Advanced Mathematics 2012 2013 Week 1 - Week 8 - 21 Jan 2013

7/26/2019 Advanced Mathematics 2012 2013 Week 1 - Week 8 - 21 Jan 2013

1/306

Advanced Mathematics

2012/2013

Version 21 January 2013

Utrecht UniversitySchool of Economics

Prof. dr. Wolter Hassink

Mathijs Janssen


2/306

ADVANCED MATHEMATICS CONTENTS

2

General information 3

Week 1 Introductory material 5Lecture 5Technical tutorial 30

Broad tutorial 35Take home assignments 40

Week 2 Linear algebra (I) 46Lecture 46Technical tutorial 72Broad tutorial 83Take home assignments 87

Week 3 Linear algebra (II) 91Lecture 91

Technical tutorial 118Broad tutorial 128Take home assignments 134Additional exercises 138

Week 4 Calculus (II) 139Lecture 139Technical tutorial 156Broad tutorial 166Take home assignments 170

Week 5 Optimization (I) 173Lecture 173Tutorials 190

Week 6 Optimization (II) and integrals (I) 201Lecture 198Technical tutorial 217Broad tutorial 230Additional exercises 236

Week 7 Integrals (II) and dynamic analysis (I) 238Lecture 238Technical tutorial 249Broad tutorial 272Take home assignments 274

Week 8 Dynamic analysis (II) 280Lecture 280Technical tutorial 297Broad tutorial 298Additional exercises 305


3/306

ADVANCED MATHEMATICS GENERAL INFORMATION

3

Assignments

From week 2: hand in your work. At random, three of these

assignments will be assessed.

From week 3: Two group assignments on broad economicquestions.

Assessment method

The average of the assessment of three, randomly chosenindividual assignments (24%);

Group assignment (16%);o

Assignment to team will be done by lecturers.

Individual end term exam, closed book (60%). For moreinformation about replacement and supplementary retake examssee the course manual.

Effort requirement

Six out of seven assignments should be handed in.


4/306

ADVANCED MATHEMATICS CONTENTS

4

Academic Skills

Problem solving: on average a sufficient grade (5.5 or higher) for

individual assignments.

o The justification of the method chosen, the systematically

presentation of both the steps in the problem solving stage

and the solution, and the interpretation of the results are all

aspects that count heavily. This means that solutions using the

problem solving skill extensively but with calculation errors

can be assessed sufficient.

Effective teamwork. 1) Sufficient contribution to team assignments.

2) Sufficient average grade (5.5 or higher) for team assignment. For

the assignment, for each of the team members the contribution to

the assignment must be shown in the paper.


5/306

ADVANCED MATHEMATICS LECTURE WEEK 1

5

ADVANCED MATHEMATICS SLIDES WEEK 1

Week 1 - Introductory material

Functional notation (domain, range etc.) Klein 2.1. [K 2.1.]

Graphs of univariate and multivariatefunctions

K 2.1.

Limits, continuity K 2.1.

Properties of functions: monotonous,convex, concave, injective, surjective,inverse, homogeneous

K 2.2.

Necessary conditions, sufficientconditions

K 2.2.

Discrete compounding K.3.1.

Exponential function K.2.3. K.3.2.

Rules of exponential functions K.2.3.

Multiple compounding per period K.3.2.

Continuous compounding K.3.2.

NPV K.3.2.Logarithm (as inverse of exponentialfunction)

K.3.3.

Rules of logarithms K.3.3.

Summations Supplementalmaterial


6/306


6

Sets

Definition:Set: collection of elements

Example 1:let

W= : the set of all non-negative integers {0,1,2...}W=

Set of positive integers: {1,2,...}W= W= : the set of all integers {..., 2, 1,0,1,2,...}W=

W= : the set of all rationalnumbers of the form /a b , where aandbare both integers

W= : the set of all realnumbers (includes both rational andirrational numbers).

Definition: Irrationalnumbers, e.g.:1

3, , ,2

e

Example 2:

let {0,1,2...,,10}S=

An element may belong to a set (or: may be a member of a set). Thusx S

Example 3:

integer 1 S integer 1 Irrational number (=3.141...) and e(=2,71...) However, 0.5 S and and e

Definition:Sub-set

Example 4: let {0,1}T=

Tis a sub-set of S: T S or : S T : inclusion symbol


7/306


7

Definition: union of sets

Example 5: S T S = All elements that belong to Sor T(or both).

