Basis Existence of Internal n-Graph Space


CAUCHY –Jurnal Matematika Murni dan Aplikasi 
Volume 6(2) (2020), Pages 58-65 
p-ISSN: 2086-0382; e-ISSN: 2477-3344 

Submitted: November 16, 2019 Reviewed: April 03, 2020 Accepted: May 31, 2020 
DOI: http://dx.doi.org/10.18860/ca.v6i2.8002   

Basis Existence of Internal n-Graph Space 

Ahmad Lazwardi  

University of Muhammadiyah Banjarmasin 
Email: lazwardiahmad@gmail.com 

ABSTRACT 

Graph of real valued continuous function with special addition and multiplication has already 
proven that is isomorphic to real number system. Furthermore, the graph of continous real valued 
function forms a field. The aim of this research was to generalize such concept to its n-tuple 
Cartesian Product and to prove that interchange of basis still able to be executed. The result of this 
research is n-tuple Cartesian Product of graph function forms a vector space over ℝ and 
interchange of basis still able to be executed. 

Keywords: n-Graph Space; General Vector Space; Internal n-Graph Space 

INTRODUCTION 

By associating the special operation to a graph of continuous function, such graph 
can be claimed as a vector space. This follows from the fact that every graph of continuous 
real valued function has a bijection to its domain i.e. the real number system. 
Furthermore, they are homeomorphic. 

Graph of real valued continuous function has a unique characteristic. It has 
continuous shape of curve along real line ℝ. More detail will be described as follows [1] 
 Definition 1. Let real valued function 𝑓: ℝ → ℝ. The graph of f is defined as 
 ℝ𝑓 = {(𝑥, 𝑓(𝑥)): 𝑥 ∈ ℝ}                                                  (1) 

 
Continuity of f indicates that ℝ𝑓 topologically equivalent to ℝ. Its described briefly as 

follows; 

Definition 2 [2]. For each 𝑈 ⊂ ℝ. Define the image of 𝑈 over ℝ𝑓 as  

 𝑈 × 𝑓(𝑈) = {(𝑥, 𝑓(𝑥)): 𝑥 ∈ 𝑈}                                               (2) 
define the associated topology for ℝ𝑓 as 

 𝜏𝑓 = {𝑈 × 𝑓(𝑈): 𝑈 𝑜𝑝𝑒𝑛}                                                      (3) 

 

It can be shown trivially that such topology implies ℝ𝑓 and ℝ are homeomorphics . The 

fact that ℝ𝑓 and ℝ are homeomorphics describes that even though their graph 

geometrically has different shapes, but they still have similarity in views of topology[3]. 
It motivates us to explore more special properties of ℝ𝑓.  

 
Now let see it further.  
Definition 3. Let real valued function 𝑓: ℝ → ℝ. Define the addition for ℝ𝑓 as follow 

http://dx.doi.org/10.18860/ca.v6i2.8002
mailto:lazwardiahmad@gmail.com


Basis Existence in Graph of Real Valued Continous Functions 

Ahmad Lazwardi 59 

𝑥𝑓⨁𝑦𝑓 = (𝑥 + 𝑦, 𝑓(𝑥 + 𝑦)) 

for any 𝑥𝑓, 𝑦𝑓 ∈ ℝ𝑓 . 

Define the scalar multiplication as follow 

(𝛼)𝑥𝑓 = 𝛵𝑓 (𝛼𝛵
−1

𝑓(𝑥, 𝑓(𝑥))) 

= (𝛼𝑥, 𝑓(𝛼𝑥)) 

For any 𝑥𝑓 ∈ ℝ𝑓 and scalar  , and Tf  is natural bijection generated by real valued function 

f  such as 𝛵𝑓(𝑥) = (𝑥, 𝑓(𝑥)).  

 
By associating ℝ𝑓 with those operations, ℝ𝑓 become real vector space. Moreover ℝ𝑓 and 

ℝ are isomorphics [4]. 
 

