ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.)
ARS MATHEMATICA CONTEMPORANEA 20 (2021) 261–274
https://doi.org/10.26493/1855-3974.2248.d3f
(Also available at http://amc-journal.eu)
A double Sylvester determinant*
Darij Grinberg †
Drexel University, Korman Center, 15 S 33rd Street, Philadelphia PA, 19104, USA
Received 7 February 2020, accepted 22 August 2020, published online 18 November 2021
Abstract
Given two (n+ 1) × (n+ 1)-matrices A and B over a commutative ring, and some
k ∈ {0, 1, . . . , n}, we consider the
(
n
k
)
×
(
n
k
)
-matrix W whose entries are (k + 1)×(k + 1)-
minors of A multiplied by corresponding (k + 1)× (k + 1)-minors of B. Here we require
the minors to use the last row and the last column (which is why we obtain an
(
n
k
)
×
(
n
k
)
-
matrix, not a
(
n+1
k+1
)
×
(
n+1
k+1
)
-matrix). We prove that the determinant detW is a multiple of
detA if the (n+ 1, n+ 1)-th entry of B is 0. Furthermore, if the (n+ 1, n+ 1)-th entries
of both A and B are 0, then detW is a multiple of (detA) (detB). This extends a previous
result of Olver and the author.
Keywords: Determinant, compound matrix, Sylvester’s determinant, polynomials.
Math. Subj. Class. (2020): 15A15, 11C20
1 Introduction
Let n and k be nonnegative integers, and let A = (ai,j)1≤i≤n+1, 1≤j≤n+1 be an (n+ 1)×
(n+ 1)-matrix over some commutative ring. Let Pk be the set of all k-element subsets
of {1, 2, . . . , n}. For any such subset K ∈ Pk, let K+ denote the subset K ∪ {n+ 1}
of {1, 2, . . . , n+ 1}. If U and V are two subsets of {1, 2, . . . , n+ 1}, then subVU A shall
denote the |U |×|V |-submatrix of A containing only the entries au,v with u ∈ U and v ∈ V.
Let WA be the Pk × Pk-matrix1 whose (I, J)-th entry (for all I ∈ Pk and J ∈ Pk) is
det
(
subJ+I+ A
)
.
*The author would like to thank Christian Krattenthaler, Peter Olver and Victor Reiner for enlightening dis-
cussions, and Peter Olver for the joint work that led to this paper. The SageMath computer algebra system [14]
has been used for experimentation leading up to some of the results below.
†Homepage: http://www.cip.ifi.lmu.de/˜grinberg/
E-mail address: darijgrinberg@gmail.com (Darij Grinberg)
1This means a matrix whose rows and columns are indexed by the k-element subsets of {1, 2, . . . , n}. If you
pick a total order on the set Pk , then you can view such a matrix as an
(n
k
)
×
(n
k
)
-matrix.
cb This work is licensed under https://creativecommons.org/licenses/by/4.0/
262 Ars Math. Contemp. 20 (2021) 261–274
(Thus, the entries of WA are all (k + 1) × (k + 1)-minors of A that use the last row and
the last column.) A particular case of a celebrated result going back to Sylvester [15] (see
[12, §2.7] or [13, Teorema 2.9.1] or [10] for modern proofs) then says that
det (WA) = a
p
n+1,n+1 · (detA)
q
, where p =
(
n− 1
k
)
and q =
(
n− 1
k − 1
)
.
Now, consider a second (n+ 1) × (n+ 1)-matrix B = (bi,j)1≤i≤n+1, 1≤j≤n+1 over
the same ring. Let WA,B (later to be just called W ) be the Pk ×Pk-matrix whose (I, J)-th
entry (for all I ∈ Pk and J ∈ Pk) is
det
(
subJ+I+ A
)
det
(
subJ+I+ B
)
.
What can be said about det (WA,B)? In general, very little2. However, under some as-
sumptions, it splits off factors. Namely, we shall show (Theorem 2.1) that det (WA,B) is
a multiple of detA if bn+1,n+1 = 0. We shall then conclude (Theorem 2.2) that if both
an+1,n+1 and bn+1,n+1 are 0, then det (WA,B) is a multiple of (detA) (detB). In either
case, the quotient (usually a much more complicated polynomial3) remains mysterious; our
proofs are indirect and reveal little about it. Our second result generalizes a curious prop-
erty of
(
n
2
)
×
(
n
2
)
-determinants [6, Theorem 10] that arose from the study of the n-body
problem (see Example 2.4 for details).
2 The theorems
Let us first introduce the standing notations.
Let N = {0, 1, 2, . . .}. Let K be a commutative ring. If a and b are two elements of K,
then we write a | b when b is a multiple of a (that is, b ∈ Ka).
If m ∈ N, then [m] shall mean the set {1, 2, . . . ,m}.
Fix an n ∈ N. If K is any subset of [n], then K+ shall mean the subset K ∪ {n+ 1}
of [n+ 1].
