Transcript MAT 322 LINEAR ALGEBRA (revision class)

MAT 322: LINEAR ALGEBRA
REVISION CLASS
AND
LAST CONTINUOUS
ASSESSMENT
TEST
1
MAT 322: LINEAR ALGEBRA
OBJECTIVES
The aim of this course is that students should be able to solve problems on
algebra of matrices as well as calculate determinants of matrices of sizes up to
4 using various methods and apply them in solving various mathematical
problems.
The objectives include fundamental operations of:
 Determinants
 System of linear equations, change of basis, equivalence and similarities
 Eigen values (latent roots) and given vectors (latent vectors)
 Minimum and characteristic polynomials of a linear transformation (matrix)
 Vector space over the real field, sub-space, linear independence, basis and
dimension
 Linear transformations and their representational matrices; range, null
space, rank, singular and non-singular transformation and matrices.
 Cayley – Hamilton theorem
2
Introduction
The study of Linear Algebra essentially
begins with the Mathematical array of
numbers called MATRIX.
The design of this scheme assumes that
students have good knowledge of matrices
and matrix algebra. It is therefore necessary
to assist the students by first giving a quick
revision of matrix and matrix algebra.
Linear algebra is the study of vectors and linear functions.
3
MATRIX
A matrix is an array of numbers
 a11 a12 a13
A

a
21
a
22
a
23


Denoted with a Capital letter
Every matrix has an order (or dimension):
that is, the number of rows  the number
of columns. So, A is 2 by 3 or (2  3).
4
MATRIX EQUALITY
Two matrices are equal if and only if;
a. they both have the same number of rows and
the same number of columns
b. their corresponding elements are equal
SQUARE MATRIX
A square matrix is a matrix that has the same
number of rows and columns (n  n)
5
DIAGONAL MATRIX
A diagonal matrix is a square matrix that has values on the
diagonal with all off-diagonal entities being zero.
0
0
a11 0
0 a

0
0
22


0
0 a33 0 


0
0
0
a
44 

IDENTITY MATRIX
An identity matrix is a diagonal matrix where the diagonal
elements all equal one.
1
0
I 
0

0
0
1
0
0
0
0
1
0
0
0
0

1
IA A
6
Null Matrix
A square matrix where all elements equal zero.
0
0

0

0
0
0
0
0
0
0
0
0
0
0
0

0
VECTOR
A single row or column of numbers denoted with bold small letters
row vector
column vector
a  1 2 3 4 5
1 
2
 
b  3
 
4
5 
7
MAIN (PRINCIPAL) DIAGONAL
The elements a11, a22, a33, ... constitute the main or principal diagonal of the
matrix A = [aij], if it is square. Eg.
TRIANGULAR MATRIX
A matrix in which all the entries below or above the main diagonal are zeros
are called upper and lower triangular matrices respectively. If only the
main diagonal is non-zero, then it is simply triangular matrix
a11
 0
Triangular  
 0

 0
0
a22
0
0
0
0
a33
0
 a11
a
Lower triangular   21
 a31

a41
0 
a11
 0
0 
, Upper Triangular  
 0
0 


a44 
 0
0
a22
0
0
a32
a42
a33
a43
0 
0 
0 

a44 
a12
a22
a13
a23
0
0
a33
0
a14 
a24 
,
a34 

a44 
8
Matrix Operations
Transposition
Addition and Subtraction
Multiplication
Inversion
9
TRANSPOSE OF A MATRIX
The transpose of a matrix is a new matrix that is
formed by interchanging the rows as columns.
The transpose of A is denoted by A' or (AT). If
 a11 a12 
A  a21 a22 
a a 
 31 32 
 a11 a21 a31 
A'  
a12 a22 a32 
Notice that the first and the last elements are always the same.
EXAMPLE
Given that A =
then A‘ =
10
A Symmetric matrix is when all the entries of MT is the
same as all the entries of M. In other words, every aij of
M is equal to every entry aij of MT
A SKEW-Symmetric matrix is when all the entries of
MT is the same as the negative of all the entries of M. In
other words, every aij of M is equal to every entry -aij of
MT
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Addition and Subtraction
Two matrices may be added (or subtracted) iff
they are the same order.
Simply add (or subtract) the corresponding
elements. So, A + B = C yields. Eg.
a11 a12  b11 b12  c11 c12 
a
  b
  c

a
b
c
 21 22   21 22   21 22 
a31 a32  b31 b32  c31 c32 
a11  b11  c11
where
a12  b12  c12
a21  b21  c 21
a22  b22  c 22
a31  b31  c31
a32  b32  c32
12
MATRIX SCALAR MULTIPLICATION
To multiply a scalar and a matrix, simply multiply
each element of the matrix by the scalar
quantity e.g. If k is a scalar and
 a11
A
a21
kA 
a12 
a22 
 a11 a12   ka11 ka12 
k


a21 a22  ka21 ka22 
13
MATRIX MULTIPLICATION
In order to multiply two matrices say AB, they
must be CONFORMABLE that is, the number
of columns in A must equal the number of
rows in B. So,
A  B = C
(m  n)  (n  p) = (m  p)
(m  n)  (p  n) = cannot be done
(1  n)  (n  1) = a scalar (1x1)
14
Matrix Multiplication (cont.)
Thus
where
a11 a12 a13  b11 b12  c11 c12 
a a
 x b b   c

a
c
 21 22 23   21 22   21 22 
a31 a32 a33  b31 b32  c31 c32 
c11  a11b11  a12b21  a13b31
c12  a11b12  a12b22  a13b32
c 21  a21b11  a22b21  a23b31
c 22  a21b12  a22b22  a23b32
c31  a31b11  a32b21  a33b31
c32  a31b12  a32b22  a33b32
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Matrix Multiplication- an example
Thus
where,
1 4 7 1 4 c11 c12  30 66
2 5 8 x 2 5  c
  36 81
c

 
  21 22  

3 6 9 3 6 c31 c32  42 96
c11  1 * 1  4 * 2  7 * 3  30
c12  1 * 4  4 * 5  7 * 6  66
c 21  2 * 1  5 * 2  8 * 3  36
c 22  2 * 4  5 * 5  8 * 6  81
c31  3 * 1  6 * 2  9 * 3  42
c32  3 * 4  6 * 5  9 * 6  96
16
AB does not necessarily equal BA
(BA may even be an impossible operation)
Eg.,
A 
B = C
(2  3)  (3  2) = (2  2)
B 
A = D
(3  2)  (2  3) = (3  3)
Matrix multiplication is Associative
A(BC) = (AB)C
Multiplication and transposition
(AB)' = B'A'
17
Note that a row matrix multiplied by a column matrix when
conformable gives a scalar
e

e' e
 e1 
e 
 2

 
en 

e1
e2
 en 
 e1 
e 
  2

 
en 

n
2
e
i
i 1
18
RANK
The rank of a matrix is defined as
rank(A) = number of linearly independent rows
= the number of linearly independent columns.
A set of vectors is said to be linearly dependent if scalars c1,
c2, …, cn (not all zero) can be found such that: c1a1 + c2a2
+ … + c nan = 0
Example:
a = [1 21 12] and b = [1/3 7 4] are linearly dependent
A matrix A of dimension n  p (p < n) is of rank p. Then A has
maximum possible rank and is said to be of full rank.
In general, the maximum possible rank of an n  p matrix A is
min(n,p).
TRACE
The trace of a matrix is the sum of the main diagonal
elements of the a square matrix
19
The Inverse of a Matrix (A-1)
a) For an n  n matrix A, there may be a B such that
AB = I = BA.
b) The inverse is comparable to a reciprocal
c) A matrix which has an inverse is nonsingular.
d) A matrix which does not have an inverse is singular.
e) An inverse exists only if A  0
PROPERTIES OF INVERSE
 AB 1
 A' 
1
A 
-1 1
 B -1 A-1


