Partial-Sum Matrix and its Rank
THITARIE RUNGRATGASAME1, PUNNA CHARRASANGSAKUL2
1,2Department of Mathematics, Srinakharinwirot University,
Sukhumvit 23, Bangkok 10110, THAILAND
Abstract: A partial-sum matrix is a matrix whose entries are partial sums of a seires associate with a sequence.
The rank of a partial-sum matrix associate with any recurrence sequence can be related to the rank of an associate
recurrence matrix, a matrix whose entries are from the same recurrence sequence. In particular, we find ranks of
partial-sum matrices associated with arithmetic series and geometric series.
Key-Words: Partial Sum, Matrix, Rank, recurrence relation.
Received: June 3, 2023. Revised: September 2, 2023. Accepted: October 1, 2023. Published: October 20, 2023.
1 Introduction
A matrix is one of the well-known tools that has been
used to solve a system of linear equations in many ar-
eas, for example, economics, statistics, engineering,
or computational science, e.g., see [1], [2], to find
values of unknown varibales. The basic approach for
finding a solution of a system of linear eqautions is a
Gaussian elimination which is an algorithm to modify
a corresponding augmented matrix of the system to
be in a reduced row echelon form. However, a com-
plication of coefficients of linear equations can lead
to a difficulty in solving a system of the linear equa-
tions. In such cases, in stead of getting a solution, the
direction is changed to determine only the existence
of solutions. Finding a rank of an augmented matrix
and a coefficient matrix then plays an important role
in verifying whether the system has a solution. This
shows that a rank of a matrix is an interesting topic
to be studied. In particular, a rank of a matrix is one
of fundamental characteristics of the matrix itself. Its
importance leads to many research studies on rank of
a matrix and its applications, e.g., see [3], [4], [5]. In
this work, we are interested in finding a rank of some
special matrices which can be associated with a sys-
tem of linear equations with coefficients in the form
of partial sums of series.
A partial-sum matrix is a matrix whose en-
tries are partial sums of a series (read left-to-right,
row-by-row). For example, the repunit sequence
{1,11,111, ...}, which can be considered as a se-
quence of partial sums of a geometric series with the
common ratio 10, forms a 4×3partial-sum matrix as
follows:
1 11 111
1111 11111 111111
1111111 11111111 111111111
1111111111 11111111111 111111111111
.
The rank of this partial-sum matrix is equal to 2. Gen-
erally speaking, it can be shown that any m×npartial-
sum matrix of the repunit sequence is equal to 2for
any m, n ≥2. In general, the rank of a partial-sum
matrix of any recurrence sequence can be related to
the rank of an associate recurrence matrix, a matrix
whose entries are from the same recurrence sequence.
Our approach is to apply row operations on a partial-
sum matrix to determine its rank.
Definition 1.1. For a sequence (aj)of complex num-
bers where j∈N, a partial sum Sjassociated with
(aj)is defined by Sj=
j
i=1
aifor any j∈N. For
m, n ∈N, we define an m×npartial-sum matrix of
(aj), written by Smn(aj), to be the matrix
S1S2S3. . . Sn
Sn+1 Sn+2 Sn+3 . . . S2n
S2n+1 S2n+2 S2n+3 . . . S3n
.
.
..
.
..
.
..
.
..
.
.
S(m−1)n+1 S(m−1)n+2 S(m−1)n+3 . . . Smn
.
To study the rank of a partial-sum matrix, we use
the fact that the rank of any matrix is the same as
the rank of its transpose, see [6] (p.114). Given a
complex-valued sequence (aj), the transpose of an
m×npartial-sum matrix is denoted by ST
mn((aj))
such that
ST
mn((aj)) =
S1Sn+1 S2n+1 . . . S(m−1)n+1
S2Sn+2 S2n+2 . . . S(m−1)n+2
S3Sn+3 S2n+3 . . . S(m−1)n+3
.
.
..
.
..
.
..
.
..
.
.
SnS2nS3n. . . Smn
.
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Since Sj+1 =Sj+aj+1, we apply the row oporations
by adding each of the jth row to the multiple of the
(j+ 1)th row (multiplied by −1) starting from j=
n−1, n −2, n −3, ..., 1, respectively to obtain that
ST
mn((aj)) is row equivalent to
S1Sn+1 S2n+1 . . . S(m−1)n+1
a2an+2 a2n+2 . . . a(m−1)n+2
a3an+3 a2n+3 . . . a(m−1)n+3
.
.
..
.
..
.
..
.
..
.