Fix k ∈ {0, 1, . . . , n}. Let Pk be the set of all k-element subsets of [n]. This is a finite
set; thus, any Pk × Pk-matrix (i.e., any matrix whose rows and columns are indexed by
k-element subsets of [n]) has a well-defined determinant4. Such matrices appear frequently
in classical determinant theory (see, e.g., the “k-th compound determinants” in [11] and in
[12, §2.6], as well as the related “Generalized Sylvester’s identity” in [12, §2.7] and [13,
Teorema 2.9.1] and [10]).
If A ∈ Ku×v is a u× v-matrix, and if I ⊆ [u] and J ⊆ [v], then subJI A shall mean the
submatrix of A obtained by removing all rows whose indices are not in I and removing all
columns whose indices are not in J . (Rigorously speaking, if A = (ai,j)1≤i≤u, 1≤j≤v and
I = {i1 < i2 < · · · < ip} and J = {j1 < j2 < · · · < jq}, then subJI A is defined to be the
matrix (aix,jy )1≤x≤p, 1≤y≤q .) When |I| = |J |, then the submatrix sub
J
I A is square; its
determinant det(subJI A) is called a minor of A.
2For example, if n = 3 and k = 2, then det
(
WA,B
)
is an irreducible polynomial in the (altogether
2 (n+ 1)2 = 32) variables ai,j and bi,j with 110268 monomials.
3Again, irreducible in the case when n = 3 and k = 2.
4Here, we are using the concepts of P × P -matrices (where P is a finite set) and their determinants. Both of
these concepts are folklore; a brief introduction can be found in [5, §1].
D. Grinberg: A double Sylvester determinant 263
Our main two results are the following:
Theorem 2.1. Let
A = (ai,j)1≤i≤n+1, 1≤j≤n+1 ∈ K
(n+1)×(n+1) and
B = (bi,j)1≤i≤n+1, 1≤j≤n+1 ∈ K
(n+1)×(n+1)
be such that bn+1,n+1 = 0. Let W be the Pk × Pk-matrix whose (I, J)-th entry (for all
I ∈ Pk and J ∈ Pk) is
det
(
subJ+I+ A
)
det
(
subJ+I+ B
)
.
Then detA | detW .
Theorem 2.2. Let
A = (ai,j)1≤i≤n+1, 1≤j≤n+1 ∈ K
(n+1)×(n+1) and
B = (bi,j)1≤i≤n+1, 1≤j≤n+1 ∈ K
(n+1)×(n+1)
be such that an+1,n+1 = 0 and bn+1,n+1 = 0. Define the Pk × Pk-matrix W as in
Theorem 2.1. Then (detA) (detB) | detW .
Example 2.3. For this example, set k = 1. Then Pk = P1 = {{1} , {2} , . . . , {n}}. Thus,
the map
[n] → Pk, i 7→ {i}
is a bijection. Use this bijection to identify the elements 1, 2, . . . , n of [n] with the elements
{1} , {2} , . . . , {n} of Pk. Thus, the Pk×Pk-matrix W in Theorem 2.1 becomes the n×n-
matrix(
det
(
sub
{j}+
{i}+ A
)︸ ︷︷ ︸
=ai,jan+1,n+1
−ai,n+1an+1,j
det
(
sub
{j}+
{i}+ B
)︸ ︷︷ ︸
=bi,jbn+1,n+1
−bi,n+1bn+1,j
)
1≤i≤n, 1≤j≤n
=
(
(ai,jan+1,n+1 − ai,n+1an+1,j) (bi,j bn+1,n+1︸ ︷︷ ︸
=0
−bi,n+1bn+1,j)
)
1≤i≤n, 1≤j≤n
= ((ai,jan+1,n+1 − ai,n+1an+1,j)(−bi,n+1bn+1,j))1≤i≤n, 1≤j≤n.
This is the matrix obtained from (ai,jan+1,n+1 − ai,n+1an+1,j)1≤i≤n, 1≤j≤n by multiply-
ing the i-th row with −bi,n+1 for all i ∈ [n] and multiplying the j-th column with bn+1,j
for all j ∈ [n]. Thus, the claim of Theorem 2.1 follows from the classical fact that
det
(
(ai,jan+1,n+1 − ai,n+1an+1,j)1≤i≤n, 1≤j≤n
)
= an−1n+1,n+1 · detA.
This fact is known as Chio pivotal condensation (see, e.g., [7, Theorem 0.1]), and is a
particular case of Sylvester’s identity ([12, §2.7]).
264 Ars Math. Contemp. 20 (2021) 261–274
Example 2.4. For this example, set k = 2, and consider the situation of Theorem 2.1
again. Then Pk = P2 = {{i, j} | 1 ≤ i < j ≤ n}. If {i, j} ∈ P2 and {k, l} ∈ P2 satisfy
i < j and k < l, then the ({i, j} , {k, l})-th entry of W is
det
(
sub
{k,l}+
{i,j}+ A
)
det
(
sub
{k,l}+
{i,j}+ B
)
= det
 ai,k ai,l ai,n+1aj,k aj,l aj,n+1
an+1,k an+1,l an+1,n+1
 det
 bi,k bi,l bi,n+1bj,k bj,l bj,n+1
bn+1,k bn+1,l 0
 .