A 
-1
'
A
20
CALCULATION OF INVERSE OF A MATRIX
If
a b 
A  

c
d


A
-1

and |A|  0
1
det( A)
 d  b
 c a 


ADJOINT
To get the adjoint of a matrix;
i. Substitute each element by its cofactor taking note of
the signs.
ii. Transpose the resulting matrix in (i) above.
21
MINORS OF A DETERMINANT
The Minor of a determinant is formed by deleting the row and column of a
particular element of a determinant of order n which is a determinant of order n-1.
𝑎11 𝑎12 𝑎13
e.g. for the determinant 𝑎21 𝑎22 𝑎23 , we have the minors of the elements as
𝑎31 𝑎32 𝑎33
follows:
𝑎11 𝑎12
Elements
a23
𝑎31
𝑎32
a11
Minors
𝑎22 𝑎23
𝑎32 𝑎33
𝑎23
𝑎33
a31
𝑎13
𝑎23
a12
𝑎21
𝑎32
𝑎12
𝑎22
𝑎22
𝑎32
a32
𝑎13
𝑎23
a13
𝑎21
𝑎31
𝑎11
𝑎21
𝑎13
𝑎33
𝑎11
𝑎21
𝑎12
𝑎22
a21
𝑎12
𝑎32
a33
a22
𝑎12
𝑎32
𝑎13
𝑎33
22
Cofactors
The minor of each element is called a
cofactor if a +ve or –ve sign is assigned
to it. The sign is determined by the
position of the element. If the sum i+j of
the element is even, the sign is +ve. It is
–ve if i+j of the element is odd.
23
Find the adjoint of A given that
EXAMPLE
3 −1 4
7
3
5
(i) A= 1 −2 1 (ii) A = 2 1 5
1 2 3
−2 4 −3
SOLUTION
𝑐11 𝑐12 𝑐13
I) let B = 𝑐21 𝑐22 𝑐23 be the matrix of cofactors.
𝑐31 𝑐32 𝑐33
Then;
𝑐11 = 2, 𝑐12 = 1, 𝑐13 = 0, 𝑐21 = 29, 𝑐22 = -11, 𝑐23 = 34, 𝑐31 = 13, 𝑐32 = -2, 𝑐33 = -17.
2
1
0
2 29
13
T
Therefore B = 29 −11 34 so that the adj A = B = 1 −11 −2 .
13 −2 −17
0 34 −17
24
• Ex 2:
2
 1 3
A   0  2 1 
 1
0  2
(a) Find the adjoint matrix of A
(b) Use the adjoint matrix of A to find A–1
Sol:
 Cij  (1)i  j M ij
2 1
 C11  
 4,
0 2
0 1
C12  
 1,
1 2
C13  
0 2
 2,
1 0
3 2
C21  
 6,
0 2
1 3
1 2
 3,
C22  
 0, C23  
1 0
1 2
3 2
C31  
 7,
2 1
1 2
C 32  
 1,
0 1
1 3
C33  
 2.
0 2
3.25

cofactor matrix of A

 4 1 2
Cij  6 0 3
7 1 2
 

adjoint matrix of A
4 6 7
T
adj( A)  Cij   1 0 1 
 2 3 2
inverse matrix of A
1
A 
adj( A) (det  A
det  A
1
4 6 7  43 2
 13 1 0 1   13 0

 
2 3 2  23 1

Check:
AA1  I

1
3

2

3
7
3
※ The computational effort of this
method to derive the inverse of a
matrix is high (especially to
 3) compute the cofactor matrix for a
higher-order square matrix)
※ However, for computers, it is easy.
since it is not necessary to judge
which row operation should be
used and the only thing needed to
do is to calculate determinants of
matrices
3.26
Echelon Matrix
When the number of zeros preceding the first
non-zero entry of a row increases row by row
until only zero rows remain; that is if there
exists non-zero entries, such a matrix is called
echelon matrix or is in echelon form
27
ECHELON MATRICES
Examples;
0
1 2 3
0
0 0 4
,
0
0 0 0
0
0 0 0
1
0
0
0
3
0
0
0
0
1
0
0
0
0
1
0
4
−3
2
0
0
2
0
0
,
0
0
0
0
2
0
0
0
2
7
0
0
0
1
0
0
4
−3
0
0
5 −6
2 0
.
6 2
0 0
DISTINGUISHED ELEMENTS
The first non-zero in each row. Thus in eg.1 above 1 and 4 are the
distinguished elements and in eg.2 the DE are 1,1,1 in eg.3 the
DE are 2,7,6).
The 3rd example above is an example of a row reduced echelon
matrix. The zero matrix irrespective of the number of rows or
columns is a row echelon matrix.
28
ELEMENTARY ROW OPERATIONS
Two matrices A and B are said to be row
equivalence if B can be obtained from A by a
finite sequence of any or all the following
operations called
Elementary Row Operation;
• Interchange the i-th row and the j-th row
• Multiply the i-th row by a non zero scalar say k
• Replace the i-th row by k times the j-th row plus the i-th
row
• Replace the i-th row by k’ times the j-th row plus k
(non-zero) times the i-th row.
29
EXAMPLES
1 2 −3 0
1) Transform 2 4 −2 2 the matri to echelon form by elementary row
3 6 −4 3
operation
Solution 1
Take the following steps:
Transform R2 by -2R1+R2 to give 0 0 4 2
Transform R3 by -3R1+R3 to give 0 0 5 3 then the new R3 by -5R2+4R3 to have
1 2 −3 0
0 0 0 2 the transformed matrix becomes 0 0 4 2 .
0 0 0 2
2 3 4 5 6
2) Transform 0 0 3 2 5 echelon matrices to row reduced matrices.
0 0 0 0 2
Solution
Transform R1 by -4R2+3R1 to get 6 9 0 7 -2 then new R1 by R3+R1 to get 69070
and R2 by -5R3+2R2 to get 0 0 6 4 0. Note that the 3rd row is echelon compliant.
6 9 0 7 0
The echelon matrix becomes; 0 0 6 4 0 .
0 0 0 0 2
Next R1 by 1/6, R2 by 1/6 and R3 by ½ to get The final canonical form as
1 3/2 0 7/6 0
0
0
1 2/3 0 .
0
0
0
0
1
30
EXERCISES
Transform to Echelon form
1
1. 2
3
−2
−1
1
3
2
2
−1
1
2 2. 2
3
3
2
4
6
−1
1
2
2
−2
−6
1
3
5
Transform to row reduced form
1
3. 0
0
2
9
0
21
−5
−26 4. 0
0
0
0
21
0
0
0
7
41
−12
−10
Reduce the following first to their echelon form then to row canonical form (i.e. row reduced
echelon form)
1 3 −1 2
0 1 3 −2
1 2 1
6 3 −4
0 11 −5 3
0 4 −1 3
5. 2 4 3 6. −4 1 −6 7.
8.
2 −5 3 1
0 0 2
1
3 6 5
1 2 −5
4 1
1 5
0 5 −3 4
31
DETERMINANT OF A MATRIX
The determinant of a matrix A is denoted by |A| (or det(A)).
Determinants exist only for square matrices.
DETERMINANT OF 2X2 MATRIX
 a11 a12 
If A  
 then
a
a
22 
 21
A  a11a22  a12 a21
Examples
Evaluate the following determinants
7 3
1.
= 7x2 – 4x3 = 14 – 12 = 2
4 2
2.
−3
2
3.
5
3
1
= -3x4 -2x1 = -12 -2 = -14
4
−1
= 5x2 – 3x (-1) = 10 + 3 =13
2
32
𝑎11 𝑎12 𝑎13
Determinant of a 3x3 matrix A of the form 𝑎21 𝑎22 𝑎23 the
𝑎31 𝑎32 𝑎33
number;
𝑎11 𝑎12 𝑎13
𝑎21 𝑎22
𝑎21 𝑎23
𝑎23
𝑎
𝑎21 𝑎22 𝑎23 = a11 22
𝑎32 𝑎33 - a12 𝑎31 𝑎33 + a13 𝑎31 𝑎32
𝑎31 𝑎32 𝑎33
= a11(a22.a33 – a32a23) – a12(a21a33 – a31a23) + a13(a21a32 – a31a22); is called
the determinant denoted detA.
Example
3 4 6
Evaluate 2 1 −1
−1 3 5
Solution
3(5+3)-4(10-1)+6(6+1) = 24-36+42 = 30
33
• Ex 3: The determinant of a square matrix of order 3
0 2 1 
A   3 1 2   det( A)  ?
 4 0 1 
Sol:
11
C11  (1)
1 2
 1
0 1
1 3
C13  (1)
1 2
C12  (1)
3 2
 (1)(5)  5
4 1
3 1
4
4 0
 det( A)  a11C11  a12C12  a13C13
 (0)( 1)  (2)(5)  (1)( 4)
 14
3.34
• Alternative way to calculate the determinant of a square matrix of order 3:
Subtract these three products
 a11 a12
A   a21 a22
 a31 a32
a13 
a23 
a33 
a11
a21
a31
a12
a22
a32
a13
a23
a33
a11
a21
a31
a12
a22
a32
Add these three products
 det( A) | A | a11a22 a33  a12 a23a31  a13a21a32  a31a22 a13
 a32 a 23 a11  a33a21a12
3.35
• Ex: Recalculate the determinant of the square matrix A in Ex 3
–4 0 6
0 2 1  0 2
A   3 1 2  3 1
 4 0 1  4 0
0 16 0
 det( A) | A | 0  16  0  (4)  0  6  14
※ This method is only valid for matrices of order 3
3.36
• Ex 4: The determinant of a square matrix of order 4
 1 2
 1 1
A
0
2

4
3
3 0
0 2 
 det( A)  ?

