Application of linear transformation in computer

LINEAR TRANSFORMATION AND ITS
APPLICATIONS IN COMPUTER
CHUKWUEDO FAVOUR
EU/SC/MTH/13/007
A PROJECT SUBMITTED TO THE DEPARTMENT OF
PHYSICAL SCIENCES COLLEGE OF SCIENCE EVANGEL
UNIVERSITY AKAEZE IN PARTIAL FULFILMENT OF
THE REQUIREMENTS FOR THE AWARD OF BACHELOR
OF SCIENCE DEGREE IN MATHEMATICS
SEPTEMBER, 2017

CERTIFICATION
CHUKWUEDO FAVOUR, an Undergraduate in the DEPARTMENT of
PHYSICAL SCIENCES (MATHEMATICS OPTION) with registration number
EU/SC/MTH/13/007 has satisfactorily completed the requirements for research work for
the degree BACHELOR of SCIENCE in MATHEMATICS. The work embodied in this
project report is original and has not been submitted in part or full for any other diploma
or degree of this or any other university.
————————————— ————————————–
MR A. C. ONAH PROF.G. A. AGBO
Project Supervisor Head of Department
———————————————-
PROF. OYESANYA, M.O
External Examiner
i

DEDICATION
To
Almighty God.
ii

ACKNOWLEDGEMENT
How can one ever acknowledge academic debts satisfactorily? Knowledge is a result of a
cumulative process spanning over many years during which an individual passes through
many people, circumstances and ideas. It is very difficult to categorise them because where
the influence of one stops, that of the other begins. But all the same, I wish to express
my gratitude to God for making all things well and to a number of people who in one way
or the other have been instrumental to the success of this project.
I offer a deserving bow to my supervisor , Mr A.C. Onah for his assistance, invaluable
suggestions, painstaking guidance, devotion, constant motivation, patience and loving
kindness. He deserves more gratitude from me than I can express here. I am also very
grateful to the Members of Staff of the Department of Mathematics, Evangel University
Akaeze, Ebonyi State. Prof. U. A. Osisiogu, Dr. J. Ezeora, Mr. R.C Ogbonna, Mr. J.
Ofoma and Mr. C. Achudume for their contributions that guided me in the course of this
work My special appreciation goes to my parents Mr and Mrs Francis Chukwuedo and
my siblings Esther, Gideon, Goodness and Samuel for their financial Support, love, un-
derstanding and prayers that created an emotional and conducive surrounding for positive
work.
There is no way that I can adequately acknowledge the influence of my course mates
Uwaoma David, Nwachi Promise, Wonders Abbah, Okwese Peter and Ogueri Chimezie
and my friends for their encouragement.
I do hope that you will tolerate my excesses in the likes of overstatements, understate-
ments, omissions and commissions, some of which hinge on inadequacy of information and
personal limitations at a particular time. There is no doubt that I have bitten more than I
can chew. In spite of this, I still insist on calling this project ”our project”, but the lapses
remain solely my responsibility.
iii

ABSTRACT
Cryptography, the science of encrypting messages in secret codes, has played an impor-
tant role in securing information since the emergence of computers. The basic idea of
cryptography is that information can be encoded using an encryption scheme and can be
decoded by anyone who knows about the scheme. There are lots of encryption schemes
ranging from very simple to very complex. Most of them are mathematical in nature.
Since matrices together with the linear transformation they represent have unique and
very powerful concept such as inverses which can be easily understood, it could be applied
as an efficient way for encrypting and storing text. This project work describes some of
the techniques of cryptography using matrices together with linear transformation repre-
sented. The technique is very simple and can be easily used for encryption of messages
confidentially but also not so easy to break if someone does not know the encryption key.
The encryption system uses different type of matrices to store the text entered by the
sender in the form of their positions and their inverses for decoding the encrypted data
into plain-text. Singh et al (2016), considered using an improved matrix cryptography to
solve this challenge of security. It requires the key matrix and its inverse in encryption
and decryption respectively.
iv

Contents
Certiﬁcation i
Dedication ii
Acknowledgement iii
Abstract iv
1 Introduction 3
1.1 Background of the Study . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
1.2 Deﬁnition Of Terms And Introduction Of Basic Concepts . . . . . . . . . . 4
1.2.1 Plain-text . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
1.2.2 Cipher-text . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
1.2.3 Encryption . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
1.2.4 Decryption . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
1.2.5 Cipher . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
1.2.6 Axioms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
1.2.7 Axiom relating Addition and Multiplication . . . . . . . . . . . . . 6
1.3 Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
2 Literature Review 8
2.1 Vector Space and Subspace of Vector Space . . . . . . . . . . . . . . 9
2.1.1 Vector Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
2.1.2 Examples of vector spaces . . . . . . . . . . . . . . . . . . . . . . . 9
1

2.1.3 Subspace of Vector Space . . . . . . . . . . . . . . . . . . . . . . . 13
2.1.4 Examples of Subspaces of vector Spaces . . . . . . . . . . . . . . . 13
2.2 Linear Combination . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
2.2.1 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
2.3 Linear Independent and Linear Dependent Vectors: . . . . . . . . 15
2.4 Spanning Sets and Bases: . . . . . . . . . . . . . . . . . . . . . . . . . 15
2.4.1 Basis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
2.4.2 Examples of Basis . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
3 Linear transformations 21
3.1 Properties of linear transformations . . . . . . . . . . . . . . . . . . . . . . 22
3.2 Algebra Of Linear Transformation . . . . . . . . . . . . . . . . . . . . 23
3.3 Range Space and Null Space of a Linear Transformation . . . . . . . . . . 26
3.3.1 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
3.4 REPRESENTATION OF LINEAR TRANSFORMATION BY MA-
TRICES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
3.4.1 Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
3.5 Linear Transformation Given By Matrices . . . . . . . . . . . . . . . . . . 29
4 Application And Conclusion 30
2

