Trace of  Positive Integer Power of  Squared Special Matrix


CAUCHY –Jurnal Matematika Murni dan Aplikasi 
Volume 6(4) (2021), Pages 200-211 
p-ISSN: 2086-0382; e-ISSN: 2477-3344 

Submitted: September 15, 2020 Reviewed: January 04, 2021 Accepted: January 27, 2021 
DOI: http://dx.doi.org/10.18860/ca.v6i4.10312   

Trace of  Positive Integer Power of  Squared Special Matrix 

Rahmawati1, Aryati Citra2, Fitry Aryani3, Corry Corazon Marzuki4, Yuslenita Muda5 
 

1,2,3,4,5 Department of Mathematics, Faculty of Science and Technology,  
State Islamic University of  Sultan Syarif Kasim Riau 

St. HR. Soebrantas No. 155 Simpang Baru, Panam, Pekanbaru, 28293 
 

Email: 1rahmawati@uin-suska.ac.id , 2aryaticitra1@gmail.com, 3khodijah_fitri@uin-suska.ac.id,  
4corry@uin-suska.ac.id, 5yuslenita.muda@uin-suska.ac.id 

ABSTRACT  

The rectangle matrix to be discussed in this research is a special matrix where each entry in each 
line has the same value which is notated by 𝐴𝑛. The main aim of this paper is to find the general 
form of the matrix trace of 𝐴𝑛 powered positive integer 𝑚 or notated by 𝑇𝑟(𝐴𝑛)

𝑚. To prove 
whether the general form of the matrix trace of 𝐴𝑛 powered positive integer can be confirmed, 
mathematics induction and direct proof are used. The main results present the general formula 
of (𝐴𝑛)

𝑚 and 𝑇𝑟(𝐴𝑛)
𝑚 with observing the pattern of power matrix for 2 ≤ 𝑚 ≤ 11,𝑛 ≥ 2, and 

𝑚 ∈ ℤ+.  

 

Keywords: direct proof; mathematics induction; matrix trace;  squared matrix 

INTRODUCTION 

The calculation of trace of power of square matrix has become attention. 
According to Brezinski [1], trace of power of matrix is often used in some fields of 
mathematics, especially Network Analysis, Number Theory, Dynamic Systems, Matrix 
Theory, and Differential Equations. The discussion about trace matrix has been widely 
studied by several researchers before. Datta et.al [2], has obtained algorithm of trace of 
power of squared matrix 𝑇𝑟(𝐴𝑘), with 𝑘 is an integer and 𝐴 is Hassenberg matrix with a 
codiagonal unit. There is also discussion of trace in several applications in matrix theory 
and numerical linear algebra. For example in determining the eigenvalue of a symmetric 
matrix, the basic procedure in estimating a trace (𝐴𝑛) and trace (𝐴−𝑛) with 𝑛 integers, 
this is explained in Pan [3]. Chu. Mt [4] discussed symbolic calculations on the power of 
squared tridiagonal of matrix trace. For example, 𝐴 a symmetric positive definite matrix, 
and for example {𝜆𝑘} notated its eigen value. For 𝑞 ∈  ℝ, 𝐴

𝑞 also symmetric definite 
matrices, and are listed in Hignam [5] with formula  

𝑇𝑟(𝐴𝑞) = ∑𝜆𝑘
𝑞

𝑘

 . 

According to Zarelua [6] in quantum and combinatorial theory, the trace matrix 
is a whole number in relation to the Euler equations 

𝑇𝑟(𝐴𝑝
𝑟
) = 𝑇𝑟(𝐴𝑝

𝑟−1
) 𝑚𝑜𝑑(𝑝𝑟) 

http://dx.doi.org/10.18860/ca.v6i4.10312
mailto:rahmawati@uin-suska.ac.id
mailto:aryaticitra1@gmail.com
mailto:khodijah_fitri@uin-suska.ac.id
mailto:corry@uin-suska.ac.id
mailto:yuslenita.muda@uin-suska.ac.id


Trace of  Positive Integer Power of  Squared Special Matrix 

Rahmawati 201 

For all matrix A integers, p is the prime number and r original number. Then this article 
also discuss about invariant in dynamic system which is illustrate as form of trace of 
integer squared matrix, for example the number Lefschetz. Next, Pahade and Jha [7], 
discuss about the formation of general form of trace matrix ordo 2 × 2 square with 
powered positive integer. In that article there are two general forms of order trace 2× 2 
with integer square n. First, the general form of order trace matrix trace 2 x 2 with even 
number square 𝑛, is 

𝑇𝑟 (𝐴𝑛) = ∑  
(−1)𝑟

𝑟 !

𝑛 2⁄
𝑟=0  𝑛 [𝑛  − (𝑟 + 1)] [𝑛  − (𝑟 + 2)] ⋯ [𝑛 − (𝑟 +(𝑟 −

1))] (Det (𝐴))
𝑟 
(𝑇𝑟 (𝐴))

𝑛−2𝑟
. 

Second, the main form of trace matrix 2 × 2 with odd number square 𝑛, is 

𝑇𝑟 (𝐴𝑛) = ∑  
(−1)𝑟

𝑟 !

(𝑛−1) 2⁄
𝑟=0  𝑛 [𝑛 − (𝑟 + 1)] [𝑛  − (𝑟 + 2)] ⋯ [𝑛 − (𝑟 + (𝑟 −

1))] (Det (𝐴))
𝑟 
(𝑇𝑟 (𝐴))

𝑛−2𝑟
. 