Note that bn+1,n+1 = 0. If we furthermore assume that
an+1,n+1 = 0, and
an+1,i = ai,n+1 = 1 for all i ∈ [n] , and
bn+1,i = bi,n+1 = 1 for all i ∈ [n] ,
then this entry rewrites as
det
ai,k ai,l 1aj,k aj,l 1
1 1 0
 det
bi,k bi,l 1bj,k bj,l 1
1 1 0

= (aj,k + ai,l − ai,k − aj,l) (bj,k + bi,l − bi,k − bj,l) .
Hence, [6, Theorem 10] can be obtained from Theorem 2.2 by setting k = 2 and A = CS
and B = CT (and observing that the matrix W then equals to WS,T ).
3 The proofs
Our proofs of Theorem 2.1 and Theorem 2.2 will rely on some basic commutative algebra:
the notion of a unique factorization domain (“UFD”); the concepts of coprime, prime and
irreducible elements; the localization of a commutative ring at a multiplicative subset. This
all appears in most textbooks on abstract algebra; for example, [8, Sections VIII.4 and
VIII.10] is a good reference5.
The content of a polynomial p over a UFD is defined to be the greatest common divisor
of the coefficients of p. For example, the polynomial 4x2 + 6y2 ∈ Z [x, y] has content
gcd (4, 6) = 2. (Of course, in a general UFD, the greatest common divisor is defined only
up to multiplication by a unit.) The following known facts are crucial to us:
Proposition 3.1. A polynomial ring over Z in finitely many indeterminates is always
a UFD. □
Proposition 3.1 appears, e.g., in [8, Remark after Corollary 8.21]. For a constructive
proof of Proposition 3.1, we refer to [9, Chapter IV, Theorems 4.8 and 4.9] or to [2, Es-
say 1.4, Corollary of Theorem 1 and Corollary 1 of Theorem 2].
Proposition 3.2. Let p be an irreducible element of a UFD K. Then the quotient ring
K/ (p) is an integral domain.
5We call “multiplicative subset” what Knapp (in [8, Section VIII.10]) calls a “multiplicative system”.
D. Grinberg: A double Sylvester determinant 265
Proof of Proposition 3.2. First of all, we recall that any irreducible element of a UFD is
prime (indeed, this follows from [8, Proposition 8.13]). Thus, the element p of K is prime.
Hence, [8, Proposition 8.14] shows that the ideal (p) of K is prime. Therefore, the quotient
ring K/ (p) is an integral domain. This proves Proposition 3.2.
We shall furthermore use the following properties of contents (whose proofs are easy):
Proposition 3.3. Let U be a UFD. Let p ∈ U [x1, x2, . . . , xm] be a polynomial over U.
Assume that the content of p is 1. Also assume that p is irreducible when considered as
a polynomial in F [x1, x2, . . . , xm], where F is the field of fractions of U. Then p is also
irreducible when considered as a polynomial in U [x1, x2, . . . , xm].
Proposition 3.4. Let U be a UFD. Let p, q ∈ U [x1, x2, . . . , xm] be two polynomials over
U. Assume that both p and q have content 1, and assume furthermore that p and q don’t
have any indeterminates in common (i.e., there is no i ∈ [m] such that degxi p > 0 and
degxi q > 0). Then p and q are coprime.
The next simple fact states that for any positive integer n, the determinant of the
“generic n × n-matrix” (i.e., of the n × n-matrix whose n2 entries are distinct indeter-
minates in a polynomial ring over Z) is irreducible as a polynomial over Z:
Corollary 3.5. Let n be a positive integer. Let G be the multivariate polynomial ring
Z
[
ai,j | (i, j) ∈ [n]2
]
. Let A ∈ Gn×n be the n × n-matrix (ai,j)1≤i≤n, 1≤j≤n. Then the
element detA of G is irreducible.
Proof of Corollary 3.5. A well-known fact (e.g., [1, Lemma 5.12]) shows that detA is
irreducible as an element of Q
[
ai,j | (i, j) ∈ [n]2
]
. This yields (using Proposition 3.3) that
detA is irreducible as an element of Z
[
ai,j | (i, j) ∈ [n]2
]
as well, since the polynomial
detA has content 1. This proves Corollary 3.5.
An element a of a commutative ring A is said to be regular if every b ∈ A satisfying
ab = 0 must satisfy b = 0. (Regular elements are also known as non-zero-divisors.) In a
polynomial ring, each indeterminate is regular; hence, each monomial (without coefficient)
is regular (since any product of two regular elements is regular).
