Maximum Absolute Determinants of Upper Hessenberg Bohemian Matrices
aa r X i v : . [ c s . S C ] M a y MAXIMUM ABSOLUTE DETERMINANTS OF UPPER HESSENBERG BOHEMIANMATRICES
JONATHAN P. KEATING ∗ AND
AHMET ABDULLAH KELES¸ † Abstract.
A matrix is called Bohemian if its entries are sampled from a finite set of integers. We determine the maximumabsolute determinant of upper Hessenberg Bohemian Matrices for which the subdiagonal entries are fixed to be 1 and uppertriangular entries are sampled from { , , . . . , n } , extending previous results for n = 1 and n = 2 and proving a recent conjectureof Fasi & Negri Porzio [8]. Furthermore, we generalize the problem to non-integer-valued entries. Key words.
Upper Hessenberg, Bohemian Matrix, Maximum Absolute Determinant.
AMS subject classifications.
1. Introduction.
Matrices whose entries are from a small subset of the integers are said to be Bo-hemian, an abbreviation of
BOunded HEight Matrix of integers . These matrices appear in many differentcontexts, including adjacency matrices of graphs [10], Hadamard matrices [1, 2], random discrete matrices[4] and alternating sign matrices [3]. They have been studied for over a century and remain a subject ofactive research. The website [7] provides a comprehensive overview of recent results and open problems.Recently, Chan et al. investigated several properties of the characteristic polynomials of upper Hessen-berg Bohemian matrices [5], and Thornton et al. obtained a number of results concerning the distribution oftheir eigenvalues, characteristic heights, and maximum absolute determinant values [6]. These papers stateseveral conjectures on the values of the determinants of Bohemian matrices [7], many of which have recentlybeen solved and generalised by Massimiliano Fasi and Gian M. N. Porzio [8].One of these conjectures, which is a refinement of a result of Li Ching [9], was until now lacking asolution that could be generalised. We here provide a generalisable solution for that problem and explore anextension of it. Specifically, we focus on the maximum absolute determinant of upper Hessenberg matriceswith fixed subdiagonal entries and a given population [0 , t ] of upper triangular entries. Our calculationsprovide a more comprehensive understanding of the behaviour of the determinants of these kind of matricesand include special cases that had previously been solved by other approaches.
2. Results.
In 1993, Ching proved the following theorem.
Theorem
The maximum absolute determinant of an n × n upper-Hessenberg matrix with uppertriangular entries from { , } and subdiagonal entries fixed at is given by the Fibonacci sequence. [9] Recently Fasi and Porzio have established the following theorem, proving a result conjectured by Thorn-ton [7]:
Theorem
The maximum absolute determinant of an n × n upper-Hessenberg matrix with uppertriangular entries from { , , } and subdiagonal entries fixed at is given by the following sequence. Let ∗ Mathematical Institute, University of Oxford, Andrew Wiles Building, Oxford, OX2 6GG, UK ([email protected]) † Department of Mathematics, Bilkent University, 06800 Bilkent, Ankara, Turkey ([email protected])1 n denote the maximum absolute determinant among these n × n matrices, then M = 2 , M = 4 , and M n = 2 · M n − + M n − for all n ≥ . [8] It is a natural question what happens if the upper triangular population is { , , . . . , n } . Fasi and Porziostated the following conjecture in this context. Conjecture
The maximum absolute determinant of an n × n upper-Hessenberg matrix with uppertriangular entries from { , , . . . , d } and subdiagonal entries fixed at is given by the following generalizedFibonacci sequence. Let M n denote the maximum absolute determinant among these n × n matrices, then M = d , M = d , M n = d · M n − + M n − for all n ≥ . [8] We discuss the following, further generalisation of this problem.
Problem
What is the maximum absolute determinant of an n × n upper-Hessenberg matrix withupper triangular entries drawn from [0 , t ] , t > , and subdiagonal entries taking a fixed value s ∈ R ? If s is negative the problem is relatively straightforward and the result can be stated as the followingtheorem, whose proof may be found in [8]. Theorem
The maximum absolute determinant of an n × n upper-Hessenberg matrix with uppertriangular entries from [0 , t ] and subdiagonal entries fixed at s < is given by t · ( t − s ) n − for all n ∈ N . However the proof of this theorem does not extend to s > s >
0. The first case we consider is when s ≤ t , thesecond is t · (1 + ǫ ) ≥ s ≥ t (where ǫ is a sufficiently small positive number depending on the dimension ofthe matrix) and the third case is s > t · · n . Our main results are contained in the following theorems: Theorem
The maximum absolute determinant of an n × n upper-Hessenberg matrix with entriesfrom the interval [0 , t ] and subdiagonal entries fixed at s , such that t ≥ s > , is given by the followingsequence. Let M n denote the maximum absolute determinant among these n × n matrices; then M = t , M = t and M n = tM n − + s M n − for all n > in Z . Remark { , , . . . , d } and [0 , d ] are equivalent. So, if we set t = d ∈ N and s = 1, then we prove Conjecture 2.3 asa corollary of Theorem 2.6. Theorem
The maximum absolute determinant of an n × n upper-Hessenberg matrix with entriesfrom the interval [0 , t ] , subdiagonal entries fixed at s and n ≥ such that t · (1 + ǫ ( n )) ≥ s ≥ t where ǫ ( n ) is a sufficiently small positive number depending on n is given by the following sequence. Let M n denotethe maximum absolute determinant among these n × n matrices; then M = 3 s t , M = s t + 4 s t and M n = tM n − + s M n − for all n > in Z . Theorem
The maximum absolute determinant of an n × n upper-Hessenberg matrix with en-tries from the interval [0 , t ] and subdiagonal entries fixed at s such that st > · n is given by s n − t + j n kj n − k s n − t . Throughout this paper we use the following definitions and notation. efinition n × n upper Hessenberg matrices whose subdiagonal entries are fixedat s and upper triangular entries are from the set P is denoted by G n × ns ( P ). Definition n × n matrix A , denote the determinant of the bottom-right ( n + 1 − k ) × ( n + 1 − k ) part of A by H k ( A ), and for convenience set H n +1 ( A ) := 1 for any n × n matrix A . Notation A , when referring to the matrix itself we use square brackets andwhen referring to the determinant of A we use straight brackets. For instance, A = (cid:20) (cid:21) , det( A ) = | A | = (cid:12)(cid:12)(cid:12)(cid:12) (cid:12)(cid:12)(cid:12)(cid:12) = −