In the network analysis field, especially on triangle counting in a graph, based on 
Avron [8], when analyzing a complex network, the important problem is calculating the 
total numbers of triangle on the simple connected graph. This number is equal to 
𝑇𝑟(𝐴3) 6⁄ , where 𝐴 is adjacency matrix from the graph. Then, in 2017, Pahade and Jha 
[9] discuss about trace of squared adjacency matrix on positive integers. In the paper, 
there is A symmetrical adjacency matrix on a complete simple graph with vertex n, for 
even number k is formulated 

𝑇𝑟(𝐴𝑘) = ∑𝑠(𝑘,𝑟)𝑛(𝑛 − 1)𝑟(𝑛 − 2)𝑘−2𝑟

𝑛
2

𝑟=1

 

and for odd number k is formulated 

             𝑇𝑟(𝐴𝑘) = ∑ 𝑠(𝑘,𝑟)𝑛(𝑛 − 1)𝑟(𝑛 − 2)𝑘−2𝑟
𝑛−1

2
𝑟=1

  

with ),( rks is a number thats depend on k and r, and defined as  

          𝑠(𝑘,1) = 1,𝑠(𝑘,
𝑘

2
) = 1,𝑠(𝑘,

𝑘−1

2
) =

𝑘−1

2
, and 𝑠(𝑘,𝑟) = 𝑠(𝑘 − 1,𝑟)+ 𝑠(𝑘 − 2,𝑟 − 1). 

Next, by this research, it will be decided the trace of rectangle matrix with the real 
number entries which for every entry row has an equal value. In this research, there are 
some related definitions and theorems.  
 
Definition 1.1 (Anton [10]) If 𝐴 is a rectangle matrix, then the definition of squared of 
powered non negative integers of 𝐴 is 

𝐴0  =  𝐼 , 𝐴𝑛  = 𝐴𝐴…𝐴⏟    
𝑛 𝑓𝑎𝑘𝑡𝑜𝑟

 (𝑛  >  0). 

Next, If 𝐴 is invertible, then the definition of squared of powered negative integers of 𝐴 
is 

𝐴−𝑛  = (𝐴−1)𝑛  = 𝐴−1𝐴−1 …𝐴−1⏟        
𝑛 𝑓𝑎𝑘𝑡𝑜𝑟

. 

Theorem 1.1 (Andrilli, [11]) If 𝐴 is a rectangle matrix, and if 𝑟 and 𝑠 are nonnegative 
integers, then 

1. 𝐴𝑟 𝐴𝑠  = 𝐴𝑟 + 𝑠 
2. (𝐴𝑟)𝑠  = 𝐴𝑟𝑠 = (𝐴𝑠)𝑡 



Trace of  Positive Integer Power of  Squared Special Matrix 

Rahmawati 202 

Definition 1.2 [10] If 𝐴 is a rectangle matrix, then the trace of 𝐴 which is stated as 
𝑇𝑟(𝐴), is defined as the total entries on main diagonal of 𝐴. Trace from 𝐴 cannot be 
defined when 𝐴 is not a rectangle matrix 

                                                     𝑇𝑟 (𝐴) = 𝑎11 + 𝑎22 + ⋯+ 𝑎𝑛𝑛 = ∑  𝑎𝑖𝑖
𝑛
𝑖=1 .                        (1.1)                                     

Theorem 1.2 [12] If 𝐴 and 𝐵 are rectangle matrix in the same order and 𝑐 is r scale, then 

apply: 

a. 𝑇𝑟(𝐴) =  𝑇𝑟(𝐴𝑇), 
b. 𝑇𝑟(𝑐𝐴) =  𝑐 𝑇𝑟(𝐴),                                (1.2) 
c. 𝑇𝑟(𝐴 +  𝐵) =  𝑇𝑟(𝐴) + 𝑇𝑟(𝐵), 
d. 𝑇𝑟(𝐴𝐵) =  𝑇𝑟(𝐵𝐴). 

 

METHODS  

The method used in order to reach the aim of this paper is using literature study 
or conceptual foundation by following steps. 

 Finding the general formula of power matrix (𝐴𝑛)
𝑚 with 𝑚  ∈ ℤ+ and proof it 

using mathematical induction, 
 Determining trace matrix (𝐴𝑛)

𝑚, notated by 𝑇𝑟 (𝐴𝑛)
𝑚, finding the general 

formula and using mathematical induction, we proof the formula obtained.  
 

RESULTS AND DISCUSSION  

This research is going to discuss about positive integers squared trace of m from 
special matrix of 𝑛 𝑥 𝑛 order with the entries of real numbers where each entry has the 
same value in a row, which is noted with matrix 𝐴𝑛

𝑚. The research started by deciding 
the general form of matrix square of 𝐴𝑛

𝑚 by calculating matrix square in order of 2 x 2 to 
order of 5 x 5 squared by m positive integers. After the general matrix of 𝐴𝑛

𝑚 is formed, 
then this research is continued by looking for 𝑇𝑟(𝐴𝑛

𝑚). 
 

Special Matrix Order of  𝒏 ×  𝒏 (𝒏  ≥  𝟐) Squared by 𝒎 Positive Integers 
This part is going to explain about squaring of special matrix order of 𝑛  × 𝑛 ,𝑛 ≥

2 with the real number entries where each entry has the same value in a row, this matrix 
is formulated as follows 

𝐴𝑛  = 

[
 
 
 
 
 
𝑎1 𝑎1 ⋯ 𝑎1
𝑎2 𝑎2 ⋯ 𝑎2
⋮ ⋮ ⋮ ⋮
𝑎𝑖 𝑎𝑖 ⋯ 𝑎𝑖
⋮ ⋮ ⋮ ⋮
𝑎𝑛 𝑎𝑛 ⋯ 𝑎𝑛]

 
 
 
 
 

 , 𝑎𝑖  ∈  ℝ ;  𝑖  =  1, 2, … ,  𝑛.                                       (2.1) 

It is special matrix in order of 22   to 55   which is formulated as follows. 