Definition:intersection of sets:Example 6: S T T = Elements that belong to both Sand T.

Definition:empty set

Example 7: {9,10}V=

V T = The sets Vand Thave no elements in common. They are disjoint.


8/306


8

Functions

Definition: Function, mapping (or transformation): element of setXinto set Y

Function: :f X Y

SetX: domainSet Y: rangeNote: range can be broad.


9/306


9

Univariate functions

Definition: one member of domain is related to one member of range( )y f x=

Definition:xis argument of the functionf(x)

Example 8:2 3y x= +

Domain: Range:

Example 9:2 3 | |y x= +

Domain: Range:[2, )

Example 10:

2 3y x= +

Domain:[0, ) Range:[2, )

Example 11:

32y

x= +

Domain: (0, )

Range:(2, )

Example 12:

yx

= +

Domain: (0, )

Range:( , )

Definition: and are parameters


10/306


10

Multivariate functions

Definition: different independent variables and one dependentvariables

1 2( , ,..., )ny f x x x=

Subscript ofxrefers to the variable name.


11/306


11

Limits and continuity

Definition:Left-hand limit

lim ( )x a

f x

existsand is equal to LL for any arbitrarily small number thereexists a small number such that

| ( ) |Lf x L < for a x a < <

Definition:Right-hand limit

lim ( )x a

f x+

existsand is equal to RL for any arbitrarily small number thereexists a small number such that

| ( ) |Rf x L < for a x a < < +

Example 13

0lim ( )x

m k x mk

+ =

Example 14

lim 0x

k

mx h=

+

if 0m


12/306


12

Continuity

Definition:a function ( )f x is continuous atx=aif

lim ( ) lim ( ) lim ( )x a x a x a

f x f x f x +

= =

and the limit equals the value of the function at that point.

Example 15

Let 1

( )8

f xx

=

The left-hand and right-hand limit are unequal

8

1lim

8x x=

and

8

1lim

8x x+=

so that the function is discontinuous atx=8

Example 16

Let2

1( )

( 8)f x

x=

28

1lim

( 8)x x=

and

28

1lim

( 8)x x+=

So that the function has the same left-hand and right-hand limit at x=8.Howeverthe functionf(.) is not defined atx=8

Definition:the function has a vertical asymptote atx=8

It is not defined atx=8

The function value approaches plus infinity or minus infinity at thatpoint.


13/306


13

Properties of functions

Definition:

Lets have ( )f x and B Ax x>

( )f x is increasing if ( ) ( )B Af x f x

( )f x is strictly increasing if ( ) ( )B Af x f x>

( )f x is decreasing if ( ) ( )B Af x f x<

( )f x is strictly decreasing if ( ) ( )B Af x f x

Definition:a monotone function is either increasing or decreasing.Definition:a strict monotone function is either strictly increasing or

strictly decreasing.

Example 17

The function 2( ) 3( 1)f x x= + is not a monotone function. However, the

function ( ) 3( 1)f x x= + is a monotone function

Definition: Strictly monotone functions are one-to-one functions (orinjective functions).

If ( ) ( )A Bf x f x= then A Bx x=

Other formulation: ( ) ( )A Bf x f x whenever A Bx x

Definition:Any monotone function has an inverse function. Notation:

( )y f x= has the inverse function 1( )y f x=

Example 183( 1)y x= +

can be rewritten as1

( 3)3

x y=

Thus the function1

( 3)3

y x=

is the inverse function of 3( 1)y x= +


14/306


14

Definition: Composite function

Argumentxof the function ( )y f x=

is also a function ( )x g z=

so that ( ( ) )y f g z=

Property1( ( ))f f x x = and 1( ( ) )f f x x =

Example 19

3( 1)y x= + and1

( 3)3

y x= are inverse functions.

thus 13[ ( 3) 1]3

x x + =

and1

[3( 1) 3]3

x x+ =


15/306


15

Extreme values

Definition:Global maximum: largest value of function over range

Global minimum: smallest value of function over range

Example 20:

Minimum:

It says that we minimize the function 25 ( 8)x+ with respect tox. Theminimum function value (at the argumentx=8) is equal to 5.

2min 5 ( 8) 5x

x+ =

The minimum function value is at the argumentx=8:2arg min 5 ( 8) 8

x

x+ =

Maximum:

It says that we maximize the function 23 2( 9)x with respect tox.The function value (atx=9) is equal to 3.