Chapter 1
Introduction
1.1 Background of the Study
People all over the world are engaged in communication through internet every day. It is
very important to protect our essential data from unauthorized users. The main challenge
in data communication is how to keep data secure against unlawful interference often called
hacking or eavesdropping. One of the common serious attacks occurs when an unautho-
rized party can access to read and in some cases, modify an important data. The data
transferred from one system to another system over the public network can be protected
by means of encryption. Each encryption creates cipher text that can be decrypted into
plain-text. Raja and Chakravarthy 2011 used Hilbert matrices to encrypt secret messages
and its inverse to decrypt the message. The idea they had behind choosing the Hilbert
matrices is that they are always invertible and have integer inverses .
Vector spaces are one of the two main ingredients making up the foundation of this
project work, the other being linear transformations. Linear transformations are functions
that send, or map, one vector to another vector. Linear refers to the fact that the transfor-
mation preserves vector addition and scalar multiplication. This means that if T is a linear
transformation sending a vector v to T(v), then for any vectors v and w, and any scalar c,
the transformation must satisfy the properties T(v+w) = T(v)+T(w) and T(cv) = cT(v).
When doing computations, linear transformations are treated as matrices. The appli-
cation of linear transformation in computer (An approach to Cryptography - a data security
technique) as described in this project work stands ﬁrm on the basis of the Theorem of
The Matrix Representation of a linear transformation.
3

”Let V be a finite-dimensional vector space over the field F and let {vj}n
j=1 be an
ordered basis for V . Let W be a vector space over the same field and let {vj}n
j=1 be any
given vectors in W . Then there exists a unique linear transformation T from V into W
such that Tvj = wj, j = 1, 2, 3, ..., n”
1.2 Definition Of Terms And Introduction Of Basic
Concepts
1.2.1 Plain-text
The message or information that is being encrypted.
Example: Any written word
1.2.2 Cipher-text
The message or information that is created after the cipher has been used.
1.2.3 Encryption
Encryption is the process of converting original plain text (data) into cipher text (data).
1.2.4 Decryption
Decryption is the process of converting the cipher text (data) to the original plain text(data)(Wu
T. M., 2005).
1.2.5 Cipher
A procedure that will render a message unintelligible to the recipient. Used to also recreate
the original message.
Example:
plaintext: A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
ciphertext: X Y Z A B C D E F G H I J K L M N O P Q R S T U V W
4

In this example, the message:
WHAT KIND OF CAKE SHOULD WE HAVE? ALICE.
will be rendered as follows:
TEXQ HFKA LC ZXHB PELRIA TB EXSB? XIFZB.
1.2.6 Axioms
Laws governing the way numbers combine together are called axioms. Any particular
axiom might be true in some number systems but not in others.
Axioms for Addition
Let S be a number system;
A1. α + β = α + β ∀ α , β ∈ S.
A2. (α + β) + γ = α + (β + γ) ∀ α, β, γ ∈ S.
A3. There is a number 0 ∈ S such that α + 0 = 0 + α = α ∀ α ∈ S.
A4. For each number α ∈ S ∃ a number −α ∈ S such that α + (−α) = (−α) + α = 0.
These axioms may or may not be satisﬁed by a given number system S. For example,
in N, A1 and A2 hold but A3 and A4 do not hold. A1 - A4 all hold in Z , Q , R , C.
5

Axioms for Multiplication
Let S be a number system;
M1. α.β = β.α ∀ α, β ∈ S.
M2. (α.β).γ = α.(β.γ) ∀ α, β, γ ∈ S.
M3. There is a number 1 ∈ S suchthat α.1 = 1.α = α∀α ∈ S.
M4. For each number α ∈ S with α = 0, ∃ a number α−1
∈ S such that α.α−1
=
α−1
.α = 1.
In N and Z, M1-M3 hold but M4 does not hold. M1-M4 all hold in Q , R and C.
1.2.7 Axiom relating Addition and Multiplication
D. (α + β).γ = α.γ + β.γ ∀ α, β, γ ∈ S
1.3 Field
Definition. A field is a non-empty set, on which two binary operations + and • called
addition and multiplication, respectively, are defined.
Concurrently, a set S on which addition and multiplication are defined is called a field if
it satisfies each of the axioms A1, A2, A3, A4, M1, M2, M3, M4, D, and if, in addition,
1 = 0.
Roughly speaking, S is a field if addition, subtraction, multiplication and division(except
by zero) are all possible in S. Elements of a field are often called Scalars.
6

Examples of ﬁelds include
(a) The set of all real numbers R with the usual addition and multiplication.
(b) The set of all rational numbers Q with the usual addition and multiplication on R.
(c) The set of all complex numbers C with the usual addition and multiplication on
complex numbers.
7

Chapter 2
Literature Review
During the last decades, information security has become a major issue (Obaida Moham-
mad Awad Al-Hazaimeh 2013). With the rapid development of network and multimedia
technologies, the digital information has been applied to many areas in real-world appli-
cations. Communication has become a very important aspect in today’s life. So, security
plays an important role in transferring the data. One such way to secure information
is cryptography. In this parlance, the theorem of the matrix representation of a linear
transformation becomes very useful. In cryptography we hide the information from unau-
thorized users by employing various techniques, encryption is one such technique where
we transform the data into a form understandable only by the authorized users. We need
to hide the data for privacy purpose and for ensuring data received at the authenticated
user end is not modiﬁed .We have several encryption and decryption algorithms for en-
crypting the data at sender end and decrypting the same at receiver side ensuring secure
data transfer.
8

