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What is Mathematics?
What is Algebra?
Introduction
History of Linear Algebra
Why Linear Algebra is imporatnt?
Linear System
Types of Solutions
Objective
Refrences

Lecture No. 1

By
Aniqa Naeem

Linear Algebra

February 7, 2021

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Main Books and Reference Books

Linear Algebra and its Applications by David C. Lay (3rd Ed)

Advance Engineering Mathematics by Peter,O Neil(EM)
Elementary Linear Algebra by Howard Anton (8th Edition)
Linear Algebra with Applications Third Edition by W. Keith
Nicholoson

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Course Outline

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Detailed Topics
System of Linear Equations and Matrices
Representation in matrix form
Matrices
Operations on Matrices
Echelon and Reduced Echelon Form
Inverse of a matrix(By elementary row operation)
Solution of Linear System
Gauss-Jordan Method
Gaussian Elimination Method
Determinants
Permutations of order two and three and definitions of
determinants of the same order.
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Computing of determinants
Definition of higher order determinants
Properties
Expansion of determinants
Vector Spaces
Definition and examples
Subspaces
Linear Combination and Spanning set
Linearly Independent sets
Finitely Generated Vector spaces
Bases and dimension of a Vector Space
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Operations on Subspaces, Intersections, Sums and Direct
Sums of Subspaces
Quotient Spaces
Linear Mappings
Definition and examples
Kernel and Image of a Linear Mapping
Rank and Nullity
Reflections, Projections and Homotheties
Change of Basis
Eigen-values and Eigen-vectors
Theorem of Hamilton Cayley
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Inner Product Spaces

Definition and Examples
Properties, Projections
Cauchy Inequality
Orthhogonal and Orthonormal Basis
Gram-Schmidt Proces
Diagonalization

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What is Mathematics?
Mathematics is an art of discovering the real facts about
imaginary objects.

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What is Algebra?
Algebra
Algebra (from Arabic al-jebr meaning ”reunion of broken parts”) is
the branch of mathematics concerning the study of the rules of
operations and relations, and the constructions and concepts
arising from them, including terms, polynomials, equations and
algebraic structures.

Linear Algebra
A branch of mathematics that is concerned with mathematical
structures closed under the operations of addition and scalar
multiplication. It includes the theory of systems of linear equations,
matrices, determinants, vector spaces, and linear transformations 9 / 50
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Introduction

When you take a digital photo with your phone or transform

the image in Photoshop, when you play a videogame or watch
a movie with digital effects, when you do a web search or
make a phone call, you are using technologies that build upon
linear algebra.
Linear algebra in turn is built on two basic elements, the
matrix and the vector.

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History of Linear Algebra

The study of linear algebra first emerged from the
introduction of determinants.
Determinants were considered by Leibniz in 1693, and
subsequently, in 1750, Gabriel Cramer used them for giving
solutions of linear systems, now called Cramer’s Rule.
Gauss further developed the theory of solving linear systems
by using Gaussian Elimination.
The study of linear algebra first emerged in England in the
mid-1800’s
Linear algebra first appearedin American graduate textbooks
in 1940’s and in undergraduate textbooks in 1950’s.
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Why Linear Algebra is important?

Linear Algebra is vital in multiple areas of computer science

because linear equations are easy to solve.
It converts the large number of problems to matrix and thus
we solve the matrix.

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Uses of Linear Algebra in CSE

Linear Algebra in computer science can broadly divided into two

categories:
Linear Algebra for Spatial Quantities:
Here you’re dealing with 2-, 3- or 4- dimensional vectors and you’re
concerned with rotations, projections, and other matrix operations
that have some spatial interpretation. This is the kind of linear
algebra that comes up,for example, in computer graphics and
physics simulations.

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Linear Algebra for Statistics:

Here you’re dealing with vectors in high-dimensional spaces that
have no particular spatial interpretation and you’re intrested in
matrix decompositions and so on. This domain includes signal
processing, statitical machine learning, and compression.

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Importance of Linear Algebra in various category

Linear ALgebra is very important in Computer vision for:

Camera modeling;
Calibration and Self-calibration;
Epipolar geometry;
Pose estimation;
Structure-from-motion;
And many other things.

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Importance of Linear Algebra in various category

Important in Machine Learning. For example:

Dimensionality reduction(e.g. Principal component analysis);
Clustering;
Classification;
Prediction;
Recommender systems (e.g. Collaborative filtering) etc.

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Linear Algebra is crucial to:

Linear Algebra is very important in Computer vision for:

Audio, video and image compression, including MP3, JPEG,
and MPEG video.
Modulation and coding, including convolutional codes and
Wi-fi, Gigabit Ethernet, HDTV, and the GPs.
Signal processing, including the Fast Fourier Transform and
autotune!
Statistics, and machine learning, including something as far a
field like automated trading in the financial markets.

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Linear Equation

Linear Equation
A linear equation in the variable x1 , x2 ,...,xn is an equation that
can be written in the form
a1 x1 +a2 x2 +...+an xn = b
where b and the coefficients a1 ,a2 ,...,an are real or complex
numbers.