We recall a few standard concepts from commutative algebra. Let K be a commutative
ring. A multiplicative subset of K means a subset S of K that contains the unity 1K of K
and has the property that every a, b ∈ S satisfy ab ∈ S.
If S is a multiplicative subset of K, then the localization of K at S is defined as follows:
Let ∼ be the binary relation on the set K× S defined by
((r, s) ∼ (r′, s′)) ⇐⇒ (t (rs′ − sr′) = 0 for some t ∈ S) .
Then it is easy to see that ∼ is an equivalence relation. The set L of its equivalence
classes [(r, s)] can be equipped with a ring structure via the rules [(r, s)] + [(r′, s′)] =
[(rs′ + sr′, ss′)] and [(r, s)] · [(r′, s′)] = [(rr′, ss′)] (with zero element [(0, 1)] and unity
[(1, 1)]). The resulting ring L is commutative, and is known as the localization of K at S.
(This generalizes the construction of Q from Z known from high school.)
The element [(r, s)] of L is denoted by rs . There is a canonical ring homomorphism
from K to L that sends each r ∈ K to [(r, 1)] = r1 ∈ L.
266 Ars Math. Contemp. 20 (2021) 261–274
When all elements of the multiplicative subset S are regular, the statement “t(rs′ −
sr′) = 0 for some t ∈ S” in the definition of the relation ∼ can be rewritten in the
equivalent (but much simpler) form “rs′ = sr′” (which is even more reminiscent of the
construction of Q).
The following fact is easy to see:
Proposition 3.6. Let K be a commutative ring. Let S be a multiplicative subset of K such
that all elements of S are regular. Let L be the localization of the ring K at S. Then:
(a) The canonical ring homomorphism from K to L is injective. We shall thus consider
it as an embedding.
(b) If K is an integral domain, then L is an integral domain.
(c) Let a and b be two elements of K. Then we have the following logical equivalence:
(a | b in L) ⇐⇒ (a | sb in K for some s ∈ S) .
Matrices over arbitrary commutative rings can behave a lot less predictably than matri-
ces over fields. However, matrices over integral domains still show a lot of the latter good
behavior, such as the following:
Proposition 3.7. Let P be a finite set. Let M be an integral domain. Let W ∈ MP×P be
a P × P -matrix over M. Let u ∈ MP be a vector such that u ̸= 0 and Wu = 0. Here, u
is considered as a “column vector”, so that Wu is defined by
Wu =
∑
q∈P
wp,quq

p∈P
, where W = (wp,q)(p,q)∈P×P and u = (up)p∈P .
Then detW = 0.
Proof of Proposition 3.7. Let m = |P |. Then we can view the P × P -matrix W as an
m × m-matrix (by “numerical reindexing”, as explained in [5, §1]), and we can view the
vector u as a column vector of size m. Let us do this from here on.
Let F be the quotient field of the integral domain M. Thus, there is a canonical embed-
ding of M into F. Hence, we can view the matrix W ∈ Mm×m as a matrix over F, and we
can view the vector u ∈ Mm as a vector over F. Let us do so from here on. We are now
in the realm of classical linear algebra over fields: The vector u ∈ Fm is nonzero (since
u ̸= 0) and belongs to the kernel of the m × m-matrix W ∈ Fm×m (since Wu = 0).
Hence, the kernel of the matrix W is nontrivial. In other words, this matrix W is singular.
Thus, detW = 0 by a classical fact of linear algebra. This proves Proposition 3.7.
Let us next recall an identity for determinants (a version of the Cauchy–Binet formula):
Lemma 3.8. Let n ∈ N, m ∈ N and p ∈ N. Let A ∈ Kn×p be an n × p-matrix. Let
B ∈ Kp×m be a p×m-matrix. Let k ∈ N. Let P be a subset of [n] such that |P | = k. Let
Q be a subset of [m] such that |Q| = k. Then
det
(
subQP (AB)
)
=
∑
R⊆[p];
|R|=k
det
(
subRP A
)
· det
(
subQR B
)
.
D. Grinberg: A double Sylvester determinant 267
Lemma 3.8 is [4, Corollary 7.251] (except that we are using the notation subJI C for
what is called subw(J)w(I) C in [4]). It also appears in [3, Chapter I, (19)] (where it is stated
using p-tuples instead of subsets).
The next lemma is just a particular case of Theorem 2.1, but it is a helpful stepping
stone on the way to proving the latter theorem:
Lemma 3.9. Let
A = (ai,j)1≤i≤n+1, 1≤j≤n+1 ∈ K
(n+1)×(n+1) and
B = (bi,j)1≤i≤n+1, 1≤j≤n+1 ∈ K
(n+1)×(n+1)
be such that bn+1,n+1 = 0. Assume further that
an+1,j = 0 for all j ∈ [n] . (3.1)
Define the Pk × Pk-matrix W as in Theorem 2.1. Then detA | detW .
The following proof is inspired by [6, proof of Theorem 10].