0 3

0  2
3.37
Sol:
det( A)  (3)(C13 )  (0)(C23 )  (0)(C33 )  (0)(C43 )
 3C13
1 1 2
 3(1)13 0 2 3
3 4 2
2
2

2 1 1
22  1
23  1 1 
 3(0)( 1)
 (2)( 1)
 (3)( 1)

4

2
3

2
3
4


 30  ( 2)(1)( 4)  (3)( 1)( 7)
 (3)(13)
 39
※ By comparing Ex 4 with Ex 3, it is apparent that the computational effort for
the determinant of 4×4 matrices is much higher than that of 3×3 matrices. In
3.38
the next section, we will learn a more efficient way to calculate the determinant
Determinant of a Triangular Matrix
If A is an n  n triangular matrix (upper triangular, lower triangular, or diagonal),
then its determinant is the product of the entries on the main diagonal. That is
det( A) | A | a11a22 a33 ann
※ Based on the above, it is straightforward to obtain that
det( I )  1
※ On the next slide, only take the case of upper triangular matrices for
example to prove . It is straightforward to apply the following proof
for the cases of lower triangular and diagonal matrices
3.39

Ex 6: Find the determinants of the following triangular matrices
(a)
0
2
 4 2
A 
 5 6
1
5

0
0
1
3
0
0

0
3
Sol:
3.40
(a)
|A| = (2)(–2)(1)(3) = –12
(b)
|B| = (–1)(3)(2)(4)(–2) = 48
(b)
 1
 0

B 0

 0
 0
0 0 0
3 0 0
0 2 0
0 0 4
0 0 0
0
0
0

0
 2
3.2 Evaluation of a Determinant Using Elementary Row Operations

The computational effort to calculate the determinant of a square matrix with a large
number of n is unacceptable. In this section, we show how to reduce the
computational effort by using elementary operations
Note: Elementary row operations and determinants
Let A and B be square matrices
(a) B  I i , j ( A)
 det( B)   det( A)
(b) B  M i( k ) ( A)  det( B)  k det( A)
(c) B  Ai(,kj) ( A)

 det( B)  det( A)
Notes: The above three properties remains valid if elementary column operations are
performed to derive column-equivalent matrices (This result will be used in Ex 5 on
Slide 3.25)
3.41

Ex: (check the characteristics of determinants
1 2 3 
A  0 1 4  det( A)  2
1 2 1 
 4 8 12 
A1  0 1 4   det( A1 )  8
1 2 1 
0 1 4 
A2  1 2 3  det( A2 )  2
1 2 1 
1 2 3
A3   2 3 2  det( A3 )  2
 1 2 1 
A1  M1(4) ( A)
 det( A1)  4det( A)  (4)(2)  8
A2  I 1, 2( A)
 det( A2)   det( A)  (2)  2
( 2)
A3  A1,2
( A)
 det( A3)  det( A)  2
3.42
• NOTE: Conditions that yield a zero determinant
If A is a square matrix and any of the following conditions is true, then det(A) = 0
(a) An entire row (or an entire column) consists of zeros
(Perform the cofactor expansion along the zero row or column)
(b) Two rows (or two columns) are equal
(c) One row (or column) is a multiple of another row (or column)
(For (b) and (c), based on the mathematical induction , perform the cofactor
expansion along any row or column other than these two rows or columns)
Notes: For conditions (b) or (c), you can also use elementary row or column operations
to create an entire row or column of zeros and obtain the results.

※ Thus, we can conclude that a square matrix has a determinant of zero if and only if it
is row- (or column-) equivalent to a matrix that has at least one row (or column)
consisting entirely of zeros
3.43

Ex:
1 2 3
0 0 0 0
4 5 6
1 4 2
1 5 2 0
1 6 2
1 4 0
2 5 0 0
3 6 0
1
2
3
4
5
6 0
2 4 6
1 1 1
2 2 2 0
4 5 6
1 8 4
2 10 5  0
3 12 6
3.44
PROPERTIES OF DETERMINATES
Determinants have several mathematical properties which
are useful in matrix manipulations.
a) |A|=|A'|.
b) If a row or column of A = 0, then |A|= 0.
c) If every value in a row or column is multiplied by k, then |A|
= k|A|.
d) If two rows (or columns) are interchanged the sign, but not
value, of |A| changes.
e) If two rows or columns are identical, |A| = 0.
f) If two rows or columns are linear combination of each other,
|A| = 0
g) |A| remains unchanged if each element of a row or each
element multiplied by a constant, is added to any other row.
h) 8. |AB| = |A| |B|
i) 9. Det of a diagonal matrix = product of the diagonal elements
45
Determinant of a matrix product
det(AB) = det(A) det(B)
• Notes:
(1)
det( A1 A2
(Verified by Ex 1 on the next slide)
An )  det( A1 ) det( A2 )
det( An )
det(Ei , j A)  det(Ei , j )det( A) and det(Ei , j A)   det( A)  det(Ei , j )  1

(2) det(Ei( k ) A)  det(Ei( k ) )det( A) and det(Ei( k ) A)  k det( A)  det(Ei( k ) )  k
 det(E ( k ) A)  det(E ( k ) )det( A) and det(E ( k ) A)  det( A)  det(E ( k ) )  1
i, j
i, j
i, j
i, j

(3) det( A  B)  det( A)  det( B)
(4)
a11
a12
a21  b21 a22  b22
a31
a32
a11 a12
a13
a23  b23  a21 a22
a31 a32
a33
a13 a11 a12
a23  b21 b22
a33 a31 a32
a13
b23
a33
(There is an example to verify this property on Slide 3.33) (Note that this property
3.46
is also valid for all rows or columns other than the second row)

Ex 1: The determinant of a matrix product
1  2 2
A  0 3 2
1 0 1
1
2 0
B  0  1  2
3 1  2
Find |A|, |B|, and |AB|
Sol:
1 2 2
| A | 0 3 2  7
1 0 1
3.47
2 0
1
| B | 0  1  2  11
3 1 2
1  8 4
1 
1  2 2 2 0
AB  0 3 2 0  1  2  6  1  10
1 0 1 3 1  2 5 1  1 
8 4
1
 | AB | 6 1 10  77
5 1 1

Check:
|AB| = |A| |B|
3.48

Ex:
1 2 2 1 2 2 1 2 2
?
A  0 3 2  1 1 2  1 2 0  B  C
1 0 1 1 0 1 1 0 1
Pf:
| A | 0(1)
2 1
| B | 1(1)
| C | 1(1)
3.49
2 2
0
2 1
2 1
1
 3(1)
2 2
0
1
2 2
0
1
2 2
 1(1)
 2(1)
1 2
1 1
2 2
2 2
 2(1)
1 2
1 1
1 2
1 1
23
 2(1)
 0(1)
1 2
1
23
23
0
1 2
1
0
1 2
1
0

Theorem 3.6: Determinant of a scalar multiple of a matrix
If A is an n × n matrix and c is a scalar, then
det(cA) = cn det(A)
(can be proven by repeatedly use the fact that if B  M i( k ) ( A)  B  k A )
• Ex 2:
1 2 4
 10 20 40
A   30
0 50  , if 3 0 5  5, find |A|
 20 30 10 
2 3 1
Sol:
1 2 4
 1  2 4
3
0 5  (1000)(5)  5000
A  10 3
0 5  A  10 3