Example
√
4x1 − 5x2 =2 and 2x1 + x2 − x3 = 2 6
are the linear equations.

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Linear System

Linear System
A system of linear equations (or a linear system) is a collection of
one or more linear equations involving the same variables-say
x1 ,x2 ,...,xn .

Example
2x1 − x2 + 1.5x3 = 8
x1 − 4x3 = −7
This is a linear system.

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We will study the following questions about linear systems:

Are there any solutions?
Does the system have no solution, a unique solution or an
infinite number of solutions?
How can we find all the solutions, if exist?
Is there some sort of structure to the solutions?

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Solution of Linear Equation

Linear Equation
A solution of the system is a list (s1 , s2 , ..., sn ) of numbers that
makes each equation a true statement with the values
(s1 , s2 , ..., sn ) are substituted for (x1 , x2 , ..., xn ), respectively.

Example
(5, 6.5, 3) is a solution of the system given in above example,
because when these values are substituted in the equation for x1 ,
x2 , x3 , respectively,

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Check the solutions by putting the values of x1 , x2 , x3 in the above

system.
2(5) − 6.5 + 1.5(3) = 8
=⇒ 10 − 6.5 + 4.5 = 8 =⇒ 10 − 2 = 8 =⇒ 8 = 8
5 − 4(3) = −7 =⇒ 5 − 12 = −7 =⇒ −7 = −7

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System of Linear Equations

Consider a system of linear equations 2x + 2y = 3 , 2x + y = 2.5

Fig. 1: Its point of intersection is (1, 0.5). It has a unique solution

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Consider a 3d system x + y + 2z = 0

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Example: Solve the linear system.
2x + 3y = 6
4x + 6y = 9
Solution:

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To escape that difficulty, we integrate (3.1.1) over the
rectangle[xl , xl+1 ] × [t m , t m+1 ] in command to frail the situation,
this contributes the formulation of particular non-homogeneous
hyperbolic- equation and is specified as
I Z t m+1 Z xl+1
bdx − f (b)dt = Q(b)dtdx, (1)
∂ω tm xl
where
Z t m+1 Z xl+1 Z t m+1 Z xl+1
Q(b)dxdt = − f (bl (t))dt + b(t m+1 , κ)dκ
tm xl tm xl
Z t m+1 Z xl+1
+ f (bl+1 (t))dt − b(t m , κ)dκ.
tm xl 27 / 50
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Dividing the above equation by 4x yields

Z t m+1 Z xl+1 Z t m+1

1 1
Q(b)dxdt = − f (bl (t))dt+
4x tm xl 4x t m
Z xl+1 Z t m+1
1 m+1 1
b(t , κ)dκ + f (bl+1 (t))dt−
4x xl 4x t m
Z xl+1
1
b(t m , κ)dκ
4x xl
(3)

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Now multiplying and dividing the above equation by 4t, we have

Z t m+1 Z t m+1
m+1 m 1 1
b̄l+ 1 = b̄l+ 1 (t ) + λ( f (bl (ς))dς − f (bl+1 (ς))dς)
2 2 4t t m 4t t m
Z t m+1 Z xl+1
1 1
+ 4t( Q(b)dxdt) (4)
4t t m 4x xl

with the help of rectangle rule we obtain

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m+1 1 m m
b̄l+ 1 = (b̄ + b̄l+1 ) + λ(f (blm − f (bl+1
m
))
2 2 l
Z t m+1 Z xl+1
1 1
+ 4t( Q(b)dxdt) (5)
4t t m 4x xl
The approximation of integral with respect to space variable by
Trapezoidal rule followed by rectangle rule for time variable gives
Z t m+1 Z xl+1 Z t m+1
1 1 1 1
4t( Q(b)dxdt) = 4t( ((Qlm )
4t t m 4x xl 4t t m 2
m 4t
+ (Ql+1 )dt) = ((Qlm ) + (Ql+1
m
)).
2
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Using previous equation in (16) , we get a first-order Lax-Friedrich

(LxF) scheme

m+1 1 m m 4t
b̄l+ 1 = (b̄ + b̄l+1 ) + λ(f (blm ) − f (bl+1
m
)) + ((Qlm ) + (Ql+1
m
)),
2 2 l 2
(7)
4t
where blm := b(t m , xl ) = b̄lm , and λ = 4x .Here b̄lm symbolizes the
cell averaged values individually. The piecewise invariable cells in
every phase are varied with respect to those in preceding stage.

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Objective

Numerical Case Studies

In this section, we will discuss the different numerical test problems

for one dimensional models, with the help of our proposed
numerical scheme that is CE/SE method, and after this we will
compare our results with central scheme:

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Test problem-01:
Let us consider one-dimensional hyperbolic system with the
following initial conditions:

2 /0.01
u(x, 0) = (0.025π)e x ,

(
1.0 if − 0.5 ≤ x ≤ 0.5,
c(x, 0) =
0.125 otherwise

v (x, 0) = 0.0

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Results:Test problem-01

Fig. 2: CE/SE and Central schemes on 100 grid points time=0.0001.