1. To avoid confusion, we use abs( · ) for absolute value sometimes.
3. Case I.
Firstly we deal with the case t ≥ s > M := 1. Lemma M n ≥ t · M n − , ∀ n ∈ Z + .Proof. Suppose that maximum absolute determinant for ( n − × ( n −
1) matrices is attained by amatrix B ( n − × ( n − . Then the following inequality holds trivially M n ≥ abs (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) t · · · s B ...0 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ! = t · M n − . For the next lemma, we start by writing the determinant for any matrix A ∈ G n × ns ([0 , t ]) using Laplaceexpansion twice, firstly for the first column of the matrix A and secondly for the first rows of the resultingtwo matrices. | A | = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) a a a · · · a n − a n s a a · · · a n − a n s a · · · a n − a n s . .. a n − a n ... . .. .. . . .. .. . ...0 0 0 · · · s a nn (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = a (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) a a · · · a n − a n s a · · · a n − a n s ... a n − a n ... . .. ... . .. ...0 0 · · · s a nn (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) − s · (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) a a · · · a n − a n ) s a · · · a n − a n s .. . a n − a n ... . .. .. . . .. ...0 0 · · · s a nn (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) == ( a a − s · a ) · (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) a · · · a n − a n s .. . a n − a n ... .. . . .. ...0 · · · s a nn (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + ( − ( a a − s · a ) · (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) s a a · · · a n a a · · · a n s a . .. a n ... . .. .. . . .. ...0 0 · · · s a nn (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ++ ( − ( a a − s · a ) · (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) s a a · · · a n s a · · · a n a .. . a n ... .. . . .. .. . ...0 0 · · · s a nn (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + · · · + ( − n − ( a a n − s · a n ) · (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) s a a · · · a n − s a · · · a n − s · · · a n − ... ... . .. . .. ...0 0 · · · s (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = (cid:20) ( a a − s · a ) H ( A ) − ( a a − s · a ) H ( A ) · s (cid:21) + (cid:20) ( a a − s · a ) H ( A ) · s − ( a a − s · a ) H ( A ) · s (cid:21) + ······ + (cid:20) ( a a n − − s · a n − ) H n ( A ) · s n − − ( a a n − s · a n ) H n +1 ( A ) · s n − (cid:21) (3.1)(The last term depends on the parity of n , if n is even, it is (cid:20) ( a a n − s · a n ) H n +1 ( A ) · s n − (cid:21) .)Now we state a new lemma which is inspired by the previous expansion. Lemma
For all k in { , , . . . , n − } we have the following inequality: (3.2) (cid:12)(cid:12)(cid:12)(cid:12) ( a a k − s · a k ) · H k +1 ( A ) − ( a a k +1) − s · a k +1) ) · s · H k +2 ( A ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ t · M n − k + ts · M n − k − and also, for the case when n is even: (3.3) (cid:12)(cid:12)(cid:12)(cid:12) ( a a n − s · a n ) · H n +1 ( A ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ t · M Proof.