METHODS 

The method of this research is done by following method: first was to prove that  ℝ𝑓 has 

dimension 1 and isomorphics to ℝ. Next step was to analyize more general space i.e. n 
tuple Cartesian Product of ℝ𝑓 denoted by ℝ𝑓

𝑛. In generalization of ℝ𝑓
𝑛 was to prove that 

such space is a vector space over ℝ and was able to change of bases. Finally, was to find 
out the briefly method to change of bases as well as in real vector spaces. 

RESULTS AND DISCUSSION  

It was proven previously that for each continuous function 𝑓: ℝ → ℝ,  the space ℝ𝑓 

generated by f forms a field.  This fact becomes the basic to generalize the idea by 
constructing new n-tuple Cartesian Product of ℝ𝑓 which preserves vector space 

properties [5]. Before any further discussion, first define some necessary terms in order 
to help in generalization. We mean linear combination is 𝑧𝑓 = (∝)𝑥𝑓 + (𝛽)𝑦𝑓 for each 𝑧𝑓 ∈

ℝ𝑓 and some scalars 𝛼, 𝛽 ∈ ℝ [6]. The set of linear combinations of 𝑥𝑓, 𝑦𝑓 is named as Span 

{𝑥𝑓, 𝑦𝑓 }  [7]. The set 𝑈 ⊂ ℝ𝑓 is said to be linearly independent if none of its members is 

able to expressed as linear combination of other members. Here is definition of basis: 
 

Definition 4. The set },...,,{
21 n

fff
xxx  is called basis of subspace 𝑈 ⊂ ℝ𝑓 if },...,,{

21 n

fff
xxx are 

linearly independent and  Span },...,,{
21 n

fff
xxx =U. 

Recall that, dimension of 𝑈 is defined as base cardinality of 𝑈. 
Next theorem is an important result. 
Theorem 1. ℝ𝑓 has dimension 1. 

Proof: 
Chose 1𝑓 ∈ ℝ𝑓. For each 𝑥𝑓 ∈ ℝ𝑓, it’s obvious that  

x𝑓 = (𝑥, 𝑓(𝑥)) 

= 𝛵𝑓(𝑥) 

   = 𝛵𝑓(𝑥. 1) 

 = (𝑥. 1, 𝑓(𝑥. 1)). 

 = (𝑥) × 1𝑓 

By last equation, we conclude that 𝑆𝑝𝑎𝑛{1𝑓} = ℝ𝑓. 

Furthermore, based on the above results, vector space theory of  ℝ𝑓  can be developed: 



Basis Existence in Graph of Real Valued Continous Functions 

Ahmad Lazwardi 60 

Definition 5. Let 𝑆 = {𝑓1, 𝑓2, … , 𝑓𝑛} be a finite collection of real valued continuous 
functions, define n-graph space as  
 ℝ𝑓𝑝

𝑛 = ∏ ℝ𝑓𝑘
𝑛
𝑘=1 = {�⃗�: {1,2, … , 𝑛} → ⋃ ℝ𝑓𝑘: �⃗�(𝑖) ∈ ℝ𝑓𝑖

𝑛
𝑘=1 }.            (4) 

Definition 6. Let 𝑆 = {𝑓1, 𝑓2, … , 𝑓𝑛} be a finite collection of real valued continuous 
functions. Define addition on ℝ𝑓𝑝

𝑛  as 

. �⃗�⨁�⃗⃗⃗� = (�⃗�(1)⨁�⃗⃗⃗�(1), �⃗�(2)⨁�⃗⃗⃗�(2), . . . , �⃗�(𝑛)⨁�⃗⃗⃗�(𝑛))                     (5) 
and scalar multiplication as 
. (∝)�⃗� = ((∝)�⃗�(1), (∝)�⃗�(2), . . . , (∝)�⃗�(𝑛)) (6) 

for each 𝛼 ∈ ℝ. 
 
