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MA1506
Mathematics II
Chapter 5
Matrices and their uses
Chew T S MA1506-14 Chapter 5 1
This chapter consists of two parts
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In part one, we shall study
Chew T S MA1506-14 Chapter 5 2
Matrix operations
Some special matrices
Inverse matrix and unique solution of AX=B
Determinant and inverse matrix
Leontief Input-Output model
PART ONE
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5.1 What is a Matrix?
Chew T S MA1506-14 Chapter 5 3
rewritten as
2x1 matrix2x2 Matrix
2 7 3x y+ =
4 8 11x y+ =
2 7 3
4 8 11
x
y
=
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3x3 Matrix
m x n Matrix: m rows, n columns
Entry at i-th row j-th column
5.1 What is a Matrix?
Chew T S MA1506-14 Chapter 5 4
11 1
1
n
m mn
a a
a a
We may write
11 12 13
21 22 23
31 32 33
a a a
a a aa a a
( )ijA a=
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5.2 Matrix operations
Chew T S MA1506-14 Chapter 5 5
Matrix addition
Scalar multiplication
Matrix multiplication
Matrix transposition
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Matrix Addition
Chew T S MA1506-14 Chapter 5 6
m x nmatrices
Term by term addition
5.2 Matrix operations
( )ijA a=( )ijB b=
( )ij ijA B a b+ = +
1 2 7 3 8 5
4 8 6 9 10 17
+ =
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Scalar multiplication
Chew T S MA1506-14 Chapter 5 7
m x nmatrix
Term by term multiplication
real or complex number
5.2 Matrix operations
( )ijA a=c
( )ijcA ca=1 2 3 6
34 8 12 24
=
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Matrix Multiplication
Chew T S MA1506-14 Chapter 5 8
Not term by term but row to column
m x nmatrix n x pmatrix
5.2 Matrix operations
( )ijA a= ( )ijB b=
m x pmatrixAB C=( )ijC c=
1 1 2 2ij i j i j in njc a b a b a b= + + +
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Example
Chew T S MA1506-14 Chapter 5 9
21
32
11
654
321
++++
++++=
)2(63514)1(62514
)2(33211)1(32211
= 78
12
5.2 Matrix operations
Animation slide
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Chew T S MA1506-14 Chapter 5 10
In general
Non commutative
5.2 Matrix operations
AB BA
2 7 1 3 16 14 8 2 1 20 4
AB = =
1 3 2 7 14 312 1 4 8 0 6
BA = =
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Matrix transposition
Chew T S MA1506-14 Chapter 5 11
m x nmatrixswap rows with columns
5.2 Matrix operations
( )ijA a=
( )T jiA a= n x mmatrix
1 2 4
6 8 9
T
=
1 6
2 8
4 9
1 7 9
6 8 2
4 10 12
T =
1 6 4
7 8 10
9 2 12
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Chew T S MA1506-14 Chapter 5 12
5.2 Matrix operations
( )T
TA A=
( )T T TA B A B+ = +
( )T TcA cA=
( )T T TAB B A=
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Symmetric matrix
Chew T S MA1506-14 Chapter 5 13
An n x nmatrix is symmetricif
5.3 Special matrices
TA A=
1 7 9
7 8 2
9 2 12
0 0 4
0 8 0
4 0 1
1 0
0 1
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Anti-Symmetric matrix
Chew T S MA1506-14 Chapter 5 14
An n x nmatrix is anti-symmetricor skewsymmetricif
5.3 Special matrices
TA A=
0 7 9
7 0 2
9 2 0
0 22 0
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Properties of Symmetric matrices
Chew T S MA1506-14 Chapter 5 15
IfA is symmetricand Bis any square matrix
T
B B+
TBAB
is symmetric
is symmetric
The above property will be used in T8 Q4
5.3 Special matrices
( ) ( )T T T T T T B B B B B B+ = + = +
Hence
( ) ( )T T T T T T T BAB B A B BAB= =
Hence
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Properties of anti-Symmetric matrices
Chew T S MA1506-14 Chapter 5 16
IfA is anti-symmetricand Bis any square matrix
TBAB
TB B is anti-symmetric
is anti-symmetric
The above property will be used in T8 Q4
5.3 Special matrices
( ) ( ) ( )T T T T T T B B B B B B = =
Hence
Hence
( ) ( )T T T T T T T BAB B A B BAB= =
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Identity matrix