For (3.2), it is enough to show the following:(3.4) max x ∈ [ − st,t ] ,y ∈ [ − st ,s t ] | x · H k +1 ( A ) + y · H k +2 ( A ) | ≤ t M n − k + ts M n − k − It suffices to check four extreme cases of x and y , i.e,( x, y ) ∈ { ( − st, s t ) , ( − st, − st ) , ( t , s t ) , ( t , − st ) } Using t ≥ s >
0, the triangle inequality, the definition of M n , and Lemma 3.1 we get the followinginequalities for these four cases: ( x, y ) = ( − st, s t ): | ( − st ) H k +1 ( A ) + ( s t ) H k +2 ( A ) | ≤ ( st ) | H k +1 ( A ) | + ( s t ) | H k +2 ( A ) | ≤≤ ( st ) M n − k + ( s t ) M n − k − ≤≤ t M n − k + s tM n − k − X ( x, y ) = ( − st, − st ): | ( − st ) H k +1 ( A ) + ( − st ) H k +2 ( A ) | ≤ ( st ) | H k +1 ( A ) | + ( st ) | H k +2 ( A ) | ≤≤ ( st ) M n − k + ( st ) M n − k − == ( st ) M n − k + ts ( t − s ) M n − k − + ts M n − k − ≤≤ ( st ) M n − k + s ( t − s ) M n − k + ts M n − k − == (2 ts − s ) M n − k + ts M n − k − ≤≤ t M n − k + ts M n − k − X ( x, y ) = ( t , s t ): | ( t ) H k +1 ( A ) + ( s t ) H k +2 ( A ) | ≤ t | H k +1 ( A ) | + s t | H k +2 ( A ) | ≤≤ t M n − k + s tM n − k − X . ( x, y ) = ( t , − st ): | ( t ) H k +1 ( A ) + ( − st ) H k +2 ( A ) | == t · abs (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) a ( k +1)( k +1) a ( k +1)( k +2) · · · a ( k +1) n s a ( k +2)( k +2) · · · a ( k +2) n ... . . . . . . ...0 · · · s a nn (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) − s · (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) a ( k +2)( k +2) a ( k +2)( k +3) · · · a ( k +2) n s a ( k +3)( k +3) · · · a ( k +3) n ... . . . . . . ...0 · · · s a nn (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ! == t · abs (cid:0) a ( k +1)( k +1) − s (cid:1) H k +2 ( A ) − s · (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) a ( k +1)( k +2) a ( k +1)( k +3) · · · a ( k +1) n s a ( k +3)( k +3) · · · a ( k +3) n ... . . . . . . ...0 · · · s a nn (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ! ≤≤ t · abs (cid:0) a ( k +1)( k +1) − s (cid:1) H k +2 ( A ) ! + t s · abs (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) a ( k +1)( k +2) a ( k +1)( k +3) · · · a ( k +1) n s a ( k +3)( k +3) · · · a ( k +3) n ... . . . . . . ...0 · · · s a nn (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ! ≤≤ t · abs (cid:0) a ( k +1)( k +1) − s (cid:1) H k +2 ( A ) ! + t s · M n − k − == t · (cid:12)(cid:12) a ( k +1)( k +1) − s (cid:12)(cid:12) · (cid:12)(cid:12) H k +2 ( A ) (cid:12)(cid:12) + t s · M n − k − ≤≤ t · (cid:12)(cid:12) a ( k +1)( k +1) − s (cid:12)(cid:12) · M n − k − + t s · M n − k − ≤≤ t · max { t − s, s } · M n − k − + t s · M n − k − If t − s = max { t − s, s } : | t H k +1 ( A ) − st H k +2 ( A ) | ≤ t · ( t − s ) · M n − k − + t s · M n − k − == t M n − k − ≤≤ t M n − k ≤≤ t M n − k + ts M n − k − X If s = max { t − s, s } : | t H k +1 ( A ) − st H k +2 ( A ) | ≤ t · s · M n − k − + t s · M n − k − == (2 t s − ts ) M n − k − + ts M n − k − ≤≤ (2 ts − s ) M n − k + ts M n − k − ≤≤ t M n − k + ts M n − k − X Hence (3.4) is done. And (3.3) is trivial.Using the triangle inequality and Lemma 3.2 in (3.1) yields the following inequality for M n :(3.5) M n ≤ t M n − + ts M n − + t s M n − + ts M n − + t s M n − + ts M n − + · · · Note as well that we have M = 1, M = t , M = t . efine a new sequence ( K n ) n ≥ in the following way: K = 1, K = t , K = t and K n = tK n − + s K n − , for all n ≥
3. We state two simple lemmas concerning this sequence.
Lemma K n = tK n − + ts K n − + ts K n − + · · · for all n ≥ .Proof. We use induction on n . For n = 1 and n = 2, the equality is trivial. It is straightforward toverify that if the statement holds for n ∈ { , , . . . , k } and then it holds for n = k + 1 where k ≥
2. By usingthe induction assumption: K k +1 = tK k + s K k − = tK k + ts K k − + ts K k − + ts K k − + · · · Lemma K n = t K n − + ts K n − + t s K n − + ts K n − + · · · for all n ≥ .Proof. For n = 2 it is clear. Using Lemma 3.3, for any n ≥ K n = tK n − + s K n − == t · (cid:0) tK n − + ts K n − + ts K n − + · · · (cid:1) + s · (cid:0) tK n − + ts K n − + ts K n − + · · · (cid:1) == t K n − + ts K n − + t s K n − + ts K n − + · · · Lemma K n ≥ M n for all n ∈ N .Proof. Notice that we already know, by Lemma 3.4 and (3.5), that K = M , K = M , K = M K n = t K n − + ts K n − + t s K n − + ts K n − + · · · M n ≤ t M n − + ts M n − + t s M n − + ts M n − + · · · By induction, it is clear that K n ≥ M n , ∀ n ∈ N . Lemma K n ≤ M n for all n ∈ N .Proof. It suffices to give an example A ∈ G n × ns ([0 , t ]) for which the absolute determinant value is equalto K n . Define an n × n matrix as follows:(3.6) U n ( s, t ) = U n := t t · · · · · · s t t · · · · · · s t · · · · · · s t .. . · · · ... .. . . .. . .. .. . ...0 0 0 0 .. . t
00 0 0 0 · · · s t n × n i.e., a ij = t if j ≥ i and j − i is even; a ij = 0 if j > i and j − i is odd. Note that | U | = t = K , | U | = t = K and by using Laplace expansion it is easy to see that(3.7) | U n | = t · | U n − | + s | U n − | which is the same recurrence relation as for ( K n ) n ≥ . Hence, for all positive integers n , we have K n = | U n | ≤ M n . s a corollary of Lemma 3.5 and 3.6, M n = | U n | = K n . This completes the proof of Theorem 2.6.QED.In the next section we discuss what happens when s ≥ t >
