Volume 10 (2009), Issue 3, Article 63, 9 pp.
AN UPPER BOUND FOR THE DETERMINANT OF A MATRIX WITH GIVEN
ENTRY SUM AND SQUARE SUM
ORTWIN GASPER, HUGO PFOERTNER, AND MARKUS SIGG
WALTROP, G ERMANY
M UNICH , G ERMANY
hugo@pfoertner.org
F REIBURG , G ERMANY
mail@MarkusSigg.de
Received 05 March, 2009; accepted 15 September, 2009
Communicated by S.S. Dragomir
A BSTRACT. By deducing characterisations of the matrices which have maximal determinant in
the set of matrices with given entry sum and square sum, we prove the inequality | det M | ≤
|α|(β − δ)(n−1)/2 for real n × n-matrices M , where nα and nβ are the sum of the entries and
the sum of the squared entries of M , respectively, and δ := (α2 − β)/(n − 1), provided that
α2 ≥ β. This result is applied to find an upper bound for the determinant of a matrix whose
entries are a permutation of an arithmetic progression.
Key words and phrases: Determinant, Matrix Inequality, Hadamard’s Determinant Theorem, Hadamard Matrix.
2000 Mathematics Subject Classification. 15A15, 15A45, 26D07.
1. I NTRODUCTION
Let n ≥ 2 be a positive integer and a = (a1 , . . . , an2 ) a vector of real numbers. What is
the maximal determinant D(a) of a matrix whose elements are a permutation of the entries of
a? The answer is unknown even for the special case a := (1, . . . , n2 ) if n > 6, see [4]. By
computational optimisation using algorithms like tabu search, we have found matrices with the
following determinants, which thus are lower bounds for D(1, . . . , n2 ):
n lower bound for D(1, . . . , n2 )
2
10
3
412
4
40 800
5
6 839 492
6
1 865 999 570
7
762 150 368 499
8
440 960 274 696 935
9
346 254 605 664 223 620
10 356 944 784 622 927 045 792
064-09
2
O RTWIN G ASPER , H UGO P FOERTNER , AND M ARKUS S IGG
It would be nice to also have a good upper bound for D(1, . . . , n2 ). We will show in this
article how to find an upper bound by treating the problem of determining D(a) as a continuous
optimisation task. This is done by maximising the determinant under two equality contraints:
by fixing the sum and the square sum of the entries of the matrix.
Our result is a characterisation of the matrices with maximal determinant in the set of matrices
with given entry sum and square sum, and a general inequality for the absolute value of the
determinant of a matrix.
For the problem of finding D(1, . . . , n2 ), the upper bound derived in this way turns out to be
quite sharp. So here we have an example where analytical optimisation gives valuable information about a combinatorial optimisation problem.
2. C ONVENTIONS
Throughout this article, let n > 1 be a natural number and N := {1, . . . , n}. Matrix always
means a real n × n matrix, the set of which we denote by M.
For M ∈ M and i, j ∈ N we denote by Mi the i-th row of M , by M j the j-th column of M ,
and by Mi,j the entry of M at position (i, j). If M is a matrix or a row or a column of a matrix,
then by s(M ) we denote the sum of the entries of M and by q(M ) the sum of their squares.
The identity matrix is denoted by I. By J we name the matrix which has 1 at all of its fields,
while e is the column vector in Rn with all entries being 1. Matrices of the structure xI + yJ
will play an important role, so we state some of their properties:
Lemma 2.1. Let x, y ∈ R and M := xI + yJ. Then we have:
(1) det M = xn−1 (x + ny)
(2) M is invertible if and only if x 6∈ {0, −ny}.
y
(3) If M is invertible, then M −1 = x1 I − x(x+ny)
J.
Proof. Since J = eeT , it holds that
M e = (xI + yeeT )e = (x + yeT e)e = (x + ny)e and M v = (xI + yeeT )v = xv
for all v ∈ Rn with v ⊥ e. Hence M has the eigenvalue x with multiplicity n − 1 and the simple
eigenvalue x + ny. This shows (1). (2) is an immediate consequence of (1). (3) can be verified
by a straight calculation.
3. M AIN T HEOREM
Let α, β ∈ R with β > 0 and Mα,β := {M ∈ M : s(M ) = nα, q(M ) = nβ}. Furthermore,
let
α2 − β
δ :=
.
n−1
In the proof of the following lemma, matrices are specified whose determinants will later turn
out to be the greatest possible:
Lemma 3.1.
(1) Mα,β 6= ∅ if and only if α2 ≤ nβ. If α2 ≤ nβ, then there exists an M ∈ Mα,β with
det M = α(β − δ)
n−1
2
.
n
(2) If α2 ≤ β, then there exists an M ∈ Mα,β with det M = β 2 .
(3) There exists an M ∈ Mα,β with det M 6= 0 if and only if α2 < nβ.
J. Inequal. Pure and Appl. Math., 10(3) (2009), Art. 63, 9 pp.
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A N U PPER B OUND FOR THE D ETERMINANT OF A M ATRIX
3
2
Proof. (1) Suppose Mα,β 6= ∅, say M ∈ Mα,β . Reading M and J as elements of Rn , the
Cauchy inequality shows that
!2
n
X
1
α2 = 2
Mi,j
n i,j=1
1
hM, Ji2
n2
n
X
1
2
≤ 2 kM k22 kJk22 =
Mi,j
= nβ.
n
i,j=1
=
1
For the other implication suppose α2 ≤ nβ, i. e. β ≥ δ, and set γ := (β − δ) 2
γI + n1 (α − γ)J. Then M ∈ Mα,β , and by Lemma 2.1
n−1
det M = γ n−1 γ + n n1 (α − γ) = γ n−1 α = α(β − δ) 2 .
1 √
3α
2
(2) Let α ≤ β. First suppose α ≥ 0, so γ := 2
− 1 gives γ 2 ≤ 1. Set
β
p

