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proceedings of the
american mathematical
Volume 79, Number 2, June 1980
society
POSriTVE DEFINITE MATRICES AND CATALANNUMBERS
FRANK THOMSON LEIGHTON AND MORRIS NEWMAN1
Abstract. It is shown that the number of n x n integral triple diagonal matrices
which are unimodular, positive definite and whose sub and super diagonal elements
are all one, is the Catalan number (^)/(n + 1). More generally, it is shown that if
A is a fixed integral symmetric matrix and d is a fixed positive integer, then there
are only finitely many integral diagonal matrices D such that A + D is positive
definite and det(/4 + D) - d.
1. Introduction. It is well known and easy to prove that any integral symmetric
matrix A is integrally congruent to a triple diagonal matrix (see [1, p. 60], for
example). Joan Birman asked the second author whether it is always possible to
make all the elements on the sub and super diagonals one, provided that det(v4) =
1. It is readily shown that this is not possible. For example, if A is positive definite
and is not integrally congruent to the identity, then A cannot be integrally
congruent to a matrix of the type desired, since all such matrices are integrally
congruent to the identity (this fact is proved in Theorem 2). The question raised by
Professor Birman suggested that it would be of interest to study the class of n X n
integral triple diagonal matrices which are unimodular, positive definite and whose
sub and super diagonal elements are all one. We will show that this class is finite
and that its cardinality is just the Catalan number (^)/(/i + 1). This supplies a
rather surprising new interpretation for the Catalan numbers. In addition, we will
prove a general theorem of this type for arbitrary integral symmetric matrices.
If positive definiteness is not required, then the class of matrices under consideration need not be finite. For example, each of the matrices
-1
10]
1-11,
0
1 a\
\-a
1
10
0 1
0
1
a - 1
is unimodular for all integral values of a.
2. A general result. We first prove a result for arbitrary integral symmetric
matrices. In the following lemma, EtJ denotes the matrix with a one in position (/',/)
and zeros elsewhere.
Received by the editors March 20, 1978 and, in revised form, May 1, 1979; presented to the Society,
April 23, 1977.
AMS (A/OS) subjectclassifications(1970).Primary 15A36,05A15.
Key words and phrases. Catalan number, congruence, determinant, integer matrix, positive definite
matrix, triple diagonal matrix, unimodular matrix.
'The work of this author was supported by NSF grant MCS76-8293.
© 1980 American Mathematical
0002-9939/8O/0000-O254/$02.25
177
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Society
178
F. T. LEIGHTON AND MORRIS NEWMAN
Lemma 1. If B = [b¡j] is a symmetric positive definite matrix, c is a positive
number, and B' = B + cEMfor some value of j, then det(B') > det(B).
Proof. If we compute det(B') by minors of the y'th row, we find that det(B') =
det(B) + cdj where dj is the cofactor of bM,and hence positive, since B is positive
definite.
□
We now prove
Theorem 1. If A is a fixed n X n integral symmetric matrix and d is a fixed
positive integer, then the number of integral diagonal matrices D such that A + D is
positive definite and det(A + D) = d is finite.
Proof. We may assume without loss of generality that A has zeros on the
diagonal. Then if A + D is positive definite, D must have positive diagonal
elements. Let tf) be the class of all D such that A + D is positive definite and
det(A + D) = d. From repeated application of Lemma 1, it is clear that if F = [/•,]
and G = [g¡\ are different elements of 6D, then there exist k and m such that
fkk > Skk and fmm < Emm-Otherwise it would not be possible for det(y4 + F) to
equal det(A + G). Thus ''D forms an antichain (order-independent set) in N". Since
all antichains in N" are known to be finite [2], there can be only a finite number of
elements in fy. □
It is clear that generalizations of this result are possible (the diagonal matrices D
may be replaced by symmetric positive definite matrices, for example) but we do
not consider these here.
3. The basic framework. Define
1
A = td(ax, a2,...,
a„) =
(1)
1
fl»-i
1
1
a.
Let Dk be the determinant of the minor matrix of A obtained by striking out the
last n — k rows and columns, and let Ek be the determinant of the minor matrix of
A obtained by striking out the first n — k rows and columns. In addition, define
D_x = 0,
D0=l,
E_x = 0,
E0=l.
(2)
Then the Dk's and the Eks satisfy the recurrences
Dk = akDk_x
D,k-2>
Ek = an
1 < k < n,
1 < k < n.