2 3 1
 2  3 1
3.50
(Determinant of an invertible matrix)
( )
A square matrix A is invertible (nonsingular) if and only if det(A)  0
If A is invertible, then AA–1 = I. , we can have |A||A–1|=|I|. Since |I|=1, neither |A|
nor |A–1| is zero
Suppose |A| is nonzero. It is aimed to prove A is invertible.
( )
By the Gauss-Jordan elimination, we can always find a matrix B, in reduced
row-echelon form, that is row-equivalent to A
1. Either B has at least one row with entire zeros, then |B|=0 and thus |A|=0 since
|Ek|…|E2||E1||A|=|B|. →←
2. Or B=I, then A is row-equivalent to I, and by Theorem 2.15 (Slide 2.59), it can
be concluded that A is invertible
3.51
•
Ex 3: Classifying square matrices as singular or nonsingular
0 2  1
A  3  2 1 
3 2  1
0 2  1
B  3  2 1 
3 2
1 
Sol:
A 0

B  12  0

A has no inverse (it is singular)
B has inverse (it is invertible/nonsingular)
3.52
Determinant of an inverse matrix
1
If A is invertible, then det( A ) 
det( A)
1
1
1
(Since AA  I , then A A
 1)
Determinant of a transpose
If A is a square matrix, then det( AT )  det(A)
(Based on the mathematical induction , compare the cofactor expansion along the row of A
and the cofactor expansion along the column of AT)
• Ex 4:
Sol:
1 0 3
A  0  1 2
2 1 0
1 0 3
 | A | 0  1 2  4
2 1 0
(a)
A 1  ?
(b)
1 1

A 4
AT  A  4
AT  ?
 A1 
3.53

Equivalent conditions for a nonsingular matrix:
If A is an n × n matrix, then the following statements are equivalent
(1) A is invertible
(2) Ax = b has a unique solution for every n × 1 matrix b
(3) Ax = 0 has only the trivial solution
(4) A is row-equivalent to In
(5) A can be written as the product of elementary matrices
(6) det(A)  0
※ The statements (1)-(5) are collected in Theorem 2.15, and the
statement (6) is from Theorem 3.7
3.54
• Ex 5: Which of the following system has a unique solution?
(a)
2 x2

1

4
3x1
 2 x2

3x1
 2 x2
 x3
 4
3x1
3x1
2 x2
 2 x2
 2 x2



 1

4
 4
(b)
3.55
 x3
x3
x3
x3
x3
Sol:
(a) Ax  b (the coefficient matrix is the matrix A in Ex 3)

(b)
A  0 (from Ex 3)
This system does not have a unique solution
Bx  b (the coefficient matrix is the matrix B in Ex 3)
B  12  0 (from Ex 3)

This system has a unique solution
3.56
MATRICES AND SYSTEMS OF LINEAR EQUATIONS
(CRAMERS’ RULE)
Consider the following equations;
a11x + a12y = p1 ……………………….. (1)
a21x + a22y = p2 ………………………... (2)
By extracting the coefficients of the unknowns in form of matrix we have
𝑎11 𝑎12
𝑎11 𝑎12
𝑎21 𝑎22 whose determinant is 𝑎21 𝑎22 = a11a22 - a2la12.
We can solve the equations simultaneously by elimination to obtain;
𝑝 𝑎 −𝑝 𝑎
x = 1 22 2 12
𝑎 11 𝑎 22 −𝑎 12 𝑎 21
y =
𝑝 2 𝑎 11 −𝑝 1 𝑎 21
……………… provided that a11a22 - a2la12 ≠ 0
𝑎 11 𝑎 22 −𝑎 12 𝑎 21
We also observe that the numerators of X and Y can also be written as determinant
so that as the common denominator;
𝑎11 𝑎12
∆= 𝑎
𝑎22 = a11a22 - a2la12.
21
𝑝1 𝑎12
For numerator of x we write ∆𝟏 = 𝑝 𝑎 = p1a22 - p2a12.
2
22
𝑎11
For numerator of y we write ∆𝟐 = 𝑎
21
∆1
∆2
Thus x = , y = ∆ ≠ 0.
∆
∆
𝑝1
𝑝2 = a11p2 - a2lp1.
57
Next consider a third order determinant with the system of equations
a11x + a12y + a13z = p1 ……………………….. (1)
a21x + a22y + a23z = p2 ………………………... (2),
a31x + a32y + a33z = p3 ………………………... (3),
𝑎11 𝑎12 𝑎13
𝑝1 𝑎12 𝑎13
𝑎11 𝑝1 𝑎13
∆ = 𝑎21 𝑎22 𝑎23 , ∆1 = 𝑝2 𝑎22 𝑎23 , ∆2 = 𝑎21 𝑝2 𝑎23 , ∆3 =
𝑎31 𝑎32 𝑎33
𝑝3 𝑎32 𝑎33
𝑎31 𝑝3 𝑎33
𝑎11 𝑎12 𝑝1
𝑎21 𝑎22 𝑝2 =
𝑎31 𝑎32 𝑝3
Each of the∆ ∆1 , ∆2 and ∆3 can be evaluated as in the definition of determinants
above.
EXAMPLES
Use determinants to solve the simultaneous equations
𝑥+ 𝑦+𝑧 =6
2𝑥 + 3𝑦 = 12
(1)
(2) 𝑥 − 𝑦 + 𝑧 = 2
𝑥 − 5𝑦 = −7
2𝑥 + 3𝑦 + 𝑧11
2𝑥 − 𝑦 + 𝑧 = 0
(3) 𝑥 + 3𝑦 + 2𝑧 = 16
3𝑥 − 𝑦 + 𝑧 = 3
58
∆1
∆
, y = 2 if . ∆ ≠ 0.
∆
∆
2 3
12 3
2 12
Where ∆ =
= -13, ∆1 =
= -39, ∆2 =
= -26;
1 −5
−7 −5
1 −7
−39
−26
then x =
= 3, y =
= 2.
1. From x =
−13
2. x =
∆1
∆
,y=
1 6
∆2 = 1 2
2 11
3. x =
∆1
∆
,y=
2 0
∆2 = 1 16
3 3
x=
−15
−5
−13
1 1 1
6
1 1
, z = if. ∆ ≠ 0. Where ∆ = 1 −1 1 = 2, ∆1 = 2 −1 1 = 2,
∆
∆
2 3 1
11 3 1
1
1 1
6
2
4
6
1 = 4 and ∆3 = 1 −1 2 = 6. Then ∴ x = 2 = 1, y = 2 = 2, z = 2 = 3 .
2 3 11
1
∆2
∆3
2 −1 1
0
,z=
if . ∆ ≠ 0. Where ∆ = 1 3 2 = -5, ∆1 = 16
∆
∆
3 −1 1
3
1
2 −1 0
2 = -25 and ∆3 = 1 3 16 = 5. Then
1
3 −1 3
∆2
= 3, y =
∆3
−25
−5
= 5, z =
5
−5
−1 1
31 2 = -15,
−1 1
= −1
59
Use Cramer’s rule to solve the system of linear equation
 x  2 y  3z  1
2x

z  0
3x  4 y  4 z  2
Sol:
1 2  3
det( A)  2
0
1  10
3 4 4
1 2 3
det( A1 )  0 0
1 8
2 4 4
1 1  3
det( A2 )  2 0 1  15,
3 2 4
1 2 1
det( A3 )  2
0 0  16
3 4 2
det( A1 ) 4
x

det( A) 5
det( A2 )  3
y

det( A)
2
det( A3 )  8
z

det( A)
5
3.60
EIGEN VALUES AND EIGEN VECTORS
An important mathematical formulation is the characteristic equation of a square matrix. If C is an n
by n covariance matrix, the characteristic equation is:
|C - I| = 0
where is a scalar and I is the identity matrix. Solving this equation for reveals that the equation is
an nth degree polynomial of . That is, there are as many s as there are variables in the covariance
matrix C. The n s that are the roots of this polynomial are the eigenvalues of C. Because C is
symmetric, all the s will be real numbers, although some of the s may be equal to or less than 0. The
s can be solved for in any order, but it is customary to order them from largest to smallest.
To examine what is meant here, let C denote a two by two correlation matrix that has the form:
1 p
p 1
Then the quantity C - I (where I is the identity matrix) may be written as:
1 p
1−ʎ
p
ʎ 0
C I 



p 1
p
1−ʎ
0 ʎ
The determinant is:
C I 12 2 ……………… Characteristic Polynomial
So the equation that requires solution is:
12 2 0. ………………… Characteristic Equation
NB: This is a quadratic in . If we had three variables, it would be a cubic; and if there were
four variables, it would be a quartic, etc.
Solving for the quadratic gives:
1
The largest root depends on the sign of . For > 0, then 1 = 1 + and 2 = 1 - .
61
EIGEN VECTORS
Each eigen value has an associated eigen vector thus, for each , one can define a nonzero vector a, such that:
C Ia 0.
The 0 to the right of the equal sign denotes a vector filled with 0s. Any number in a that satisfies this equation is called
latent vector or eigenvector of matrix C. Each eigenvector is associated with its own . Note that the solution to a is no
unique because if a is multiplied by any scalar, the above equation still holds. Thus, there are an infinite set of values fo
a, although each solution will be a scalar multiple of any other solution.
𝑎1
1−ʎ
𝜌
1 − ʎ 𝑎1 + 𝜌𝑎2
0
=
=