Chew T S MA1506-14 Chapter 5 17
n x nidentity matrix
sometimes denoted by
In
5.3 Special matrices
1 0 00 1 0
0 0 0 1
I=
AI IA A= =
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Vectors as special matrices
Matrices containing only one column are often
called column vectors or vectors
Matrices containing only one row are often
called row vectors or vectors
Chew T S MA1506-14 Chapter 5 18
1
2
1
2
3
[ ]1 2 [ ]1 2 3
5.3 Special matrices
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Vectors
Chew T S MA1506-14 Chapter 5 19
=1
0
0
1
=
1
0
0
0
1
0
0
0
1
== =
(0,1)
(1,0)
(1,0,0) (0,1,0)
(0,0,1)
5.3 Special matrices
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Chew T S MA1506-14 Chapter 5 20
a
b
1 0
0 1a b
= +
1 0 0
0 1 0
0 0 1
a
b a b c
c
= + +
=
=
5.3 Special matrices
ai bj+
ai bj ck + +
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Rotation
Chew T S MA1506-14 Chapter 5 21
Rotation (anti-clockwise)
through angle
Let
5.3 Special matrices
cos sin( )
sin cosR
=
cos sinRi i j = +
sin cosRj i j = +
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Rotation
Chew T S MA1506-14 Chapter 5 22
Rotation (anti-clockwise)
through angle
R transforms
5.3 Special matrices
cos sinRi i j = +
sin cosRj i j = + sin cosi j +
cos sini j +
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Chew T S MA1506-14 Chapter 5 23
cos sin( )
sin cos
R
=
[ ]
15061506 cos sin
( ) sin cos
cos(1506 ) sin(1506 )
sin(1506 ) cos(1506 )
R
=
=
5.3 Special matrices
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Orthogonal matrix
Chew T S MA1506-14 Chapter 5 24
Ann x nmatrix, Bis orthogonal if
is orthogonal
TBB I=
cos sin
sin cos
cos sin cos sin
sin cos sin cos
2 2
2 2
cos sin 0
0 sin cosI
+= =
+
5.3 Special matrices
Check
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Chew T S MA1506-14 Chapter 5 25
cos sin
sin cos
( )cos ,sin
( )sin ,cos
( )cos , sin
( )sin , cos
An nn matrix A isorthogonal iff its columns
are perpendicular unitvectors.
An nn matrix A isorthogonal iff its rows areperpendicular unitvectors.
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Chew T S MA1506-14 Chapter 5 26
Involutory matrix ( a matrix that is its own inverse)
AA I=21
a b
A aab
=
Example
Then
AA I=
= 1 2
or
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Shear angle and parallel to x-axis
Chew T S MA1506-14 Chapter 5 27
Let S=
1 tan 1
0 1 0
Si
=
1
0
i
= =
1 tan 0
0 1 1Sj
=
5.3 Special matrices
Then
So
tan tan
1
i j
= = +
1 tan
0 1
Si i= tansj i j= +
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Chew T S MA1506-14 Chapter 5 28
5.3 Special matrices
Si i= tansj i j= +
tan i j +
multiplication multiplication
j becomes ,turn angle clockwise,measurefrom
tan i j +
j
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Chew T S MA1506-14 Chapter 5 29
(S transforms)
5.3 Special matrices
Si i= tansj i j= +
tan i j +
S maps
We may say that S maps i ito
mapsj to
tan i j +
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Si =1
0
Sj=1 tan0 1
S
=
tan
1
1 tan1 2
0 1
= +
1 2 tan
1 0 2 1
+ = +
1 2tan
2
+ =
1
2
1 2tan
2
+
or 1 1 tan 1 1 2tan
2 0 1 2 2S
+ = =
1 (1 2 ) 1 22
S S i j Si Sj = + = +
30Chew T S MA1506-14 Chapter 5
5.3 Special matrices
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Si =1
0
Sj=1 tan0 1
S
=
tan
1
1 tan( 1) 2
0 1
= +
1 2 tan
( 1) 0 2 1
+ = +
1 2 tan
2
+ =
1
2
1 2 tan
2
+
or 1 1 tan 1 1 2 tan
2 0 1 2 2S
+ = =
1 (( 1) 2 ) ( 1) 22
S S i j Si Sj = + = +
31Chew T S MA1506-14 Chapter 5
5.3 Special matrices
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Question 7 (a) [5 marks] (2007 Exam )In two dimensions, we perform a shear
[shearing angle 45 degrees] by means offorces parallel to an axis which makes anangle of 30 degrees with respect to thepositive x-axis. Find the final location of
the point which was originally at (x, y)coordinates (0, 1).