2.1 Vector Space and Subspace of Vector Space
2.1.1 Vector Space
A vector space over a field K is a set V which has two basic operations, addition and scalar
multiplication, satisfying certain requirements. Thus for every pair u, v ∈ V, u + v ∈ V is
defined, and for every α ∈ K, αv ∈ V is defined. For V to be called a vector space, the
following axioms must be satisfied for all α, β ∈ K and all u, v ∈ V. In this project, unless
otherwise stated, scalars are chosen from the set of real numbers.
Vector addition satisfies axioms A1, A2, A3 and A4 above.
α(u + v) = αu + αv;
(α + β)v = αv + βv;
(αβ)v = α(βv);
1v = v.
If V is a vector space over the field of real numbers, R , then V is called a real vector
space. V is called a complex vector space if V is a vector space (over C) the field of
complex numbers.
2.1.2 Examples of vector spaces
Examples of Vector Spaces
1. Let f be any field and fn
= f × f × ... × f (n factors of f)
x ∈ f ⇒ x = (x1, x2, ..., xn); xi ∈ f, ∀i = 1, 2, ..., n)i.e.
fn
= {(x1, x2, ..., xn) : xi ∈ f, ∀i = 1, 2, ..., n)}
define addition and scalar multiplication on fn
by
x + y = (x1, x2, ..., xn) + (y1, y2, ..., yn)
= (x1 + y1, x2 + y2, ..., xn + yn)
αx = α(x1, x2, ..., xn) = (αx1, αx2, ..., αxn)
9

for all x = (x1, x2, ..., xn) ; y = (y1, y2, ..., yn) ∈ fn
and α ∈ f
With this addition and scalar multiplication, fn
is a vector space.
Veriﬁcation:
Let x, y, z ∈ fn
and α, β ∈ f; be arbitrary such that
x = (x1, x2, ..., xn); y = (y1, y2, ..., yn); z = (z1, z2, ..., zn)
Closure under addition
(i) x + y = (x1, x2, ..., xn) + (y1, y2, ..., yn)
= x + y = (x1 + y1, x2 + y2, ..., xn + yn) ∈ fn
, since each xi, yi ∈ f and so is xi + yi
(ii) Closure under scalar multiplication
αx = α(x1, x2, ..., xn) = (αx1, αx2, ..., αxn) ∈ fn
, since αxi ∈ f for each i
(iii) Commutativity under addition
x + y = (x1, x2, ..., xn) + (y1, y2, ..., yn)
= (x1 + y1, x2 + y2, ..., xn + yn)
= (y1 + x1, y2 + x1, ..., yn + xn)
= (y1, y2, ..., yn) + (x1, x2, ..., xn)
= y + x
(iv) Associativity under addition.
(x + y) + z = [(x1, x2, ..., xn) + (y1, y2, ..., yn)] + (z1, z2, ..., zn)
= [(x1 + y1, x2 + y2, ..., xn + yn)] + (z1, z2, ..., zn)
= [x1 + y1 + z1, x2 + y2 + z2+, ..., +xn + yn + zn]
= [x1 + (y1 + z1), x2 + (y2 + z2)+, ..., +xn + (yn + zn)]
= (x1, x2, ..., xn) + [y1 + z1, y2 + z2, ..., yn + zn]
= x + (y + z)
(v) Existence of additive identity.
There exist 0 = (0, 0, ..., 0) ∈ fn
such that
0 + x = (0, 0, ..., 0) + (x1, x2, ..., xn)
10

= (0 + x1, 0 + x2, ..., 0 + xn)
= (x, x, ..., x) = x
(vi) Existence of additive inverse
Let x = (x1, x2, ..., xn) ∈ fn
be arbitrary, then, by property of ﬁeld, there exist
−x = (−x1, −x2, ..., −xn) ∈ fn
, such that
x + (−x) = (x1, x2, ..., xn) + (−x1, −x2, ..., −xn)
= (x1 − x2, x2 − x2, ..., xn − xn)
= (0, 0, ..., 0) = 0
(vii) : α(x + y) = α(x1 + y1, x2 + y2, ..., xn + yn)
= [α(x1 + y1), α(x2 + y2), ..., α(xn + yn)]
= (αx1 + αy1, αx2 + αy2, ..., αxn + αyn)
= (αx1, αx2, ..., αxn) + (αy1, αy2, ..., αyn)
= α(x1, x2, ..., xn) + α(y1, y2, ..., yn)
= αx + αy
(viii) : (α + β)x = (α + β)(x1, x2, ..., xn)
= ((α + β)x1, (α + β)x2, ..., (α + β)xn)
= (αx1 + βx1, αx2 + βx2, ..., αxn + βxn)
= (αx1, αx2, ..., αxn) + (βx1, βx2, ..., βxn)
= α(x1, x2, ..., xn) + β(x1, x2, ..., xn) = αx + βx
(ix) : (αx)β = [α(x1, x2, ..., xn)β]
= (αx1, αx2, ..., αxn)β
= (αβx1, αβx2, ..., αβxn)
= α(βx1, βx2, ..., βxn)
= α[β(x1, x2, ..., xn)] = α(βx)
(x) : ∃1 ∈ f 1 • x = 1 • (x1, x2, ..., xn)
= (1 • x1, 1 • x2, ..., 1 • xn)
= (x1, x2, ..., xn) = x
... fn
is a vector space over f
This example shows that the cartesian products R , Q and C are vector spaces
over the ﬁelds R, Q and C respectively.
11

2.1.3 Subspace of Vector Space
Many interesting examples of vector spaces are subsets of a given vector space V that are
vector spaces in their own right.
Let V be a vector space over the field K. Certain subsets of V have the nice property
of being closed under addition and scalar multiplication; that is, adding or taking scalar
multiples of vectors in the subset gives vectors which are again in the subset. We call such
a subset a subspace:
Definition. A subspace of V is a non-empty subset W ⊆ V such that
i W is closed under addition: u, v ∈ W =⇒ u + v ∈ W.
ii W is closed under scalar multiplication: v ∈ W, α ∈ K → αv ∈ W.
These two conditions can be replaced with a single condition
u, v ∈ W, α, β ∈ K → αu + βv ∈ W.
A subspace W is itself a vector space over K under the operations of vector addition
and scalar multiplication in V . Notice that all vector space axioms of W hold automati-
cally. (They are inherited from V .)
2.1.4 Examples of Subspaces of vector Spaces
1. Let S be any real vector space. Then, {0} is a subspace of S, usually called the
trivial subspace of any vector space.
Verification: Given, V = {0},then, x, y ∈ V ; ⇒ x = y = 0 and so
∀α, β ∈ ; αx + βy = α(0) + β(0) = 0 + 0 = 0 ∈ V
12