𝑎
𝜌
1−ʎ
1 − ʎ 𝑎2 + 𝜌𝑎1
0
2
or, by carrying out the multiplication, we find
1a1 a2 0
1a2 a1 0
Now take the largest eigenvalue, l = 1 + r, and substitute. This gives
a2 a1)0
a1 a2 0
(C ʎI)a 
EXAMPLE: Given the matrix; A =
Solution
−5
If A =
2
−5
2
2
, Find the eigenvalues of A
−2
2
we obtain the Characteristic polynomial
−2
From det.(A - ʎI) where ʎ is a scalar and I is identity matrix;
1 0
−5 2
−5 − ʎ
2
i.e. det
−ʎ
=
0 1
2 −2
2
−2 − ʎ
∴ the characteristic equation is; ʎ2 + 7ʎ + 6 = 0
Solving the quadratic equation gives the eigen values:
62
LINEAR TRANSFORMATION
It is one of the interesting applications of matrices in algebra. It adopts the principle of 1-1 mapping of an of
a point (object) to another point (image) in a plane. Linear transformation can take any of the following;
- Rotation
- Shear or
- Enlargement/Reduction
- Stretch.
- Reflection
As our basis of starting, let us consider a unit square OABC on a Cartesian plane with vertices as follows O(0,0),
A(1,0), B(1,1) and C(0,1).
C(0,1)
O(0,0)
B(1,1)
A(1,0)
ROTATION
For OABC above a rotation with axis at origin for 90o in the positive direction (anti-clockwise) will give the following
figure with vertices as O(00), A/(01), B/ (-1,1) and C/(-1,0) as in the figure below.
If we denote this mapping (rotation) as T it means that T(A) → A/, etc. If we denote the coordinates of the vertices
of OABC as a column matrix each we shall have
0
1
0
1
O=
A=
B=
and C =
0
1
1
0
63
FOR ROTATION THROUGH 90O ABOUT THE ORIGIN
We also consider the unit square in the following diagram.
B’(-1,1)
C’(-1,0)
Under the transformation T, T
A(0,1)
O’(0,0)
1
−𝟏
𝟎
0
=
;T
=
.
0
𝟏
1
𝟎
0 −1
called the matrix of transformation has the effect of turning every
1 0
point through 90o about the origin.
FOR ROTATION THROUGH 180O ABOUT THE ORIGIN
We also consider the unit square in the following diagram.
Y
The matrix M=
-X
A’(-1,0)
O’(0,0)
X
C’(0,-1)
B’(-1,-1)
-Y
64
1
−𝟏
−1
0
𝟎
=
;T
=
. If we take the matrix of transformation to be M then M =
0
𝟎
1
−𝟏
0
of turning every point through 180o about the origin.
FOR ROTATION THROUGH 270O ABOUT THE ORIGIN
We also consider the unit square in the following diagram.
Under the transformation T, T
0
has the effect
−1
Y
-X
C’(1,0)
O’(0,0)
X
B’(1,-1)
A’(0,-1)
-Y
1
𝟏
0 1
𝟎
0
Under the transformation T, T
=
;T
=
. If we take the matrix of transformation to be M then M =
has the effect of
0
−𝟏
1
𝟎
−1 0
turning every point through 270o about the origin.
FOR ROTATION THROUGH 360O ABOUT THE ORIGIN
1
𝟏
0
𝟎
We also consider the unit square under the transformation T, T
=
;T
=
. If we take the matrix of transformation to be M then
0
𝟎
1
𝟏
1 0
M=
which is an identity matrix has the effect of turning every point through 360o about the origin (i.e. restoring every figure back to
0 1
its original position).
FOR ROTATION THROUGH + 𝜽O ABOUT THE ORIGIN
We also consider the unit square in the following diagram.
Y
B’
C’
A’
-X
o
O
O’
X
-Y
1
𝒄𝒐𝒔𝜽
0
−𝒔𝒊𝒏𝜽
𝑐𝑜𝑠𝜽
Under the transformation T, T
=
;T
=
. If we take the matrix of transformation to be M then M =
0
𝒔𝒊𝒏𝜽
1
𝒄𝒐𝒔𝜽
𝑠𝑖𝑛𝜽
o
the effect of turning every point through 𝜽 about the origin.
−𝑠𝑖𝑛𝜽
has
𝑐𝑜𝑠𝜽
65
EXAMPLES:
1. Find the vertices of the image of the triangle ABC which undergoes a rotation of 90 o in the anti-clockwise sense if the vertices of ABC are
A(0,0), B(1,1) and C(1,4)
Solu:
0 1 1
The vertices of ABC can be written as a single matrix P =
.
0 1 4
0 −1
The matrix of transformation of every point through 90o is M =
.
1 0
Let Q be the matrix of the vertices of the image of the triangle after transformation then;
Q
= MP
0 −1 0 1 1
0 −1 −4
=
=
1 0
0 1 4
0 1
1
∴ the vertices of the image of the triangle are A’(0,0), B/(-1,1), C/(-4,1).
2. Find the matrix P which transforms the triangle ABC with vertices A(0,0), B(3,2), C(1,5) onto a triangle with vertices A /( 0,0 B/
5
/ 1
5
3
5
3
2
1
− 3, −
2
o
3 and C − 3,
+ . Show that the transformation is equivalent to a rotation of 60 about the origin in the anti-clockwise sense.
2
2
2
2
Solution
𝑎 𝑏
Let P =
be the matrix of the transformation.
𝑐 𝑑
0 3 1
let us write the vertices of ABC as a single matrix A =
.
0 2 5
3
1
5
0
− 3
− 3
2
2
2
/ / /
Similarly the vertices of A B C can be written as a single matrix B =
.
1
5
3
5
0
− 3
+
2
2
2
2
=> P:A → B i.e. P(A) = B.
2
𝑎
=>
𝑐
𝑏
𝑑
0
0
3
2
0
1
=
5
0
3
2
1
2
− 3
−
5
2
2
3
3
3
−
2
3𝑎 + 2𝑏
3𝑐 + 2𝑑
we get the values as; a = ½ , b= ∴P=
½
3
2
−
½
3
2
− 3
5
2
+
1
3
5
2
5
− 3
𝑎 + 5𝑏
2
2
2
=
1
5
3
5
𝑐 + 5𝑑
0
− 3
+
2
2
2
2
By equating corresponding elements and solving them simultaneously:
0
=>
0
0
1
,c=
3
2
and d = ½ .
3
2
66
REFLECTION
(a) ABOUT X-AXIS
Here x-axis is taken as the line of symmetry (mirror). Consider the diagram below
Y
B(1,1)
C(0,1)
-X
A(1,0)
X
O(00)
C’(0,-1)
B’(1,-1
-Y
Let the transformation be T, then under T
1
𝟏
0
𝟎
T
=
;T
=
0
𝟎
1
−𝟏
Let M be the matrix of this transformation (reflection) then M =
𝟏
𝟎
𝟎
which has the effect of reflecting every point about x-axis.
−𝟏
(b)
ABOUT Y-AXIS
Here y-axis is taken as the line of symmetry (mirror). Consider the diagram below
Y
B’(1,1)
-X
A’(1,0)
C(0,1)
O(00)
B(1,1)
A(1,0)
X
-Y
Let the transformation be T, then under T
1
−𝟏
0
𝟎
T
=
;T
=
0
𝟎
1
𝟏
Let M be the matrix of this transformation (reflection) then M =
−𝟏
𝟎
𝟎
which has the effect of reflecting every point about y-axis.
𝟏
(c) Reflection about the line y = xtan𝜃
The matrix of transformation M is given by M =
𝑐𝑜𝑠2𝜽
𝑠𝑖𝑛2𝜽
𝑠𝑖𝑛2𝜽
−𝑐𝑜𝑠2𝜽
67
EXAMPLE
1. Triangle ABC has vertices A(1,1), B(3,2), C(2,4), find the vertices of the image of triangle
ABC if it undergoes a reflection about the line through the origin making +45o with the
positive x-axis
Solution
1 3 2
If we take P as the coordinates of the triangle ABC then P =
.
1 2 4
The matrix of transformation of reflection about the line through the origin making an angle
𝜃with the x-axis in the anti-clockwise is given by
𝑐𝑜𝑠2𝜃 𝑠𝑖𝑛2𝜃
M=
.
𝑠𝑖𝑛2𝜃 −𝑐𝑜𝑠2𝜃
𝑐𝑜𝑠90 𝑠𝑖𝑛90
0 1
If 𝜃 = 45o then M =
=> M =
.
𝑠𝑖𝑛90 −𝑐𝑜𝑠90
1 0
Let Q be the matrix of the vertices of the image of the triangle after transformation then;