(0,1)
3045
Rotate 30
Shear45
Rotate 30
parallel to x axis
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cos sin( )
sin cosR
=
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Chew T S MA1506-14 Chapter 5 34
Summary We have learnt
symmetric matrix A
anti symmetric matrix A
identity matrix
orthogonal matrix B
shear matrix
vector
rotation matrix
TA A=
TA A= I
TBB I=
1
2
[ ]1 21 tan
( ) 0 1S
= cos sin
( )sin cos
R
=
5.3 Special matrices
Involutory matrix A 2A I=
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The inverse matrix of A is denoted by1
A
WARNING: We only define inverse for
n n matrices, i.e. square matrices.
Let A=11 1
1
n
n nn
a a
a a
be an n n matrix
5.4 Inverse matrix and unique solution
Chew T S MA1506-14 Chapter 5 36
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B=
1
2
n
b
b
b
X=
1
2
n
x
x
x
Consider a linear system AX=B
11 1
1
n
n nn
a a
a a
1
2
n
x
x
x
1
2
n
b
b
b
i.e., =
Chew T S MA1506-14 Chapter 5 37
5.4 Inverse matrix and unique solution
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which is equivalent to
11 1 12 2 1 1... n na x a x a x b+ + + =
21 1 22 2 2 2... n na x a x a x b+ + + =..............................................
1 1 2 2 ...n n nn n na x a x a x b+ + + =
Chew T S MA1506-14 Chapter 5 38
5.4 Inverse matrix and unique solution
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Theorem 1The linear system of equations AX=B
has a unique solution if and only if A has an
inverse 1A
How to find 1A if1
A
exists?
We can perform elementary rowoperations to A to get 1A
Chew T S MA1506-14 Chapter 5 39
Remark: B can be any n-dim vector
Proof omitted
5.4 Inverse matrix and unique solution
See Appendix
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5.5 Determinants and inverse
Chew T S MA1506-14 Chapter 5 40
In this section, we introduce determinants,
which help us to determine whether ann n matrix has an inverse or not.
Consider11 12
21 22
a aa a
1
2
x
x
1
2
b
b
=
Then
11 1 12 2 1
a x a x b+ =
21 1 22 2 2a x a x b+ =
(1)
(2)
5 5 D t i t d i
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Hence 22 12(1) (2)a a get
11 22 12 21 1 1 22 12 2( )a a a a x b a a b = Similarly
21 12 22 11 2 1 21 11 2( )a a a a x b a a b =
Therefore the linear system of equations
has unique solution if and only if
5.5 Determinants and inverse
Chew T S MA1506-14 Chapter 5 41
11 22 12 21 0a a a a
5 5 Determinants and inverse
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We define the determinant of
11 12
21 22
a aa a
to be 11 22 12 21a a a a
We write
det11 12
21 22
a a
a a
11 22 12 21a a a a
or 11 1221 22
a aa a
11 22 12 21a a a a
=
=
5.5 Determinants and inverse
Chew T S MA1506-14 Chapter 5 42
5 5 Determinants and inverse
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Let A=11 12 13
21 22 23
31 32 33
a a a
a a a
a a a
Then we define the determinant of A,
(denoted by det(A) or )A
to bedet(A)
11a2322
32 33
aa
a a
12a2321
31 33
aa
a a
13a+ 21 22
31 32
a a
a a
=
How to get this? see next slide
5.5 Determinants and inverse
Chew T S MA1506-14 Chapter 5 43
5 5 Determinants and inverse
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called cofactorexpansion
det A
5.5 Determinants and inverse
Chew T S MA1506-14 Chapter 5 44
cofactorcofactorcofactor
Animation slide
5 5 Determinants and inverse
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Cofactor expansion can be done about any
row or column
5.5 Determinants and inverse
Chew T S MA1506-14 Chapter 5 45
2x2 matrix
5 5 Determinants and inverse
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Examples
Chew T S MA1506-14 Chapter 5 46
5.5 Determinants and inverse
Animation slide
5 5 Determinants and inverse
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Cofactor expansion can be used for any
n x ndeterminant
5.5 Determinants and inverse
Chew T S MA1506-14 Chapter 5 47
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2ndmethod (cofactor expansion) finding Inverse
Chew T S MA1506-14 Chapter 5 48