0. We are able to say something in tworegimes.
4. Case II.
We consider first the case when s & t , i.e., s is slightly greater than t .Define permutation matrices P ( r ) n and P ( c ) n in the following way: Let P ( r ) n be obtained by interchangingthe first two rows of the n × n identity matrix and P ( c ) n be obtained by interchanging the last two columnsof the n × n identity matrix. And then define U ( r ) n ( s, t ) = U ( r ) n , U ( c ) n ( s, t ) = U ( c ) n , U ( rc ) n ( s, t ) = U ( rc ) n in G n × ns ( { , t } ) for n ≥ P ( r ) n · U ( r ) n = s t · · · · · · t t t · · · · · · s t · · · · · · s t ... · · · ... .. . .. . . .. ... ...0 0 0 0 ... t
00 0 0 0 · · · s t , U ( c ) n · P ( c ) n = t t · · · · · · s t t · · · · · · s t · · · · · · s t .. . · · · ... .. . .. . . .. .. . ...0 0 0 0 .. . t
00 0 0 0 · · · t s (4.1) and P ( r ) n · U ( rc ) n · P ( c ) n = s t · · · · · · t t t · · · · · · s t · · · · · · s t .. . · · · ... .. . . .. .. . .. . ...0 0 0 0 .. . t
00 0 0 0 · · · t s
We can determine the relation between the determinants of these matrices and the determinant of U n in the following way.Using the Laplace expansion for the first column of P ( r ) n · U ( r ) n we obtain(4.2) abs( | U ( r ) n | ) = abs (cid:0)(cid:12)(cid:12) P ( r ) n · U ( r ) n (cid:12)(cid:12)(cid:1) = s · | U n − | + ts · | U n − | and similarly(4.3) abs( | U ( c ) n | ) = s · | U n − | + ts · | U n − | for all n ≥
4. And again using the Laplace expansion for both the first column and the last row of P ( r ) n · U ( rc ) n · P ( c ) n we get | U ( rc ) n | = (cid:12)(cid:12) P ( r ) n · U ( rc ) n · P ( c ) n (cid:12)(cid:12) = s · | U n − | + 2 ts · | U n − | + t s · | U n − | (4.4)for all n ≥
4, defining | U | := 0 for convenience. roposition For the statement of Theorem 2.6, is the exact upper bound for st . More explicitly,for any s > t and n ≥ , the matrix that gives the maximum absolute determinant is not U n .Proof. If s > t , by (3.7) and (4.4) we have the following when n ≥ | U ( rc ) n | = s | U n − | + 2 ts | U n − | + t s | U n − | > s | U n − | + ts | U n − | + (cid:0) t | U n − | + t s | U n − | (cid:1) == s | U n − | + ts | U n − | + t | U n − | == s | U n − | + t | U n − | = | U n | (4.5)That means Theorem 2.6 is not valid for any s > t and n ≥ Proposition
For any n ≥ , s, t > consider the matrix in G n × ns ([0 , t ]) which has the maximumabsolute determinant among this set. We call it maximizing matrix. By Theorem 2.6 we know the maximizingmatrix for < st ≤ . Proposition 4.1 shows that the maximizing matrix changes at st = 1 . We claim thatthe maximizing matrix changes from U n to U ( rc ) n at st = 1 .Proof. Suppose that for an n ≥ U n to a matrix R n = R n ( s, t ) ∈G n × ns ([0 , t ]) at st = 1. Then the absolute value of the determinant of R n (1 ,
1) must be equal to the absolutevalue of the determinant of U n (1 , | R n (1 , | ) = | U n (1 , | if and only if R n (1 , ∈ (cid:8) U n (1 , , U ( r ) n (1 , , U ( c ) n (1 , , U ( rc ) n (1 , (cid:9) . Hence R n ∈ (cid:8) U n , U ( r ) n , U ( c ) n , U ( rc ) n (cid:9) .Furthermore, we have the following inequalities for 2 t > s > t with n ≥ | U n − | (2 t − s )( s − t ) s + s ( s − t ) | U n − | > ⇒ ts | U n − | + ( s + t s ) | U n − | > ( s + 2 st ) | U n − | + 2 ts | U n − |⇒ s | U n − | + 2 ts | U n − | + t s | U n − | > s | U n − | + ts | U n − |⇒ | U ( rc ) n | > abs( | U ( c ) n | ) = abs( | U ( r ) n | )(4.6)by (3.7), (4.2), (4.3) and (4.4).So, in (4.5) and (4.6) we have shown that for 2 > st > U ( rc ) n among these four matrices. Therefore R n = U ( rc ) n .In view of the fact that U ( rc ) n has become an important matrix in our investigation, we establish therecurrence relation satisfied by its determinant in the following proposition. Proposition | U ( rc )4 | = 3 s t , | U ( rc )5 | = s t + 4 s t and (cid:0) | U ( rc ) n | (cid:1) n ≥ satisfies the same recurrencerelation as M n : (4.7) | U ( rc ) n | = t · | U ( rc ) n − | + s · | U ( rc ) n − | for any n ≥ .Proof. By substituting (4.4) and using (3.7) | U ( rc ) n | = s | U n − | + 2 ts | U n − | + t s | U n − | == t (cid:0) s | U n − | + 2 ts | U n − | + t s | U n − | (cid:1) + s (cid:0) s | U n − | + 2 ts | U n − | + t s | U n − | (cid:1) == t · | U ( rc ) n − | + s · | U ( rc ) n − | emark U ( rc ) n has the maximum absolute determinant value inthe case 1 ≤ st ≤ ǫ ( n ) for sufficiently small ǫ ( n ) > n . So, Propositions 4.2 and 4.3 finishthe proof of Theorem 2.8.