Theorem 2. Let 𝑆 = {𝑓1, 𝑓2, … , 𝑓𝑛} be a finite collection of real valued continuous function. 
n-graph space ℝ𝑓𝑝

𝑛  is a vector space over ℝ under operation ⨁ and scalar multiplication 

(∝)�⃗�. 
In special case which 𝑓1 = 𝑓2 = 𝑓3 = ⋯ = 𝑓𝑛, graph space ℝ𝑓𝑝

𝑛  is called internal n-graph 

space i.e. Cartesian Product n-tuple of ℝ𝑓 itself. One can write 

ℝ𝑓𝑝
𝑛 = ℝ𝑓

𝑛. 

Euclidean Space ℝ𝑛 is one of finest example of graph space which f is defined as identity 
mapping.  
 
One of the most important tools to analyze relation between two internal n-graph space 
is linear transformation[8], here we still able to define linear transformation as well as 
done on commonly vector spaces. 
Definition 7. Let two graph spaces ℝ𝑓

𝑚, ℝ𝑓
𝑛. Mapping 𝐿: ℝ𝑓

𝑚 → ℝ𝑓
𝑛 is linear transformation 

if the following properties hold  
𝐿(�⃗�⨁�⃗⃗⃗�) = 𝐿(�⃗�)⨁𝐿(�⃗⃗⃗�),      ∀�⃗�, �⃗⃗⃗� ∈ ℝ𝑓

𝑚 

𝐿((∝)�⃗�) = (∝)𝐿(�⃗�),        ∀∝∈ ℝ 

The set of all linear transformation from ℝ𝑓
𝑚 to ℝ𝑓

𝑛 is denoted as  𝐿𝑖𝑛(ℝ𝑓
𝑚, ℝ𝑓

𝑛) [7].  One of 

necessary example of linear transformation is linear mapping from ℝ𝑓
𝑚 to its coordinate 

i.e. 
𝑣𝑖⃗⃗⃗ ⃗: ℝ𝑓

𝑚 → ℝ 

by specific formula  
𝑣𝑖⃗⃗⃗ ⃗(�⃗�) = �⃗�(𝑖),        𝑖 = 1,2,3, … , 𝑚.   
 
The term of linear transformation is very useful in constructing theory change bases in 
graph space[9]. On internal graph space ℝ𝑓

𝑚 change of bases still able to be constructed.  

Definition 8. Let �⃗�1, �⃗�2, . . . , �⃗�𝑘 ∈ ℝ𝑓
𝑚 are said to be linearly independent if for each  �⃗�𝑖

⃗⃗⃗ ⃗ can’t 

be expressed as linear combination of some others. 
The term of linear combination refers to equation 
 �⃗� = (∝1)𝑣1⃗⃗⃗⃗⃗⨁ . . . ⨁ (∝𝑘)𝑣𝑘⃗⃗⃗⃗⃗ (7) 
 
 
Another way to express coordinate transformation is by defining linear transformation 
which maps a vector to its scalars corresponding to the basis used to. 
 



Basis Existence in Graph of Real Valued Continous Functions 

Ahmad Lazwardi 61 

Definition 9. Let V be a nontrivial vector space over field F. Let 𝑆 = {𝑠1⃗⃗⃗⃗ , 𝑠2⃗⃗⃗⃗ , 𝑠3⃗⃗⃗⃗ , … , 𝑠𝑚⃗⃗ ⃗⃗⃗} be a 
basis of V. Define coordinate transformation as ∅𝑆: 𝑉 → 𝐹

𝑚 such that for each  �⃗� = 𝑣1𝑠1⃗⃗⃗⃗ +
𝑣2𝑠2⃗⃗⃗⃗ + ⋯ + 𝑣𝑚𝑠𝑚⃗⃗ ⃗⃗⃗, the map 

∅𝑆(�⃗�) = [�⃗�]𝑆 = [

𝑣1
𝑣2
⋮

𝑣𝑚

] . 