p
γ
1 − γ2
p
2
p
α
β
−
α
p
and B := β − 1 − γ 2
A :=
γ
− β − α2
α
0
0
and M :=

0
0 .
1
3
Then s(A) = 2α, q(A) = 2β, det A = β, s(B) = 3α, q(B) = 3β, det B = β 2 . In the case of
n = 2k with k ∈ N, use k copies of A to build the block matrix


A
..
,
M := 
.
A
which has the required properties. In the case of n = 2k + 1 with k ∈ N, use k − 1 copies of A
to build the block matrix


A
...


,
M := 


A
B
which again fulfills the requirements.
n
In the case of α < 0, an M 0 ∈ M−α,β with det M 0 = β 2 exists. For even n, the matrix
M := −M 0 ∈ Mα,β has the requested determinant, while for odd n swapping two rows of −M 0
gives the desired matrix M .
(3) If α2 < nβ, then the existence of an M ∈ Mα,β with det M 6= 0 is proved by (1) in the case
of α 6= 0 and by (2) in the case of α = 0. For α2 = nβ and M ∈ Mα,β , the calculation in (1)
shows that hM, Ji = kM k2 kJk2 . However, this equality holds only if M is a scalar multiple
of J, so we have det M = 0 because of det J = 0.
For α2 ≤ β we have given two types of matrices in Lemma 3.1, the first one having the
n−1
n
determinant α(β − δ) 2 , the second one with the determinant β 2 . The proof of Theorem 3.3
below will use the fact that for α2 < β the determinant of the first type is strictly smaller than
that of the second type. Indeed, the following stronger statement holds:
Lemma 3.2. Let α2 ≤ nβ. Then |α|(β − δ)
J. Inequal. Pure and Appl. Math., 10(3) (2009), Art. 63, 9 pp.
n−1
2
n
≤ β 2 with equality if and only if α2 = β.
http://jipam.vu.edu.au/
4
O RTWIN G ASPER , H UGO P FOERTNER , AND M ARKUS S IGG
Proof. This is obvious for α = 0, so let α 6= 0. With f (x) := x
have
r n−1
n
2
−
|α|(β − δ) 2 β 2 = f αβ .
n−x n−1
n−1
for x ∈ [0, n] we
The proof is completed by applying the AM-GM inequality to f (x)1/n :
n−1 ! n1
x + (n − 1) n−x
1
n
−
x
n−1
n
=1
≤
f (x) = x
n
n−1
with equality if and only if x =
n−x
,
n−1
i. e. if and only if x = 1.
If α2 < nβ, then by Lemma 3.1 there exists an M ∈ Mα,β with det M 6= 0, and, by possibly
swapping two rows of M , det M > 0 can be achieved. As Mα,β is compact, the determinant
function assumes a maximum value on Mα,β . The next theorem, which is essentially due to
O. Gasper, shows that this maximum value is given by the determinants noted in Lemma 3.1:
Theorem 3.3. Let (
α2 < nβ and M ∈ Mα,β with maximal determinant. Then
(1) M M T = βI
if α2 ≤ β:
n
(2) det M = β 2

j

(3) s(Mi ) = s(M ) = α for all i, j ∈ N
if α2 ≥ β: (4) M M T = (β − δ)I + δJ

n−1

(5) det M = |α|(β − δ) 2
Proof. From Lemma 3.1, we know that det M > 0. The matrix M solves an extremum problem
with equality contraints