.
ana2na3n. . . amn
.(1.1)
Let Amn((aj)) be the matrix
a2a3a4. . . an
an+2 an+3 an+4 . . . a2n
.
.
..
.
..
.
..
.
..
.
.
a(m−1)n+2 a(m−1)n+3 a(m−1)n+4 . . . amn
and let S1((aj)) denote the matrix
S1Sn+1 S2n+1 . . . S(m−1)n+1.
The row-reduced matrix in (1.1) implies that the rank
of Smn((aj)) is the same as the rank of S1((aj))
AT
mn((aj)).
In particular, if the row of S1((aj)) is linearly in-
dependent to all rows of AT
mn((aj)), then
rank Smn((aj)) = rank AT
mn((aj)) + 1
=rank Amn((aj)) + 1.(1.2)
Then finding the rank of Amn((aj)) becomes the key
to our goal. However, the matrix Amn((aj)) is related
to the matrix of the sequence (aj), as defined in [7]
and [8], for which we shall explain.
Definition 1.2. For a sequence (aj)of complex num-
bers where j∈Nand m, n ∈N, we define an m×n
matrix of the sequence (aj), denoted by Mmn(aj), to
be the matrix
a1a2a3. . . an
an+1 an+2 an+3 . . . a2n
a2n+1 a2n+2 a2n+3 . . . a3n
.
.
..
.
..
.
..
.
..
.
.
a(m−1)n+1 a(m−1)n+2 a(m−1)n+3 . . . amn
.
For any k= 1,2,3, ..., n, let Ck((aj)) be the ma-
trix
ak
an+k
a2n+k
.
.
.
a(m−1)n+k
.Then we rewrite Mmn((aj)) to
be Mmn((aj)) = [C1((aj)) Amn((aj))]
where
Amn((aj)) = [C2((aj)) C3((aj)) ... Cn((aj))].
Clearly, rank(Mmn((aj))) is the same as rank
[Amn((aj)) C1((aj))].Moreover, we can consider
[Amn((aj)) C1((aj))]as the augmented matrix of
the system of the following linear equations
a2x2+a3x3+· · · +anxn=a1
an+2x2+an+3x3+· · · +a2nxn=an+1
.
.
. (1.3)
a(m−1)n+2x2+a(m−1)n+3x3+· · · +amnxn
=a(m−1)n+1.
Notice that the system (1.3) has a solution if and only
if the rank of [Amn((aj)) C1((aj))]is the same as
the rank of Amn((aj)).That is, the system (1.3) has a
solution if and only if Mmn((aj)) has the same rank
as Amn((aj)).
From the above result and (1.2) ,
if S1((aj)) is linearly independent to
C2((aj)), C3((aj)), ..., Cn((aj)) and the system
(1.3) has a solution, then
rank Smn((aj)) = rank Mmn((aj)) + 1.(1.4)
We will apply (1.4) to study the rank of some partial-
sum matrices associated with special series in the next
section.
2 Rank of partial-sum matrices in
some types
2.1 Partial-sum matrices of arithmetic
series
Let (aj)be an arithmetic sequence with an initial
value a1and a nonzero common difference d. We
have that aj+k=ak+ (j−1)dfor all j, k ∈N.The
system (1.3) can be written in the following system
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a1(x2+x3+· · · +xn)
+d(x2+ 2x3+· · · + (n−1)xn) = a1
an+1(x2+x3+· · · +xn)
+d(x2+ 2x3+· · · + (n−1)xn) = an+1
.
.
. (2.1)
a(m−1)n+1(x2+x3+· · · +xn)
+d(x2+ 2x3+· · · + (n−1)xn) = a(m−1)n+1
which is equivalent to the system
x2+x3+· · · +xn= 1
x2+ 2x3+· · · + (n−1)xn= 0.(2.2)
The system (2.2) is a consistent system. Therefore,
(2.1) is also consistent. We can conclude that rank
Amn((aj)) = rank Mmn((aj)). C. Lee and V. Peter-
son showed in [8] that rank Mmn((aj)) = 2 when
m, n ≥2, and so rank Amn((aj)) = 2 when m, n ≥
2. Next, we will show that the row of S1((aj)) is lin-
early independent to all rows of AT
mn((aj)).
In this case, Amn((aj)) is column equivalent to
a1+d d 0. . . 0
a1+ (n+ 1)d d 0. . . 0
.
.
..
.
..
.
..
.
.
a1+ ((m−1)n+ 1)d d 0. . . 0
.