2max 3 2( 9) 3x

x =

The argument at which the function has a maximum:2arg max 3 2( 9) 9

x

x =


16/306


16

Average rate of change of function

Definition: The average rate of change of the function ( )y f x=

over the closed interval [ , ]A B

x x is

( ) ( )B A

B A

f x f xy

x x x

=

Definition:

Secant line: line between the points ( , )A Ax y and ( , )B Bx y

where ( )A Ay f x= and ( )B By f x=

( ) ( )' ( ' )B AA A

B A

f x f xy y x x

x x

=

For any point ( ', ')x y on this line, 'x is within [ , ]A Bx x and 'y is

within [ , ]A By y


17/306


17

Concavity and convexity

Definition:A function is strictly concave in an interval if for any distinct points

Ax and Bx in that interval, and for all values in the open interval

(0,1)

( (1 ) ) ( ) (1 ) ( )A B A B

f x x f x f x + > +

Definition:A function is strictly convex in an interval if for any distinct points

Ax and Bx in that interval, and for all values in the open interval(0,1)

( (1 ) ) ( ) (1 ) ( )A B A B

f x x f x f x + < +


18/306


18

Necessary and sufficient conditions: some logic

Definition:WheneverPis true, Qis necessarily true.

Qis a necessary condition forP:Pis a sufficient condition for Q:

P Q

Read: It means that ifPthen Qor Qis the consequence ofP

Example 21:Xis a square Xis a rectangleA sufficient condition forXto be a rectangle is thatXbe a square.orA necessary condition forXto be a square is thatXbe a rectangle.

Example 22:

Person is healthy Person breathes without difficulty

Wrong implications(reverse implication of above):A person breathes without difficulty is necessarily healthy.andBreathing is a sufficient condition for a person to be healthy.

Example 23:25 25x x> >

25 25x x< >

Example 24:

0 0xy x= = or 0y=

Example 25:

0x= or 0y= 0xy =


19/306


19

Definition:

Pis a necessary and sufficient condition for Q:

P Q

and Q P

Read: It means that Pif and only if QP Q

Example 26:

0x= or 0y= 0xy =

Example 27:

5x< or 5x> 2 25x >


20/306


20

More on logic: Structure of a mathematical proof

Method 1 (direct proof):P Q

P: set of propositions. Also referred to as premise (what we know)Q: set of propositions. Also referred to as conclusion (what we wantto know)

Method 2 (indirect proof):Alternative structure of a proof:

P Q is equivalent to not notQ P

P Q

is equivalent to Q P

Example 28:Direct structure: If it is raining, the grass is getting wet.Indirect structure: If the grass is not getting wet, then it is not raining.


21/306


21

A menu of functions: power function

Definition:power function

( ) py f x kx= =

for whichpis referred to as the exponent of the function

Rules of exponents:0 1x = 1x x=

1pp

xx

=

/ nm n m

x x= a b a bx x x

+ = a

a b

b

xx

x

=

( )b

a abx x=

( )aa ax y xy=

aa

ax xy y

=

( 0)y

Definition:polynomial function2

0 1 2( ) ... n

ny f x a a x a x a x= = + + + +

Degree of the polynomial function: highest exponent of the function(=n)


22/306


22

A menu of functions: exponential function

Definition:exponential function

( ) xy f x kb= =

b: base of the function

Example 29:

lim 0xx

b

= if | | 1b <

and lim 1xx

b

= if | | 1b =

and lim xx

b

= if | | 1b >

Example 30:

lim 0xx

b

=If | | 1b >

Which is equivalent to1

lim 0xx b

= If | | 1b >


23/306


23

Summations

Definition:

The sum from i= 1 to i= 5 of ix :5

1 2 3 4 5

1

i

i

x x x x x x=

= + + + +

i: summation index (an integer).