Work out the matrix of cofactor of each entry
Take transpose
Divide by determinant
5.5 Determinants and inverse
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Chew T S MA1506-14 Chapter 5 49
5.5 Determinants and inverse
5.5 Determinants and inverse
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Chew T S MA1506-14 Chapter 5 50
=
1
=
1
=
1
det
=1
Let
Then
5.5 Determinants and inverse
Inverse of 2x2 matrix
5.5 Determinants and inverse
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Important Properties of Determinants
Chew T S MA1506-14 Chapter 5 51
size of M
det( ) (det )(det ) det( )ST S T TS = =
det detT
M M=
det( ) detn
cM c M =
5.5 Determinants and inverse
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If A is upper (lower) triangular, thendet(A)=product of all diagonal entries of A
1 2 4
0 3 5 1 3 6 18
0 0 6
= =
4 0 0
1 5 0 4 5 6 1202 3 6
= =
Chew T S MA1506-14 Chapter 5 52
5.5 Determinants and inverse
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Theorem 2
Chew T S MA1506-14 Chapter 5 53
An nxn matrix A has an inverse if and only if det(A)0
Proof omitted
1 1 1
0 2 3
0 0 3
A
=
1 1 1
0 0 30 0 2
B
=
det 1 2 3 6A= =
det 1 0 2 0B= =
A has an inverse
B does not have aninverse
5.5 Determinants and inverse
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Theorem3
Chew T S MA1506-14 Chapter 5 54
Let A be an nxn matrix. Then AX=B has a unique solution
if and only if det(A)0Proof omitted
1
2
3
1 1 1 4
0 2 3 5
0 0 3 6
x
x
x
=
has a unique solution
1 1 1
det 0 2 3 1 2 3 6 0
0 0 3
= =
since
Remark: B can be any n-dim vector
5.5 Determinants and inverse
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Theorem 4
Chew T S MA1506-14 Chapter 5 55
Let A be nxn matrix. Then AX=0
has nontrivial (nonzero) solutionsif and only if det(A)= 0.
Proof omitted
Note that A0=0.Hence zero vector 0= is always a solution of AX=0
0
0
0
Zero vector
5.5 Determinants and inverse
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1
2
3
1 1 1 0
0 2 3 0
0 0 3 0
x
x
x
=
has a unique solution
Then the unique solution should be1
2
3
0
0
0
x
x
x
=
Chew T S MA1506-14 Chapter 5 56
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5.5 Determinants and inverse
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Remark 2:AX=0 has either unique solution
(detA0)or infinitely many solutions
(detA=0)
Remark 3:If detA =0 , then AX=B, where B 0,
has two cases: infinitely many solns or no
solns
Chew T S MA1506-14 Chapter 5 58
5 6 L i f I O M d l
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5.6 Leontief Input-Output Model
Chew T S MA1506-14 Chapter 5 59
we use this model to analyze economics of
interdependent sectors
Example ---Oil and Transportation industries
(1) Total output $y of Transportation industry requires
(i) $0.50y gasoline from the oil industry
(ii) $0.20y transportation of equipment from thetransportationindustry
5.6 Leontief Input-Output Model
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We will look at a single oil company and a single
transportation company as a closed system
Chew T S MA1506-14 Chapter 5 60
(2) The total output $x of Oil industry requires
(i) $0.12x transportation of gasoline from thetransportationindustry
(ii) $0.32x oil-based fuels for processing fromthe oil industry
5.6 Leontief Input-Output Model
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Chew T S MA1506-14 Chapter 5 61
(3)Suppose that the demand from the outsidesector of the economy (all consumers outsideof oil and transportation) is:
$15 billion for oil
$1.2 billion for transportation
5.6 Leontief Input-Output Model
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$x b = the total output from oil company=oil demand
$y b = the total output from transportation company
=transportation demand
Chew T S MA1506-14 Chapter 5 62
Internal demand External
demandFrom oilcompany Fromtransportationcompany
Oil demand
x0.32x 0.50y d1= $15b
Transportation
demand y0.12x 0.20y d2= $1.2b
.