Note that any subspace of V that contains W1 and W2 has to contain all vectors of the
form u + v for u ∈ W1, v ∈ W2. This motivates the following definition.
Definition. Let W1, W2 be subspaces of the vector space V . Then W1 +W2 is defined
to be the set of vectors v ∈ V such that v = w1 + w2 for some w1 ∈ W1, w2 ∈ W2.
Or Preferably, W1 + W2 = w1 + w2 | w1 ∈ W1, w2 ∈ W2 .
2.2 Linear Combination
Given that λ1, λ2, λ3, . . . , λnare scalars in a scalar field Fand v1, v2, v3, . . . , vnare vectors
in a vector space, V over the scalar field. Then, the linear combination of v1, v2, v3, . . . , vn
(with those scalars as coefficients) is an expression of the form
λ1v1 + λ2v2 + λ3v3 + . . . + λnvn or n
k=1 λkvk in compact form. In line with this, a given
vector v ∈ V is said to be a linear combination of a set of vectors v1, v2, v3, . . . , vnif there
exists scalars λ1, λ2, λ3, . . . , λn such that v = λ1v1 + λ2v2 + λ3v3 + . . . + λnvn.
2.2.1 Example
1. Let the field K be the set of real numbers. Let the vector space V be the Euclidean
space, 3
. Consider the vectors, e1 = (1, 0, 0), e2 = (0, 1, 0)and e3 = (0, 0, 1).
Then, any vector in 3
is a linear combination of e1, e2 and e3
Verification: Let v = (α1, α2, α3)be arbitrary. We seek scalars λ1, λ2 and λ3such that
(α1, α2, α3) = λ1e1 + λ2e2 + λ3e3. This means that,
(α1, α2, α3) = λ1e1 + λ2e2 + λ3e3
= λ1(1, 0, 0) + λ2(0, 1, 0) + λ3(0, 0, 1)
= (λ1, 0, 0) + (0, λ2, 0) + (0, 0, λ3)
= (λ1, λ2, λ3)
Which implies that λ1 = α1, λ2 = α2 and λ3 = α3.
13

2.3 Linear Independent and Linear Dependent Vec-
tors:
Let V be a real vector space. The vectors v1, v2, ..., vnin V are said to be linearly independent
over R if for t1, t2, ..., tn in R, we have
n
Σ
k=1
tkvk = 0 implies thatt1 = t2 = ... = tn = 0.
Vectors which are not linearly independent are said to be linearly dependent.
In a linearly dependent set of vectors one can express at least one of the vectors as a
linear combination of the others. While, this is not possible in a linearly independent set
of vectors. If we want to check whether a set of vectors v1, v2, ..., vn is linearly independent
or not, we form a linear combination
n
Σ
k=1
tkvkof the vectors and assume this to be zero
and then solve for t1, t2, ..., tk. If all the t’s are zero, then our set of vectors is linearly
independent, otherwise, it is linearly dependent.
Example
1. In the vector space R , the following set of vectors
S = {ei = (0, 0, ..., 0, 1, 0, 0, ..., 0); i = 1, 2, ..., n}, (with 1 in the ith entry) is linearly
independent .
Veriﬁcation: Without loss of generality, in R ,
S = {(1, 0, 0) , (0, 1, 0) , (0, 0, 1)}.
Let v1 = (1, 0, 0) , v2 = (0, 1, 0) , v3 = (0, 0, 1) ; t1, t2, t3 ∈ R
Then, 3
k=1 tkvk = 0, implies that
t1 (1, 0, 0) + t2 (0, 1, 0) + t3 (0, 0, 1) = (0, 0, 0)
⇒ (t1, t2, t3) = (0, 0, 0)
⇒ t1 = 0; t2 = 0; t3 = 0
2.4 Spanning Sets and Bases:
Let V be a real vector space and S = {v1, v2, ..., vn} a subset of V. The span of S denoted
by span[s] is the set of all ﬁnite linear combinations of the vectors, v1, v2, ..., vn, i.e.
Span[s] = {v ∈ V : v =
n
Σ
k=1
tkvk; tk ∈ R, k = 1, 2, ..., n}
14

2.4.1 Basis
A basis for a vector space V over R is a linearly independent subset of V which spans
V.
2.4.2 Examples of Basis
• Let V = R3
and S = {(1, 2, 0), (−1, 1, 0)}. What is span[S] ?
• Find the span of the vector (−1, 1).
• Find the span of S = {(1, 1, 0), (−1, 0, 0)}
Solution
The solution to the above can only be completed by the preceding theorem.
Theorem
Let V be a vector space over R, and let S = {v1, v2, ..., vn} be a subset of V. Then
• Span[S] is a subspace of V.
• If S is linearly independent, then every vector in span[S] can be written in only one
way as a linear combination of the vectors in S.
• If S is linearly independent and y is not in span[S] then the set obtained by adding
y to S is linearly independent.
A set of vectors S = {v1, v2, ..., vn} in V is said to generate or span V if every vector in V
can be expressed as a linear combination of v1, v2, ..., vn, i.e. if v is an arbitrary element
of V, then there exist t1, t2, ..., tk ∈ R such that v = n
k=1 tkvk, i.e , V = span[S]. In
other words, a subset of a vector space is said to span (or generate) the vector space if for
every vector in the vector space, one can ﬁnd suitable scalars such that the vector can be
expressed as alinear combination of the vectors in the subset.
A basis for a vector space V over R is a linearly independent subset of V which spans
V.
Examples
• In the vector space R , the set of vectors S = {ei = (0, 0, ..., 0, 1, 0, 0, ..., 0); i = 1, 2, ..., n},
(with 1 in the ith entry) is a basis for R (why?). The set of vectors S = {e1, e2, ..., en}
is called the usual or standard basis for R .
15

