The Newton-Raphson Method
1
Introduction
The Newton-Raphson method, or Newton Method, is a powerful technique
for solving equations numerically. Like so much of the dierential calculus,
it is based on the simple idea of linear approximation. The Newton Me

Exercise
1. Determine the inverse of each of the following 3 X 3
matrices, if it exists, using the method of Gauss Jordan
elimination.
1 2 3
a. 0 1 2
4 5 3
1 2 3
b. 1 2 1
5 2 3
2. Determine the inverse of each of the following 4 X 4
matrices, if it exists

1.7
Matrix Representations of Linear Transformation
In this module we introduce a way of representing a
linear transformation between general vector spaces by a
matrix. We lead up to this discussion by looking at the
information below that is necessary to

Problem 4: Consider the linear operator , = 2, +
on 2 . Find the matrix of with respect to the standard
basis = 1,0 , 0,1
of 2 . Use the transformation
= 1 to determine the matrix with respect to the
basis = 2,3 , 1, 1 .
Solution: The effect of on the v

Problem 3: Consider the linear transformation : 3 2 ,
defined by , , = ( + , 2). Find the matrix of with
respect to the bases cfw_1 , 2 , 3 and cfw_1 , 2 of 3 and 2 ,
where
1 = 1,1,0 , 2 = 0,1,4 , 3 = 1,2,3 1 = 1,0 , 2 = (0,2).
Use this matrix to find t

Problem 4: Consider the linear operator , = 2, +
on 2 . Find the matrix of with respect to the standard
basis = 1,0 , 0,1
of 2 . Use the transformation
= 1 to determine the matrix with respect to the
basis = 2,3 , 1, 1 .

Exercise
1) Let : be a linear transformation. Let be
defined relative to bases cfw_1 , 2 , 3 and cfw_1 , 2 , 3 of
and as follows:
1 = 1 + 2 + 3
2 = 31 22
3 = 1 + 22 3 .
Find the matrix of with respect to these bases. Use
this matrix to find the imag

Problem 1: Consider the linear transformation T : R3 R2
defined as follows on basis vectors of R 3 . Find T 1, 2,3.
T 1,0,0 3, 1 , T 0,1,0 2,1 , T 0,0,1 3,0
Solution: Since T is defined on basis vectors of R 3 , it is
defined on the whole space. To find,

Problem 2: Let : be a linear transformation. T is
defined relative to bases = cfw_1 , 2 , 3 and = 1 , 2 of
and as follows
1 = 21 2
2 = 31 + 22
3 = 1 42
Find the matrix representation of with respect to these
bases and use this matrix to determine the

Problem 3: Consider the linear transformation : 3 2 ,
defined by , , = ( + , 2). Find the matrix of with
respect to the bases cfw_1 , 2 , 3 and cfw_1 , 2 of 3 and 2 ,
where
1 = 1,1,0 , 2 = 0,1,4 , 3 = 1,2,3 1 = 1,0 , 2 = (0,2).
Use this matrix to find t

Problem 2: Let : be a linear transformation. T is
defined relative to bases = cfw_1 , 2 , 3 and = 1 , 2 of
and as follows
1 = 21 2
2 = 31 + 22
3 = 1 42
Find the matrix representation of with respect to these
bases and use this matrix to determine the

1.6
LINEAR TRANSFORMATIONS
A vector space has two operations defined on it, namely,
addition and scalar multiplication. Linear transformations
between vectors spaces are those functions that preserve
these linear structures in the following sense.
DEFINIT

Problem 2: Let be the vector space of real polynomial
functions of degree n. Show that the following
transformation : 2 1 is linear.
T (ax2+bx+c)= (a+b)x+c
Solution:
Let ax2+bx+c and px2+qx+r be arbitrary
elements of P2. Then
T (ax2+bx+c)+(px2+qx+r) = T (