5.6 Leontief Input-Output Model
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Setting up the demand equations
The total output of each company will equal thesum of the internal and external demands:
Expressed as a matrix equation:
Chew T S MA1506-14 Chapter 5 63
X MX D= +
1
2
0.32 0.50
0.12 0.20
x x y d
y x y d
= + +
= + +
1
20.32 0.500.12 0.20
dx xdy y
= +
5.6 Leontief Input-Output Model
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Solving the demand equations
Solve forX
:
In our example:
Chew T S MA1506-14 Chapter 5 64
IX MX D =
called technologymatrix
X MX D= +1( )X I M D=
1
2
0.32 0.50
0.12 0.20
dx x
dy y
= +
( )I M X D =
0.32 0.50
0.12 0.20M
=
5.6 Leontief Input-Output Model
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Chew T S MA1506-14 Chapter 5 65
( ) 1 1.65 1.03
0.25 1.40I M
=
0.68 0.50
0.12 0.80I M =
5.6 Leontief Input-Output Model
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The solution
Putting it all together:
In order to meet the demand the companies need
to produce$26.0 billion of oil
$5.4 billion of transportation
Chew T S MA1506-14 Chapter 5 66
1
2
151.2
dDd = =
( )1 1.65 1.03 15 26.0
0.25 1.40 1.2 5.4X I M D
= = =
5.6 Leontief Input-Output Model
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Chew T S MA1506-14 Chapter 5 67
0.32 0.50 15
0.12 0.20 1.2
x x y
y x y
= + +
= + +Can you solve the above without using matrices ?
Sure, you can
However if it involves many variables, then solving
it by matrices is easier.
End of part I
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Chapter 5
PART TWO
Chew T S MA1506-14 Chapter 5 68
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In part II, we shall study
Eigenvalues and eigenvectors---very important
concepts
Chew T S MA1506-14 Chapter 5 69
Diagonalization of matrix
Weather forecasting model
Trace of a matrix
5 7 Eigenvalues and eigenvectors
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parallel
=
=
Vectors are in differentdirections
Vectors are in samedirection
Consider
5.7 Eigenvalues and eigenvectors
Vectors are in different
directions
Chew T S MA1506-14 Chapter 5 70
Rewrite5.7 Eigenvalues and eigenvectors
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Let T=
T
Then
= =T
So we are interested in the last case
2 221 1
=
T multiplying21
by
is the same as multiplying2
1
by 2
Rewrite
Chew T S MA1506-14 Chapter 5 71
Eigenvector T i t i5.7 Eigenvalues and eigenvectors
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Eigenvector
Chew T S MA1506-14 Chapter 5 72
After multiplication (transformation),
does not change the direction
eigenvalue eigenvector
u
Tu
u
Matrix multiplication
is the same as scalar multiplication
which is much easier
T is nxn matrix
suppose We may use
u uorTu u=
5.7 Eigenvalues and eigenvectors
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How to find Eigenvalues and Eigenvectors
Chew T S MA1506-14 Chapter 5 73
We want to find nonzero eigenvectors
Tu u= How to find eigenvalue and eigenvector u
First note that 0 0T =Hence zero vector 0 is always an eigenvector
5.7 Eigenvalues and eigenvectors
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Chew T S MA1506-14 Chapter 5 74
we want nonzero vector
By Thm 4 in section 5.5, we have
Hence we can find nonzero eigenvaluesusing above equation
T-isnot meaningful
How to find nonzeroeigenvectors
First note that
Tu u Iu = =
( ) 0T I u =
det( ) 0T I =
Example5.7 Eigenvalues and eigenvectors
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Example
Chew T S MA1506-14 Chapter 5 75
Find eigenvalues of
Cont.5.7 Eigenvalues and eigenvectors
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Cont.