 
It’s easy to check that coordinate transformation is an isomorphism [10]. The concept of 
coordinate mapping above is used to construct the concept of change of basis. First, we 
have to decide standard basis which lies on ℝ𝑓

𝑚.  

Standard basis of ℝ𝑓
𝑚 is 

𝑣𝑖⃗⃗⃗ ⃗ = (1𝑓, 0𝑓, 0𝑓, 0𝑓, . . . , 0𝑓) 

𝑣2⃗⃗⃗⃗⃗ = (0𝑓, 1𝑓, 0𝑓, 0𝑓, . . . , 0𝑓) 

𝑣𝑚⃗⃗⃗⃗⃗⃗ = (0𝑓, 0𝑓, 0𝑓, 0𝑓, . . . , 1𝑓) 

Now let’s see how the change of basis works: 
For each �⃗� ∈ ℝ𝑓

𝑚. By previous definition, one can write  

 �⃗� = (�⃗�(1), �⃗�(2), . . . , �⃗�(𝑚))                                               (8) 

which �⃗�(𝑖) ∈ ℝ𝑓   for each i = 1,2, . . ., m. Therefore 

�⃗� = �⃗�(1)�⃗�1⨁�⃗�(2)�⃗�2⨁ . . . ⨁�⃗�(𝑚)�⃗�𝑚 
for corresponding coordinate �⃗�(𝑖) ∈ ℝ𝑓.    

This is how the above coordinate and corresponding scalars related, let’s see it as follows: 
�⃗�(1). 1𝑓 = �⃗�(1) = 𝛵𝑓(𝛵𝑓

−1(�⃗�(1))) 

 = 𝛵𝑓(𝛵𝑓
−1(�⃗�(1). 1) 

 = (𝛵𝑓
−1(�⃗�(1). 1), 𝑓(�⃗�(1). 1)) 

 = (𝛵𝑓
−1(�⃗�(1). 1) . 1𝑓 

Hence 

 �⃗� = (𝛵𝑓
−1(�⃗�(1)) . 𝑣1⃗⃗⃗⃗⃗⨁ (𝛵𝑓

−1(�⃗�(2)) . 𝑣2⃗⃗⃗⃗⃗⨁ . . . ⨁ (𝛵𝑓
−1(�⃗�(𝑚)) . 𝑣𝑚⃗⃗⃗⃗⃗⃗  (9) 

for some (𝛵𝑓𝑘
−1(�⃗�(1)) ∈ ℝ. 

The last equation can be expressed in term of more visual vector addition as follows: 

�⃗� = (𝛵𝑓
−1(�⃗�(1)) . 𝑣1⃗⃗⃗⃗⃗⨁. . . ⨁ (𝛵𝑓

−1(�⃗�(𝑚)) . 𝑣𝑚⃗⃗⃗⃗⃗⃗  

 = (𝛵𝑓
−1(�⃗�(1)) .

[
 
 
 
1𝑓
0𝑓
⋮
0𝑓]

 
 
 

⨁. . . ⨁ (𝛵𝑓
−1(�⃗�(𝑚))

[
 
 
 
0𝑓
0𝑓
⋮
1𝑓]

 
 
 

 

 =

[
 
 
 
1𝑓 0𝑓
0𝑓
⋮
0𝑓

…
…
0𝑓

     

… 0𝑓
…
…
…

0𝑓
⋮
1𝑓]

 
 
 

[

𝛵𝑓
−1(�⃗�(1))

…
…

𝛵𝑓
−1(�⃗�(𝑚))

] = 𝐼𝑓𝛵𝑓
−1(�⃗�) 

The above equation gives a consequence that ℝ𝑓
𝑚 is isomorphic  to ℝ𝑚. 

 
The next theorem will ensure that ℝ𝑓

𝑚 has more than just standard basis. 