5. Case III.
The third case we consider is when s ≫ t >
0, i.e. s is sufficiently larger than t , dependingon n .We start with an important observation which illustrates the reason why we are interested in s ≫ t instead of s ≥ t . Observation
The case when s ≥ t > is excessively general to reach a conclusion on M n . Moreexplicitly, for this case, there is no fixed matrix structure that gives the maximum absolute determinant forall values of s > t .Proof. Consider the case when s = 100 and t = 1 and n = 6. By using MATLAB and testing all possible2 cases, it is not difficult to check that the maximum absolute determinant for this case is 10006000000and this value is attained by only two matrices: A = , A = If there is a unique matrix that maximizes the absolute determinant value for all s > t in G × s ( { , t } ),then it must be the matrix U ( rc )6 by Proposition 4.2. Hence U ( rc )6 (100 ,
1) must be the same with either A or A , but clearly it is not. So, the maximizing matrix still depends on the ratio st .The next theorem describes the case s ≫ t > Theorem
The maximum absolute determinant of an n × n upper-Hessenberg matrix with entriesfrom the interval [0 , t ] and subdiagonal entries fixed at s such that s ≫ t > (i.e. s can be taken sufficientlylarger than t for any case) is given by s n − t + j n kj n − k s n − t . (We shall go on afterwards to establisha precise lower bound for st such that this statement holds.) Define M n := max A ∈G n × ns ([0 ,t ]) abs( | A | ) again. The proof is in two parts; the first is giving an example toshow that M n ≥ s n − t + j n kj n − k s n − t , and the second showing that this example has the maximumabsolute determinant. We start with the first part. Define a matrix V n for n ≥ (5.1) V n := t t t t · · · t s · · · t s · · · t s · · · t ... ... . .. ... .. . ... ...0 0 0 0 · · · t · · · s t n × n roposition | V n | = ( − n ( n − t s n − for all n ≥ .Proof. For n = 2, it is clear. We use induction on n , specifically, we assume that the statement is validfor n = 2 , , . . . , k − n = k . By using Laplace expansion for the last row,and the induction assumption: | V k | = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) t t t t · · · t s · · · t s · · · t s · · · t ... ... . .. .. . .. . ... ...0 0 0 0 · · · t · · · s t (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) k × k = t · (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) t t t t · · · t ts · · · s · · · s · · · · · · · · · s (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ( k − × ( k − − s · | V k − | == t · ( ts k − ( − k − ) − s · (( − k − ( k − t s k − ) == t s k − ( − k + ( k − t s k − ( − k == ( − k ( k − t s k − Now, we define a new sequence of matrices ( W n ) n ≥ depending on the parity of n . (5.2) W k +1 := k z }| { t t · · · t k z }| { · · · ts · · · t t · · · t s · · · t t · · · t · · · s t t · · · t
00 0 · · · s · · · t · · · s .. . 0 t ... .. . . .. .. . ... ...0 0 · · · · · · · · · s t W k +2 := k z }| { t t · · · t k+1 z }| { · · · ts · · · t t t · · · t s · · · t t t · · · t · · · s t t t · · · t
00 0 · · · s · · · t · · · s t ... . .. .. . . .. ... ...... . .. .. . . .. ... ...0 0 · · · · · · · · · s t Notice that W k +1 contains a k × k block full of t ’s whereas W k +2 contains a k × ( k + 1) block; and notethat if we define W k +2 such that it would contain a ( k + 1) × k block of t ’s instead of a k × ( k + 1) block itwould not change the determinant value. For later convenience we denote the alternative version W ′ k +2 Proposition | W n | = j n − k ( − s ) n − t − s | W n − | for all n ≥ .Proof. We need to show | W k +1 | = kt ( − s ) k − − s | W k | and | W k +2 | = kt ( − s ) k − − s | W k +1 | . Weare going to show only one, because the proofs in both cases are essentially identical.Using Laplace expansion for the last row of W k +2 (and by iteratively – exactly k times – doing it tothe last row of each the resulting matrices), and by using Proposition 5.3 we get: | W k +2 | = t · s k · ( − k · | V k +1 | − s | W k +1 | == t · s k · ( − k · ( − k +1 kt s k − − s | W k +1 | == kt ( − s ) k − − s | W k +1 | roposition | W n | = ( − n − (cid:16) s n − t + j n kj n − k s n − t (cid:17) for all n ≥ .Proof. Using the previous proposition, this follows straightforwardly by induction.As a corollary of the last proposition, we have(5.3) M n ≥ abs( | W n | ) = s n − t + j n kj n − k s n − t for all n ≥ G n × ns ( { , t } ) instead of G n × ns ([0 , t ]).Notice that the determinant value of any A ∈ G n × ns ( { , t } ) is a polynomial in s and t . More explicitly,the determinant of A is an element of the following set:( − n − s n − t ·{ , } +( − n − s n − t · n , , . . . , (cid:18) n − (cid:19)o +( − n − s n − t · n , . . . , (cid:18) n − (cid:19)o + · · · + t n ·{ , } Recall that we assumed s ≫ t and already know (5.3). Then M n ≥ s n − t + j n kj n − k s n − t > s n − t · (cid:18) n − (cid:19) + s n − t · (cid:18) n − (cid:19) + s n − t · (cid:18) n − (cid:19) + · · · == abs (cid:16) ( − n − s n − t · (cid:18) n − (cid:19) + ( − n − s n − t · (cid:18) n − (cid:19) + · · · (cid:17) (5.4)This means that the sign of the determinant which gives the maximum absolute determinant cannot be( − n − . So, its sign is ( − n − .By (5.3) and using s ≫ t again,(5.5) M n ≥ s n − t + j n kj n − k s n − t > s n − t · (cid:18) n − (cid:19) + s n − t · (cid:18) n − (cid:19) + · · · So, the maximum absolute determinant must contain the term s n − t , i.e., the maximum absolute deter-minant is of the form s n − t + · · · .Moreover, we also have the following M n ≥ s n − t + j n kj n − k s n − t >> s n − t + ( − s n − t + s n − t · (cid:18) n − (cid:19) + s n − t · (cid:18) n − (cid:19) + · · · (5.6)This means that the maximum absolute determinant cannot contain the term s n − t , i.e., the maximumabsolute determinant is of the form 1 · s n − t + 0 · s n − t + · · · . gain by s ≫ t and (5.3), we can state: M n ≥ s n − t + j n kj n − k s n − t >> s n − t + (cid:16)j n kj n − k − (cid:17) s n − t + s n − t · (cid:18) n − (cid:19) + s n − t · (cid:18) n − (cid:19) + · · · (5.7)Therefore, the coefficient of s n − t in the maximum absolute determinant must be at least j n kj n − k . Lemma
For any matrix A ∈ G n × ns ( { , t } ) , if the coefficient of s n − t in | A | is zero, then the absolutevalue of the coefficient of s n − t is at most j n kj n − k .Proof. Consider the determinant as a polynomial with variable s . It is easy to see that for A = a a a · · · a n − a n s a a · · · a n − a n s a · · · a n − a n s . .. a n − a n ... . .. .. . . .. .. . ...0 0 0 · · · s a nn , the determinant of A can be expressed as follows: | A | =( − n − s n − · a n + ( − n − s n − · (cid:0) a a n + a a n + a a n + · · · + a n − a nn (cid:1) ++ ( − n − s n − · (cid:16) X ≤ i 2) chessboardby observing that setting a i = 0 corresponds to colouring the i th row (from the top) black, and setting a jn = 0 corresponds to colouring the ( n − j + 1) st column (from the right) black for i ∈ { , , . . . , n − } and j ∈ { , , . . . , ( n − } . (Notice that we do not colour for the cases a = 0 and a nn = 0) (See Figure 1). a a a a a a a a a a a a a a Figure 1 . Case n = 9 and a = a = a = a = a = 0 The number of black squares is less than or equal to the number of zero terms in the expansion (5.11).We know that the total number of colored rows and columns is at least ( n − 3) since k + k ≥ n − 3. Itis clear that to minimize the number of the black squares, coloured columns must be the leftmost ones andcoloured rows must be lowermost ones, and | k − k | must be 0 or 1 (depending on the parity of n ). SeeFigure 2 for the minimizing example in the case n = 9.It is easy to calculate that the minimum number of black squares is at least(5.12) 2 · n − · n − · 12 = n − n + 34if n is odd, and(5.13) n − · n − · 12 + n − · n · 12 = n − n + 44if n is even.Hence, in the expansion (5.11), we know that at least n − n + 34 or n − n + 44 terms are zero. Andthe number of terms in (5.11), is (cid:0) n − (cid:1) = n − n + 22 . a a a a a a a a a a a a a Figure 2 . Case n = 9 and a = a = a = a = a = a = 0 Therefore, by (5.12) and (5.13) the number of nonzero terms in the (5.11) is less than or equal to(5.14) n − n + 22 − n − n + 34 = n − n + 14 = j n kj n − k if n is odd, and(5.15) n − n + 22 − n − n + 44 = n − n j n kj n − k if n is even. As a consequence, by (5.8), (5.14) and (5.15) the absolute value of the coefficient of s n − t isat most j n kj n − k .Hence, the maximum absolute determinant is1 · s n − t + 0 · s n − t + j n kj n − k s n − t + · · · by (5.6), the previous lemma and (5.7).Now we are going to consider the entries of the matrix that makes the coefficient of s n − t equal to j n kj n − k .If n is odd, say n = 2 k + 1, we have a (2 k − × (2 k − 1) half chessboard, and we colour 2 k − j n kj n − k , we colour exactly ( k − 1) rows (the lowermostones) and ( k − 1) columns (the leftmost ones). This means we set(5.16) a k − = a k − = · · · = a k +1) = 0(5.17) a k +1) = a k +1) = · · · = a ( k +1)(2 k +1) = 0 dditionally all the nonzero terms in (5.11) must be t , which means(5.18) a = a = · · · = a k = t (5.19) a ( k +2)(2 k +1) = a ( k +3)(2 k +1) = · · · = a (2 k +1)(2 k +1) = t (5.20) a ij = t for all ( i, j ) ∈ { , , . . . , ( k + 1) } × { ( k + 1) , ( k + 2) , . . . , (2 k ) } because there is a term a i − · a ij · a ( j +1)(2 k +1) that corresponds to a white square in the chessboardrepresentation for each ( i, j ) ∈ { , , . . . , ( k + 1) } × { ( k + 1) , ( k + 2) , . . . , (2 k ) } .Moreover, recall the second equation at (5.10) and we already have a = a (2 k +1)(2 k +1) = t by (5.18)and (5.19), then(5.21) a k +1) = a k ) = 0We already know that the coefficient of s n − in (5.8) is not zero, so(5.22) a k +1) = t So far we have arrived at the following matrix structure by (5.16)-(5.22)(5.23) k z }| { t t · · · t k z }| { · · · ts ? ? ? t t · · · t s ? ? t t · · · t · · · s t t · · · t 00 0 · · · s ? ? ? t · · · s ? ? t ... . . . . . . . . . ? ...0 0 · · · · · · · · · s t For the case when n is even, we have the same, but according to the choice of the difference betweenthe number of zero terms in the sets { a , a , . . . , a n − } and { a n , . . . , a ( n − n } , i.e. ( k − k ), as − W k +2 or W ′ k +2 as defined at (5.2). From now on, we just consider the odd case,the even case can be done in exactly the same way.Recall that we know that the maximum absolute determinant is1 · s n − t + 0 · s n − t + j n kj n − k s n − t + · · · Because of (5.3) and s ≫ t we have the following inequality, M n ≥ s n − t + j n kj n − k s n − t > > s n − t + j n kj n − k s n − t + ( − · s n − t + s n − t · (cid:18) n − (cid:19) + s n − t · (cid:18) n − (cid:19) + · · · Hence, the coefficient of s n − t must be 0 to have the maximum absolute determinant. Now we aregoing to show that this fact forces all entries with a question mark in (5.23) to be filled with 0. Proposition Recall that we have defined V n in (5.1) . If any of the entries of the triangular blockof ’s in the upper half is t instead of , then the determinant contains the term s n − t .Proof. Consider the following permutation: t · · · t · · · · · · t s ... t .. . ..... . ... ... s · · · t ... ... ... s · · · · · · t . .. ... s t This permutation gives the term s n − t , and because all the terms with s n − have the same sign, s n − t does not vanish. Proposition Consider the matrix in (5.23) , if there is t in any of the entries that are filled with aquestion mark, then the determinant has the term s n − t .Proof. Let n = 2 k + 1 be odd, the other case can be done by the same way. Suppose that the t is placedin the top-left triangular block of ?’s WLOG. Then consider the determinant as follows: (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) t t · · · t · · · ts ? ? ? t t · · · t s ? ? t t · · · t · · · s t t · · · t 00 0 · · · s ? ? ? t · · · s ? ? t ... ... ... ... ? ...0 0 · · · · · · · · · s t (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) The determinant of the upper-left ( k + 1) × ( k + 1) square contains the term t s k − by Proposition 5.7.And as we have boxed, there is a permutation that gives a ts k − term from the bottom-right square. Whenwe consider these two together we get the term t s k − which is exactly what we are looking for.As a corollary, we can state that all the entries with question mark must be filled with 0 to have themaximum absolute determinant. Therefore the matrix which has the maximum absolute determinant is thematrix W k +1 as we have defined at (5.2). QED. ow we can discuss the condition s ≫ t . We have used this in our proof in the following ways in (5.4),(5.5), (5.6), (5.7) and (5.24). We can rewrite these inequalities setting x := st : x n − + j n kj n − k x n − > x n − · (cid:18) n − (cid:19) + x n − · (cid:18) n − (cid:19) + x n − · (cid:18) n − (cid:19) + · · · (5.25) x n − + j n kj n − k x n − > x n − · (cid:18) n − (cid:19) + x n − · (cid:18) n − (cid:19) + · · · (5.26) x n − + j n kj n − k x n − > x n − · (cid:18) n − (cid:19) + x n − · (cid:18) n − (cid:19) + · · · (5.27) x n − > x n − · (cid:18) n − (cid:19) + x n − · (cid:18) n − (cid:19) + · · · (5.28) x n − > x n − · (cid:18) n − (cid:19) + x n − · (cid:18) n − (cid:19) + · · · (5.29)Note that except in the last inequality, these allow us to take x asymptotic to n . However, the last onerequires n . Fortunately we can overcome this issue using another way to tackle the problem. Recall thatwe deduced (5.29) from (5.24). Before stating (5.24), we found that the maximum absolute determinant hasthe form 1 · s n − t + 0 · s n − t + j n kj n − k s n − t + · · · and the matrix with this determinant has the form(5.23) Lemma Let A = t t · · · t · · · ts a · · · a k t t · · · t s ... ... t t · · · t ... ... a kk ... ... ... ... ... · · · s t t · · · t 00 0 · · · s a ( k +2)( k +2) · · · a ( k +2)(2 k ) t · · · s ... ... t ... ... ... ... ... a (2 k )(2 k ) ... · · · · · · · · · s t , where A ∈ G n × ns ( { , t } ) . And let (5.30) abs( | A | ) = 1 · s n − t + j n kj n − k s n − t − b · s n − t + b · s n − t − b · s n − t + · · · Then, b m · n ≥ b m +1 for all m ∈ { , , . . . , n − } .Proof. Recall that in the permutation definition, if we consider the determinant as a function of s , theabsolute value of the coefficient of s n − l is:(5.31) X ≤ i < ···