Chew T S MA1506-14 Chapter 5 76
( ) 0T I u =
Now we shall find eigenvectorassociated to eigenvalue =2 u
=
Tu u=
1 2 2 0( 2 ) 2 2 2 0T I
= =
1 2
2 2T
=
2, 3=
5.7 Eigenvalues and eigenvectors
Cont.
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Chew T S MA1506-14 Chapter 5 77
two equations same
1 equation 2 unknowns
They are many solutions, i.e. many eiqenvectors
Cont.
5.7 Eigenvalues and eigenvectors
Cont.
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Chew T S MA1506-14 Chapter 5 78
choose
Eigenvectors associated to eigenvalue 2 are ofthe form
1
12 2
=
0
get eigenvector
need only to choose one
=
All eigenvectors are
parallel i.e. all arelinearly dependent
Cont.
1=
1
12
5.7 Eigenvalues and eigenvectors
Cont.
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Chew T S MA1506-14 Chapter 5 79
eigenvector associated to = -3
two eqs same
Now find
1 3 2 0( ( 3) )
2 2 3 0
T I
+ = =
+
Cont.
5.7 Eigenvalues and eigenvectors
Cont.
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Chew T S MA1506-14 Chapter 5 80
choose get eigenvector
1
2 2
=
eigenvectors associated to =-3 is of the form
0
need only to choose one
2 0 + =
=
Cont.
1=1
2
Example5.7 Eigenvalues and eigenvectors
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Example
Chew T S MA1506-14 Chapter 5 81
1
01
i
i
=
0i =
0i = Two eqs same
( ) 0T I
=
det( ) 0T I =T=
consider i=
Cont5.7 Eigenvalues and eigenvectors
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Cont.
Chew T S MA1506-14 Chapter 5 82
eigenvector
0i =0i =
Two eqs same
1
i i
=
1i
get complex eigenvectorchoose 1=
use 0i =
cont.5.7 Eigenvalues and eigenvectors
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cont.
Chew T S MA1506-14 Chapter 5 83
get eigenvector
10
1
i
i
=
0i =
0i + = Two eqs same
1
i i
=
choose get complexeigenvector
Now consider second eigenvalue
get
use 0i + =
i=
1
i
i=
5.7 Eigenvalues and eigenvectors
Recallcont.cos sin
( )R
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Chew T S MA1506-14 Chapter 5 84
When rotating through 90 degrees, everyrealvector should change direction,so
has NO real eigenvector
1
i
represents rotation through 90 degrees
are its eigenvectors which are complex
Recall
So the matrix we just mentioned
cont. ( )sin cos
R
=
1
i
5.8 Diagonalization
5.8 Diagonalization
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We say a matrix A is diagonalizable if
Matrix of eigenvectorsDiagonal matrix ofeigenvalues
ANS:If A is nxn matrix , and it has n non paralleleigenvectors, then A can be diagonalized.
Chew T S MA1506-14 Chapter 5 85
1A PDP
=
5.8 Diagonalization
When a given matrix A can be diagonalized ?
Example
5.8 Diagonalization
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Example
Chew T S MA1506-14 Chapter 5 86
has eigenvalues 2 and -3and corresponding eigenvectors
From the first example in section 5.7, we know
Then2 0
0 3
1e
2e
= =
1 2
2 2
1
12
1
2
= =
1 1
12
2
5.8 Diagonalization
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Next we can verify that
Chew T S MA1506-14 Chapter 587
1 2
2 2
P
2 0
0 3
1P
=
Example
5.8 Diagonalization
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p
Chew T S MA1506-14 Chapter 5 88
has only one eigenvalue since
It has only one eigenvector
Not possible to diagonalize
1 tan
0 1
21 tandet (1 ) 00 1
= =
1
0
1 tan
0 1
Findingn
M5.8 Diagonalization
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11 1
2
0
0
n
n n
nM PD P P P
= =
Finding M
Chew T S MA1506-14 Chapter 5 89
S
5.8 Diagonalization
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Chew T S MA1506-14 Chapter 5 90
[ ]
1506
1506cos sin
( )sin cos
cos(1506 ) sin(1506 )
sin(1506 ) cos(1506 )
R
=
=
Some special cases Notused this method
Involutory matrix A 2A I=
(1)
(2)
Then
=
5 9 Model weather forecasting
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5.9 Model weather forecasting
Today Tomorrow ProbabilityRainy Rainy
Sunny
Sunny Rainy
Sunny
Chew T S MA1506-14 Chapter 5 91
60%
70%
30%
40%
We can write the above information inthe following matrix form
5..9 Model weather forecasting
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Current : R S Next:R
columns add to 1
Chew T S MA1506-14 Chapter 5 92
called transit ion matrix
0.6 0.3
0.4 0.7
R R S RM
R S S S
= = S
5..9 Model weather forecasting
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Today is sunny, will it be rainy 2 days later?