Teorema 3. If 𝑆 = {𝒔1, 𝒔2, . . . , 𝒔𝑚} is basis for ℝ
𝑚 then  𝑇𝑓(𝑆) = {𝑇𝑓(𝒔𝟏), 𝑇𝑓(𝒔2), . . . , 𝑇𝑓(𝒔𝑚)} 

forms a basis for ℝ𝑓
𝑚. 

Bukti: 



Basis Existence in Graph of Real Valued Continous Functions 

Ahmad Lazwardi 62 

Suppose that �⃗� ∈ ℝ𝑓
𝑚. Let’s see the following fact, another fine expression for vector in ℝ𝑓

𝑚 

is  �⃗� = [

�⃗�(1)

�⃗�(2)
⋮

�⃗�(𝑚)

] . Thus we have 𝛵𝑓
−1(�⃗�) =

[
 
 
 
 
𝛵𝑓

−1(�⃗�(1))

𝛵𝑓
−1(�⃗�(2))

⋮
𝛵𝑓

−1(�⃗�(𝑚))]
 
 
 
 

 is actually lies in ℝ𝑚. Therefore 

𝛵𝑓
−1(�⃗�) =

[
 
 
 
 
𝛵𝑓

−1(�⃗�(1))

𝛵𝑓
−1(�⃗�(2))

⋮
𝛵𝑓

−1(�⃗�(𝑚))]
 
 
 
 

= (∝1) [

𝑠11
𝑠12
⋮

𝑠1𝑚

] + . . . +(∝𝑚) [

𝑠𝑚1
𝑠𝑚2
⋮

𝑠𝑚𝑚

] , by 𝒔𝑖 = [

𝑠𝑖1
𝑠𝑖2
⋮

𝑠𝑖𝑚

]  for each i. 

 

One can have  �⃗� = 𝛵𝑓 (𝛵𝑓
−1(�⃗�)) = 𝛵𝑓((𝛼1)𝒔1 + (𝛼2)𝒔2+ . . . +(𝛼𝑚)𝒔𝑚) 

 = ((𝛼1)𝒔1 + (𝛼2)𝒔2+ . . . +(𝛼𝑚)𝒔𝑚, 𝑓((𝛼1)𝒔1 + (𝛼2)𝒔2+ . . . +(𝛼𝑚)𝒔𝑚)) 

 = (𝛼1)𝛵𝑓(𝒔1)⨁(𝛼2)𝛵𝑓(𝒔2)⨁ . . . ⨁(𝛼𝑚)𝛵𝑓(𝒔𝑚). 

In other words,  {𝛵𝑓(𝒔1), 𝛵𝑓(𝒔2), . . . , 𝛵𝑓(𝒔𝑚)} spans ℝ𝑓
𝑚. The rest is to prove that 𝛵𝑓(𝒔) is 

linearly independent.  
Suppose the statement is not true, then there exist 𝛵𝑓(𝒔) which become linear 

combination of other members. Let’s assume that is 𝛵𝑓(𝒔𝒊) then we have 

𝛵−1 (𝛵𝑓(𝒔𝑖)) = 𝒔𝑖 = (𝛼1)𝒔1+ . . . +(𝛼𝑖−1)𝒔𝑖−1 +  (𝛼𝑖+1)𝒔𝑖+1+ . . . + (𝛼𝑚)𝒔𝑚 

But since {𝒔𝟏, 𝒔𝟐, . . . , 𝒔𝒎} is linearly independent, then it should be a contradiction.   
 