𝐴2  = [
𝑎1 𝑎1
𝑎2 𝑎2

] , 𝐴3  = [

𝑎1 𝑎1 𝑎1
𝑎2 𝑎2 𝑎2
𝑎3 𝑎3 𝑎3

] , 𝐴4  = [

𝑎1 𝑎1 𝑎1 𝑎1
𝑎2 𝑎2 𝑎2 𝑎2
𝑎3 𝑎3 𝑎3 𝑎3
𝑎4 𝑎4 𝑎4 𝑎4

], 

 𝐴5  = 

[
 
 
 
 
𝑎1 𝑎1 𝑎1 𝑎1 𝑎1
𝑎2 𝑎2 𝑎2 𝑎2 𝑎2
𝑎3 𝑎3 𝑎3 𝑎3 𝑎3
𝑎4 𝑎4 𝑎4 𝑎4 𝑎4
𝑎5 𝑎5 𝑎5 𝑎5 𝑎5]

 
 
 
 

 



Trace of  Positive Integer Power of  Squared Special Matrix 

Rahmawati 203 

For 𝑛 = 2, it is decided the matrix squaring values of (𝐴2)
2 to (𝐴2)

11 which are 
presented in Table 1 below. 

Table 1. Special Matrix Squaring Values of (𝐴2)
2 to (𝐴2)

11
 

No Special Matrix Squaring of 𝑨2 Matrix Squaring Values 𝑨2 
1. (𝐴2)

2 (𝑎1 + 𝑎2) 𝐴2 
2. (𝐴2)

3 (𝑎1 + 𝑎2)
2 𝐴2 

3. (𝐴2)
4 (𝑎1 + 𝑎2)

3 𝐴2 
4. (𝐴2)

5 (𝑎1 + 𝑎2)
4 𝐴2 

5. (𝐴2)
6 (𝑎1 + 𝑎2)

5 𝐴2 
6. (𝐴2)

7 (𝑎1 + 𝑎2)
6 𝐴2 

7. (𝐴2)
8 (𝑎1 + 𝑎2)

7 𝐴2 
8. (𝐴2)

9 (𝑎1 + 𝑎2)
8 𝐴2 

9. (𝐴2)
10 (𝑎1 + 𝑎2)

9 𝐴2 
10. (𝐴2)

11 (𝑎1 + 𝑎2)
10 𝐴2 

 

 

After getting the values of special matrix squaring of 𝐴2 which are in Table 1, then it can 
be predicted that the general form of the special matrix squaring based on its recursive 
pattern is (𝐴2)

𝑚  = (𝑎1 + 𝑎2)
𝑚−1 𝐴2. According to the prediction, then the general 

form of matrix squaring of 𝐴2  is presented in Theorem 2.1 below. 

 

Theorem 2.1 If given the special matrix of 𝐴2  = [
𝑎1 𝑎1
𝑎2 𝑎2

] ; 𝑎1, 𝑎2  ∈  ℝ,  then            

            (𝐴2)
𝑚  = (𝑎1 + 𝑎2)

𝑚−1 𝐴2 with 𝑚  ∈ ℤ
+.                                                                (2.2) 

Proof: Using mathematic induction. 

For example 𝑝(𝑚) : (𝐴2)
𝑚  = (𝑎1 + 𝑎2)

𝑚−1 𝐴2 

1. For  𝑚 = 1 then  
𝑝 (1) : (𝐴2)

1  = (𝑎1 + 𝑎2)
1−1 𝐴2 

                         = (𝑎1 + 𝑎2)
0 𝐴2 

= 𝐴2 
2. For  𝑚 = 𝑘 then it is assumed that 𝑝 (𝑘) is correct, which is 

𝑝 (𝑘) : (𝐴2)
𝑘  = (𝑎1 + 𝑎2)

𝑘−1 𝐴2 
 
Will be presented for 𝑚  =  𝑘 + 1 then 𝑝 (𝑘 + 1) is also correct, which is 

                         𝑝 (𝑘 + 1) : (𝐴2)
𝑘+1  = (𝑎1 + 𝑎2)

𝑘 𝐴2.                (2.3)
     

Then,  

(𝐴2)
𝑘+1  = (𝐴2)

𝑘 (𝐴2) 
                  = (𝑎1 + 𝑎2)

𝑘−1 𝐴2 𝐴2 
                  = (𝑎1 + 𝑎2)

𝑘−1 (𝐴2)
2 

                  = (𝑎1 + 𝑎2)
𝑘−1 (𝑎1 + 𝑎2) 𝐴2 

                  = (𝑎1 + 𝑎2)
(𝑘−1)+1 𝐴2 

                  = (𝑎1 + 𝑎2)
𝑘 𝐴2 

By giving attention to the Equation (2.3) then 𝑝 (𝑘 + 1) is correct. Due to Step (1) and 
(2) are presented correctly, then Theorem 2.1 is proven. ∎ 



Trace of  Positive Integer Power of  Squared Special Matrix 

Rahmawati 204 

For 𝑛 = 3, it is decided the value of matrix squaring of (𝐴3)
2 to (𝐴3)

11 which are 
presented in Table 2 below. 

Table 2. The Value of Special Matrix Squaring of (𝐴3)
2 to (𝐴3)

11
 

No Special Matrix Squaring of 𝐴3   
The Value Matrix Squaring 

of 𝐴3 

1. (𝐴3)
2   3321 Aaaa   

2. (𝐴3)
3 (𝑎1 + 𝑎2 + 𝑎3)

2 𝐴3 
3. (𝐴3)

4 (𝑎1 + 𝑎2 + 𝑎3)
3 𝐴3 

4. (𝐴3)
5 (𝑎1 + 𝑎2 + 𝑎3)

4 𝐴3 
5. (𝐴3)

6 (𝑎1 + 𝑎2 + 𝑎3)
5 𝐴3 

6. (𝐴3)
7 (𝑎1 + 𝑎2 + 𝑎3)

6 𝐴3 
7. (𝐴3)

8 (𝑎1 + 𝑎2 + 𝑎3)
7 𝐴3 

8. (𝐴3)
9 (𝑎1 + 𝑎2 + 𝑎3)

8 𝐴3 
9. (𝐴3)

10 (𝑎1 + 𝑎2 + 𝑎3)
9 𝐴3 

10. (𝐴3)
11  (𝑎1 + 𝑎2 + 𝑎3)

10 𝐴3 
 

After getting the values of the special matrix squaring of 𝐴3 which is in Table 2, then it 
can be predicted the general form of the special matrix squaring is based on its recursive 
pattern which is (𝐴3)

𝑚  = (𝑎1 + 𝑎2 + 𝑎3)
𝑚−1 𝐴3. According to the prediction, then 

the general form of matrix squaring of 3A  is presented in Theorem 2.2 below. 