We have that the partial sum is Sk=ka1+k(k−1)
2d
for any k. Then S1((aj)) is linearly independent to
both
[a1+d a1+(n+1)d ... a1+((m−1)n+1)d]and
[d d d . . . d].
As a result, rank Smn((aj)) = rank Mmn((aj)) +
1 = 3 when m, n ≥2. We have proved the following
theorem.
Theorem 2.1. Let (aj)be an arithmetic sequence
with an initial value a1and a nonzero common dif-
ference d. Then a partial-sum matrix Smn((aj)) with
m, n ≥3has rank 3.
2.2 Partial-sum matrices of geometric series
Let (aj)be a geometric sequence with a nonzero ini-
tial value a1and a nonzero common ratio r. Accord-
ingly, aj+k=rjakfor all j, k ∈N.We rewrite the
system (1.3) to be
ra1x2+r2a1x3+· · · +rn−1a1xn
=a1
ran+1x2+r2an+1x3+· · · +rn−1an+1xn
=an+1
.
.
.
ra(m−1)n+1x2+r2a(m−1)n+1x3+· · ·
+rn−1a(m−1)n+1xn=a(m−1)n+1
which is equivalent to
rx2+r2x3+· · · +rn−1xn= 1.
The latter equation always has a solution, for exam-
ple, let x2=1
r, x3=x4=· · · =xn= 0.By the
same reason as in the case of an arithmetic sequence,
we derive that Smn((aj)) = rank Mmn((aj)) + 1 =
2, referring to [8] that rank Mmn((aj)) = 1 when
m, n ≥2for a geometric sequence (aj). The follow-
ing theorem is now proved.
Theorem 2.2. Let (aj)be a geometric sequence with
a nonzero initial value a1and a nonzero common
ratio r. Then a partial-sum matrix Smn((aj)) with
m, n ≥2has rank 2.
2.3 Partial-sum matrices of linear
recurrence relations
Let (aj)be a homogeneous linear recurrence se-
quence of order ksuch that for all j≥k
aj=α1aj−1+α2aj−2+α3aj−3+· · · +αkaj−k
(2.3)
with initial values a1, a2,· · · , akand α1, α2,· · · , αk
where αk= 0. An geometric sequence is an example
of a homogeneous linear recurrence sequence of order
1because aj=raj−1. Since αk= 0, we arrange
(2.3) to be that for any j∈N
aj=1
αk
aj+k−α1
αk
aj+k−1−α2
αk
aj+k−2− · · ·
−αk−1
αk
aj+1.(2.4)
Let m, n ≥k+ 1. We derive the following equations
afterward.
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a1=1
αk
ak+1 −α1
αk
ak−α2
αk
ak−1− · · · −
αk−1
αk
a2
an+1 =1
αk
an+1+k−α1
αk
an+k−α2
αk
an+k−1−
· · · − αk−1
αk
an+2
.
.
.
a(m−1)n+1 =1
αk
a(m−1)n+1+k−α1
αk
a(m−1)n+k−
α2
αk
a(m−1)n+k−1− · · · −
αk−1
αk
a(m−1)n+2.
Therefore, the system (1.3) has a solution, that is,
rank Mmn((aj)) = rank Amn((aj)).In this linear re-
currence sequence, the row R1of Smn((aj)) is lin-
early independent to all rows of AT
mn((aj)), and so
Smn((aj)) = rank Mmn((aj)) + 1.Here Mmn((aj))
must have a rank drop. To determine the rank of
Mmn((aj)), we shall recall the occurrence of a rank
drop in a recurrence matrix, which is first mentioned
in [8] and further studied in[7].
2.3.1 Order rank drops
The order rank drops in Mmn((aj)) occurs when
(aj)also satisfies a recurrence relation with order
less than k. For example, the recurrence sequence
aj= 4aj−1−3aj−2with a1= 1 and a2= 3. Thus
aj= 3j−1, and hence aj= 3aj−1. By Theorem 2.2,
rank Smn((aj)) = 2 since Mmn((aj)) = 1. This
means that Smn((aj)) has a rank drop if m, n ≥3.
We also call that Smn((aj)) has an order rank drop.
To clarify the order of a recurrence sequence in this
case, let the minimal order of (aj)be the smallest or-
der satisfied by (aj).
2.3.2 Width rank drops
Let (aj)be a recurrence sequence satisfying aj=
aj−3with a1= 1, a2= 0, a3= 0.