The above theorem indirectly explains that the internal graph space ℝ𝑓

𝑚 has infinitely 

many vectors that can form a basis for ℝ𝑓
𝑚. The next theorems will be discussing about 

how the basis related each other. 
Let 𝑆 = {𝑠1⃗⃗⃗⃗ , 𝑠2⃗⃗⃗⃗ , . . . , 𝑠𝑚⃗⃗ ⃗⃗⃗} be a basis for ℝ𝑓

𝑚. Then for each �⃗� ∈ ℝ𝑓
𝑚, one can write  

 �⃗� = (𝛼1
𝑠)𝑠1⃗⃗⃗⃗ + (𝛼2

𝑠)𝑠2⃗⃗⃗⃗ + . . . +(𝛼𝑚
𝑠 )𝑠𝑚⃗⃗ ⃗⃗⃗   dengan (𝛼𝑖

𝑠) ∈ ℝ.  (10) 

 = (𝛼1
𝑠)

[
 
 
 
 𝑠1(1)
⃗⃗⃗⃗⃗⃗⃗⃗ ⃗⃗⃗

𝑠1(2)
⃗⃗⃗⃗⃗⃗⃗⃗ ⃗⃗⃗

⋮

𝑠1(𝑚)
⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗]

 
 
 
 

+ (𝛼2
𝑠)

[
 
 
 
 𝑠2(1)
⃗⃗⃗⃗⃗⃗ ⃗⃗ ⃗⃗⃗

𝑠2(2)
⃗⃗⃗⃗⃗⃗ ⃗⃗ ⃗⃗⃗

⋮

𝑠2(𝑚)
⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗ ⃗⃗⃗]

 
 
 
 

+…+(𝛼𝑚
𝑠 )

[
 
 
 
 𝑠𝑚(1)
⃗⃗⃗⃗ ⃗⃗ ⃗⃗ ⃗⃗ ⃗⃗

𝑠𝑚(2)
⃗⃗⃗⃗ ⃗⃗ ⃗⃗ ⃗⃗ ⃗⃗

⋮

𝑠𝑚(𝑚)
⃗⃗⃗⃗⃗⃗ ⃗⃗⃗⃗ ⃗⃗ ⃗⃗ ]

 
 
 
 

 

 =

[
 
 
 
 𝑠1(1)
⃗⃗⃗⃗⃗⃗⃗⃗ ⃗⃗⃗

𝑠1(2)
⃗⃗⃗⃗⃗⃗⃗⃗ ⃗⃗⃗

⋮

𝑠1(𝑚)
⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗

𝑠2(1)
⃗⃗⃗⃗⃗⃗ ⃗⃗ ⃗⃗⃗

𝑠2(2)
⃗⃗⃗⃗⃗⃗ ⃗⃗ ⃗⃗⃗

⋮

𝑠2(𝑚)
⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗ ⃗⃗⃗

…

𝑠𝑚(1)
⃗⃗⃗⃗ ⃗⃗ ⃗⃗ ⃗⃗ ⃗⃗

𝑠𝑚(2)
⃗⃗⃗⃗ ⃗⃗ ⃗⃗ ⃗⃗ ⃗⃗

⋮

𝑠𝑚(𝑚)
⃗⃗⃗⃗⃗⃗ ⃗⃗⃗⃗ ⃗⃗ ⃗⃗ ]

 
 
 
 

[

(𝛼1
𝑠)

(𝛼2
𝑠)
⋮

(𝛼𝑚
𝑠 )

] 

 =[𝑀]𝑆,𝑓[𝛼]𝑆  

by [𝑀]𝑆,𝑓 is matrics which for each entry lies in ℝ𝑓
𝑚 and [𝛼]𝑆  is real scalar vector.  