Proof of Lemma 3.9. We WLOG assume that K is the polynomial ring over Z in n2 +
(n+ 1) + ((n+ 1)
2 − 1) indeterminates
ai,j for all i ∈ [n] and j ∈ [n] ;
ai,n+1 for all i ∈ [n+ 1] ;
bi,j for all i ∈ [n+ 1] and j ∈ [n+ 1] except for bn+1,n+1.
And, of course, we assume that the entries of A and B that are not zero by assumption are
these indeterminates.6
The ring K is a UFD (by Proposition 3.1).
We WLOG assume that n > 0 (otherwise, the result follows from detW =
det
(
0
)
= 0).
The set Pk is nonempty (since k ∈ {0, 1, . . . , n}); thus, |Pk| ≥ 1.
Let A be the n × n-matrix (ai,j)1≤i≤n, 1≤j≤n ∈ K
n×n. Then, because of (3.1), we
have
detA = an+1,n+1 · detA (3.2)
(by [4, Theorem 6.43], applied to n+ 1 instead of n).
The matrix A is a completely generic n× n-matrix (i.e., its entries are distinct indeter-
minates); thus, its determinant detA is an irreducible polynomial in the polynomial ring
Z
[
ai,j | (i, j) ∈ [n]2
]
(by Corollary 3.5). Hence, detA also is an irreducible polynomial
in the ring K (since K differs from Z
[
ai,j | (i, j) ∈ [n]2
]
only in having more variables,
which clearly cannot contribute any factors to detA). Thus, Proposition 3.2 (applied to
p = detA) shows that the quotient ring K/(detA) is an integral domain.
Let M be the quotient ring K/(detA). Then M is an integral domain (since K/(detA)
is an integral domain). All monomials in the variables bi,j (with (i, j) ̸= (n+ 1, n+ 1))
are nonzero in M. Likewise, an+1,n+1 ̸= 0 in M.
6These assumptions are legitimate, because if we can prove Lemma 3.9 under these assumptions, then the
universal property of polynomial rings shows that Lemma 3.9 holds in the general case.
268 Ars Math. Contemp. 20 (2021) 261–274
Let w be the element
∏
j∈[n] bn+1,j ∈ M. (Strictly speaking, we mean the canonical
projection of
∏
j∈[n] bn+1,j ∈ K onto the quotient ring M.) Then, w is a nonzero element
of the integral domain M (since bn+1,j ̸= 0 in M for all j ∈ [n]).
For each i ∈ [n], we define zi ∈ M by zi =
∏
j∈[n]; j ̸=i bn+1,j (projected onto M).
This is a nonzero element of M. In M, we have
bn+1,izi = bn+1
∏
j∈[n];
j ̸=i
bn+1,j =
∏
j∈[n]
bn+1,j = w (3.3)
for all i ∈ [n].
We need another piece of notation: If M is a p × q-matrix, and if u ∈ [p] and v ∈ [q],
then M∼u,∼v denotes the (p− 1)× (q − 1)-matrix obtained from M by removing the u-th
row and the v-th column.
The matrix A∼1,∼(n+1) has determinant 0 (because (3.1) shows that its last row consists
of zeroes). In other words, det
(
A∼1,∼(n+1)
)
= 0.
Also, due to (3.1), we see that each i ∈ [n] satisfies
det (A∼1,∼i) = an+1,n+1 · det
(
A∼1,∼i
)
(3.4)
(by [4, Theorem 6.43], applied to A∼1,∼i instead of A), because the last row of the matrix
A∼1,∼i is (0, 0, . . . , 0, an+1,n+1).
For each i ∈ [n+ 1], we define an element ui ∈ M by
ui =
{
zi (−1)i det (A∼1,∼i) , if i ∈ [n] ;
1, if i = n+ 1.
Claim 1. All these n+ 1 elements u1, u2, . . . , un+1 of M are nonzero.
Proof of Claim 1. Let i ∈ [n]. Then, det
(
A∼1,∼i
)
̸= 0 in M because det
(
A∼1,∼i
)
is a
polynomial of smaller degree than detA, and thus is not a multiple of detA. Now,
ui = zi (−1)i
=an+1,n+1·det(A∼1,∼i) (by (3.4))︷ ︸︸ ︷
det (A∼1,∼i)
= zi︸︷︷︸
̸=0 in M
̸=0 in M︷ ︸︸ ︷
(−1)i an+1,n+1︸ ︷︷ ︸
̸=0 in M
·
̸=0 in M︷ ︸︸ ︷
det
(
A∼1,∼i
)
̸= 0 in M (since M is an integral domain).
Thus, u1, u2, . . . , un are nonzero. Moreover, un+1 is nonzero (since un+1 = 1). Thus, we
are done.
Let u = (uJ)J∈Pk ∈ M
Pk be the vector defined by
uJ =
∏
j∈J
uj .