• The set of vectors S = {(1, 1, 1), (0, 1, 2), (0, 0, 1)}is a basis for R (why?).
• The set of vectors S = {(1, 1, 1), (0, 1, 2), (0, 0, 1)} is a basis for R (see example (i)).
• In R any set consisting of just one non-zero number is a basis.
• The set {(1, 0, 0), (0, 1, 0), (1, 1, 0)}is not a spanning set of R instead its span is the
space of all vectors in R whose last component is zero.
• In R , the vectors v1 = (1, 1) and v2 = (−1, 2)is a basis for R .
16

To illustrate with example (iv), we have to prove that these two vectors form a basis for
2
. To this end, it suffices to prove that these vectors are linearly independent and that
they generate (or span) 2
.
Part I: Recall that two vectors v1 and v2 are linearly independent if λ1v1 + λ2v2 = 0
with (λ1 and λ2 being scalars) implies that λ1 = 0 and λ2 = 0. Hence, to prove that
(1, 1) and (−1, 2) are linearly independent, we assume that there exists scalars λ1 and λ2
such that λ1(1, 1) + λ2(−1, 2) = (0, 0)and strive to show that λ1 = λ2 = 0. But,
λ1(1, 1) + λ2(−1, 2) = (0, 0)
⇒ (λ1, λ1) + (−λ2, 2λ2) = (0, 0)
⇒ (λ1 − λ2, λ1 + 2λ2) = (0, 0)
⇒ λ1 − λ2 = 0 and λ1 + 2λ2 = 0
On solving, we have thatλ1 = 0 and λ2 = 0. Therefore, (1, 1) and (−1, 2) are linearly
independent.
Part II: Recall that two vectors v1 and v2 are said to generate or span a vector space, V ,
over a scalar field F if for every vector, v ∈ V , it is possible to find λ1 and λ2in Fsuch
that v = λ1v1 + λ2v2. Hence, to prove that (1, 1) and (−1, 2) span 2
, we assume that
v = (x, y) is an arbitrary vector in 2
and seek λ1 and λ2in Fsuch that λ1v1 +λ2v2 = (x, y)
But,
λ1v1 + λ2v2 = (x, y)
⇒ λ1(1, 1) + λ2(−1, 2) = (x, y)
⇒ (λ1, λ1) + (−λ2, 2λ2) = (x, y)
⇒ (λ1 − λ2, λ1 + 2λ2) = (x, y)
⇒ λ1 − λ2 = x and λ1 + 2λ2 = y
On solving for λ1 and λ2 by subtracting the last two equations, we have that 3λ2 =
y−xwhich in turn implies that λ2 = 1
3
(y−x)and on substitution followed by simplification,
we have that λ1 = 1
3
(y + 2x). Therefore, (1, 1) and (−1, 2) span 2
because for any vector
v = (x, y) ∈ 2
, we can find λ1 = 1
3
(y + 2x) and λ2 = 1
3
(y − x)both in F such that
(x, y) = 1
3
(y + 2x)(1, 1) + 1
3
(y − x)(−1, 2).
Theorem
Let V be a vector space that has a basis consisting of n vectors. Then,
i any set with more than n vectors is linearly independent
ii any spanning set has at least n vectors
iii any linearly independent subset of V has at most n vectors
17

iv every basis of V contains exactly n vectors
18

Proof
Let V be a real vector space. We say that the dimension of V is n , dim(V ) = n, if
V has a basis of n vectors. If V is the vector space that consists of the zero vector only,
we define the dimension of V to be 0. If dim(V ) = n, we call V an n-dimensional vector
space.
If for any n, V has a set of n linearly independent vectors, we say that the dimension of
V is infinite. The set, of all polynomials with coefficients from R is infinite dimensional.
The set of all real-valued functions defined on [a, b] is also infinite dimensional.
19

Chapter 3
Linear transformations
Definition. Let V and W be real vector spaces. A Linear transformation from V into W
is a function T : V −→ W which satisfies the following properties.
• T(x + y) = Tx + Ty; ∀x, y ∈ V ; Property 1
(We say that T preserves additivity.)
• T(ax) = aTx; ∀x ∈ V ; ∀a ∈ R. Property 2
(We say that T preserves scalar multiplication.)
If both properties itemised above are satisfied, then, ∀x, y ∈ V ; ∀a, b ∈ R , we have:
T(ax + by) = T(ax) + T(by) = aTx + bTy; ∀x, y ∈ V ; a, b ∈ R
Conversely, if
T(ax + by) = aTx + bTy; ∀x, y ∈ V ; a, b ∈ R,
then by taking a = b = 1 in the equation above we obtain ( Property 1 ) and taking
b = 0 we obtain ( Property 2 ).
Thus Property 1 and Property 2 can be replaced by :
20