 
Then we have  
 
 

 𝑇𝑓
−1(�⃗�) =

[
 
 
 
 𝑇𝑓

−1(𝑠1(1)
⃗⃗⃗⃗⃗⃗⃗⃗ ⃗⃗⃗)

𝑇𝑓
−1(𝑠1(2)

⃗⃗⃗⃗⃗⃗⃗⃗ ⃗⃗⃗)

⋮

𝑇𝑓
−1(𝑠1(𝑚)

⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗)

𝑇𝑓
−1(𝑠2(1)

⃗⃗⃗⃗⃗⃗ ⃗⃗ ⃗⃗⃗)

𝑇𝑓
−1(𝑠2(2)

⃗⃗⃗⃗⃗⃗ ⃗⃗ ⃗⃗⃗)

⋮

𝑇𝑓
−1(𝑠2(𝑚)

⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗ ⃗⃗⃗)

…

𝑇𝑓
−1(𝑠𝑚(1)

⃗⃗⃗⃗ ⃗⃗ ⃗⃗ ⃗⃗ ⃗⃗ )

𝑇𝑓
−1(𝑠𝑚(1)

⃗⃗⃗⃗ ⃗⃗ ⃗⃗ ⃗⃗ ⃗⃗ )

⋮

𝑇𝑓
−1(𝑠𝑚(𝑚)

⃗⃗⃗⃗⃗⃗ ⃗⃗⃗⃗ ⃗⃗ ⃗⃗ )]
 
 
 
 

[

(𝛼1
𝑠)

(𝛼2
𝑠)
⋮

(𝛼𝑚
𝑠 )

] 



Basis Existence in Graph of Real Valued Continous Functions 

Ahmad Lazwardi 63 

Therefore, 
   𝑇𝑓

−1(�⃗�)=[𝑇𝑓
−1(𝑆)][𝛼]𝑆 (11) 

 
The above equation explains how the basis S in internal graph space ℝ𝑓

𝑚 mapped to ℝ𝑚. 

It will make us easier to change basis from the old to the new one. Let’s pay attention to 
the following discussion; 
Let 𝑆 = {𝑠1⃗⃗⃗⃗ , 𝑠2⃗⃗⃗⃗ , . . . , 𝑠𝑚⃗⃗ ⃗⃗⃗} be a basis of ℝ𝑓

𝑚 and 𝑊 = {𝑤1⃗⃗ ⃗⃗⃗, 𝑤2⃗⃗⃗⃗⃗⃗ , . . . , 𝑤𝑚⃗⃗⃗⃗⃗⃗⃗} another basis. We want 

to change the old basis S into the new one W. We have 
�⃗� = [𝑀]𝑆,𝑓[𝛼]𝑆 

On the other hand 
�⃗� = [𝑀]𝑊,𝑓[𝛼]𝑊 

Therefore 
 𝑇𝑓

−1(�⃗�) =[𝑇𝑓
−1(𝑆)][𝛼]𝑆 

 )(
1

)(
1 vT fST f


 

 
=[𝛼]𝑆 

There exists transition metrics   )(),( 11 WTST ffM  such that the following works  

 )(
1

)(
1 vT fWT f


  =    SWTST ffM )(),( 11   

     = [𝛼]𝑊. 
 
Now we send it back through the map  

 
)(

1

)(

1

)(
11 vT fWTWT ff


    = [𝑇𝑓

−1(𝑊)][𝛼]𝑊. 

Hence 
�⃗�= [𝑀]𝑆,𝑓[𝛼]𝑆 = [𝑀]𝑊,𝑓[𝛼]𝑊. 

That’s how the basis change. For more understanding, let’s see the example 
 
Let  𝑓(𝑥) = 𝑒𝑥. The corresponding isomorphism is 𝑇𝑓(𝑥) = (𝑥, 𝑒

𝑥) for each 𝑥 ∈ ℝ. 