Jk-2>
t + lEtk-l
Furthermore, expanding det(/4) by minors of the kth row of A, we find that
det(yi) = akDk_xE„_k
= £>kEn-k -
(3)
- Dk_2En_k - Dk_xE„_k_x
r)k-iE„_k_x,
1 < k < n.
(4)
We are interested in the matrices A given by (1) which are unimodular and
positive definite. A necessary and sufficient condition for this to happen is that
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POSITIVE DEFINITE MATRICES AND CATALANNUMBERS
179
ax, a2, . . . , a„ be positive integers such that
Dk > 0,
1 < k < n - 1, and Dn= I.
(5)
The set of such matrices A will be denoted by S„, and will be the object of our
study. We commence with some elementary lemmas.
Lemma 2. If A = td(ax, . . . , a„) E Sn, then ak = 1 for some k, 1 < k < n.
Proof.
Assume that a¡ > 2 for 1 < / < n. Then D0 =1,
Dx = ax > 2, and
Dk = akDk_x - Dk_2 > 2Dk_x - Dk_2 for 2 < k < n. Thus Dk - Dk_x > Dk_x
— Dk_2 for 2 < k < n and £>t > k + 1 for 1 < k < w. In particular, this means
that Dn > n + 1 > 1 for n > 1, which is a contradiction.
□
In the following, we write [1] + C to denote the direct sum of the matrices [1]
and C. In addition, we use the sequence notation ak, . . . , ar — I to denote the
sequence ak, ak+x, . . ., ar_x, ar — 1 when r > k, the integer ak — 1 when r = k,
and the empty sequence when r < k. Similar interpretations are understood for
ak, . . ., ar + 1, for ak — 1, . . ., ar, and for ak + I, . . . , ar. We are now ready to
prove
Lemma 3. Let ax, . . ., an and k be integers such that ak = 1 and I < k < n. Then
A = td(a„ . . . ,an) is integrally congruent to [1] 4- td(a,, . . ., ak_x — 1, ak+x —
1, . . . , a„).
Proof. Perform the following congruence transformations on A. Subtract row k
from rows k - 1 and k + 1 (as far as present) and then do the same with the
columns. Multiply the first k — 1 rows and columns by -1. Now permute the rows
and columns of the matrix according to the permutation (123 • • • k). It is not
difficult to check that the resulting matrix is [1] + td(ax, . . ., ak_x — 1, ak+x —
l,...,an).
D
As an important consequence of Lemma 3, we note without proof:
Lemma 4. td(ax, . . . , an) E S„ if and only if td(ax, . . ., ak_x + 1, 1, ak +
1, . . . , an) G Sn+xfor
every integer k, 1 < k < n + 1.
□
4. The class Sn. We are now prepared to prove the main results of this paper.
Theorem
2. Every element of Sn is integrally congruent to the identity.
Proof. The result is trivial for n = 1. Suppose the theorem is true for n — 1
where n > 2. Let A be any element of S„. From Lemmas 2, 3 and 4, we know that
there is a matrix B such that A is integrally congruent to [1] + B and B G Sn_,. By
the induction hypothesis, B, and thus A, is congruent to the identity. □
Theorem 3. The set S„ has cardinality (^)/(n
+ 1).
Proof. The proof consists of two parts. In the first, we show that for every r,
1 < r < n, any element of Sr_x may be "composed" with any element of S„_r to
determine in a unique fashion an element of S„. In the second part of the proof, we
show that, conversely, every element of Sn is uniquely "decomposable" into an
element of Sr_x and an element of S„_r for exactly one value of r, 1 < r < n. It
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180
F. T. LEIGHTON AND MORRIS NEWMAN
will then immediately follow that
|$J-
2 \Sr-i\ \S„-r\,
n>l,
\S0\=l.
(6)
r~\
Since the recurrence relation (6) defines the Catalan numbers, we will thus have
shown that \Sn\ = ft)/(n
+ 1).
If A = td(iz„ . . . , a„) E Sn, then for each r, 1 < r < n, ar is uniquely determined
by the other a„ 1 < / < n, i =£ r. This follows from formulas (2)-(4) and the fact
thatdet(,4) = 1. In fact,
a, - 0 + 0,-23,-,
+ Dr-xEn-,-x)/Dr-xEH-r-
(?)