S
S
R
S
R
S
R
0.7
0.7
0.3
0.3 0.4
0.6
0.39
0.610.12
0.18
0.49
0.21
Chew T S MA1506-14 Chapter 5 93
0.6 0.3
0.4 0.7
R R S RM
R S S S
= =
Observe that
5..9 Model weather forecasting
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Chew T S MA1506-14 Chapter 5 94
In fact
2 0.6 0.3 0.6 0.3
0.4 0.7 0.4 0.7
M =
0.6 0.6 0.3 0.4 0.3 0.7 0.6 0.3
0.4 0.6 0.7 0.4 0.7 0.7 0.4 0.3
+ + = + +
Observe that
2 22
2 2
R R S RM
R S S S
=
2
2
0.48 0.39
0.52 0.61
S R
S S
= =
Summary of the previous slide5..9 Model weather forecasting
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Today is sunny, will it besunny 2 days later?
Chew T S MA1506-14 Chapter 5 95
0.6 0.3
0.4 0.7
R R S RM
R S S S
= =
Today is rainy, will it besunny 2 days later?
Probability=0.52 Probability=0.61
2 22
2 2
0.48 0.39
0.52 0.61
R R S RM
R S S S
= =
Today is rainy, will it berainy 2 days later?
Probability=0.48
Today is sunny, will it berainy 2 days later?
Probability=0.39
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T d i R i il l i t b i 30 d l t ?
5..9 Model weather forecasting
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Eigenvaluesof
Chew T S MA1506-14 Chapter 5 97
0.6 0.3
0.4 0.7M
=
are
Corresponding eigenvectors are
1 0.3 = 2 1 =
1
1
1
4
3
Today is Rainy, will i t be rainy 30 days later?
Find
30
M Should use30 30 1
M PD P
=
1 1
4 3
7 7
5..9 Model weather forecasting
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Chew T S MA1506-14 Chapter 5 98
When use this method?suggest: n>4
41
3
P =
1 7 7
3 3
7 7
P
=
30 630 0.3 0 2 10 0
0 1 0 1D
=
630 30 1
4 31 1
2 10 0 7 74
3 31 0 13 7 7
M PD P
=
3 3
7 7
4 4
7 7
=
5.10 Trace of a Matrix
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Chew T S MA1506-14 Chapter 5 99
Let Mbe a square matrix.
The trace of M, denoted Tr (M),is the sum of thediagonal entries
( ) ( )Tr MN Tr NM=5.10 Trace of a Matrix
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( ) ( )Tr MN Tr NM =
1
( ) ( ) ( )Tr M Tr PDP Tr D
= =Given matrix
Chew T S MA1506-14 Chapter 5 100
For a diagonalizable matrix M,Tr(M)= Tr(D)=sum of its eigenvalues.
Example 1
5.10 Trace of a Matrix
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Use this to check your calculations ofeigenvalues and to find the remainingeigenvalue
Chew T S MA1506-14 Chapter 5 101
0.6 0.3
0.4 0.7M
=
eigenvalue 1 0.3 =
Tr(M)=0.6+0.7=1.3
Hence the 2ndeigenvalue is2 1 =
a p e
Tr(M)= Tr(D)=sum of its eigenvalues.