This summation is the same as5

1 2 3 4 5

1

j

j

x x x x x x=

= + + + +

Example 31:6

2 2 2 2 2 2 2

1

1 2 3 4 5 6 91i

i=

= + + + + + = 2

0

1 1 1 1 21

( 1)( 3) 3 8 15 40j j j== + + =

+ +


24/306


24

Summations: properties

Additive property:

1 1 1( )

n n n

i i i i

i i ia b a b

= = =+ = +

Homogeneity property:

1 1

n n

i i

i i

ca c a= =

=

So that:

1

n

i

c nc=

=


25/306


25

Double summations

Property:

1 2

1 1 1 1 1 1 1...

m n m n n n n

ij ij j j mj

i j i j j j ja a a a a

= = = = = = =

= = + + +

or

1 2

1 1 1 1 1 1 1

...n m n m m m m

ij ij i i in

j i j i i i i

a a a a a= = = = = = =

= = + + +

thus

1 1 1 1

m n n m

ij ij

i j j i

a a= = = =

=


26/306


26

Calculating growth (1)

Growth from period tto period t+1: 1 (1 )t tX r X+ = +

Growth from period tto period t+n: (1 )

n

t n tX r X+ = +

Subscript: discrete time period

A variable X growths by the rate rcompounded ktimes during aperiod:

1 1

k

t t

rX X

k+

= +

Definition:Exponential e(irrational number)

1lim 1 2.71828182845...

k

ke

k

= + =

Example 32:With an annual interest rate of 100 percent, the value of $100 after oneyear of continuously compounded interest is

1lim 1 $100 $271.83

k

k k

+ =


27/306


27

Calculating growth (2)

Consequence:/ /

11 1 1/

r rk k r k r

r rk k k r

+ = + = +

/

/

1 1lim 1 lim 1 (2.71828...)

/

r rk r m

r r

k r me

k r m

+ = + = =

Thus:

( 1) ( )rX t e X t+ =

tis denoted in parentheses: moments measured in continuous time.

( ) ( )rnX t n e X t+ =

n: any real number

Present value: see book and tutorial


28/306


28

Logarithmic functions

Definition:xy b=

has a point in logarithmic form: log ( )by x=

Example 33log ( )b xb x=

Rule 1:

log ( ) log ( ) log ( )b b bxy x y= +

Rule 2:

log ( / ) log ( ) log ( )b b bx y x y=

Rule 3:

log ( ) log ( )b bx x =

Property:

loglog

log

WW

J

HJ

H=


29/306


29

Natural Logarithms

Rule 1: ln( )ze z=

Rule 2:

lnx

e x=

Rule 3: ln( ) ln( ) ln( )xy x y= +

Rule 4: ln( / ) ln( ) ln( )x y x y=

Rule 5: ln( ) ln( )zx z x=


30/306

ADVANCED MATHEMATICS TUTORIALS WEEK 1

30

ADVANCED MATHEMATICS TUTORIAL WEEK 1

Advanced Mathematics Tutorials week 1 - Solutions week 1

Exercises with an asterisk (

*) are meant to deepen the knowledge of the material of that

week. Hence, these exercises are less likely to be asked on the exam.

Technical tutorial (Wednesday)

Exercise 1.Consider the following graph of a functionf.

Graphically assess the domain and the range off, its limits 0, , ,x x A x B x C and its

continuity at these points. By graphically assessing, we mean that the exact function values donot matter, but the procedure followed should be clear.

Solution:The domain is the set on which the function is defined.fappears to be defined everywhere,except on the interval [-1,-0.5] and atx=0, wherefhas no values. So its domain is

/ ([ 1, .5] {0}) (this means the set , which denotes the real numbers, with the interval

[-1,-0.5] and the point x=0 cut out).The range is the set of outcomes of the function. In this case all numbers from seem to be

reached except for the small interval at about [ ]2, 1 . So the range of the function is/ ([ 2, 1]) . Note that we cant tell from the picture whether this is equal to the co-domain

off, i.e., if :f X Y , whether ( )Y range f = . For instance, Yhere could be the whole of ,

or it could be / ([ 2, 1]) . In the latter case, the function would be called onto, or surjective.

A function which is onto or surjective has range equal to its codomain.


31/306


31

For the limits and continuity we start with point C. Recall that a function is continuous if youcan draw it without lifting you pencil from the paper. Clearly around point Cthis is possible,so f is continuous at C. Continuity at a point means that the function value at that point is

equal to its limit, so lim ( ) ( ) 0.9x C

f x f C

= , where the last approximate equality is our guess

looking at the graph.

We turn to pointB. The function has a jump here, so it is not continuous at B. Furthermore,from the left the function tends to a different value than from the right (from the right it goesto, say, 2, whereas from the left it appears to go to something like 1.2). Therefore a limit atBis not defined.