Because of (7), we can also say that if td(ax, . . . , ar_x) G Sr_x and td(ar+1, . . . ,
an) E Sn_r for any integers ax,. . . , ar_x, ar+x, . . ., a„, then there is a unique
integer ar such that td(o,, . . . , an) E Sn. In fact, since Dr_x — En_r = 1, ar is given
by
ar = l + Dr_2 + E„_r_x.
(8)
This completes the first part of the proof.
Now suppose that td(a,, . . ., an) G S„. We say that a, is a breakpoint
td(o,, . . . ,ar_x)
E Sr_x and
td(ar+1, . . . , an) E S„_r.
In the following,
if
we will
show that every element of S„ possesses a breakpoint. This is trivially true for
n = 1. Assume that it is true for n — 1 where n > 2. Let A = td(ax, . . ., a„) be any
element of S„. We know by Lemmas 2 and 4 that there is a k such that 1 < k < n,
ak = 1 and B = td(a„ . . . , ak_x — I, ak+x - I, . . . , an) E S„_x. By the induction
hypothesis, this element has a breakpoint which divides the sequence of diagonal
elements into two subsequences of lengths r — 1 and n — r — I, say, and which
yield elements of Sr_x and Sn_r_x. The subsequence that does not contain
ak_x — 1 or ak+x — 1 is contained identically in the original sequence of diagonal
elements ax, . . . ,an. The other subsequence is empty or contains one or both of
ak_x — 1 and ak+x — 1. In any case, we know by Lemma 4 that we can insert a 1
and add a 1 to the adjoining number(s) to get the complementary part of
ax, .. ., an, and that this new sequence will be the sequence of diagonal elements of
a member of Sr or of 5n_r. Thus A has a breakpoint,
We now show that the location of the breakpoint
for n = 1. Assume that it is true for n — 1 where n
S„, and suppose that A has two breakpoints a, and
and the induction is complete.
is unique. This is certainly true
> 2. Let A = td(ax, . . ., an) E
a}, i i*j. By (8), we must have
ar = 1 + Dr_2 + En_r_x for r = i,j. It follows that if n > 1, then a¡ > 1 and
a, > 1. Thus there is a ac different from / and/ such that ak = 1. But this means
that B = td(a,, . . ., ak_x — 1, ak+x — 1, . . ., an) E Sn_x. Thus a, and a, are also
breakpoints of B, which contradicts the inductive hypothesis. Finally, formula (8)
insures the uniqueness of the value of ar at the breakpoint. This completes the
induction and establishes the uniqueness of the decomposition.
□
5. Additional results. In this section, we list some miscellaneous results, some of
which have not been proved completely.
(9) If td(a„ . . ., an) E S„ and D0, . . . , Dn are as in formulas (2)-(5), then
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POSITIVE DEFINITE MATRICES AND CATALAN NUMBERS
181
(i) a¡ < n for 1 < í < n,
(ii)2n - 1 < 2"_,a, < 3n - 3,
(iii) Dk/Dk+X < k + 1 for 0 < k < n,
(iv) Dk/Dk_x < n - k + 1 for 0 < k < n.
(10) The number of elements of S„ with trace t is given by
/,_
n(3»-,-i)(3,,-,-2)
lf2/I_1<i<3n_3)
\ n-2
I
n- 1
0 otherwise.
(11) The number of elements of S„ with exactly i diagonal elements equal to one
is given by
2"~"^l(n
(n -2i+
- 1)!
!)!/!(/-
1)!
if 0 < i <
w+ 3
2 J'
0 otherwise.
(9) is straightforward but (10) and (11) seem more difficult. We do not believe
them to offer any fundamental problems however, and the reader is invited to
supply proofs.
We are indebted to the referee for his careful reading of the paper and for his
numerous valuable comments.
References
1. Morris Newman, Integral matrices, Academic Press, New York, 1972.
2. Richard Stanley, Elementary problems and solutions problem E2546, Amer. Math. Monthly 83 (10)
(1976),813-814.
National
Bureau of Standards, Operations Research Division, Washington, D.C 20234
Institute for the Interdisciplinary Applications of Algebra and Combinatorics, Department
of Mathematics, University of California, Santa Barbara, California 93106 (Current address of
Morris Newman)
Current address (F. T. Leighton): Applied Mathematics Department, Massachusetts Institute of
Technology, Cambridge, Massachusetts 02139
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