p. 59] or [7, Conjecture 11]. Conjecture. If A ∈ Dn is a nonsingular matrix, then
−1 A 2n , F n+1 where the equality holds if and only if A is an S-matrix. This conjecture arose from a problem in spectroscopy. A detailed discussion of its applications in spectroscopy can be found in [2]. The conjecture has been proved in recent papers [9,8,3] for some special matrices. Apparently, the authors of these papers were not aware of the fact that for odd dimensions the conjecture has already been proved in [1], while for even dimensions a slightly weaker lower bound for the norm has been derived; see [1, Corollary 3.4]. The proof is based on the celebrated equivalence theorem due to Kiefer and Wolfowitz [5] that connects the problem with the approximate theory of optimal design. For an extensive treatment of this theory we refer to [4]. In this note we give a short and transparent proof of the conjecture when n is odd, while for even n our method gives the same (weaker) lower bound as in [1, Corollary 3.4]. Theorem. Let A ∈ Dn be a nonsingular matrix. If n 3 is an odd integer, then
−1 A 2n , F n+1
(1)
where the equality holds if and only if A is an S-matrix. If n 4 is an even integer, then
√ 2 −1 A > 2 n − 2n + 2 . F
(2)
n
If n = 2 then
√ −1 A 2, F
(3)
where the equality holds if and only if A is either the identity matrix or
01 10
.
Proof. Let e = (1, 1, 1, . . . , 1) T ∈ Rn and J = ee T . We divide the proof into three cases. Case 1. n 3 is an odd integer, so that n = 2k − 1 for some k ∈ N. Define the matrices M and N of order n + 1 by
M=
1 eT e −k A −1
Since
MNT =
we have
n+1
∗
and
N=
∗ , (n + 1) I
1 eT e −(2I − 1k J ) A T
.
R. Drnovšek / Linear Algebra and its Applications 439 (2013) 3555–3560
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M , N = tr M N T = (n + 1)2 . By the Cauchy–Schwarz inequality, we then obtain
2 (n + 1)4 = M , N M 2F · N 2F
2 1 2 −1 2 = 1 + 2n + k A 1 + 2n + 2I − J A T . F k
(4)
F
If we show that
2I − 1 J A T n, k
(5)
F
then (4) gives the inequality
2 (n + 1)2 1 + 2n + k2 A −1 F , and so
−1 2 n 2 A = F 2
2
2n n+1
k
completing the proof of (1). To show (5), we determine the maximum of the function f defined on Dn by
f ( A) = 2I −
2
1 = tr A 2I − J J AT k k F
1
= 4 tr A A =4
T
n n
−
a2i , j
2
AT
2k + 1 T = tr 4 A A T − ( Ae )( Ae ) 2 k
2k + 1
( Ae )T ( Ae ) n n 2k + 1
k2
−
k2
i =1 j =1
i =1
2 ai , j
,
j =1
where A = [ai , j ]ni, j =1 ∈ Dn . Since f is a continuous function on a compact set, it attains its maximum
at some matrix B = [b i , j ]ni, j =1 ∈ Dn . Assume that 0 < b i , j < 1 for some i , j ∈ {1, 2, . . . , n}. Then we have
∂f ( B ) = 0 and ∂ ai , j
∂2 f ( B ) 0. ∂ a2i , j
However,
n ∂f 2k + 1 ( A ) = 8ai , j − 2 ai ,l , ∂ ai , j k2 l =1
and so
∂2 f 2(2k + 1) 2(4k2 − 2k − 1) ( A) = 8 − = >0 2 2 k k2 ∂ ai , j for all A = [ai , j ]ni, j =1 ∈ Dn . Therefore, we conclude that B is necessarily a {0, 1}-matrix. Let p i be the number of ones in the i-th row of B. Then
f (B) = 4
n i =1
2k + 1 n
pi −
k2
i =1
2k + 1 n
p 2i = −
k2
i =1
pi −
2k2 2k + 1
2 +
4k2 2k + 1
(2k − 1).
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R. Drnovšek / Linear Algebra and its Applications 439 (2013) 3555–3560 2
Since k − 2k2k+1 = 2kk+1 ∈ (0, 12 ), we have |m − that p i = k for all i. It follows that
f ( B ) = 4nk −
2k + 1 k2
2k2 | 2k+1
> |k −
2k2 | 2k+1
for all m ∈ {1, 2, . . . , n} \ {k}, implying
nk2 = n(4k − 2k − 1) = n2 .
This completes the proof of the inequality (5). Assume that the equality holds in (1). Then there are equalities in (4) and (5), that is, M = N and A is an invertible {0, 1}-matrix with Ae = ke. It follows that k A −1 = (2I − 1k J ) A T = 2 A T − J , and
so e = k A −1 e = 2 A T e − (2k − 1)e implying that A T e = ke. Therefore, we have
NNT = MNT =
1 eT e −k A −1
This means that
NT =
eT J − 2A
1 e
=
n+1 0 0 (n + 1) I
.
eT J − 2A
1 e
is a Hadamard matrix, and so A is an S-matrix. As the equality holds in (1) when A is an S-matrix, the proof is complete for odd dimensions. Case 2. n 4 is an even integer, so that n = 2k for some integer k 2. Define the matrices M and N of order n + 1 by
M=
0 e
Then
√e
MNT =
2k
∗
T
√k k k −1
A −1
∗
k(2k−1) I k −1
and
N=
0 e
√
eT
k√−1 k(2k−1) ( k −1 I k k
− J)A
T
.