At pointAthe function also has a jump, so it is again discontinuous. However from the left

and from the right it tends towards the same value, so it does have a limit: lim ( ) 1x A

f x

. Do

note that lim ( ) ( )x A

f x f A

.

Finally, at point 0 something strange happens. The function is not defined at this point.However, it does have a left-hand and a right-hand limit. From the left it goes to , from theright to . These values are different, so the limit at 0 is not defined. Even if it were, wecould not decide whether f is continuous at 0, because the condition

0lim ( ) (0)x

f x f

= is not

well-defined: f(0) does not exist.Hopefully this rather convoluted example displays all the conceptual pitfalls of limits andcontinuity.

Exercise 2.

LetAandBbe two sets and A B . We wonder ifx B . Would x A be a sufficient

condition for that? And a necessary condition? And if we knew x B and wonderedx A ?

Solution:

Ifx A , then certainly x B , so it is sufficient. However, it is not necessary, because xcouldbe inBwithout being inA.

Ifx B , then it might still be that x A , so it is not a sufficient condition. However, it is anecessary condition, for ifx B , then certainlyx A .

Exercise 3.Exercise 2.2.2. from Klein.Which of the following functions are one-to-one:

a)

A function relating countries to their citizensb)

A function relating street addresses to zip codesc) A function relating library call numbers to books.d)

A function relating a students identification number to a course grade in a specificclass.

Solution:

a) Not, because Pierre and Jacques are both from France, but not the same person.b) Notc) Don't know, I don't know what a library call number is. If there is only one call number per

book (and one book per library call number), then it will be one-to-one.d) Not, if two different students obtain the same grade.


32/306


32

Exercise 4.Exercise 3.1.4. from Klein.Suppose you are a farmer that stores grain in a leaky silo so that you lose 2% of your cropeach year as a result of dampness and rot. To obtain the future value ofX, you still obtain the

same formula (1 )nt n tX X r+ = +

If r= -2%, what is the value of t nX + as napproaches infinity? That is, what is the value of

lim lim (1 )nt n tn n

X X r+

= + . Will the value of t nX + ever equal zero?

Solution:

lim (0.98) 0ntn

X

= , however, 0 is never actually reached.

Exercise 5.Exercise 3.1.6. from Klein.Assume a firms net profits are $50 million in 2000 and are expected to grow at a steady stat

of 6% per year through the end of the next decade. How much would you expect the firm toearn in 2001? In 2003? Now assume that the firms profits have been growing at 6% since1997. If a negative value of ncan be interpreted as the number of time periods before t, howmuch did the company earn in 1998? Graph the path of income growth between 1998 and2003 and explain why the curve gets steeper over time.

Solution:

In 2001: 11.06 50 53 = In 2003:

31.06 50 59.5508... =

In 1998: 21.06 50 44.4998... =

Graph:

This actually looks very much like a straight line, so for sake of clarity, we also draw a graphup to 2030:


33/306


33

The line gets steeper, because at each point growth is 6% of profits, and profits are alwaysincreasing. Therefore growth is always increasing.

Exercise 6.Exercises 3.2.1. and 3.2.5. from Klein.

3.2.1.Using the formula of multiple compounding in one period1 (1 )

k

t t

rX X

k+ = + . Calculate

the value of 1tX + for the following values of rand k. Assume that 20tX = .

a)

r= 8%, k= 4b) r= 0.5%, k= 2c)

r= 10%, k= 365

3.2.5.Assuming thatX(t) = 75, determine the value ofX(t + n) for the following values of kand nusing the continuous compounding formula:

1

r n

t tX X e

++ =

a) n= 3, r= 9%b) n= 0.5, r= 2.5%c) n= -2, r= 11%d)

n= 0.25, r= 6%e) n= 3, r= 9%f)

n= 0.75, r= -3%

Solution:We do one from each, the point should be clear:

3.2.1. a) 41 (1 ) 20 (1.02) 21.649k

t t

rX X

k+ = + = =

3.2.5. a)0.27( ) ( ) 75 98.2473r nX t n X t e e+ = = =

Exercise 7.Exercise 3.3.1 from Klein.

a) 10log (100)10

b) ln( )ln x xe e

c)10 5

1log ( )x


34/306


34

d)2log ( )a b+

e) ln( )a bx cz e + +

f) 3ln(4 )x

e) 25

1ln [ ]x y

e

Solution:

a) 10log (100)10 100= b)

ln( )ln( ) 0x xe e x x = =

c)