Example 25.10 Trace of a Matrix
4 4 matrixA = entries= {1 2 3 16}
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Chew T S MA1506-14 Chapter 5 102
entries= {1,2,3...16}
All rows, all columns,all diagonalshave the same sum
T A BExample 3
5.10 Trace of a Matrix
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Chew T S MA1506-14 Chapter 5 103
( ) ( )( ) ( )
Tr T Tr A BTr A Tr B
= += +
T A B = +Suppose that
Find1 2 5 6 A=3 4 7 8
T =
[ ] [ ]1 1( ) ( ) ( ) 5 13Tr B Tr T Tr A
= =
( )Tr B
Solution Tr is linear
We need thisproperty inTutorial 9 Q3
R k
5.10 Trace of a Matrix
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Chew T S MA1506-14 Chapter 5 104
Remark:
det = det(1
)= det(1) = det
= product of eigenvalues
0.6 0.3
0.4 0.7M
=
1 0.3 = 2 1 =eigenvalues
EndChapter 5
DetM=(0.3)(1)=0.3
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cont. cossin ( sin )
d d
dt dt
= =
Appendix
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Chew T S MA1506-14 Chapter 5 106
( )dt dt
is constant
Composing two shears Appendix
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Chew T S MA1506-14 Chapter 5 107
S:shear degrees parallel to x axis
S: shear degrees parallel to x axis
Still a shear but note that
Rotation in 3D Appendix
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Chew T S MA1506-14 Chapter 5 108
Rotate 90 degrees (anticlockwise) about z-axisRotate 90 degrees (anticlockwise) about x-axis
About z-axis About x-axis
Cont. Appendix
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Chew T S MA1506-14 Chapter 5 109
Rotate 90 degrees
(anticlockwise) about z-axis
Rotate 90 degrees
(anticlockwise) about x-axis
D t i t f O th l M t i M
Appendix
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Determinant of Orthogonal Matrix
Chew T S MA1506-14 Chapter 5 110
M
Examples
1P
Appendix
P P exits
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are parallel iff
11 22 12 21det P P P P P=
11 22 12 21det 0P P P P P= =
11 21
12 22
P P
P P= =iff
11
21
P
u P
=
12
22
Pv
P
=
12
22
Pv
P
= =
det 0P=
not parallel iff det 0P
So if not parallel then1
P
exits
Chew T S MA1506-14 Chapter 5 111
Hence
11 12
21 22
P PP
P P
=
,u v
,u v
,u v
Row operations to find inverse
Appendix
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Row operations to find inverse
Chew T S MA1506-14 Chapter 5 112
Refer to Textbook Section 3.3 for more details
Remark: 2
nd
method finding inverse see slide 46
Three row operations on a matrix
multiply a constant to a row
switch two rows add a multiple of another row
(see example)
AppendixCont.
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Example
Let A=
1 4 2
2 8 3
0 1 1
Suppose that we know that A has an inverse.
Now we shall do the following to get 1A
Chew T S MA1506-14 Chapter 5 113
First writeAppendixCont.
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1 4 2 1 0 0
2 8 3 0 1 00 1 1 0 0 1
Try to get zero as many as possible
for the lower triangular part.
1 22R R+
1 4 2 1 0 0
0 0 7 2 1 00 1 1 0 0 1
Chew T S MA1506-14 Chapter 5 114
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1 4 0 3 7 2 7 0
AppendixCont.
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3 12R R +1 4 0 3 7 2 7 0
0 1 0 2 7 1 7 1
0 0 1 2 7 1 7 0
2 14R R +1 0 0 11 7 2 7 4
0 1 0 2 7 1 7 10 0 1 2 7 1 7 0
When we reach this step,we get
Chew T S MA1506-14 Chapter 5 116
1A 11 7 2 7 4
2 7 1 7 1
2 7 1 7 0
=
Consider linear system AX=B where
AppendixCont.
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Consider linear system AX B where
1 4 2
2 8 3
0 1 1
A= X=1
2
3
x
x
x
B= 232
1
Now the inverse of A exists. Hence we have
Chew T S MA1506-14 Chapter 5 117
AX B=1 1A AX A B =
1IX A B
=1X A B=
Therefore
AppendixCont.
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Therefore
1
2
3
x
x
x
11 7 2 7 42 7 1 7 1
2 7 1 7 0
232
1
=
2
3
4
=
(11/ 7)( 2) (2 / 7)(32) ( 4)(1) 2 + + =
( 2 / 7)( 2) ( 1/ 7)(32) (1)(1) 3 + + =
(2 /7)( 2) (1/ 7)(32) (0)(1) 4 + + =
Endof