 

Theorem 2.2 If given the special matrix of 𝐴2  = [

𝑎1 𝑎1 𝑎1
𝑎2 𝑎2 𝑎2
𝑎3 𝑎3 𝑎3

] ; 𝑎1, 𝑎2, 𝑎3  ∈  ℝ,  then  

          (𝐴3)
𝑚  = (𝑎1 + 𝑎2 + 𝑎3)

𝑚−1 𝐴3 with 𝑚  ∈ ℤ
+.               (2.4) 

Proof: Applying the same steps with Theorem 2.1, then this theorem is proved. ∎ 

 

For 𝑛 = 4, it is decided the value of matrix squaring of (𝐴4)
2 to (𝐴4)

11 which are 
presented in Table 3 below. 

Table 3. The Value of Special Matrix Squaring of (𝐴4)
2 to (𝐴4)

11
 

No Special Matrix Squaring of 𝐴4 Matrix Squaring Value of𝐴4 

1. (𝐴4)
2  (𝑎1 + 𝑎2 + 𝑎3 + 𝑎4) 𝐴4 

2. (𝐴4)
3  (𝑎1 + 𝑎2 + 𝑎3 + 𝑎4)

2 𝐴4 
3. (𝐴4)

4  (𝑎1 + 𝑎2 + 𝑎3 + 𝑎4)
3 𝐴4 

4. (𝐴4)
5  (𝑎1 + 𝑎2 + 𝑎3 + 𝑎4)

4 𝐴4 
5. (𝐴4)

6   (𝑎1 + 𝑎2 + 𝑎3 + 𝑎4)
5 𝐴4 

6. (𝐴4)
7  (𝑎1 + 𝑎2 + 𝑎3 + 𝑎4)

6 𝐴4 
7. (𝐴4)

8        (𝑎1 + 𝑎2 + 𝑎3 + 𝑎4)
7𝐴4 

8. (𝐴4)
9 (𝑎1 + 𝑎2 + 𝑎3 + 𝑎4)

8𝐴4 
9. (𝐴4)

10 (𝑎1 +𝑎2 +𝑎3 + 𝑎4)
9𝐴4 

10. (𝐴4)
11 (𝑎1 + 𝑎2 + 𝑎3 + 𝑎4)

10𝐴4 
After getting the values of special matrix squaring of 𝐴4 which is in Table 3, then it can 
be predicted that the general form of the special matrix squaring is based on its 
recursive pattern which is (𝐴4)

𝑚  = (𝑎1 + 𝑎2 + 𝑎3 + 𝑎4)
𝑚−1 𝐴4. According to the 



Trace of  Positive Integer Power of  Squared Special Matrix 

Rahmawati 205 

prediction, then the general form of matrix squaring of 𝐴4 is presented in Theorem 2.3 
below. 
 
Theorem 2.3 If given the special matrix of  

                              𝐴4  = [

𝑎1 𝑎1 𝑎1 𝑎1
𝑎2 𝑎2 𝑎2 𝑎2
𝑎3 𝑎3 𝑎3 𝑎3
𝑎4 𝑎4 𝑎4 𝑎4

];  𝑎1, 𝑎2, 𝑎3, 𝑎4  ∈  ℝ,  

then   

                       (𝐴4)
𝑚  = (𝑎1 + 𝑎2 + 𝑎3 + 𝑎4)

𝑚−1 𝐴4  with 𝑚  ∈ ℤ
+.              (2.5) 

Proof: Adopting the proof in Theorem 2.1, then this theorem is proven as well. ∎ 

 

For 𝑛 = 5, it is decided the value of matrix squaring of (𝐴5)
2 to (𝐴5)

11 which is 
presented in the Table 4 below. 

Table 4. The Value of Special Matrix Squaring of (𝐴5)
2to (𝐴5)

11
 

No 
Special Matrix 
Squaring of 𝐴5 

Matrix Squaring Value of 𝐴5 

1. (𝐴5)
2 (𝑎1 + 𝑎2 + 𝑎3 + 𝑎4 + 𝑎5) 𝐴5 

2. (𝐴5)
3 (𝑎1 + 𝑎2 + 𝑎3 + 𝑎4 + 𝑎5)

2𝐴5 
3. (𝐴5)

4 (𝑎1 + 𝑎2 + 𝑎3 + 𝑎4 + 𝑎5)
3𝐴5 

4. (𝐴5)
5 (𝑎1 + 𝑎2 + 𝑎3 + 𝑎4 + 𝑎5)

4𝐴5 
5. (𝐴5)

6 (𝑎1 + 𝑎2 + 𝑎3 + 𝑎4 + 𝑎5)
5𝐴5 

6. (𝐴5)
7 (𝑎1 + 𝑎2 + 𝑎3 + 𝑎4 + 𝑎5)

6𝐴5 
7. (𝐴5)

8 (𝑎1 + 𝑎2 + 𝑎3 + 𝑎4 + 𝑎5)
7𝐴5 

8. (𝐴5)
9 (𝑎1 + 𝑎2 + 𝑎3 + 𝑎4 + 𝑎5)

8𝐴5 
9. (𝐴5)

10 (𝑎1 + 𝑎2 + 𝑎3 + 𝑎4 + 𝑎5)
9𝐴5 

10. (𝐴5)
11 (𝑎1 + 𝑎2 + 𝑎3 + 𝑎4 + 𝑎5)

10𝐴5 
 
After getting the values of matrix squaring of 𝐴5 which is in Table 3, then in can be 
predicted that the general form of the special matrix squaring is based on its recursive 
pattern which is (𝐴5)

𝑚  = (𝑎1 + 𝑎2 + 𝑎3 + 𝑎4 + 𝑎5)
𝑚−1𝐴5. According to the 

prediction, then the general form of matrix squaring of 𝐴5 is presented in Theorem 2.4 
below. 