D. Grinberg: A double Sylvester determinant 269
Then the entries of the vector u are nonzero (because they are products of the nonzero
elements u1, u2, . . . , un+1 of the integral domain M). Since the vector u has at least one
entry (because |Pk| ≥ 1), we thus conclude that u ̸= 0.
Let ∆ be the diagonal matrix ∆ = diag (u1, u2, . . . , un+1) ∈ M(n+1)×(n+1).
Let x ∈ Mn+1 be the column vector defined by
x =
(
(−1)1 det (A∼1,∼1) , (−1)2 det (A∼1,∼2) , . . . , (−1)n+1 det(A∼1,∼(n+1))
)T
.
Let (e1, e2, . . . , en+1) be the standard basis of the free M-module Mn+1. Thus, for any
(n+ 1)× (n+ 1)-matrix C ∈ M(n+1)×(n+1) and any j ∈ {1, 2, . . . , n+ 1}, we have
(the j-th column of the matrix C) = Cej . (3.5)
Now, using Laplace expansion, it is easy to see that
Ax = − detA · e1. (3.6)
To prove Equation (3.6), consider the adjugate adjA of the matrix A. A standard fact ([4,
Theorem 6.100]) says that A ·adjA = detA ·In+1. But the definition of adjA reveals that
the first column of the matrix adjA is −x. Hence, the first column of the matrix A · adjA
is A · (−x) = −Ax. On the other hand, the first column of the matrix A ·adjA is detA ·e1
(since A · adjA = detA · In+1). Comparing the preceding two sentences, we conclude
that −Ax = detA · e1, so that Ax = −detA · e1. This proves Equation (3.6).
Also, Equation (3.5) (applied to C = BT and j = n+ 1) yields
BT en+1 =
(
the (n+ 1)-st column of the matrix BT
)
= (bn+1,1, bn+1,2, . . . , bn+1,n+1)
T
.
Hence,
∆BT en+1 = ∆(bn+1,1, bn+1,2, . . . , bn+1,n+1)
T
= (u1bn+1,1, u2bn+1,2, . . . , un+1bn+1,n+1)
T
(3.7)
(since ∆ = diag (u1, u2, . . . , un+1)).
Claim 2. We have
uibn+1,i = w · (−1)i det (A∼1,∼i) for each i ∈ [n+ 1] . (3.8)
Proof of Claim 2. Let i ∈ [n+ 1]. If i = n+ 1, then both sides of (3.8) are zero (because
bn+1,n+1 = 0 and det
(
A∼1,∼(n+1)
)
= 0). If i ̸= n+1, then i ∈ [n] and thus the definition
of ui yields ui = zi(−1)i det(A∼1,∼i). Hence,
uibn+1,i = zi (−1)i det (A∼1,∼i) bn+1,i = bn+1,izi︸ ︷︷ ︸
=w (by (3.3))
(−1)i det (A∼1,∼i)
= w · (−1)i det (A∼1,∼i) .
Hence, Equation (3.8) is proven in both cases.
270 Ars Math. Contemp. 20 (2021) 261–274
Now, (3.7) becomes
∆BT en+1
= (u1bn+1,1, u2bn+1,2, . . . , un+1bn+1,n+1)
T
=
(
w · (−1)1 det (A∼1,∼1) , w · (−1)2 det (A∼1,∼2) , . . . ,
w · (−1)n+1 det
(
A∼1,∼(n+1)
) )T
(by (3.8))
= w ·
(
(−1)1 det (A∼1,∼1) , (−1)2 det (A∼1,∼2) , . . . , (−1)n+1 det
(
A∼1,∼(n+1)
))T
︸ ︷︷ ︸
=x (by the definition of x)
= wx.
Hence,
A∆BT en+1 = Awx = w ·
=− detA·e1 (by (3.6))︷︸︸︷
Ax = −w · detA︸ ︷︷ ︸
=an+1,n+1·detA (by (3.2))
· e1
= −w · an+1,n+1 · detA︸ ︷︷ ︸
=0 (since we are in M)
· e1 = 0.
In other words, the (n+ 1)-st column of the matrix A∆BT is 0 (since the (n+ 1)-st col-
umn of the matrix A∆BT is A∆BT en+1 (by (3.5), applied to C = A∆BT and j = n+1)).
Now, fix I ∈ Pk. Then, the last column of the matrix subI+I+(A∆BT ) is 0 (because
this column is a piece of the (n+ 1)-st column of the matrix A∆BT , but as we have just
shown the latter column is 0). Thus, det
(
subI+I+(A∆B
T )
)
= 0.
But Lemma 3.8 (applied to M, n + 1, n + 1, n + 1, ∆BT , k + 1, I+ and I+ instead
of K, n, m, p, B, k, P and Q) yields
det
(
subI+I+(A∆B
T )
)
=
∑
R⊆[n+1];
|R|=k+1
det
(
subRI+ A
)
det
(
subI+R (∆B
T )
)
.