Problem 1: Prove that the following
: 2 2 is linear. T(x,y) = (2x, x+y)
transformation
Solution: We first show that T preserves addition. Let
(x1,y1) and (x2,y2) be elements of 2 . Then
T(x1,y1)+ (x2,y2) = T (x1+x2,y1+y2) by vector addition
= (2x1+2x2, x1

Problem 3: Solve the system of equations
1 2 23 = 1
21 32 53 = 3
1 + 32 + 53 = 2
Solution: This system can be written in the following
matrix form:
1
2
1
1
3
3
1
1
2 = 3
3
2
2
5
5
If the matrix of coefficients is invertible, the unique
solution is
1
1
2 =

1.8
The Inverse of a Matrix
In this module we introduce the concept of the matrix
inverse. We will see how an inverse can be used to solve
certain systems of linear equations, and we will see an
application of matrix inverse in cryptography, the study of

Lecture 2
The rank of a matrix
Eivind Eriksen
BI Norwegian School of Management
Department of Economics
September 3, 2010
Eivind Eriksen (BI Dept of Economics)
Lecture 2 The rank of a matrix
September 3, 2010
1 / 24
Linear dependence
Linear dependence
To

1.10
Eigenvalues and Eigenvectors
Definition: Let A be an matrix. A scalar is called an
eigenavlue of A if there exists a nonzero vector in such
that = . The vector x is called an eigenvector
corresponding to .
Let us look at the geometrical significance

Problem 6: Find
3
of the matrix 1
1
values.
the
4
2
1
3
Solution: Let = 1
1
sum and product of the eigen values
4
4 without actually finding the eigen
3
4
2
1
4
4
3
Sum of the eigen values = trace of = 3 + 2 + 3 = 4.
Product of the eigen values = .
3
Now,

3
Problem 5: Find the eigen values of when = 5
3
5
Solution: The characteristic equation of
3 94 ) 1 = 0.
Hence the eigen values of are 3, 4, 1.
The eigen values of 5 are 35 , 45 , 15 .
0
4
6
0
0 .
1
is obviously

3
Problem 4: If the given values of = 2
3
1
2
2, 2, 3 find the given values of and .
10
3
5
5
4
7
are
Solution: Since 0 is not an eigen value of , is a non
singular matrix and hence 1 exists.
Eigen values of 1 are
22 , 22 32 .
1 1 1
, ,
2 2 3
and eigen va

Problem 3: Let A be an matrix A with eigenvalues
1 , . . and corresponding eigenvectors 1 , . . . Prove
that if 0, then the eigenvalues of cA are 1 , . . with
corresponding eigenvectors 1 , . . .
Solution: Let be one of the eigenvalues of A with
correspon

Problem 1: Find the eigenvalue and eigenvectors of the
matrix
4
3
=
6
5
Solution: Let us first derive the characteristic polynomial
of A. We get
2 =
4
3
6
1
5
0
0
4
=
1
3
6
5
Note that the matrix 2 is obtained by subtracting
from the diagonal elemen

Problem 2: Find the eigenvalues and eigenvectors of the
5 4 2
matrix 4 5 2
2 2 2
Solution: The matrix 3 is obtained by subtracting
from the diagonal elements of A. Thus
5
3 =
4
2
4
5
2
2
2
2
The characteristic polynomial of A is 3 . Using row
and col

Problem 3: Let A be an matrix A with eigenvalues
1 , . . and corresponding eigenvectors 1 , . . . Prove
that if 0, then the eigenvalues of cA are 1 , . . with
corresponding eigenvectors 1 , . . .

1.9
The Rank of a Matrix
In this module the student is introduced to the concept of
the rank of a matrix. Rank enables one to relate matrices to
vectors, and vice versa. Rank is a unifying tool that enables
us to bring together many of the concepts discus

Problem 3: Find a basis for the subspace of 4 spanned
by the vectors
1,2,3,4 , 1, 1, 4, 2 , 3,4,11,8
Solution: We construct a matrix having these vectors as
row vectors.
1
= 1
3
2
1
4
3
4
4 2
11 8
Determine a reduced echelon form of . We get
1
1
3
2
1
4