Theorem 2.4 If given the special matrix of 𝐴5  = 

[
 
 
 
 
𝑎1 𝑎1 𝑎1 𝑎1 𝑎1
𝑎2 𝑎2 𝑎2 𝑎2 𝑎2
𝑎3 𝑎3 𝑎3 𝑎3 𝑎3
𝑎4 𝑎4 𝑎4 𝑎4 𝑎4
𝑎5 𝑎5 𝑎5 𝑎5 𝑎5]

 
 
 
 

; 

𝑎1,𝑎2,𝑎3,𝑎4,𝑎5 ∈ ℝ  then 

             (𝐴5)
𝑚  = (𝑎1 + 𝑎2 + 𝑎3 + 𝑎4 + 𝑎5)

𝑚−1𝐴5 with 𝑚  ∈ ℤ
+.                          (2.6) 

 

Proof: It is clear from above theorems. ∎ 

By giving attention to the recursive pattern of Equation (2.2), Equation (2.4), 
Equation (2.5) and Equation (2.6) which are 



Trace of  Positive Integer Power of  Squared Special Matrix 

Rahmawati 206 

(𝐴2)
𝑚  = (𝑎1 + 𝑎2)

𝑚−1 𝐴2 
(𝐴3)

𝑚  = (𝑎1 + 𝑎2 + 𝑎3)
𝑚−1 𝐴3 

(𝐴4)
𝑚  = (𝑎1 + 𝑎2 + 𝑎3 + 𝑎4)

𝑚−1 𝐴4 
(𝐴5)

𝑚  = (𝑎1 + 𝑎2 + 𝑎3 + 𝑎4 + 𝑎5)
𝑚−1 𝐴5. 

It can be predicted that the general form of the special matrix squaring in order of 𝑛 ×𝑛,
𝑛 ≥ 2 is equal to the Equation (2.1), which is

 
(𝐴𝑛)

𝑚  = (𝑎1 + 𝑎2 + … + 𝑎𝑛)
𝑚−1 𝐴𝑛 

               = (∑ 𝑎𝑖

𝑛

𝑖=1

)

𝑚−1

𝐴𝑛

 According to the prediction, then the general form of special matrix squaring in order of 
𝑛  × 𝑛 ,𝑛 ≥ 2 is equal to Equation (2.1) is presented in the Theorem 2.5 below. 

 

Theorem 2.5 If given the special matrix in order 𝑛 ×𝑛 , 𝑛 ≥ 2 which is 

 𝐴𝑛  = 

[
 
 
 
 
 
𝑎1 𝑎1 ⋯ 𝑎1
𝑎2 𝑎2 ⋯ 𝑎2
⋮ ⋮ ⋮ ⋮
𝑎𝑖 𝑎𝑖 ⋯ 𝑎𝑖
⋮ ⋮ ⋮ ⋮
𝑎𝑛 𝑎𝑛 ⋯ 𝑎𝑛]

 
 
 
 
 

 ; 𝑎𝑖  ∈  ℝ ,   𝑖  =  1,  2, … ,  𝑛 

then 
  

 (𝐴𝑛)
𝑚  = (∑  𝑎𝑖

𝑛
𝑖=1 )

𝑚−1𝐴𝑛 ,  𝑑𝑒𝑛𝑔𝑎𝑛 𝑚  ∈ ℤ
+. 

 
Proof: Again, by using mathematic induction, for example 𝑝 (𝑚) : (𝐴𝑛)

𝑚  =
 (∑  𝑎𝑖

𝑛
𝑖=1 )

𝑚−1𝐴𝑛 , with 𝑚 ∈ ℤ
+ 

1. For  𝑚  =  1 then 

𝑝 (1) : (𝐴𝑛)
1  = (∑ 𝑎𝑖

𝑛

𝑖=1

)

1−1

𝐴𝑛 

                             = (∑ 𝑎𝑖

𝑛

𝑖=1

)

0

𝐴𝑛 

                                                                               = 𝐴𝑛         

 
2. For  km   is assumed that 𝑝 (𝑘) is correct, which is 

𝑝 (𝑘) : (𝐴𝑛)
𝑘  = (∑  𝑎𝑖

𝑛
𝑖=1 )

𝑘−1𝐴𝑛, with 𝑚  ∈ ℤ
+.  

 
Will be presented for 𝑚  =  𝑘 + 1 then 𝑝 (𝑘  +  1) is also correct, which is 

𝑝 (𝑘  +  1) : (𝐴𝑛)
𝑘+1  = (∑ 𝑎𝑖

𝑛

𝑖=1

)

(𝑘+1)−1

𝐴𝑛 

                                                                                  =  (∑  𝑎𝑖
𝑛
𝑖=1 )

𝑘𝐴𝑛              (2.7)

 

The proof is below 

 

 (𝐴𝑛)
𝑘+1  = (𝐴𝑛)

𝑘 (𝐴𝑛) 



Trace of  Positive Integer Power of  Squared Special Matrix 

Rahmawati 207 

                  = (∑ 𝑎𝑖

𝑛

𝑖=1

)

𝑘−1

𝐴𝑛𝐴𝑛 

                 = (∑ 𝑎𝑖

𝑛

𝑖=1

)

𝑘−1

[
 
 
 
 
 
𝑎1 𝑎1 ⋯ 𝑎1
𝑎2 𝑎2 ⋯ 𝑎2
⋮ ⋮ ⋮ ⋮
𝑎𝑖 𝑎𝑖 ⋯ 𝑎𝑖
⋮ ⋮ ⋮ ⋮
𝑎𝑛 𝑎𝑛 ⋯ 𝑎𝑛]

 
 
 
 
 

[
 
 
 
 
 
𝑎1 𝑎1 ⋯ 𝑎1
𝑎2 𝑎2 ⋯ 𝑎2
⋮ ⋮ ⋮ ⋮
𝑎𝑖 𝑎𝑖 ⋯ 𝑎𝑖
⋮ ⋮ ⋮ ⋮
𝑎𝑛 𝑎𝑛 ⋯ 𝑎𝑛]

 
 