The graph is ℝ𝑓 = {𝑎𝑓 = (𝑎, 𝑒
𝑎): 𝑎 ∈ ℝ}. By applying the method to find the 

corresponding addition, we have  
 𝑎𝑓⨁𝑏𝑓 = (𝑎 + 𝑏, 𝑒

𝑎+𝑏) ,   𝑎, 𝑏 ∈ ℝ (12) 

And the scalar multiplication 
 (𝛼)(𝑥𝑓) = (𝛼𝑥, 𝑒

𝛼𝑥), 𝑥 ∈ ℝ  (13) 

Now the internal graph space dimension 2 has the form 

 ℝ𝑓
2 = {(

𝑥𝑓
𝑦𝑓

) : 𝑥𝑓, 𝑦𝑓 ∈ ℝ𝑓} (14) 

Now let see how the basis change ℝ𝑓
2. Choose a basis   

𝑆 = {(
(2, 𝑒2)
(0,1)

) , (
(0,1)

(2, 𝑒2)
)} 

and 

𝑊 = {(
(1, 𝑒)
(0,1)

) , (
(1, 𝑒)

(−1, 𝑒−1)
)}. 

Suppose �⃗� = (
(5, 𝑒5)

(2, 𝑒2)
). It will be shown that �⃗� is linear combination of S. 

Let’s pay attention to this 

�⃗� = (𝛼) (
(2, 𝑒2)
(0,1)

) ⨁(𝛽) (
(0,1)

(2, 𝑒2)
) 



Basis Existence in Graph of Real Valued Continous Functions 

Ahmad Lazwardi 64 

 =(
(2𝛼, 𝑒2𝛼)

(0,1)
) ⨁ (

(0,1)

(2𝛽, 𝑒2𝛽)
) 

 = (
(2𝛼, 𝑒2𝛼)

(2𝛽, 𝑒2𝛽)
) 

 =(
(5, 𝑒5)

(2, 𝑒2)
) 

hence  2𝛼 = 5, 2𝛽 = 2, 𝑒2𝛼 = 𝑒5, 𝑒2𝛽=𝑒2. Those imply  𝛼 = 5
2
 and 𝛽 = 1. 

Therefore 

 �⃗� = [
2𝑓 0𝑓
0𝑓 2𝑓

] [
5

2

1
] 

 =[𝑀]𝑆,𝑓[𝛼]𝑆 

On the other hand,  

�⃗� = (𝛼) (
(1, 𝑒)
(0,1)

) ⨁(𝛽) (
(1, 𝑒)

(−1, 𝑒−1)
) 

 =(
(𝛼, 𝑒𝛼)
(0,1)

) ⨁ (
(𝛽, 𝑒𝛽)

(−𝛽, 𝑒−𝛽)
) 

 = (
(𝛼 + 𝛽, 𝑒𝛼+𝛽)

(−𝛽, 𝑒−𝛽)
) 

 =(
(5, 𝑒5)

(2, 𝑒2)
) 

We have 𝛼 + 𝛽 = 5, −𝛽 = 2, 𝑒𝛼+𝛽 = 𝑒5, 𝑒−𝛽=𝑒2. Those imply 𝛼 = 7 and 𝛽 = −2. 
Therefore 

 �⃗� = [
1𝑓 1𝑓
0𝑓 −1𝑓

] [
7

−2
] 

 =[𝑀]𝑊,𝑓[𝛼]𝑊. 

Its already shown how the vector �⃗� to be expressed as linear combination of S and W. To 
change basis from S to W, we must transfer all members of S to the ℝ2 i.e. 𝑇𝑓

−1(𝑆) =

{(
2
0
) , (

0
2
)} and 𝑇𝑓

−1(𝑊) = {(
1
1
) , (

0
−1

)} . By elementary calculation we have 

 (
2
0
) = 2 (

1
1
) + 2 (

0
−1

) 

 (
0
2
) = 0 (

1
1
) − 2 (

0
−1

) 

thus, the transition metrics is 

 )(
1

)(
1 vT fWT f


  =[

2 0
2 −2

] [
5
2

1
] 

 =[
7

−2
]. That’s the new coordinate. 

CONCLUSIONS 

For each real valued function f, the corresponding internal graph space forms a vector 
space over ℝ. The concept of linear mapping and linear combinations still can be adapted 
from the graph space and still well defined.  
  



Basis Existence in Graph of Real Valued Continous Functions 

Ahmad Lazwardi 65 

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