T(ax + by) = T(ax) + T(by) = aTx + bTy; ∀x, y ∈ V ; a, b ∈ R
It is important to observe that if a = 0 in ( Property 2 ), then we obtain T(0) = 0 .
Thus, if T : V −→ W is a Linear Transformation, then T must take the zero element of
V to the zero element of W
The following examples copiously explains the concept.
Example
Let V be a given real vector space. Then the identity operator, T : V → V deﬁned for
any given x ∈ V by Tx = x is a linear transformation, since
T(ax + by) = ax + by ( from definition of T)
= aTx + bTy; ∀x, y ∈ V ; a, b ∈ R.
Example
The zero transformation, T : V → W deﬁned for all x ∈ V by Tx = 0 is a linear
transformation, since
T(x + y) = 0 ( from definition of T) = 0 + 0 = Tx + Ty, and T(ax) = 0 = a0 = aTx.
3.1 Properties of linear transformations
Theorem. Let V and W be two vector spaces. Suppose T : V −→ W is a linear
transformation. Then
1. T(0) = 0.
2. T(−v) = −T(v) ∀ v ∈ V
3. T(u − v) = T(u) − T(v) ∀ u, v ∈ V
4. If v = c1v1 + c2v2 + ... + cnvn then,
T(v) = T(c1v1 + c2v2 + ... + cnvn) = c1T(v1) + c2T(v2) + ... + cnT(vn).
21

3.2 Algebra Of Linear Transformation
Basically this explains some ways of combining linear transformation to get another linear
transformation.
Theorem
Let V and W be real vector spaces and let T1 and T2 be linear transformation from
V into W. Then the function (T1 + T2) defined for all x ∈ V by (T1 + T2)(x) = T1x + T2x
is a linear transformation.
If c is any element of R, then the function (cT1) : V → W defined for each x ∈ V by
(cT1)(x) = cT1x is a linear transformation from V into W. Furthermore, the set of all
linear transformations defined from V into W with the addition and scalar multiplication
defined above is a vector space.
22

Proof
Let x, y ∈ V ; a, b ∈ R be arbitrary. Then,
(T1 + T2)(ax + by) = T1(ax + by) + T2(ax + by)
= aT1x + bT1y + aT2x + bT2y
= a(T1 + T2)x + b(T1 + T2)y
Furthermore,
(cT)(ax + by) = cT(ax) + cT(by) = a(cT)(x) + b(cT)(y)
Let L(V, W) denote the set of all linear transformations from V into W. we prove that
L(V, W) is a vector space. Observe that closure under addition and scalar multiplication
has been shown above. The additive identity is the zero transformation which transforms
every vector in V into the zero-vector in W. the rest of the properties corresponding prop-
erties of the operations in the space W.
Theorem
Let U, V and W be real vector spaces, and let T1 : U → V ; T2 : V → W be linear
transformations. Then, the function T2oT1 = T2T1 : U → W and
T1oT2 = T1T2 : V → V are linear transformations(if U ∩ V = ∅).
Proof: Let x, y ∈ V and a, b ∈ R be arbitrary. Then,
T2oT1(ax + by) = T2[T1(ax + by)]
= T2[aT1x + bT1y]
= aT2(T1x) + bT2(T1y)
= a(T2T1)(x) + b(T2T1)(y)
The proof that T1oT2 is a linear transformation follows similarly.
Invertible Linear Transformation
Theorem
Let V, W be real vector spaces, and T : V → W a linear transformation. T is said
to be invertible if there exist a function denoted by T−1
deﬁned from W into V such
that ToT−1
is the identity transformation on V which sent every element of V to itself
and ToT−1
is the identity element of W which sends every element of W to itself, T−1
is
called the inverse of T.
23

Theorem:
If T : V → W is invertible, then T−1
: W → V is a linear transformation.
Proof Let y1, y2 ∈ W and a, b ∈ R be, arbitrary. Then we show thatT−1
(ay1 + by2) =
aT−1
y1 + bT−1
y2. Since,y1, y2 ∈ W, and T is invertible, then there exists x1, x2 ∈ V
such that Tx1 = y1, and Tx2 = y2. Hence, x1 = T−1
y1 and x2 = T−1
y2. Then,
T−1
(ay1 + by2) = T−1
(aTx1 + bTx2)
= T−1
[T(ax1) + T(bx2)]
= T−1
[T(ax1 + bx2)]
= ax1 + bx2
= aT−1
y1 + bT−1
y2
Let T : V → W be a linear transformation. Then T is invertible (one-to-one) if and
only if N(T) = {0}
Proof: Suppose T is 1-1, we prove that N(T) = {0}. Since T is 1-1, Tx = Ty implies
thatx = y. Let v ∈ N(T) be arbitrary, then, Tv = 0 = T0, so that Tv = T0 = 0. Hence,
v = 0, and consequently N(T) = {0}.
Conversely, suppose that N(T) = {0}, we prove T is 1-1.Let x, y ∈ V be arbitrary and
suppose Tx = Ty, then Tx − Ty = 0. Hence, T(x − y) = 0, so that (x − y) ∈ N(T) = {0}
from which it follows that x = y, completing the proof.
3.3 Range Space and Null Space of a Linear Trans-
formation
Deﬁnition
Let V and W be real vector spaces, and let T : V → W be a linear transformation.
The range of T which we shall denote by R(T) is the set:
R(T) = {y ∈ W : Tx = y, for − some, x ∈ V } = {Tx : x ∈ V }
The Null Space (or the Kernel) of T which we shall denote by N(T)
[or, Ker(T)] is the set
N(T) = Ker(T) = {v ∈ V : Tv = 0}
3.3.1 Example
Let V = R3
and deﬁne T : V → V by
24

T(x, y, z) = (x, y, 0); ∀(x, y, z) ∈ V . Then T is a linear transformation and
R(T) = T(x, y, z) : (x, y, z) ∈ R3
R(T) = {T(x, y, z) : (x, y, z) ∈ R3
}
= {(x, y, 0) : x, y ∈ R}
N(T) = {(x, y, z) ∈ V : T(x, y, z) = (x, y, 0) = (0, 0, 0)}
= {(0, 0, z) : z ∈ R} = z − axis
Example
let V = R2
and let A =
1 2
−1 1
Deﬁne T : V → V by Tx = Ax; ∀x =
x
y
∈ V .
Then T is a linear transformation (from above example)
N(T) = x ∈ V : Ax =
0
0
.
Observe that Ax = 0 implies that
1 2
−1 1
x
y
=
0
0
Solving this system of equations now yields:
x
y
=
0
0
(multiply LHS and
apply equality of matrices).
Hence,N(T) = {0}, (the set containing only the zero vector in v). In this case,
R(T) = V.
Example
Let V = R2
and let A =
1 2
1 2
, Deﬁne T : V → V by Tx = Ax; ∀x =
x
y
∈ V .
Then T is a linear transformation
N(T) = Span[
−2
1
]
25