0 −1 1 3 2
1 2 4
Q = MP =>
=
1 0
1 2 4
1 3 2
∴ the vertices of the image of the triangle ABC after reflection are:
A’(1,1), B/(2,3), C/(4,2)
68
ENLARGEMENT
Consider the transformation of the unit square under the enlargement of the scale factor 2 as seen
below:
Y
B’(2,2)
C’(0,2)
C(0,1)
-X
O(00)
B(1,1)
A(1,0)
A(2,0)
X
-Y
Let the transformation be T hence under T;
1
𝟐
0
𝟎
T
=
;T
=
0
𝟎
1
𝟐
Let M be the matrix of this transformation (reflection) then M =
𝟐
𝟎
𝟎
.
𝟐
𝒌 𝟎
.
𝟎 𝒌
If k > 1, the matrix has the effect of enlarging/ magnifying the original object by scale factor of K.
If 0<k < 1 the matrix has the effect of reducing the original object by scale factor of K.
Now let A1 be the area of the unit square and let A2 be the area of the image of the unit square under
𝒌 𝟎
the transformation M =
then
𝟎 𝒌
A1 = 1 sq unit and
A2 = K2 sq units. Then
In general we can consider a scale factor k and we have M =
𝐴2
𝐴1
= 𝐾 2 . = det. M …………….. (verify)
69
EXAMPLE
1. Find the effect of the matrix R =
𝟑 𝟎
on the square whose vertices are O(0,0) A(1,0) B(1,1) C(0,1).
𝟎 𝟑
Solution:
If P is the matrix of the vertices of the square and Q is the matrix of the vertices of the image of the square, then
Q = RP
𝟑 𝟎 𝟎 𝟏 𝟏 𝟎
𝟎 𝟑 𝟑 𝟎
=> Q =
=
𝟎 𝟑 𝟎 𝟎 𝟏 𝟏
𝟎 𝟎 𝟑 𝟑
NB the unit square is magnified by a scale factor of 3 to a square whose vertices are A/(0,0), B/(3,0) C/(3,3) and
D/(0,3)
𝑎 𝑏
2. The matrix P =
transforms a triangle whose vertices are O(0,0), A(3,4), B(5,2) into a triangle with
𝑐 𝑑
vertices O/(0,0), A/(-6,8), B/(-10,4). Determine P.
Solution:
Let R be the matrix of the coordinates of the vertices of ∆OAB and Let S be the matrix of the coordinates of the
vertices of ∆OAB / ; then
0 −6 −10
𝑎 𝑏 0 3 5
PR = S =>
=
0 8
4
𝑐 𝑑 0 4 2
=>
0 3𝑎 + 4𝑏
0 3𝑐 + 4𝑑
5𝑎 + 2𝑏
0 −6
=
5𝑐 + 2𝑑
0 8
−10
4
3𝑎 + 4𝑏 = −6 …………………… 1
3𝑐 + 4𝑑 = 8 ……………………… 2
5𝑎 + 2𝑏 = −10 …………………… 3
5𝑐 + 2𝑑 = 4 ……………………….. 4
By solving them simultaneously we obtain the following results:
a = -2, b = 0, c = 0 and d = 2.
−2 0
∴P=
.
0 2
Thus the geometrical effect of P on ∆OAB is an enlargement by a scale factor of 2 followed by a reflection on y-axis.
=>
70
STRETCH
We consider a stretch on a unit square along the x-axis by a scale factor of K.
Y
B(1,1)
B’(k,1)
C(0,1)
-X
O(00)
X
A’(k,0)
A(1,0)
-Y
1
0
𝟎
𝒌
=
; and T
=
0
1
𝟏
𝟎
𝒌 𝟎
Let M be the matrix of this transformation (stretch) then M =
which has the effect of stretching the figure along the axis by a scale factor of k
𝟎 𝟏
For a stretch along y-axis, we consider the diagram below.
Let the transformation be T hence under T; T
Y
C(0,L)
B’(1, L)
C(0,1)
B(1,1)
-X
O(00)
X
A(1,0)
-Y
1
𝟏
0
𝟎
Let the transformation be T hence under T; T
=
; and T
=
0
𝟎
1
𝟏
𝟏 𝟎
Let M be the matrix of this transformation (stretch) then M =
.
𝟎 𝟏
STRETCH ALONG BOTH AXES
This can take place simultaneously. If the stretch along the x-axis is by scale factor k and along y-axis it is 4, then we can consider the diagrams below if the transformation is
T:
Y
B’(k, L)
C(0,L)
C(0,1)
-X
O(00)
B(1,1)
A(1,0)
A(k,0)
X
-Y
1
0
𝟎
𝒌
𝒌 𝟎
T
=
; and T
=
. The matrix of transformation is M =
has the effect transforming a unit square.
0
1
𝑳
𝟎
𝟎 𝑳
71
SHEAR
This transformation distorts the original figure by a factor in the direction of a straight line called INVARIANT LINE. Consider
the shear of a unit square by a factor P along x-axis as below;
Y
P
B(1,1)
C(0,1)
-X
O(00)
C’(p,1)
B’(p+1,1)
X
A(1,0)
-Y
𝒑
1
𝟏
0
Let the transformation be T the T
=
; and T
=
. If the matrix of transformation is M = then
𝑳
0
𝟎
1
𝟏 𝒑
M=
which shears a unit square by a factor P in the direction of x-axis (invariant line). We note
𝟎 𝟏
that this matrix transforms a unit square into a parallelogram
SHEAR ALONG Y-AXIS by a factor q
B’(1,q+1)
Y
C(0,1)
-X
O(00)
A’(1,q)
B(1,1)
A(1,0)
q
X
-Y
𝟏
1
0
𝟎
Let the transformation be T the T
=
; and T
=
. If the matrix of transformation is M = then M=
𝒒
0
1
𝟏
𝟏 𝟎
which shears a unit square by a factor q in the direction of y-axis (invariant line). We note that this matrix
𝒒 𝟏
transforms a unit square into a parallelogram. We observe also that the determinant of the matrix of
72
EXAMPLES
1. A square OABC with vertices (0,0), (1,0), (1,1) and (0,1) is transformed into a rectangle
3 0
. Find the:
OA’B’C’ by the matrix
0 4
a. vertices of OA’B’C’
b. the area of OA’B’C’
Solution
a) Let P be the matrix of the coordinates of OABC and let Q be the matrix of the coordinates of
and let T be the matrix of the transformation; then
3 0
0 1 1 0
.
,T=
P=
0 4
0 0 1 1
0 3 3 0
3 0 0 1 1 0
.
=
But Q = TP =>
0 0 4 4
0 4 0 0 1 1
Therefore the vertices of OA’B’C’ are (0,0), (3,0), (3,4) and (0,4).
b. The area of OA’B’C’ is given by det T X area of OABC.
OABC. Is a unit square with area = 1sq unit
3 0
= 12
det T =
0 4
Therefore area of OABC = 12 X 1 = 12sq units.
73
VECTOR SPACE AND SUB-SPACE
(BASICS)
GROUPOIDS
A non-empty set which is closed wrt a binary operation
i.e if G is a non-empty set G with a binary operation * defined on it is called a groupoid (G, *) if
∀ a, b ∈ G, a ∗ b ∈ G ..... (Closure).
E.g. (N,+), (N, x), (Z, x), (R, +)
If P {i,-i,1,-1} is P a groupoid?
SEMI - GROUP
Let G be a non-empty set G with a binary operation * defined on it then is called a semi- group (G, *)
if:
i. ∀ a, b ∈ G, a ∗ b ∈ G ..... (Closure).
ii. ∀ a, b, c ∈ G, a ∗ b ∗ c = a ∗ (b ∗ c) ∈ G ....... Associative
e.g. (z,+), (N,+)
MONOID
A non-empty set G with a binary operation * defined on it is called a group (G, *) if it satisfies the
following 4 axioms;
i) Closure i.e. ∀ a, b ∈ G, a ∗ b ∈ G
ii) Associative i.e. ∀ a, b, c ∈ G, a ∗ b ∗ c = a ∗ (b ∗ c) ∈ G
iii) Existence of identity element e ∈ G ∋ ∀ a ∈ G, a ∗ e = e ∗ a = a
e.g. (z,x), (N,x)
NB: Explain additive and multiplicative identities
74
GROUPS
Defn: A non-empty set G with a binary operation * defined on it is called a group (G, *) if it
satisfies the following 4 axioms;
i) Closure i.e. ∀ a, b ∈ G, a ∗ b ∈ G
ii) Associative i.e. ∀ a, b, c ∈ G, a ∗ b ∗ c = a ∗ (b ∗ c) ∈ G
iii) Existence of identity element e ∈ G ∋ ∀ a ∈ G, a ∗ e = e ∗ a = a