det X −→ max
(P)
(X ∈ M∗ ),
s(X) = nα

q(X) = nβ
where M∗ is the set of invertible matrices. The Lagrange function of (P) is given by
L(X, λ, µ) = det X − λ(s(X) − nα) − µ(q(X) − nβ),
so there exist λ, µ ∈ R with
d
L(M, λ, µ)
dMi,j
d
det M
dMi,j
= 0 for all i, j ∈ N . It is well known that
−1
= (det M ) (M T )
i,j
−1
(see e. g. [3], 10.6), thus we get (det M ) (M T )
− λM − 2µJ = 0, i. e.
(det M )I = λM M T + 2µJM T .
(3.1)
Suppose λ = 0. Then
(det M )n = det(2µJM T ) = det(2µJ) det M = 0 det M = 0
by applying the determinant function to (3.1). This contradicts det M > 0. Hence
(3.2)
λ 6= 0.
As M M T has diagonal elements q(M1 ), . . . , q(Mn ), and JM T has diagonal elements s(M1 ),
. . . , s(Mn ), we get
n det M = λq(M ) + 2µs(M ) = λnβ + 2µnα
J. Inequal. Pure and Appl. Math., 10(3) (2009), Art. 63, 9 pp.
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A N U PPER B OUND FOR THE D ETERMINANT OF A M ATRIX
5
by applying the trace function to (3.1), consequently
det M = λβ + 2µα.
(3.3)
The symmetry of (det M )I and the symmetry of λM M T in (3.1) show that µJM T is symmetric as well. As all rows of JM T are identical, namely equal to (s(M1 ), . . . , s(Mn )), we
obtain
µs(M1 ) = · · · = µs(Mn ).
(3.4)
In the following, we inspect the cases µ = 0 and µ 6= 0 and prove:
(
µ = 0 =⇒ α2 ≤ β ∧ (1) ∧ (2),
(3.5)
µ 6= 0 =⇒ α2 ≥ β ∧ (3) ∧ (4) ∧ (5).
Case µ = 0: Then (3.3) reads det M = λβ, so taking (3.2) into account and dividing (3.1) by λ
gives βI = M M T , i. e. (1). Part (2) followsby
the determinant function to (1). Using
√ applying
β M is orthogonal and thus an isometry w.r.t.
the Cauchy inequality and the fact that 1
the euclidean norm k · k2 , we get:
(3.6)
1
α2 = 2
n
≤
=
1
n
n2
n
X
!2
s(Mi )
i=1
n
X
s(Mi )2
i=1
1
1
1
kM ek22 = βkek22 = βn = β.
n
n
n
Case µ 6= 0: Then s(M1 ) = · · · = s(Mn ) by (3.4). The identity
s(M1 ) + · · · + s(Mn ) = s(M ) = nα
shows that s(Mi ) = α for all i ∈ N . Taking into account that the determinant is invariant
against matrix transposition, this proves (3). Furthermore, JM T = αJ, and (3.1) becomes
(3.7)
λM M T = (det M )I − 2µαJ,
hence
q(Mi ) = (M M T )i,i =
1
(det M − 2µα)
λ
for all i ∈ N , and q(M1 ) = · · · = q(Mn ). With
q(M1 ) + · · · + q(Mn ) = q(M ) = nβ,
this shows that
(3.8)
(M M T )i,i = q(Mi ) = β
J. Inequal. Pure and Appl. Math., 10(3) (2009), Art. 63, 9 pp.
for all i ∈ N .
http://jipam.vu.edu.au/
6
O RTWIN G ASPER , H UGO P FOERTNER , AND M ARKUS S IGG
Let i, j ∈ N with i 6= j. Equation (3.7) gives (M M T )i,k = − λ1 2µα for all k ∈ N \ {i}, and
we get
X
β + (n − 1)(M M T )i,j = (M M T )i,i +
(M M T )i,k
k6=i
=
=
=
=
n
X
(M M T )i,k
k=1
n X
n
X
Mi,p Mk,p
k=1 p=1
n
X
Mi,p s(M p )
p=1
n
X
Mi,p α = s(Mi ) α = α2 ,
p=1
so
α2 − β
= δ.
(3.9)
(M M )i,j =
n−1
Equations (3.8) and (3.9) together prove (4). With Lemma 2.1, this yields
T
(det M )2 = det(M M T ) = (β − δ)n−1 (β − δ + nδ) = α2 (β − δ)n−1 ,
and taking the square root gives (5). Suppose that α2 < β. Then by Lemma 3.1 there exists an
n
M 0 ∈ Mα,β with det M 0 = β 2 , and by Lemma 3.2,
det M = |α|(β − δ)
n−1
2
n
< β 2 = det M 0 ,
which contradicts the maximality of det M . Hence α2 ≥ β.
We have now proved (3.5) and are ready to deduce the statements of the theorem: If α2 < β,
then (3.5) shows that µ = 0 and thus (1) and (2). If α2 > β, then (3.5) shows that µ 6= 0 and
thus (3), (4) and (5). Finally suppose that α2 = β. Then δ = 0, hence (1) ⇐⇒ (4) and (2)
⇐⇒ (5). If µ 6= 0, then (3.5) shows (3), (4) and (5), from which (1) and (2) follow. If µ = 0,
then (3.5) shows (1) and (2), from which (4) and (5) follow. It remains to prove (3) in the case
of α2 = β and µ = 0. To this purpose, look at (3.6) again, where α2 = β means equality in