Theorem
Let V and W be real vector spaces and T : V → W a linear transformation. Then,
1. R(T) is a subspace of W.
2. N(T) is a subspace of V.
Proof
Let y1,y2 ∈ R(T); a, b ∈ R be arbitrary. Then, there exist x1,x2 ∈ V such that Tx1 =
y1; Tx2 = y2. Since, ax1 + bx2 ∈ V (since V is a vector space) and
T(ax1 + bx2) = aTx1 + bTx2 = ay1 + by2. Therefore, ay1 + by2 ∈ R(T).
To prove that N(T) is a subspace of V, let v1,v2 ∈ N(T) be arbitrary. Then, Tv1 = Tv2 =
0. For any a, b ∈ R, we have T(av1 + bv2) = aTv1 + Tv2 = 0. So that
(av1 + bv2) ∈ N(T), completing the proof.
3.4 REPRESENTATION OF LINEAR TRANSFOR-
MATION BY MATRICES
Let V be an n-dimensional vector space over a field F and let W be an m-dimensional vector
space over F. Tentatively, it can be shown that every linear transformation T : V −→ W
could be represented by an m×n matrix, A. Therefore a detailed study of how such matrix
representation of T could be obtained
3.4.1 Theorem
Let V be a finite-dimensional vector space over the field F and let {vj}n
j=1 be an ordered
basis for V . Let W be a vector space over the same field and let {vj}n
j=1 be any given
vectors in W . Then there exists a unique linear transformation T from V into W such
that Tvj = wj, j = 1, 2, 3, ..., n
3.5 Linear Transformation Given By Matrices
Suppose A is a matrix of size m x n. Given a vector
26

V =





v1
v2
...
vn





∈ Rn
define T(v) = Av = A =





v1
v2
...
vn





Then T is a linear transformation from Rn
to Rm
.
Proof. From properties of matrix multiplication, for u, v ∈ Rn
and scalar c we have :
T(u + v) = A(u + v) = A(u) + A(v) = T(u) + T(v)
and
T(cu) = A(cu) = cAu = cT(u).
Completing the proof.
27

Chapter 4
Application And Conclusion
Cryptography, to most people, is concerned with keeping communications private.
Indeed, the protection of sensitive communications has been the emphasis of cryp-
tography throughout much of its history. Encryption is the transformation of data into
some unreadable form. Its purpose is to ensure privacy by keeping the information hidden
from anyone for whom it is not intended, even those who can see the encrypted data.
Decryption is the reverse of encryption; it is the transformation of encrypted data back
into some intelligible form.
Encryption and decryption require the use of some secret information, usually referred
to as a key. Depending on the encryption mechanism used, the same key might be used for
both encryption and decryption, while for other mechanisms, the keys used for encryption
and decryption might be diﬀerent.
Application 1
For the security, we ﬁrst code the alphabet as follows:
28

A B C D E F G H I J K L M
1 2 3 4 5 6 7 8 9 10 11 12 13
N O P Q R S T U V W X Y Z
14 15 16 17 18 19 20 21 22 23 24 25 26
1. Next, obtain a Cipher matrix -



−3 − 3 − 4
0 1 1
4 3 4



For this example we will use the following Plain-text -
PENGUINS ARE ONE TO ONE
2. Now we will replace each letter with its numerical representation, using 1-26 for
A-Z and 27 for a space between the words. Leaving us with :
16, 5, 14, 7, 21, 9, 14, 19, 27, 1, 18, 5, 27, 15, 14, 5, 27, 20, 15, 27, 15,
14, 5
3. Now separate the Plain-text into 3x1 vectors until the whole Plain-text is used.
[ 16 ] [ 7 ] [ 14 ] [ 1 ] [ 27 ] [ 5 ] [ 15 ] [ 14 ]
[ 5 ] [ 21 ] [ 19 ] [ 18 ] [ 15 ] [ 27 ] [ 27 ] [ 5 ]
[ 14 ] [ 9 ] [ 27 ] [ 5 ] [ 14 ] [ 20 ] [ 15 ] [ 27 ]
4. Augment these vectors into a plaintext matrix -



16 7 14 1 27 5 15 14
5 21 19 18 15 27 27 5
14 9 27 5 14 20 15 27



29

5. Multiply the Plain-text matrix with the Cipher matrix to form the encrypted
matrix -



−3 −3 −4
0 1 1
4 3 4


 ∗



16 7 14 1 27 5 15 14
5 21 19 18 15 27 27 5
14 9 27 5 14 20 15 27



6. The newly formed matrix contains the Cipher-text -



−119 −120 −207 −77 −182 −176 −186 −165
19 30 46 23 29 47 42 32
135 127 221 78 209 181 201 179



7. To decrypt the matrix back into Plain-text, multiply it by the inverse of the cipher -



1 0 1
4 4 3
−4 −3 −3


 ∗



−119 −120 −207 −77 −182 −176 −186 −165
19 30 46 23 29 47 42 32
135 127 221 78 209 181 201 179






16 7 14 1 27 5 15 14
5 21 19 18 15 27 27 5
14 9 27 5 14 20 15 27


 −→



P G N A E O N
E U S R O E
N I E N T O



Which contains the Plain-text -
PENGUINS ARE ONE TO ONE
Application 2
Suppose we want to send the following message to our friend,
MEET TOMORROW.
For the security, we ﬁrst code the alphabet as described in Application 1 above:
Thus, the code message is
MEET TOMORROW
M E E T T O M O R R O W
13 5 5 20 20 15 13 15 18 18 15 23
The sequence
30