10 105

1log ( ) 5 log ( )x

x=

d)Cannot be simplified further.

e) ln( )a bx cz e a bx cz + + = + +

f)

4

log(3 )x Unclear, has two interpretations:Either:

3log((4 ) ) 3log(4 ) 3(2log(2) log( ))x x x= = +

Or:4(log(3 ))x Which cannot be simplified further.

g)2 5 2

5

1ln( [ ] ) ln( ) ln([ ] ) 5 2(ln( ) ln( ))

5 2( ln( ) ln( ))

x y e x y x ye

x y

= + = + + =

+

*Exercise 8.Exercise 3.3.5. from Klein.The theory of consumer behaviour is one of the foundations of economic analysis. The linearlogarithmic utility function is one of the original functions developed to measure consumerutility and is still widely use by economists. It is written as

1

ln lnn

i i

i

u U q=

= =

where uis the index of utility, iq is the quantity of good i, and 0 1i< < . Transform the

function back to its original form, where Uis utility.

Solution:

1

log( )log( ) log( )log( )

1 1 1

( )

n

i i

i i i i i i

n n nqq qU

i

i i i

U e e e e q

=

= = =

= = = = =


35/306


35

Advanced Mathematics Week 1 - Broad tutorial (Friday of week 1)

Exercises with an asterisk (*) are meant to deepen the knowledge of the material of thatweek. Hence, these exercises are less likely to be asked on the exam.

*Exercise 1.Evaluate the following limit directly from the definition:

4lim ( )x

f x

Where 2( )f x x= . Would your answer change if2 if 4

( )0 if 4

x xf x

x

=

=

Solution:

Sincef(.) is clearly continuous, we want to find that 24

lim ( ) 4 16x

f x

= = . Recall the formal

definition of a limit. It should hold simultaneously that2

4lim ( ) 4 16x f x+ = = and2

4lim ( ) 4 16

xf x

= = . We consider the left-hand limit, the right-hand is similar. The idea is now

that for any number close to 16 we can find numbers close to, but smaller than 4, such that

their square is even closer to 16. That is, given a number 0> , we have to find a 0> such

that for numbersxsuch that 4 4x < < , we find2 16x < . This means that we have to

find as a function of .

If we can make this work for 1< , we can obviously also make it work for larger , so werestrict our attention to that case.

Let's try

1

8 = . Then, because 4 4x < }, and a square (2( )

A Bx x >0), so that the last term is negative.

which proves our result.

2b.Recall the definition of homogeneity:

The functionf(.) is homogeneous of degree kif, for >0, ( ) ( )k

f x f x =

We write out the definition:

2 2 2 2( ) ( ) ( )f x x x f x = = =

So we find thatf(.) is homogeneous of degree 2.

*Exercise 3.Supposef(x) andg(x) are both monotonous functions on the same domain. Show thata. f(.) +g(.) is also monotonous.

b.

And iff(x) andg(x) are both concave, can you show thatf(.) +g(.) is concave?c. And iff(x) andg(x) are both homogeneous, of degrees land mrespectively, can you then

show that ( ) ( )f x g x is homogeneous, and of what degree?

Solution:

3a. Monotonicity:

We know that, if A Bx x< , ( ) ( )A Bf x f x< and ( ) ( )A Bg x g x< , for any

, ( ) ( )A Bx x Dom f Dom g (meaning that Ax and Bx are both on the domain of bothf(.) andg(.)).

Now we have to prove that, if we let ( ) ( ) ( )h x f x g x= + that ( ) ( )A Bh x h x< .

Thus:

( ) ( ) ( ) ( ) ( ) ( )A A A B b Bh x f x g x f x g x h x= + < + =

This proves what we want.

3b. Concavity: We know that for any 0


38/306


38

and

( ) (1 ) ( ) ( (1 ) )A B A Bg x g x g x x + +

Now we want to show that, if we define again ( ) ( ) ( )h x f x g x= + , then

(1) ( ) (1 ) ( ) ( (1 ) )A B A B

h x h x h x x + + .

Thus we start with the left-hand side of equation (1) and we rewrite it in terms of functionsf(.)andg(.)

( ) (1 ) ( )

( ( ) ( )) (1 )( ( ) ( ))

( ) (1 ) ( ) ( ) (1 ) ( ) ( (1 ) ) ( (1 ) )

( (1 ) )

A B

A A B B

A B A B A B A B

A B

h x h x

f x g x f x g x

f x f x g x g x f x x g x x

h x x

+ =

+ + + =

+ + + + + + =

+

That proves what we want.