1) + ( k − ≤ n herefore, we get that the preimage of any element in B m has cardinality less than or equal to n andthis means | B m | · n ≥ | B m +1 | .Note that: b m · t m = X ≤ i < ··· − (2) · n these inequalities hold for n ≥ .Proof. Start with the first one, (5.25), we know that32 · n < − (2) · n < x , then x n − (cid:18) n − (cid:19) + x n − (cid:18) n − (cid:19) + x n − (cid:18) n − (cid:19) + · · · < x n − · n + x n − · n 3! + x n − · n 5! + · · · << · x n − + (cid:0) (cid:1) · x n − 3! + (cid:0) (cid:1) · x n − 5! + · · · << · x n − · (cid:0) · · · (cid:1) << · x n − · sinh(1) < x n − << x n − + j n kj n − k x n − X or the second one (5.26), use the first inequality (5.25), x n − + j n kj n − k x n − > x n − (cid:18) n − (cid:19) + x n − (cid:18) n − (cid:19) + x n − (cid:18) n − (cid:19) + · · · >> x n − · n · (cid:18) n − (cid:19) + x n − · n · (cid:18) n − (cid:19) + · · · >> x n − · (cid:18) n − (cid:19) + x n − · (cid:18) n − (cid:19) + · · · X For the third case (5.27), x n − · (cid:18) n − (cid:19) + x n − · (cid:18) n − (cid:19) + · · · < x n − · n 2! + x n − · n 4! + x n − · n 6! + · · · << x n − · (cosh − (2)) 2! + x n − · (cosh − (2)) 4! + x n − · (cosh − (2)) 6! + · · · << x n − · (cosh − (2)) 2! + x n − · (cosh − (2)) 4! + x n − · (cosh − (2)) 6! + · · · << x n − · (cosh − (2)) 2! + x n − · (cosh − (2)) 4! + x n − · (cosh − (2)) 6! + · · · << x n − · (cid:20) cosh (cid:0) cosh − (2) (cid:1) − (cid:21) = x n − < x n − + j n kj n − k x n − X And for the last inequality (5.28), x n − · (cid:18) n − (cid:19) + x n − · (cid:18) n − (cid:19) + · · · < x n − · n 4! + x n − · n 6! + · · · << x n − · (cosh − (2)) 4! + x n − · (cosh − (2)) 6! + · · · << x n − · (cosh − (2)) 4! + x n − · (cosh − (2)) 6! + · · · << x n − · (cosh − (2)) 4! + x n − · (cosh − (2)) 6! + · · · << x n − · (cid:20) cosh (cid:0) cosh − (2) (cid:1) − (cid:21) = x n − X Remark − (2) since we are not interested in thestrict lower bound. Thus Theorem 5.2 and Lemma 5.10 finish the proof of Theorem 2.9. Remark n for our proof to hold. 6. Concluding Remarks. We found the maximum absolute determinants (and the matrices givingthe corresponding values) for the cases st > · n (Theorem 2.9) and 0 ≤ st ≤ st ≤ t for Theorem 2.6 (Proposition 4.1), and found the first maximizing matrix after st = 1 (Proposition 4.2)with the recurrence relation satisfied by its absolute determinant value (Proposition 4.3). Nevertheless, forthe most part the problem as to what happens if 1 < st < · n still remains open. Note that as st goes to1 from 45 · n , the matrix which gives the maximum absolute determinant changes from having a localizedto a much more uniform structure. In view of these observations, further investigations might be based onthe following questions:1. Which matrices maximize the absolute determinant as st goes to 1 from n , in other words, whatare the transition forms?2. It is reasonable to work on the case s = 2 and t = 1 as a first step to understanding the transitionform. Is U ( rc ) n the maximizing matrix in this case when n ≥ n ≥ st = 1 from U n to U ( rc ) n . When does the secondtransition occur, in other words what is the maximum possible value of ǫ ( n ) in Remark 4.4?5. What should be the precise lower bound in Theorem 5.2? Is it asymptotic to n ?6. When do other transitions occur?7. Note that the number of t ’s in the matrices U n (3.6), U ( rc ) n (4.1) and W n (5.2) are equal. Is it truethat for any fixed n , all maximizing matrices have the same number of t ’s?8. Moreover, there is no alternating sign among the nonzero permutations in the determinants for anyof these three matrices. Is this condition valid for all maximizing matrices? Acknowledgements. We thank Professor N. J. Higham for drawing our attention to questions con-cerning Bohemian Matrices. The work reported here was carried out as part of an undergraduate summerresearch project by Mr Ahmet Abdullah Kele¸s at the University of Bristol, June–July 2019. Mr Kele¸s is grate-ful for financial support and the hospitality of the School of Mathematics at the University of Bristol duringhis visit. JPK was supported by a Royal Society Wolfson Research Merit Award, EPSRC Programme GrantEP/K034383/1 LMF: L -Functions and Modular Forms, and by ERC Advanced Grant 740900 (LogCorRM). REFERENCES[1] Sylvester J. J. “Thoughts on inverse orthogonal matrices, simultaneous sign successions, and tessellated pavements in twoor more colours, with applications to Newton’s rule, ornamental tile-work, and the theory of numbers.” Philos Mag .1867; 34: 461-475.[2] Hadamard J. “Resolution d’une question relative aux determinants.” Bull Sci Math . 1893; 17: 240– 246.[3] Bressoud, D. Proofs and Confirmations: The Story of the Alternating Sign Matrix Conjecture . 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