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Test problem-02:
Let us consider one-dimensional hyperbolic system with the
following initial conditions:

2 /0.0018 2 /0.0018
u(x, 0) = 1/0.0072π[e −(x−0.09) + e −(x+0.09) ] ,

(
1.0 if − 0.5 ≤ x ≤ 0.5,
c(x, 0) =
0.125 otherwise

v (x, 0) = 0.0

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Results:Test problem-02

Fig. 3: CE/SE and Central schemes on 100 grid points time=0.0001.

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Test Problem-03:
Let us consider a piecewise initial data for the given model


5.0 if 0 < x ≤ 0.2,

6.0 if

 0.2 < x ≤ 0.4,
u(x, 0) = 8.0 if 0.4 < x ≤ 0.7,



9.5 if 0.7 < x ≤ 0.9,

6.0 if 0.9 < x ≤ 1.0.


2.0 if
 0 < x ≤ 0.2,

3.0 if

 0.2 < x ≤ 0.4,
v (x, 0) = 2.0 if 0.4 < x ≤ 0.7,



3.0 if 0.7 < x ≤ 0.9, 37 / 50

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

−1.0 if 0 < x ≤ 0.2,

2.0 if 0.2 < x ≤ 0.4,



c(x, 0) = 0.0 if 0.4 < x ≤ 0.7,

−1.0

 if 0.7 < x ≤ 0.9,


2.0 if 0.9 < x ≤ 1.0.


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Results: Test Problem-03

Fig. 4: CE/SE and NT central schemes on 100 grid points for

time=0.001.

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Table 1: Comparison of errors at different grids

Methods grid Density(u) Velocity(v) cpu-
points time(sec)
CE/SE 50 0.2071 0.1616 0.1809
Central 50 0.2482 0.2178 0.1892
CE/SE 100 0.1690 0.1142 0.2465
Central 100 0.2108 0.1407 0.3132
CE/SE 200 0.1114 0.0649 0.344
Central 200 0.1388 0.0792 0.4431
CE/SE 400 0.0411 0.0242 0.4285
Central 400 0.0484 0.0296 0.6439
CE/SE 800 0.0165 0.0107 0.6698
Central 800 0.0169 0.0110 1.219 40 / 50
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Results: Test Problem-03

Fig. 5: Error graph in density and velocity at time=0.001.

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Test problem-04:
This test problem consists of piecewise initial data ,which is
defined as follows:

1.0
 if 0 ≤ x ≤ 0.4,
u(x, 0) = 1.3765 if 0.4 ≤ x ≤ 0.7,

0.138 if 0.7 ≤ x ≤ 1.0.



0.0
 if 0 ≤ x ≤ 0.4,
v (x, 0) = −0.3948 if 0.4 ≤ x ≤ 0.7,

0.0 if 0.7 ≤ x ≤ 1.0.


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
1.0 if
 0 ≤ x ≤ 0.4,
c(x, 0) = 1.57 if 0.4 ≤ x ≤ 0.7,

1.0 if 0.7 ≤ x ≤ 1.0.


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Results:Test problem-04

Fig. 6: CE/SE and Central schemes on 100 grid points time=0.0001.

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Test problem-05:
This test problem consists of piecewise initial data ,which is
defined as follows:

2.667
 if 0 ≤ x ≤ 0.3,
u(x, 0) = 1.0 if 0.3 ≤ x ≤ 0.6,

0.287 if 0.6 ≤ x ≤ 1.0.



1.479
 if 0 ≤ x ≤ 0.3,
v (x, 0) = 0.0 if 0.3 ≤ x ≤ 0.6,

0.0 if 0.6 ≤ x ≤ 1.0.


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
4.500 if
 0 ≤ x ≤ 0.3,
c(x, 0) = 1.0 if 0.3 ≤ x ≤ 0.6,

1.0 if 0.6 ≤ x ≤ 1.0.


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Results:Test problem-05

Fig. 7: CE/SE and Central schemes on 100 grid points time=0.0001.

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Test problem-06:
This test problem consists of piecewise initial data ,which is
defined as follows:
 √
 5εbK cos(√ τ x) − aK if 0 < x ≤ 0.2,
τ χD cos( τ√x̄) τD
u(x, 0) = −5εbK cos( τ x)
 √ − aK
τ χD cos( τ x̄) τD if 0.2 < x ≤ 1.0.
 √
 2εbK cos(√ τ x) − aK if 0 < x ≤ 0.5
τ χD cos( τ√x̄) τD
v (x, 0) = −2εbK cos( τ x)
 √ − aK
τ χD cos( τ x̄) τD if 0.5 < x ≤ 1.0.

c(x, 0) = 0.0
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Results:Test problem-06

Fig. 8: CE/SE and Central schemes on 100 grid points time=0.0001.

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Refrences
The numerical approximation of a one-dimensional chemotaxis
models were performed.

Second order accurate splitting based numerical schemes were

applied to solve the model equations, i.e.CE/SE and NT
central.

Several case studies were carried out. The numerical results

obtained with different values of density and concentration .

It was found that the CE/SE method is capable to capture

and resolve all discontinuities more accurately. This method
gives sharp peaks as compared to the NT central scheme. 50 / 50