3c.Homogeneity:

We know that ( ) ( )lf x f x = and ( ) ( )mg x g x = .

Now we want that, if we define ( ) ( ) ( )k x f x g x= , then we want ( ) ( )pk x k x = for somep

still to be determined.

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )l m l m l mk x f x g x f x g x f x g x k x += = = = .

So we see that the function k(.) is homogeneous of degree l+m.

Exercise 4.Write out the following:

a)4 4

2 1

*2i

i j

i= =

b)

3 22

1 1

( )i j

i j= =

+

Solution:

a)4 4 4 4 4 4 4

2 1 2 1 2 2 2

2 2 ( 2 2 2 2 ) 4( 2 ) 4 ( 2 )i i i i i i i i

i j i j i i i

i i i i i i i i= = = = = = =

= = + + + = = =

2 3 44(2 2 3 2 4 2 ) 384 + + =

We could also have shown:4 4 4 4 4 4

2 3 4

1 2 1 2 1 1

2 2 (2 2 3 2 4 2 ) 96 4 96 384i i

j i j i j j

i i= = = = = =

= = + + = = =

Thus both outcomes are equal:4 4 4 4

1 2 2 1

2 2i i

j i i j

i i

= = = =

=


39/306


39

b)

( )

( ) ( ) ( )

3 2 3 2 32 2 2 2

1 1 1 1 1

2 2 2 2 2 2

( ) ( ) ( 1) ( 2)

(1 1) (1 2) (2 1) (2 2) (3 1) (3 2)

4 9 9 16 16 25 76

i j i j i

i j i j i i= = = = =

+ = + = + + + =

+ + + + + + + + + + + =

+ + + + + =

Again, one can show that3 2 2 3

2 2

1 1 1 1

( ) ( )i j j i

i j i j= = = =

+ = +

Note that always the indices have dropped out after you have evaluated the sums. They areonly useful within the sum and for that reason are sometimes called dummies.

Exercise 5.Let kbe some constant andf(.) some function. Show, or at least make clear, that

1 1

( ) ( )n n

i i

kf i k f i= =

= and1

n

i

k nk=

= .

Solution:

1 1

( ) (1) (2) ... ( ) ( (1) (2) ... ( )) ( )n n

i i

kf i k f k f k f n k f f f n k f i= =

= + + + = + + + =

1times

( ... )n

in

k k k k nk =

= + + + = ((((


40/306

ADVANCED MATHEMATICS TAKE HOME WEEK 1

40

ADVANCED MATHEMATICS TAKE HOME ASSIGNMENT

MATERIAL OF WEEK 1

Advanced mathematics

Solutions take home assignments of week 2 (on material of week 1)

* Exercise 1.The following are all graphs of functions . Determine whether they are one-to-one(i.e. injective) and whether they are onto (i.e. surjective).

a)

Solution:We check for injectivity. Clearly the function does not take the same value twice, so thefunction is injective.

We check for surjectivity. The range for the function is only about[ 3, ) , while it is given

that Y= . So the function is not surjective.


41/306


41

b)

Solution:We check for injectivity. We see that, for instance, both atx=2 and atx=-2f(x)=1, so thefunction is not injective.

We check for surjectivity. The range for the function is only about[ 3, ) , while it is given

that Y= . So the function is not surjective.


42/306


42

c)

Solution:We check for injectivity. We see that the function does not take the same value twice, so it isinjective.We check for surjectivity. We see that the range of the function is Y= , so it is surjective.

d)

Solution:We check for injectivity. We see that for instance atx=0 and atx=40 the function takes the

valuef(x)=0, so it is not injective.We check for surjectivity. We see that the range of the function is Y= , so it is surjective.


43/306


43

Exercise 2.

Write out:2 4

1 1

(2 3 )i j

i j= =

+

Solution:2 4 4 4

1 1 1 1

(2 3 ) (2 3 ) (4 3 ) (6 6 ) (6 6) (6 12) (6 18) (6 24) 14 6 84i j j j

i j j j j= = = =

+ = + + + = + = + + + + + + + = =

Or the other way around:2 4 2 2

1 1 1 1

(2 3 ) ((2 3) (2 6) (2 9) (2 12)) (8 30) (8 30) (16 30) 84i j i i

i j i i i i i= = = =

+ = + + + + + + + = + = + + + =

Of course, not all of these brackets are necessary, they are mostly to show what comes fromwhat.