iv) ∀ a ∈ G has an inverse say 𝑎−1 ∋ 𝑎 ∗ 𝑎−1 = 𝑎−1 ∗ 𝑎 = 𝑒
Example;
a) (Z6, +) where Z6 = [0,1,2,3,4,5]
b) (Z,+) where Z = [...-3,-2,-1,0,1,2,3, .....]
c) G = {1,3,5.7} under multiplication (mod.8)
ORDER OF A GROUP:
It is the Cardinal number of its underlying set denoted by 𝐆 i.e. 𝐆 = n(G)
ORDER OF AN ELEMENT: If a is an element of G and e is an identity element, the order of a is the
least multiple of a that gives the identity element i.e.
i.
an = e the n is the order of a n∈N (multiplication
ii. na = e then n is the order of a n∈N (addition
Consider the group G = (Z6, +) = {0,1,2,3,4,5} where 0 is the identity element
Order of
0 = 1 since 1(0) = 0
1 = 6 since 6(1) = 1+1+1+1+1+1 = 0
2 = 3 since 3(2) = 2+2+2 = 0
3 = 2 since 2(3) = 3+3 = 0
4 = 3 since 3(4) = 4+4+4 = 0
5 = 6 since 6(5) = 5+5+5+5+5+5 = 0
75
1. SUB-GROUP
Let (G, *) be a group and H a subset of G such that (H, *) is a group then (H, *) is a
subgroup of (G, *).
We note that:
i. the [e,*] and (G, *) are trivial subgroups
ii. identity element of G is also the identity element of each such subgroup
iii. if H ≠ e and H ≠ G the H is called a proper sub-group of G
Examples
a) (Z, +) is a subgroup of (R, +)
b) (Q*,x) is a subgroup of (R*, x) where Q* and R* do not contain the zero element
set {0}
2. COSETS OF A GROUP
Let T be a sub-group of G. The left coset of T is denoted by
gT = {gt: g ∈ G, ∀ t ∈ T} .
(similarly right coset of T is denoted by
Tg = {tg: g ∈ G, ∀ t ∈ T}
76
Example
a) Consider the group (Z6, +) and its sub-group (T, +) where Z6 = {0,1,2,3,4,5} and T =
{0,2,4}.
Left cosets of T in Z are 0T, 1T, 2T, 3T, 4T, and 5T
0T = 0+T = {0+0, 0+2, 0+4} = {0,2,4}
1T = 1+T = {1+0, 1+2, 1+4} = {1,3,5}
2T = 2+T = {2+0, 2+2, 2+4} = {2,4,0}
3T = 3+T = {3+0, 3+2, 3+4} = {3,5,1}
4T = 4+T = {4+0, 4+2, 4+4} = {4,0,2} and
5T = 5+T = {5+0, 5+2, 5+4} = {5,1,3}.
We note that the index of T in G is 2 (i.e. the number of distinct cosets) => {0,2,4}
& {1,3,5}
b) Consider the group (Z* 8, x) and its sub-group (G, x) where Z = {1,2,3,4,5,6,7} and
G = {1,3,5,7}. The right cosets of G in Z are G1, G2, G3, G4, G5, G6 and G7 as
follows;
G1 = Gx1 = {1x1, 3x1, 5x1, 7x1} = {1,3,5,7}
G2 = Gx2 = {1x2, 3x2, 5x2, 7x2} = {2,6,2,6} = {2,6}
G3 = Gx3 = {1x3, 3x3, 5x3, 7x3} = {3,1,7,5}
G4 = Gx4 = {1x4, 3x4, 5x4, 7x4} = {4,4,4,4} = {4}
G5 = Gx5 = {1x5, 3x5, 5x5, 7x5} = {5,7,1,3}
G6 = Gx6 = {1x6, 3x6, 5x6, 7x6} = {6,2,6,2} = {2,6}
G7 = Gx7 = {1x7, 3x7, 5x7, 7x7} = {7,5,3,1}
We note that the index of G in Z is 3 (i.e. the number of distinct cosets) => {1,3,5,7},
{2,6}, &{4}
77
1. CYCLIC GROUPS
A group G is cyclic if all the elements of G can be expressed in terms of one element of the set
Example
a) (Z5, +) where Z = {0,1,2,3,4} is a cyclic group with generators as 1 or 2 or 3 or 4 since
<1> = {1, 1+1, 1+1+1, 1+1+1+1, 1+1+1+1+1} = {1,2,3,4,0} as generated by 1
<2> = {2, 2+2, 2+2+2, 2+2+2+2, 2+2+2+2+2} = {2,4,1,3,0} as generated by 2
b) (Z6, +) is cyclic generated by 1 0r 5. Show
c) (Z*5, x) where Z* = {1,2,3,4} is generated by 2 or 5. Show
Definitions
i. Lagrange’ theorem: If H is a sub-group of a group G then the order of H divides the order of G
i.e. 𝑯 / 𝑮
ii. Normal sub group or invariant sub-group: if N is a subgroup of G and left cosets are equal to right cosets of N in G
then N is normal subgroup of G
i.e. gN=Ng
iii. Maximal normal sub-group: A normal sub-group M of G is called maximal normal sub-group iff no normal sub-group
H of G exists properly contained in M and properly contained in G i.e. M⊂ H ⊂G
iv. Simple groups: A group G is called a simple group if it has no proper normal sub-group
v. Conjugate elements of a group: Let x, y ∈ G. If there exists an element g ∈ G such that y = gxg-1, then y is the
conjugate of x.
vi. Torsion groups: A group G is called Torsion if every element of G is of finite order.
i.e. ∀ g ∈ G, ∃ n ∈ N ∋ gn = e.
Note that G is torsion free if no element other than the identity element is of finite order.
(if a ∈ G, the normalizer of a in G is denoted by N{a} and given by: N{a} = (x: x ∈ G, a = xax-1) )
vii. Normaliser of a sub-group: Let G be a group and A sub-group of G. The normalizer of A in G is the set denoted by
NG(A) and given by:
NG(A) = {g: g∈ G, gAg-1 = A}
viii. Centre of a group: the centre of a group G denoted by CG is the set of all x ∈ G such that xg = gx for all g∈ G
i.e. CG = {x: x ∈ G, xg = gx ∀ g ∈ G}
This simply means that CG is the set of all elements of G that commute with every element of G
ix. Centralizer of a sub-group: Let G be any group and A a sub-group of G. The centralizer or centre of A in G is the set
denoted by CG(A) = {g: g ∈ G, ga = ag ∀ ∈ 𝐴 }
78
1. Quotient groups (*****)
Let H be a sub-group of G. The group formed by the cosets of H in G is called
quotient group of H in G or factor group of G relative to H denoted as G/H.
This exists only when H is a normal (invariant) subgroup of G
2. Direct Product Groups: Let H and K be two groups and a Cartesian product
H x K = {(h,k) ∀ h ∈ H and ∀ k ∈ K}
The group formed with respect to the binary operation H x K given by
(a,b)(x,y) = (ax,by) for all (a,b), (x,y) ∈ HxK is a Direct Product Group.
3. Symmetric group (****) A symmetric group of degree n Sn is a group of all
the n! Permutations of n objects. The zero permutation is the identity element
of symmetric group
Alternating group (****) An alternating group An of degree n is a group formed
by all the even permutations of n objects. An contains
n!
2
even permutations of n
objects with zero permutation as the identity element.
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MAPPING BETWEEN GROUPS
1. Homomorphism: A homomorphism is a mapping from one group into another group which
preserves the structure in the groups
Consider (H,*) and (K, ⨁) be two groups and let f: (H,*) → (K, ⨁) be a mapping such that:
∀ a,b ∈ H, f(a*b) = f(a) ⨁f(b)
Then f is a homomorphism.
Note: If f: X→ Y is a homomorphism then Kernel of f denoted as Ker f is the set of elements of X