Comparing this with det
(
subI+I+(A∆B
T )
)
= 0, we obtain
0 =
∑
R⊆[n+1];
|R|=k+1
det
(
subRI+ A
)
det
(
subI+R (∆B
T )
)
.
In the sum on the right hand side, all addends for which n + 1 /∈ R are zero (because if
R ⊆ [n+ 1] satisfies |R| = k + 1 and n+ 1 /∈ R, then the last row of the matrix subRI+ A
consists of zeroes (by (3.1), since n + 1 /∈ R but n + 1 ∈ I+), and therefore we have
det
(
subRI+ A
)
= 0), and thus can be discarded. Hence, we are left with
0 =
∑
R⊆[n+1];
|R|=k+1;
n+1∈R
det
(
subRI+ A
)
det
(
subI+R (∆B
T )
)
.
D. Grinberg: A double Sylvester determinant 271
But the subsets R of [n+ 1] satisfying |R| = k + 1 and n+ 1 ∈ R can be parametrized as
J+ with J ranging over Pk. Hence, this rewrites further as
0 =
∑
J∈Pk
det
(
subJ+I+ A
)
det
(
subI+J+(∆B
T )
)
.
It is easily seen that det
(
subI+J+(∆B
T )
)
= det
(
subJ+I+ B
)
uJ for each J ∈ Pk (indeed,
recall the definition of ∆ and the fact that un+1 = 1 and that det
(
CT
)
= detC for each
square matrix C). Thus, the above equality simplifies to
0 =
∑
J∈Pk
det
(
subJ+I+ A
)
det
(
subJ+I+ B
)
uJ .
Now, forget that we fixed I . We thus have proven that
0 =
∑
J∈Pk
det
(
subJ+I+ A
)
det
(
subJ+I+ B
)
uJ (3.9)
for each I ∈ Pk. This rewrites as Wu = 0 (indeed, the left hand side of (3.9) is the I-th
entry of the zero vector 0, whereas the right hand side of (3.9) is the I-th entry of Wu).
Now, consider the matrix W as a matrix in MPk×Pk . Then, Proposition 3.7 (applied to
P = Pk) yields detW = 0 in M (since u ̸= 0 and Wu = 0). In view of the definition of
M, this rewrites as detA | detW in K.
Let us consider the matrix W again as a matrix over K. Each entry of W has the form
det
(
subJ+I+ A
)
det
(
subJ+I+ B
)
for some I, J ∈ Pk.
Claim 3. det
(
subJ+I+ A
)
is a multiple of an+1,n+1 for all I, J ∈ Pk.
Proof of Claim 3. Let I, J ∈ Pk. Then, the equality (3.1) shows that the last row of the
matrix subJ+I+ A is (0, 0, . . . , 0, an+1,n+1). Hence, an application of [4, Theorem 6.43]
shows that det
(
subJ+I+ A
)
= an+1,n+1 det
(
subJI A
)
. Thus, det
(
subJ+I+ A
)
is a multiple
of an+1,n+1.
By Claim 3, all entries of W are multiples of an+1,n+1. Hence, the determinant of W
is a multiple of (an+1,n+1)
|Pk|, thus a multiple of an+1,n+1 (since |Pk| ≥ 1). In other
words, an+1,n+1 | detW in K.
Recall that K is a UFD. Also, the two polynomials an+1,n+1 and detA in K both
have content 1, and don’t have any indeterminates in common; thus, these two polynomials
are coprime (by Proposition 3.4). Hence, any polynomial in K that is divisible by both
an+1,n+1 and detA must be divisible by the product an+1,n+1 · detA as well. Thus, from
an+1,n+1 | detW and detA | detW , we obtain an+1,n+1 · detA | detW . In view of
(3.2), this rewrites as detA | detW . This proves Lemma 3.9.
We shall now derive Theorem 2.2 from Lemma 3.9, following the same idea as in [12,
§2.7] and [13, Teorema 2.9.1] and [10]:
Proof of Theorem 2.1. We WLOG assume that n > 0 (otherwise, the result follows from
detW = det
(
0
)
= 0).
272 Ars Math. Contemp. 20 (2021) 261–274
We WLOG assume that K is the polynomial ring over Z in (n+ 1)2 + ((n+ 1)2 − 1)
indeterminates
ai,j for all i ∈ [n+ 1] and j ∈ [n+ 1] ;
bi,j for all i ∈ [n+ 1] and j ∈ [n+ 1] except for bn+1,n+1.
And, of course, we assume that the entries of A and B that are not zero by assumption
are these indeterminates. Proposition 3.1 shows that the ring K is a UFD (since it is a
polynomial ring over Z).
Let S be the multiplicative subset
{
apn+1,n+1 | p ∈ N
}
of K. Then, all elements of S
are regular (since they are monomials in a polynomial ring).
Let L be the localization of the commutative ring K at the multiplicative subset S. Then,
Proposition 3.6(a) shows that the canonical ring homomorphism from K to L is injective;
we shall thus consider it as an embedding. Also, Proposition 3.6(b) shows that L is an
integral domain.