 
 
 

  

 

   = (∑  𝑎𝑖
𝑛
𝑖=1 )

𝑘−1 

[
 
 
 
 
 
𝑎1
2 + 𝑎1𝑎2 + ⋯ + 𝑎1𝑎𝑖  + ⋯ + 𝑎1𝑎𝑛 𝑎1

2 + 𝑎1𝑎2 + ⋯ + 𝑎1𝑎𝑖  + ⋯ + 𝑎1𝑎𝑛 ⋯ 𝑎1
2 + 𝑎1𝑎2 + ⋯ + 𝑎1𝑎𝑖  + ⋯ + 𝑎1𝑎𝑛

𝑎1𝑎2 + 𝑎2
2 + ⋯ + 𝑎2𝑎𝑖  + ⋯ + 𝑎2𝑎𝑛 𝑎1𝑎2 + 𝑎2

2 + ⋯ + 𝑎2𝑎𝑖  + ⋯ + 𝑎2𝑎𝑛 ⋯ 𝑎1𝑎2 + 𝑎2
2 + ⋯ + 𝑎2𝑎𝑖  + ⋯ + 𝑎2𝑎𝑛

⋮ ⋮   ⋮
𝑎1𝑎𝑖  + 𝑎2𝑎𝑖  + ⋯ + 𝑎𝑖

2 + ⋯ + 𝑎𝑖𝑎𝑛 𝑎1𝑎𝑖  + 𝑎2𝑎𝑖  + ⋯ + 𝑎𝑖
2 + ⋯ + 𝑎𝑖𝑎𝑛 ⋯ 𝑎1𝑎𝑖  + 𝑎2𝑎𝑖  + ⋯ + 𝑎𝑖

2 + ⋯ + 𝑎𝑖𝑎𝑛
⋮ ⋮ ⋮ ⋮

𝑎1𝑎𝑛  + 𝑎2𝑎𝑛  + ⋯ + 𝑎𝑖𝑎𝑛  + ⋯ + 𝑎𝑛
2 𝑎1𝑎𝑛  + 𝑎2𝑎𝑛  + ⋯ + 𝑎𝑖𝑎𝑛  + ⋯ + 𝑎𝑛

2 ⋯ 𝑎1𝑎𝑛  + 𝑎2𝑎𝑛  + ⋯ + 𝑎𝑖𝑎𝑛  + ⋯ + 𝑎𝑛
2]
 
 
 
 
 

 

 

= (∑ 𝑎𝑖

𝑛

𝑖=1

)

𝑘−1

 

[
 
 
 
 
 
(𝑎1 + 𝑎2 + ⋯ + 𝑎𝑖  + ⋯ + 𝑎𝑛) 𝑎1 (𝑎1 + 𝑎2 + ⋯ + 𝑎𝑖  + ⋯ + 𝑎𝑛) 𝑎1 ⋯ (𝑎1 + 𝑎2 + ⋯ + 𝑎𝑖  + ⋯ + 𝑎𝑛) 𝑎1
(𝑎1 + 𝑎2 + ⋯ + 𝑎𝑖  + ⋯ + 𝑎𝑛) 𝑎2 (𝑎1 + 𝑎2 + ⋯ + 𝑎𝑖  + ⋯ + 𝑎𝑛) 𝑎2 ⋯ (𝑎1 + 𝑎2 + ⋯ + 𝑎𝑖  + ⋯ + 𝑎𝑛) 𝑎2

⋮ ⋮ ⋮ ⋮
(𝑎1 + 𝑎2 + ⋯ + 𝑎𝑖  + ⋯ + 𝑎𝑛) 𝑎𝑖 (𝑎1 + 𝑎2 + ⋯ + 𝑎𝑖  + ⋯ + 𝑎𝑛) 𝑎𝑖 ⋯ (𝑎1 + 𝑎2 + ⋯ + 𝑎𝑖  + ⋯ + 𝑎𝑛) 𝑎𝑖

⋮ ⋮ ⋮ ⋮
(𝑎1 + 𝑎2 + ⋯ + 𝑎𝑖  + ⋯ + 𝑎𝑛) 𝑎𝑛 (𝑎1 + 𝑎2 + ⋯ + 𝑎𝑖  + ⋯ + 𝑎𝑛) 𝑎𝑛 ⋯ (𝑎1 + 𝑎2 + ⋯ + 𝑎𝑖  + ⋯ + 𝑎𝑛) 𝑎𝑛]

 
 
 
 
 

 

= (∑  𝑎𝑖
𝑛
𝑖=1 )

𝑘−1 (𝑎1 + 𝑎2 + ⋯ + 𝑎𝑖  + ⋯ + 𝑎𝑛) 

[
 
 
 
 
 
𝑎1 𝑎1 ⋯ 𝑎1
𝑎2 𝑎2 ⋯ 𝑎2
⋮ ⋮ ⋮ ⋮
𝑎𝑖 𝑎𝑖 ⋯ 𝑎𝑖
⋮ ⋮ ⋮ ⋮
𝑎𝑛 𝑎𝑛 ⋯ 𝑎𝑛]

 
 
 
 
 

  

 = (∑  𝑎𝑖
𝑛
𝑖=1 )

𝑘−1 (∑  𝑎𝑖
𝑛
𝑖=1 ) 𝐴𝑛 

= (∑  𝑎𝑖
𝑛
𝑖=1 )

(𝑘−1)+1 𝐴𝑛  

= (∑  𝑎𝑖
𝑛
𝑖=1 )

𝑘 𝐴𝑛  

By giving attention to the Equation (2.7) then 𝑝 (𝑘 +  1) is correct.

 

Due to step (1) and  

(2) are presented correctly, then the Theorem 2.5 is proven. ∎ 

 

Trace of Special Matrix in Order 𝒏 × 𝒏 (𝒏  ≥  𝟐) Squared by Positive Integers  

In this part is going to be given the trace of special matrix of 𝐴2
𝑚,𝐴3

𝑚,𝐴4
𝑚, and 𝐴5

𝑚 
which are contained in Theorem 3.1 to Theorem 3.4 as follows. 