the Cauchy inequality, which tells us that (s(M1 ), . . . , s(Mn )) is a scalar multiple of e, hence
s(M1 ) = · · · = s(Mn ), and (3) follows as in the case µ 6= 0.
4. A PPLICATION
The following is a more application-oriented extract of Theorem 3.3:
Proposition 4.1. Let M ∈ M, α := n1 s(M ), β := n1 q(M ) and δ :=
α2 −β
.
n−1
Then:
n
α2 < β
=⇒
| det M | ≤ β 2
α2 = β
=⇒
| det M | ≤ |α|(β − δ)
α2 > β
=⇒
| det M | ≤ |α|(β − δ)
n−1
2
n−1
2
n
= β2
n
< β2
Proof. This is clear if det M = 0. In the case of det M 6= 0, we get α2 < nβ by Lemma 3.1,
and the stated inequalities are true by Lemma 3.2 and Theorem 3.3.
J. Inequal. Pure and Appl. Math., 10(3) (2009), Art. 63, 9 pp.
http://jipam.vu.edu.au/
A N U PPER B OUND FOR THE D ETERMINANT OF A M ATRIX
7
For M ∈ M with |Mi,j | ≤ 1 for all i, j ∈ N , Proposition 4.1 tells us that
! n2
! n2
n
n
n
n
1 X 2
1 X
(4.1)
| det M | ≤ β 2 =
Mi,j
≤
1
= n2,
n i,j=1
n i,j=1
which is simply the determinant theorem of Hadamard [2]. If Mi,j ∈ {−1, 1} for all i, j ∈ N
and | det M | = nn/2 , i. e. M is a Hadamard matrix, then Proposition 4.1 shows that α2 ≤ β must
hold. For a Hadamard matrix M , the value s(M ) is called the excess of M . Since q(M ) = n2
in the case of Mi,j ∈ {−1, 1}, Proposition 4.1 yields an upper bound for the excess, known as
Best’s inequality [1]:
M is a Hadamard matrix
(4.2)
=⇒
3
s(M ) ≤ n 2
The results (4.1) and (4.2), which both can be proved more directly, are mentioned here just
as by-products of Proposition 4.1. In the following, we are interested only in the case α2 ≥ β,
where the inequality
| det M | ≤ |α|(β − δ)
n−1
2
=: g(M )
n
2
holds. Note that Lemma 3.2 states that g(M ) < β is true for α2 < β also, but | det M | is not
necessarily bounded by g(M ) in this situation:
1 0
M :=
, | det M | = 1 , g(M ) = 0.
0 −1
We are now going to apply Proposition 4.1 to the problem stated in the introduction. This
problem is a special case of finding an upper bound for the determinant of matrices whose
entries are a permutation of an arithmetic progression:
Proposition 4.2. Let p, q be real numbers with q > 0 and M a matrix whose entries are a
permutation of the numbers p, p + q, . . . , p + (n2 − 1)q. Set
r :=
p n2 − 1
+
q
2
and
% :=
n3 + n2 + n + 1
.
12
Then
n
2
| det M | ≤ n q
n
n4 − 1
r +
12
2
n2
and
2
r >%
=⇒
n n
| det M | ≤ n q |r|%
n−1
2
n
2
<n q
n
n4 − 1
r +
12
2
n2
.
Proof. For α := n1 s(M ) and β := n1 q(M ) a calculation shows that α2 −β = n(n−1)q 2 (r2 −%),
hence (α2 > β ⇐⇒ r2 > %). The bounds noted in Proposition 4.1 yield the asserted inequalities
for | det M |.
Corollary 4.3. If M is a matrix whose entries are a permutation of 1, . . . , n2 , then
3
n−1
2
n + n2 + n + 1 2
n n +1
| det M | ≤ n
.
2
12
Proof. Apply Proposition 4.2 to (p, q) := (1, 1). For r = (n2 + 1)/2 it is easy to see that r2 > %,
which yields the stated bound.
J. Inequal. Pure and Appl. Math., 10(3) (2009), Art. 63, 9 pp.
http://jipam.vu.edu.au/
8
O RTWIN G ASPER , H UGO P FOERTNER , AND M ARKUS S IGG
Comparing the lower bounds for D(1, . . . , n2 ) noted in the introduction with the upper
bounds resulting from rounding down the values given by Corollary 4.3 shows that the quality
of these upper bounds is quite convincing:
n determinant of best known matrix upper bound given by Corollary 4.3
2
10
11
3
412
450
4
40 800
41 021
5
6 839 492
6 865 625
6
1 865 999 570
1 867 994 210
7
762 150 368 499
762 539 814 814
8
440 960 274 696 935
441 077 015 225 642
9
346 254 605 664 223 620
346 335 386 150 480 625
10
356 944 784 622 927 045 792
357 017 114 947 987 625 629
These are the record matrices R(n) corresponding to the noted determinants:
R(2) =
4 2
1 3