which are mapped onto the identity in Y.
i.e. ker f = {x: f(x) = ey, where x ∈ 𝑋 and ey is the identity in Y
Example
1. The mapping (Z4,+) →(Z*5, x) given by: f(0)=1, f(1)=2, f(2)=4 and f(3)=3 is a homomorphism ker f
= {0} in that
f(0 + 1) = f(1) = 2;
f(0) x f(1) = 1 x 2 = 2
f(1+ 2) = f(3) = 3;
f(1) x f(2) = 2 x4 = 3
f(2+3) = f(1) = 2;
f(2) x f(3) = 4 x 3 = 2
f(1+1) = f(2) = 4;
f(1) x f(1) = 2 x 2 = 4
f(2 + 2) = f(0) = 1;
f(2) x f(2) = 4 x 4 = 1
f(3 +3) = f(2) = 4;
f(3) x f(3) = 3 x 3 = 4
The following are also homomorphism
2. h:(Z, +) → (R,+) given by h(x) = 3x. we show that ∀ a,b ∈Z, h(a+b) = h(a) + h(b)
i.e.
h(a+b)
= 3(a + b) = 3a + 3b; ...... i
and
h(a) + h(b)
= 3a + 3(b) .....................ii
thus h is a homomorphism since i=ii
3𝑥
3. h:(Z,+) → (Q,+) given by h(x) = .
4
+
4. h:(R,+) → (R ,x) given by h(x) = 3x.
In each case we show that h(a+b) = h(a) + h(b)
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TYPES OF HOMOMORPHISM
1. endomorphism: It is a homomorphism of a group into itself. Let G be a
group and f:G→G a mapping such that f(ab) = f(a)f(b) for all a,b∈G
2. monomorphism: It is an injective (1-1) homomorphism. Let G and H
be groups and f:G→H a mapping such that:
(i) for all a,b∈G f(ab) = f(a)f(b)
(ii) for all a,b∈G f(a) = f(b) iff a=b and f(a) ≠ f(b) iff a≠b
3. epimorphism: It is a surjective (onto) homomorphism. Let G and H be
groups and f:G→H a mapping such that:
(i) for all a,b∈G f(ab) = f(a)f(b)
(ii) for all h∈H ∃ g ∈ G: f(g) = h;
then f is an epimorphism from G onto H
4. isomorphism: It is bijective (1-1, onto) homomorphism. Let G and H
be groups and f:G→H a mapping such that:
(i) for all a, b ∈ G f(ab) = f(a)f(b)
(ii) for all a, b ∈ G f(a) = f(b) iff a=b and f(a) ≠ f(b) iff a≠b
(iii) for all h∈H ∃ g ∈ G: f(g) = h
then f is an isomorphism from G onto H
5. authomophism: It is an isomorphism of a group onto itself.
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DEFINITION: VECTOR SPACE AND SUB-SPACE
Defn: Let K be a given field (with scalar elements) and V a non-empty set
with the rules of addition and scalar multiplication which assigns to any
u,v∈V a sum u+v ∈ V and to any u ∈V and k €K a product ku ∈V. Then V is
called a vector space over K and the elements of V are vectors if the
following conditions (axioms) hold true;
a) For any vectors u,v,w ∈ V (u+v)+w = u+(v+w)
b) There exists a vector in V, denoted by 0 and called zero vector for
which u+0=u for any u∈V
c) For any vector in u∈V there is a vector in V denoted as –u for which
u+(-u)=0
d) For any vectors u,v,w∈V, u+v=v+u
e) For any scalar k∈K and any vectors u,v€V, k(u+v) = ku+kv
f) For any scalars a,b∈K, and any vector v∈V, (a+b)u = au+bu.
g) For any scalars a,b∈K, and any vector v∈V, (ab)u = a(bu)
h) For the unit scalar 1∈K, 1u=u for any u∈V
By implication of the above axioms, we observe that the left and right
cancellation laws hold also i.e. if u+w=v+w => u=v. similarly, subtraction is
defined by u-v = u+(-v).
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Some examples of vector spaces are;
1. Let K be an arbitrary field. The set of all n-tuples of elements of K with vector addition and
scalar multiplication denoted by (a1, a2, a3, ---an) + (b1, b2, b3, ---, bn)=
(a1+b1,a2+b2,a3+b3, ---,an+bn) where ais ∈K is a vector space and we denote this space by
Kn and the zero vector in Kn is the n-tuple of zeros i.e. 0=o,0,0,…,0).
2. Let V be the set of all mxn matrices with entries from an arbitrary field K. Then V is a
vector space over K with respect to the operation of matrix addition and scalar
multiplication.
3. Let V be a set of all polynomials a0+a1t+a2t2+---antn with coefficients ai from a field K. Then
V is a vector space over K w.r.t. the usual operations of addition of polynomials and
multiplication by a constant.
SUB-SPACES
Defn: Given the above definition of a vector space V over K, if W is a subset of V, then W is a
subspace of V provided that W itself is a vector space over K w.r.t. the operations of vector
addition and scalar multiplication on V.
We note that W is a subspace as V iff
i. W is non-empty
ii. W is closed under vector addition i.e. if v,w∈W, => v+w∈W
iii. W is closed under scalar multiplication i.e. if v ∈W => kv ∈W for all k in K.
The above definition can be put in other words as: “W is a subspace of V iff (i)0∈W or W≠0, and
(ii) v,w ∈W=> av+bw ∈ W for every a,b∈K.
In a special way we note that V is a subspace of itself as well as the set {0}consisting of all the
zero vectors.
Trivial result: The intersection of any number of subspaces of a vector space V is a
subspace of V.
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CAYLEY-HAMILTON THEOREM
Cayley–Hamilton theorem
states
every square matrix over a commutative
satisfies its own characteristic equation.
that
ring
If A is a given n×n matrix and In is the n×n identity
matrix, then the characteristic polynomial of A is
defined
as:
where det is
    det A  I  ,
the determinant operation and λ is a scalar element
of the base ring. Since the entries of the matrix are
polynomials in λ, the determinant is also an n-th
order monic polynomial in λ. The Cayley–Hamilton
theorem states that substituting the matrix A for λ in
this polynomial results in the zero matrix, i.e.  A  0 .
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CHECK SLIDE 90
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FINAL TEST QUESTIONS
Answer all questions. Time 30mins
1 1 2
1. Derive the characteristic polynomial of A  0 3 2 using the variable y.


1 3 9
2. Use Cramer’s rule to solve the system of linear equation :
 x  2 y  3z  1
2x

z  0
3x  4 y  4 z  2
7 3 5
3. Find the adjoint of A given that A   1 - 2 1 .
- 2 4 - 3
1 2 −3 0
4. Transform 2 4 −2 2 the matri to echelon form by elementary row operation
3 6 −4 3
5. Find the vertices of the image of the triangle ABC which undergoes a rotation of 90o in
the anti-clockwise sense if the vertices of ABC are A(0,0), B(1,1) and C(1,4).
6. State Cayley-Hamilton theorem.
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