Claim 1. We claim that
detA | detW in L. (3.10)
Proof of Claim 1. Consider A, B and W as matrices over L. The entry an+1,n+1 of A
is invertible in L (by the construction of L). Hence, we can subtract appropriate scalar
multiples7 of the (n+ 1)-st column of A from each other column of A to ensure that
all entries of the last row of A become 0, except for an+1,n+1. (Specifically, for each
j ∈ [n], we have to subtract aj,n+1/an+1,n+1 times the (n+ 1)-st column of A from the j-
th column of A.) All these column transformations preserve the determinant detA, and also
preserve the minors det
(
subJ+I+ A
)
for all I, J ∈ Pk (because when the (n+ 1)-st column
of A is subtracted from another column of A, the matrix subJ+I+ A either stays the same or
undergoes an analogous column transformation8, which preserves its determinant); thus,
they preserve the matrix W . Hence, we can replace A by the result of these transformations.
This new matrix A satisfies (3.1). Hence, Lemma 3.9 (applied to L instead of K) yields
that detA | detW in L. This proves (3.10).
But we must prove that detA | detW in K. Fortunately, this is easy: Since K embeds
into L, we can translate our result “detA | detW in L” as “detA | apn+1,n+1 detW in K
for an appropriate p ∈ N” (by Proposition 3.6(c), applied to a = detA and b = detW ).
Consider this p.
Claim 2. The polynomial an+1,n+1 ∈ K is coprime to detA.
Proof of Claim 2. The polynomial detA contains the monomial a1,n+1a2,n · · · an+1,1 =∏
i∈[n+1] ai,n+2−i, and thus is not a multiple of an+1,n+1. Hence, it is coprime to an+1,n+1
(since the only non-unit divisor of an+1,n+1 is an+1,n+1 itself, up to scaling by units).
So we know that an+1,n+1 is coprime to detA. Hence, its power a
p
n+1,n+1 is co-
prime to detA as well. Hence, we can cancel the apn+1,n+1 from the divisibility detA |
apn+1,n+1 detW , and conclude that detA | detW in K. This proves Theorem 2.1.
7The scalars, of course, come from L here.
8Here we are using the fact that n + 1 ∈ J+ (so that the matrix subJ+I+ A contains part of the (n+ 1)-st
column of A).
D. Grinberg: A double Sylvester determinant 273
Proof of Theorem 2.2. We WLOG assume that K is the polynomial ring over Z in the
((n+ 1)
2 − 1) + ((n+ 1)2 − 1) indeterminates
ai,j for all i ∈ [n+ 1] and j ∈ [n+ 1] except for an+1,n+1;
bi,j for all i ∈ [n+ 1] and j ∈ [n+ 1] except for bn+1,n+1.
And, of course, we assume that the entries of A and B that are not zero by assumption are
these indeterminates. The ring K is a UFD (by Proposition 3.1).
WLOG assume that n > 0 (otherwise, the result follows from detW = det
(
0
)
= 0).
Thus, the monomial a1,n+1a2,n · · · an+1,1 =
∏
i∈[n+1] ai,n+2−i occurs in the polynomial
detA with coefficient ±1. Hence, the polynomial detA has content 1. Similarly, the
polynomial detB has content 1.
Theorem 2.1 yields detA | detW . The same argument yields detB | detW (since
the matrices A and B play symmetric roles in the construction of W ). But Proposition 3.4
shows that the polynomials detA and detB in K are coprime (because they have content 1,
and don’t have any indeterminates in common). Thus, any polynomial in K that is divisible
by both detA and detB must be divisible by the product (detA) (detB) as well. Thus,
from detA | detW and detB | detW , we obtain (detA) (detB) | detW . This proves
Theorem 2.2.
4 Further questions
While Theorems 2.1 and 2.2 are now proven, the field appears far from fully harvested.
Three questions readily emerge:
Question 4.1. What can be said about detWdetA (in Theorem 2.1) and
detW
(detA)(detB) (in Theo-
rem 2.2)? Are there formulas?
Question 4.2. Are there more direct proofs of Theorems 2.1 and 2.2, avoiding the use of
polynomial rings and their properties and instead “staying inside K”? Such proofs might
help answer the previous question.
Question 4.3. The entries of our matrix W were products of minors of two (n+ 1) ×
(n+ 1)-matrices that each use the last row and the last column. What can be said about
products of minors of two (n+m)× (n+m)-matrices that each use the last m rows and
the last m columns, where m is an arbitrary positive integer? The “Generalized Sylvester’s
identity” in [12, §2.7] answers this for the case of one matrix. It is not quite obvious what
the right analogues of the conditions an+1,n+1 = 0 and bn+1,n+1 = 0 are; furthermore,
nontrivial examples become even more computationally challenging.
ORCID iD
Darij Grinberg https://orcid.org/0000-0002-9661-8432
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