25
7

R(5) = 
6
10
16

46
9

39

R(7) = 
26
20

4
32
42
48
11
22
8
28
16
15
36
44
17
40
19
3
,

15 9 11 4
24 14 3 17

12 23 20 5 
,
13 2 22 19
1 18 8 21
2
30
34
41
6
25
37


12 13 6 2
 3 8 16 7 

R(4) = 
14 1 9 10 ,
5 11 4 15


9 3 5
R(3) = 4 8 1 ,
2 6 7
27
7
13
47
33
38
10
24
14
29
1
23
49
35

68
67

30

56

R(9) = 
23
1

45

15
64

18
31

5

21
 ,
45

12
43
7
37
20
50
14
38
74
77
52
J. Inequal. Pure and Appl. Math., 10(3) (2009), Art. 63, 9 pp.
12
43
79
8
54
32
31
35
75

36
3

13
R(6) = 
14

20
26
24
35
7
22
4
18

1
44

57

28
R(8) = 
25

19

59
26
62
10
53
27
63
21
80
39
13
73
3
49
42
11
65
17
51
57
26
61
71
60
18
66
46
16
6
29
33
40
81
72
22
5
59
28
21
25
34
2
19
9
12
35
2
22
36
60
39
54
58
78
25
4
44
48
24
69
19
17
15
16
33
29
1
20
3
51
55
42
33
9
47
52
14
49
4
34
56
37
13
5
23
10
12
32
30
40
43
23
64
5
46
30
10

8
11

31
 ,
28

6
27
50
15
11
41
48
6
58
31
53
45
38
18
7
16
21
62

32
61

29

27
,
63

24

8
17

34
36

2

41

70

76

47

9
55
http://jipam.vu.edu.au/
A N U PPER B OUND FOR THE D ETERMINANT OF A M ATRIX

1
 53

 46

 79

 17
R(10) = 
 68

100

 21

 69
52
2
57
63
42
75
93
16
72
15
70
61
3
47
49
87
76
31
29
99
23
84
65
4
71
58
50
35
9
32
96
81
94
45
5
30
85
62
48
44
11
82
20
78
95
6
56
34
73
24
36
39
91
83
51
38
7
8
43
90
55
54
22
28
10
27
37
64
97
74
92
41
66
13
77
86
14
67
89
40
12
9

60
33

98

26

80

19

88

25

18
59
Calculating the matrix M M T for each record matrix M reveals that M M T has roughly the
structure (β − δ)I + δJ that was noted in Theorem 3.3 for the optimal matrices of the corresponding real optimisation problem.
R EFERENCES
[1] D.S. BERNSTEIN, Matrix Mathematics: Theory, Facts, and Formulas with Application to Linear
Systems Theory, Princeton University Press, 2005.
[2] M.R. BEST, The excess of a Hadamard matrix. Nederl. Akad. Wet., Proc. Ser. A, 80 (1977), 357–361.
[3] J. HADAMARD, Résolution d’une question relative aux déterminants, Darboux Bull., (2) XVII
(1893), 240–246.
[4] N.J.A. SLOANE, The Online Encyclopedia of Integer Sequences, id:A085000. [ONLINE: http:
//www.research.att.com/~njas/sequences/A085000].
J. Inequal. Pure and Appl. Math., 10(3) (2009), Art. 63, 9 pp.
http://jipam.vu.edu.au/