,
and so
2k2 (2k − 1) 2k(2k2 − 1) M , N = tr M N T = 2k + = . k−1 k−1 By the Cauchy–Schwarz inequality, we then obtain
2k(2k2 − 1)
2
k−1
2 = M , N M 2F · N 2F
k 3 −1 2 A = 4k + 4k + g ( A ) , F k−1
where g ( A ) is defined by
√
2 k − 1 k(2k − 1) g( A) = − J AT I √ . k−1 k k
If A =
F
∈ Dn then
(2k − 1)2 T 2 2 (2k − 1)2 g ( A ) = tr A I − J AT = tr A A − ( Ae ) T ( Ae ) k(k − 1) k k(k − 1) k 2k 2 2k 2k 2k 2 (2k − 1)2 2 ai , j − ai , j . = k(k − 1) k [ai , j ]ni, j=1
i =1 j =1
Since
i =1
j =1
(6)
R. Drnovšek / Linear Algebra and its Applications 439 (2013) 3555–3560
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∂2 g 2(2k − 1)2 4 2(4k2 − 6k + 3) ( A) = − = > 0, 2 k(k − 1) k k(k − 1) ∂ ai , j we conclude (similarly as in Case 1) that the maximum of the function g on Dn is attained at some matrix B = [b i , j ]ni, j =1 ∈ Dn with b i , j ∈ {0, 1} for all i and j. Let p i be the number of ones in the i-th row of B. Then
2 2k 2k 2k (2k − 1)2 (2k − 1)2 (2k − 1)4 2 2 2 g(B) = pi − pi = − pi − + . k(k − 1) k k 4(k − 1) 4(k − 1)2 i =1
Since k −
(2k−1)2 4(k−1)
=
− 4(k1−1)
∈
i =1
we obtain that p i = k for all i. It follows that
(2k − 1)2 2k(2k2 − 2k + 1) . 2k − 4k2 = k−1 k−1
g(B) =
Now, since 4k + g ( B ) =
4k +
i =1
(− 12 , 0),
2k(2k2 −1) , k−1
the inequality (6) gives
2 k 3 −1 2 A 2k(2k − 1) , F k−1 k−1
and so
−1 2 2(2k2 − 2k + 1) 4(n2 − 2n + 2) A = . F 2 2 k
(7)
n
To complete the proof of the inequality (2), we must exclude the possibility of the equality in (7). So, assume that for some matrix A ∈ Dn the equality holds in (7). Then A is a {0, 1}-matrix and M = N. Therefore, we have
k3 k−1
A
−1
k(2k − 1)
=
or
k−1
k2 2k − 1
I=
I−
I − J AT
k−1 k(2k − 1)
J A T A,
implying that
AT A =
k2
2k − 1
I−
k−1 k(2k − 1)
−1 J
=
k2 2k − 1
I+ T
k−1 k
J .
It follows that the off-diagonal entries of the matrix A A are equal to the number an integer. This is a contradiction with the fact that A is a {0, 1}-matrix. Case 3. n = 2. If
A=
a b c d
is an invertible matrix in D2 , then
A −1 =
1 ad − bc
d −c
−b a
,
and so
−1 2 a 2 + b 2 + c 2 + d 2 A = . F (ad − bc )2
k(k−1) 2k−1
that is not
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R. Drnovšek / Linear Algebra and its Applications 439 (2013) 3555–3560
Now, we have
2 (ad − bc )2 A −1 F − 2 = (a − d)2 + (b − c )2 + 2ad(1 − ad) + 2bc (1 − bc ) + 4abcd 0. We conclude that
−1 2 A 2 F and the equality holds if and only if a = d, b = c, ad ∈ {0, 1}, bc ∈ {0, 1} and abcd = 0. This implies the desired conclusions. 2 Acknowledgement The author was supported in part by the Slovenian Research Agency. References [1] C.-S. Cheng, An application of the Kiefer–Wolfowitz equivalence theorem to a problem in Hadamard transform optics, Ann. Statist. 15 (4) (1987) 1593–1603. [2] M. Harwit, N.J.A. Sloane, Hadamard Transform Optics, Academic Press, New York, 1979. [3] X. Hu, Some inequalities for unitarily invariant norms, J. Math. Inequal. 6 (4) (2012) 615–623. [4] J. Kiefer, General equivalence theory for optimum designs (approximate theory), Ann. Statist. 2 (1974) 849–879. [5] J. Kiefer, J. Wolfowitz, The equivalence of two extremum problems, Canad. J. Math. 12 (1960) 363–366. [6] N.J.A. Sloane, M. Harwit, Masks for Hadamard transform optics, and weighing designs, Appl. Optics 15 (1976) 107–114. [7] X. Zhan, Open problems in matrix theory, in: L. Ji, K. Liu, L. Yang, S.-T. Yau (Eds.), Proceedings of the 4th International Congress of Chinese Mathematicians, vol. I, Higher Education Press, Beijing, 2008, pp. 367–382. [8] L. Zou, On a conjecture concerning the Frobenius norm of matrices, Linear Multilinear Algebra 60 (1) (2012) 27–31. [9] L. Zou, Y. Jiang, X. Hu, A note on a conjecture on the Frobenius norm of matrices, J. Shandong Univ. Nat. Sci. 45 (2010) 48–50.