Then (aj) = (1,0,0,1,0,0,1,0,0, . . . )and (Sj) =
(1,1,1,2,2,2,3,3,3, . . . ). However, both
S44((aj)) =
1 1 1 2
2 2 3 3
3 4 4 4
5 5 5 6
and S55((aj)) =
11122
23334
44555
66677
78889
have rank 4. The difference between
these two matrix is that S44((aj)) does not have a drop
in rank whereas S55((aj)) does. The rank drop de-
pending on the width (or the number of columns) of a
matrix is called a width rank drop.
From the two cases of rank drops in recurrence
matrices, S. Bozlee derives the exact rank of a recur-
rence matrix by using a characteristic polynomial of
a recurrence relation. For more details, readers may
consult [7].
Theorem 2.3. [7] Let (aj)be a recurrence sequence
with the minimal order kand qdistinct eigenvalues
λ1, λ2, . . . , λqwith multiplicities k1, k2, . . . , kq, re-
spectively. If m, n ≥k, let Λnbe the set of all distinct
values taken by λn
1, λn
2, . . . , λn
q. Then
rank Mmn((aj)) =
a∈Λn
max{kl:λl=a}
Then the rank of Smn((aj)) can be derived conse-
quently.
Corollary 2.4. Let (aj)be a homogeneous recur-
rence sequence with the minimal order kand qdis-
tinct eigenvalues λ1, λ2, . . . , λqwith multiplicities
k1, k2, . . . , kq, respectively. If m, n ≥k+1, let Λnbe
the set of all distinct values taken by λn
1, λn
2, . . . , λn
q.
Then
rank Smn((aj)) =
a∈Λn
max{kl:λl=a}+ 1
In particular, Theorem 5.1 in [7] shows that
if (aj)is a homogeneous linear recurrence se-
quence of order two such that aj=αaj−1+βaj−2
with given initial values a1and a2, then for m, n ≥2,
rank Mmn((aj)) =
0if a1=a2= 0,
1if a2
2−αa1a2−βa2
1= 0,
1if α2+ 4c2
2= 0 and α+√α2+4β
α−√α2+4βn
= 1,
2otherwise
and hence, for m, n ≥3,rank Smn((aj)) =
0if a1=a2= 0,
2if a2
2−αa1a2−βa2
1= 0,
2if α2+ 4β2= 0 and α+√α2+4β
α−√α2+4βn
= 1,
3otherwise.
(2.5)
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The Fibonacci sequence (1,1,2,3,5, . . . )is ho-
mogeneous linear recurrence sequence of order two
satisfying Fj=Fj−1+Fj−2with a1= 1 and a2= 1.
This sequence does not satisfy the first three condi-
tions in (2.5). That is, an m×npartial-sum matrix of
the Fibonacci sequence has rank 3whenever m, n ≥
3. Clearly, the m×1and 1×npartial-sum matrices of
the Fibonacci sequence has rank 1. For the remaining
case, it is easy to see that m×2and 2×mpartial-
sum matrices of the Fibonacci sequence for m≥2
has rank 2, e.g., see S24((aj)) = 1 1 2 3
5 8 13 21.
In the case of an inhomogeneous recurrence rela-
tion (aj)of order ksuch that for all j≥k
aj=α1aj−1+α2aj−2+α3aj−3+· · · +αkaj−k
(2.6)
with initial values a1, a2,· · · , akand α1, α2,· · · , αk
with αk= 0.
We can rewrite (2.6) to be the homogeneous recur-
rence relation of order k+ 1
aj+1 = (α1+ 1)aj+ (α2−α1)aj−1+· · · +
(αk−αk−1)aj−k+1 +αkaj−k.
Therefore, one can also derive the rank of
Smn((aj)) by referring to the homogeneous case. We
shall leave it to reader to work on the details.
3 Conclusion
We have defined a partial-sum matrix and provide the
rank of this type of matrices of some special cases.
The process we have applied is from Linear Algebra.
However, in analysis, the sequence of partial sums of
a sequence will lead to a series, then we may question
whether a sequence of m×npartial-sum matrices (for
either fixed mor n) can be related to the associated
series to the partial sums in some ways. This can be
an open problem for further study.
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Contribution of Individual Authors to
the Creation of a Scientific Article
(Ghostwriting Policy)
The authors equally contributed in the present re-
search, at all stages from the formulation of the prob-
lem to the final findings and solution.
Sources of Funding for Research
Presented in a Scientific Article or
Scientific Article Itself
No funding was received for conducting this study.
Conflicts of Interest
The authors have no conflicts of interest to declare
that are relevant to the content of this article.
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