13 5 5 20 20 15 13 15 18 18 15 23
is the original code message. To encrypt the original code message, we can apply a
linear transformation to original code message. Let
L : R3
−→ R3
, L(x) = Ax
where
A =



1 2 3
1 1 2
0 1 2



Then, we break the original message into 4 vectors ﬁrst,



13
5
5


 ,



20
20
15


 ,



13
15
18


 ,



18
15
23


 ,
and use the linear transformation to obtain the encrypted code message
L






13
5
5





 = A



13
5
5



=



1 2 3
1 1 2
0 1 2






13
5
5



=



38
28
15


 ,
L






20
20
15





 = A



20
20
15



31

=



1 2 3
1 1 2
0 1 2






20
20
15



=



105
70
50



L






13
15
18





 = A



13
15
18



=



1 2 3
1 1 2
0 1 2






13
15
18



=



97
64
51


 ,
and
L






18
15
23





 = A



18
15
23



32

=



1 2 3
1 1 2
0 1 2






18
15
23



=



117
79
61


 .
Then, we can send the encrypted message code
28 15 105 70 50 97 64 51 117 79 61
Suppose our friend wants to encode the encrypted message code. Our friend can ﬁnd the
inverse matrix of A ﬁrst,
A−1
=



1 2 3
1 1 2
0 1 2



−1
=



0 1 −1
2 −2 −1
−1 1 1



and then
A−1



38
28
15


 =



0 1 −1
2 −2 −1
−1 1 1






38
28
15



=



13
5
5


 ,
A−1



105
70
50


 =



0 1 −1
2 −2 −1
−1 1 1






105
70
50


 =



20
20
15


 ,
A−1



97
64
51


 =



0 1 −1
2 −2 −1
−1 1 1






97
64
51


 =



13
15
18



33

and
A−1



117
79
61


 =



0 1 −1
2 −2 −1
−1 1 1






117
79
61


 =



18
15
23



Thus, our friend can ﬁnd the original message code
13 5 5 20 20 15 13 15 18 18 15 23
via the inverse matrix of A.
Similarly, if we receive the following message code from our friend
77 54 38 71 49 29 68 51 33 76 48 40 86 53 52
and we know the message from our friend transformed by the same linear transformation
L : R3
→ R3
, L (x) = Ax =



1 2 3
1 1 2
0 1 2


 x.
Thus, we ﬁrst break the message into 5 vectors,



77
54
38


 ,



71
49
29


 ,



68
51
33


 ,



76
48
40


 ,



86
53
52


 ,
and then the original message code can be obtained by
A−1



77
54
38


 =



0 1 −1
2 −2 −1
−1 1 1






77
54
38



=



16
8
15


 ,
A−1



71
49
29


 =



0 1 −1
2 −2 −1
−1 1 1






71
49
29



=



20
15
7


 ,
34

A−1



68
51
33


 =



0 1 −1
2 −2 −1
−1 1 1






68
51
33



=



18
1
16


 ,
A−1



76
48
40


 =



0 1 −1
2 −2 −1
−1 1 1






76
48
40



=



8
16
12


 ,
and
A−1



86
53
52


 =



0 1 −1
2 −2 −1
−1 1 1






86
53
52



=



1
14
19


 .
Thus, the original message from our friend is
16 8 15 20 15 7 18 1 16 8 16 12 1 14 19
P H O T O G R A P H P L A N S
35

Conclusion
As we live in a society where automated information resources are increased and cryp-
tography will continue to increase in importance as a security mechanism. Electronic
networks for banking, shopping, inventory control, beneﬁt and service delivery, informa-
tion storage and retrieval, distributed processing, and government applications will need
improved methods for access control and data security. The information security can be
easily achieved by using Cryptography technique. DES is now considered to be insecure for
some applications like banking system. there are also some analytical results which demon-
strate theoretical weaknesses in the cipher. So it becomes very important to augment this
algorithm by adding new levels of security to make it applicable. DES Encryption with
two keys instead of one key already will increase the eﬃciency of cryptography.
36

References
DOGAN-DUNLAP, H. Linear Algebra Students Modes of Reasoning: Geometric
Representations. Linear Algebra and Its Applications (LAA),432. pp. 2141-2159,
2010.
Lester S. Hill
CRYPTOGRAPHY IN AN ALGEBRAIC ALPHABET
The American Mathematical Monthly, Vol. 36, No. 6. (Jun. - Jul., 1929), pp.
306-312.
Obaida Mohammad Awad Al-Hazaimeh
A NEW APPROACH FOR COMPLEX ENCRYPTING AND DECRYPTING
DATA, International Journal of Computer Networks and Communications Vol.5,
No.2, March 2013
Raja P. V K., Chakravarthy A. S. N., a cryptosystem based on Hilbert matrix using cipher
block chaining mode, International Journal of Mathematics Trends and Technology,
Issue 2011
Singh, Kirat Pal, and Shiwani Dod. Performance Improvement in MIPS Pipeline
Processor based on FPGA. International Journal of Engineering Technology,
Management and Applied Sciences 4.1 (2016): 57-64.
Sunitha K, Prashanth K.S.
Enhancing Privacy in Cloud Service Provider Using Cryptographic Algorithm.
IOSR Journal of Computer Engineering
(IOSR-JCE) e-ISSN: 2278-0661, p- ISSN: 2278-8727Volume 12, Issue 5
(Jul. - Aug. 2013). pp. 64.
Wu T. M., Applied Mathematics and Computation,
International Journal of Industrial and Systems Engineering
Volume 169, Issue 2, 2005
37

Application of linear transformation in computer

More Related Content

What's hot (20)

Similar to Application of linear transformation in computer (20)

Recently uploaded (20)

Application of linear transformation in computer