Theorem 3.1 If it is given the special matrix of 𝐴2 = [
𝑎1 𝑎1
𝑎2 𝑎2

] ; 𝑎1,𝑎2 ∈ ℝ then 

𝑇𝑟 (𝐴2)
𝑚 = (𝑎1 + 𝑎2)

𝑚, with  𝑚 ∈ ℤ+.                                                                   (3.1)  
 

Proof. The Proof of Theorem uses direct proof. Because of the known matrix of 𝐴2 then 

𝑇𝑟(𝐴2) = 𝑎1 + 𝑎2. According to Theorem 2.1, is got Equation (2.2) which is (𝐴2)
𝑚 =

(𝑎1 + 𝑎2)
𝑚−1𝐴2. By using the Definition 1.2 and Theorem 1.2 (b), it is formulated 

𝑇𝑟 (𝐴2)
𝑚 = 𝑇𝑟((𝑎1 + 𝑎2)

𝑚−1𝐴2) 

                   = (𝑎1 + 𝑎2)
𝑚−1𝑇𝑟(𝐴2) 

                                                                         = (𝑎1 + 𝑎2)
𝑚−1(𝑎1 + 𝑎2) 

           = (𝑎1 + 𝑎2)
(𝑚−1)+1 



Trace of  Positive Integer Power of  Squared Special Matrix 

Rahmawati 208 

                                                                         = (𝑎1 + 𝑎2)
𝑚. 

According to the proof, then Theorem 3.1 is proven. ∎ 

Theorem 3.2 If it is given special matrix of 𝐴3 = [

𝑎1 𝑎1 𝑎1
𝑎2 𝑎2 𝑎2
𝑎3 𝑎3 𝑎3

] ;𝑎1,𝑎2,𝑎3 ∈ ℝ then 

𝑇𝑟 (𝐴3)
𝑚 = (𝑎1 + 𝑎2 + 𝑎3)

𝑚, with 𝑚 ∈ ℤ+.                                                      (3.2) 
 

Proof. It is clear from above theorem. ∎ 

 
Theorem 3.3 If it is given the special matrix of 

𝐴4  = [

𝑎1 𝑎1 𝑎1 𝑎1
𝑎2 𝑎2 𝑎2 𝑎2
𝑎3 𝑎3 𝑎3 𝑎3
𝑎4 𝑎4 𝑎4 𝑎4

]; 𝑎1, 𝑎2, 𝑎3, 𝑎4  ∈  ℝ,  

then 
𝑇𝑟 (𝐴4)

𝑚 = (𝑎1 + 𝑎2 + 𝑎3 + 𝑎4)
𝑚, with 𝑚 ∈ ℤ+.                                                      (3.3)  

Proof. The proof is clear. ∎ 

Theorem 3.4 If given the special matrix of 𝐴5  = 

[
 
 
 
 
𝑎1 𝑎1 𝑎1 𝑎1 𝑎1
𝑎2 𝑎2 𝑎2 𝑎2 𝑎2
𝑎3 𝑎3 𝑎3 𝑎3 𝑎3
𝑎4 𝑎4 𝑎4 𝑎4 𝑎4
𝑎5 𝑎5 𝑎5 𝑎5 𝑎5]

 
 
 
 

; 𝑎1, 𝑎2, 𝑎3,𝑎4 

and 𝑎5  ∈  ℝ then  

             𝑇𝑟 (𝐴5)
𝑚 = (𝑎1 +𝑎2 +𝑎3 +𝑎4 +𝑎5)

𝑚, with  𝑚 ∈ ℤ+.                               (3.4)  

 

Proof. Clearly proven by following Theorem 3.1.  ∎ 

 

By giving attention to the recursive pattern on Equation (3.1), Equation (3.2), Equation 

(3.3) and Equation (3.4) which are 

𝑇𝑟 (𝐴2)
𝑚  = (𝑎1 + 𝑎2)

𝑚 
𝑇𝑟 (𝐴3)

𝑚  = (𝑎1 + 𝑎2 + 𝑎3)
𝑚 

𝑇𝑟 (𝐴4)
𝑚  = (𝑎1 + 𝑎2 + 𝑎3 + 𝑎4)

𝑚 
𝑇𝑟 (𝐴5)

𝑚  = (𝑎1 + 𝑎2 + 𝑎3 + 𝑎4 + 𝑎5)
𝑚. 

It can be predicted that the general form of the trace of special matrix in order 𝑛 𝑥 𝑛,𝑛 ≥
2 is equal to Equation (2.1) squared by positive integer (nonnegative integer) which is 

𝑇𝑟 (𝐴𝑛)
𝑚  = (𝑎1 + 𝑎2 + … + 𝑎𝑛)

𝑚 = (∑ 𝑎𝑖

𝑛

𝑖=1

)

𝑚

. 

According to the prediction, then the general form of trace of special matrix in order 
𝑛 𝑥 𝑛 ,𝑛 ≥ 2 is presented in Theorem 3.5 below. 
 

  



Trace of  Positive Integer Power of  Squared Special Matrix 

Rahmawati 209 

Theorem 3.5 If given special matrix in order 𝑛 𝑥 𝑛 ,𝑛 ≥ 2  which is 

 𝐴𝑛  = 

[
 
 
 
 
 
𝑎1 𝑎1 ⋯ 𝑎1
𝑎2 𝑎2 ⋯ 𝑎2
⋮ ⋮ ⋮ ⋮
𝑎𝑖 𝑎𝑖 ⋯ 𝑎𝑖
⋮ ⋮ ⋮ ⋮
𝑎𝑛 𝑎𝑛 ⋯ 𝑎𝑛]

 
 
 
 
 

 ,  𝑎𝑖  ∈  ℝ ;  𝑖  =  1, 2, … ,  𝑛.  

 
then 

𝑇𝑟 (𝐴𝑛)
𝑚  = (∑  𝑎𝑖

𝑛
𝑖=1 )

𝑚,  with  𝑚  ∈ ℤ+. 
 

Proof: This theorem will be proven by direct proof. Because matrix 𝐴𝑛 is known, then 
𝑇𝑟 (𝐴𝑛) = ∑  𝑎𝑖

𝑛
𝑖=1 . From Theorem 2.5, obtained (𝐴𝑛)

𝑚  = (∑  𝑎𝑖
𝑛
𝑖=1 )

𝑚−1𝐴𝑛. So that by 
using Definition 1.2 and Theorem 1.2 (b) obtained 

𝑇𝑟(𝐴𝑛)
𝑚  =  𝑇𝑟 ((∑ 𝑎𝑖

𝑛

𝑖=1

)

𝑚−1

𝐴𝑛) 

                   = (∑ 𝑎𝑖

𝑛

𝑖=1

)

𝑚−1

 𝑇𝑟 (𝐴𝑛) 

                   = (∑ 𝑎𝑖

𝑛

𝑖=1

)

𝑚−1

 (∑ 𝑎𝑖

𝑛

𝑖=1

) 

                   = (∑ 𝑎𝑖

𝑛

𝑖=1

)

𝑚

. 

Based on the evidence, then Theorem 3.5 is Proven. ∎ 
 

The Application of Matrix 𝑨𝒏
𝒎 and 𝑻𝒓(𝑨𝒏

𝒎) in Examples 
The following is given the example of question related to Theorem 2.5 and Theorem 

3.5 as follows. 
 
Example 1. Consider Matrix 𝐴4 as follows 

  𝐴4 = [

3 3 3 3
12 12 12 12
25 25 25 25
10 10 10 10

] 

Determine (𝐴4)
80 and 𝑇𝑟(𝐴4)

80. 
Solution: 
By giving attention to matrix 𝐴4, value of 𝑎1 = 3,𝑎2 = 12,𝑎3 = 25, and 𝑎4 = 10. Based 
on Theorem 2.5 obtained 

(𝐴4)
80 = (∑𝑎𝑖

4

𝑖=1

)

80−1

𝐴4 

              = (𝑎1 + 𝑎2 + 𝑎3 + 𝑎4)
79𝐴4 

              = (3+ 12 + 25 + 10)79 [

3 3 3 3
12 12 12 12
25 25 25 25
10 10 10 10

] 



Trace of  Positive Integer Power of  Squared Special Matrix 

Rahmawati 210 

              = (50)79 [

3 3 3 3
12 12 12 12
25 25 25 25
10 10 10 10

] 

Based on Theorem 3.5 obtained 

𝑇𝑟(𝐴4)
80 = (∑𝑎𝑖

4

𝑖=1

)

80

 

                   = (𝑎1 + 𝑎2 + 𝑎3 + 𝑎4)
80 

                   = (3+ 12 + 25 + 10)80 
                   = (50)80 

 
Example 2. Given matrix 𝐴5 as follows 

  𝐴5 =

[
 
 
 
 

8 8 8 8 8
5/16 5/16 5/16 5/16 5/16
−12 −12 −12 −12 −12
2/3 2/3 2/3 2/3 2/3
−5/12 −5/12 −5/12 −5/12 −5/12]

 
 
 
 

 

Determine (𝐴5)
27 and  𝑇𝑟(𝐴5)

27. 
Solution : 
By giving attention to matrix ,5A value of 𝑎1 = 8,𝑎2 = 5/16,𝑎3 = −12,𝑎4 = 2/3 

and 𝑎5 = −5/12. Based on Theorem 2.5 obtained 
                       

 (𝐴5)
27 = (∑𝑎𝑖

5

𝑖=1

)

27−1

𝐴5   

  
              = (𝑎1 + 𝑎2 + 𝑎3 + 𝑎4 + 𝑎5)

26𝐴5 

      = (8 + (5/16) +(−12) +(2/3) + (−5/12))26

[
 
 
 
 

8 8 8 8 8
5/16 5/16 5/16 5/16 5/16
−12 −12 −12 −12 −12
2/3 2/3 2/3 2/3 2/3
−5/12 −5/12 −5/12 −5/12 −5/12]

 
 
 
 

 

      = (−3
7

16
)26

[
 
 
 
 

8 8 8 8 8
5/16 5/16 5/16 5/16 5/16
−12 −12 −12 −12 −12
2/3 2/3 2/3 2/3 2/3
−5/12 −5/12 −5/12 −5/12 −5/12]

 
 
 
 

 

According to Theorem 3.5 it is formulated 

𝑇𝑟(𝐴5)
27 = (∑𝑎𝑖

5

𝑖=1

)

27

 

                   = (𝑎1 + 𝑎2 + 𝑎3 + 𝑎4 + 𝑎5)
27 

                   = (8+ (5/16) + (−12) + (2/3) +(−5/12))27 

                   = (−3
7

16
)27. 

  



Trace of  Positive Integer Power of  Squared Special Matrix 

Rahmawati 211 

CONCLUSIONS 
Based on elaboration and discussion in previous part, several conclusions can be 

drawn as follows. 
1. The general form of integer of a special matrix form in order 𝑛 × 𝑛 ,𝑛 ≥ 2 in 

Equation (2.1) is as follows. 

(𝐴𝑛)
𝑚  = (∑ 𝑎𝑖

𝑛

𝑖=1

)

𝑚−1

 𝐴𝑛,  with 𝑚  ∈ ℤ
+. 

2. General form of trace in a special matrix form in order 𝑛 × 𝑛 ,𝑛 ≥ 2 in Equation (2.1) 
is as follows. 

                               𝑇𝑟 (𝐴𝑛)
𝑚  = (∑  𝑎𝑖

𝑛
𝑖=1 )

𝑚,  with 𝑚  ∈ ℤ+.  

 
ACKNOWLEDGMENTS  

We thank all authors (RR, AC, FA, CCM, and YM) for their responsibility to 
designed the research and approved the final manuscript. RR, AC wrote the manuscript, 
FA, CCM gave their suggestion and edited the manuscript and YM read, edited for the 
final content of the manuscript. None of the authors had a conflict of interest. 

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