* Exercise 3.Determine whether the following function is homogeneous. If it is, determine the degree.

3( ) ( )f x h x= , where ( )h x is homogeneous of degree 7. (Hint: if you find this confusing, first

try it with 7( )h x x= , which is a homogeneous function of degree 7.)

Solution:First the general problem:

We check if ( ) ( )mf t x t f x = for some m.3 3 3( ) (( ) ) ( )f t x h t x h t x = =

Now we know that his homogeneous of degree 7, i.e. 7( ) ( )h r y r h y = . Take 3r t= and3

y x= to find:3 3 3 7 3 21 3 21( ) ( ) ( ) ( ) ( )h t x t h x t h x t f x= = =

Sofis homogeneous of degree 21.

The hint is solved similarly, but now we take 7( )h x x= :3 3 3 7 21 3 7 21 3 21( ) (( ) ) ( ) ( ) ( ) ( )f t x h t x t x t x t h x t f x = = = = =

Which is not really easier, I suppose.

* Exercise 4.Of course you all know intuitively what the derivative of a functionf(x) is: it is the very smallchange that occurs inf(x) when you very slightly changex. The picture illustrates this. The

blue line is the graph of the functionf(x). If you take xever smaller, you will approach evermore closely the slope of the red line of the derivative.


44/306


44

For this approaching ever more closely, we naturally think of the limit (in fact, it was in thecontext of derivatives that the notion of limit was first developed).

We define:0

( ) ( )( ) lim

x

d f x x f xf x

dx x

+ =

. (Note that ( ) ( ) ( )f x x f x f x+ = .)

Now you must prove that for 2( )f x x= it indeed holds that ( ) 2d

f x xdx

= by writing out the

limit. Do this in three steps: first, before evaluating the limit, observe that2( ) ( ) 2( )f x x f x x x x+ = + . Then, still before evaluating the limit, show that

( ) ( )f x x f x

x

+

simplifies to 2x x+ . Then evaluate the limit

0lim 2x

x x

+ directly from

the definition (either doing it from the left or the right hand side is enough).If you succeed, you have proved a rule that youve already known for a long time. Isnt thatfun!

Solution:So, let the fun begin:We first write out the definition. To emphasize that x is a single number and not amultiplication of and x, I will now define x=h. (So this is just giving it a new name).

2 22

0 0

( ) ( ) ( )lim limh h

d f x h f x x h xx

dx h h

+ + = =

Now, before we touch the limit, we just apply algebra to what is inside the limit. This is

allowed, because were basically not changing the expression over which we take the limit.2 2 2 2 2 2

0 0 0 0

( ) 2 2

lim lim lim lim(2 )h h h h

x h x x xh h x xh h

x hh h h

+ + + +

= = = + Strictly speaking, for our last step, we should observe that h

0,becauseotherwiseit would


45/306


45

not be allowed to divide by h. However, if you recall the definition of a limit, you will see that

hnever actually takes the value to which it goes, i.e. 0 in this case. Therefore our last step is

valid and is obtained by dividing both the denominator and the numerator by h. For this last

expression we can now apply the definition of a limit.

How does it work again? In general, the idea is that, if

lim ( )x a

c f x=

thenf(x) will get ever closer to c, asxgets closer to a. This was formalised thus: if you say

how close to cyou want to get, then I should be able to give a distance from a so that you will

indeed get that close or closer to c. You saying how close you want to get is setting an , me

providing you with this distance is picking the .

Lets apply this to our case. The function over which we are taking a limit is 2x+h, wherexis

now just some given number. Intuitively, we would expect that as h goes to zero, this function

will just go to 2x. So lets make that our guess for the limit.

Now, since this function is very simple, we can do the right hand limit and the left hand limit

at the same time.

You provide me with >0 and I decide to pick = (Why? It turns out that it works. This is

basically backward engineering.) Then if I only look at hwhose distance to 0 is less than ,

i.e.

0 h h =

For which tis 1' 0x A x >

Does it matter for these results that A is a symmetric matrix? So that 'A A=

So, check whether the results is different for e.g.1 1

2B

t

=

?

Solution:

a) det( ) 1 2 2 4A t t= =

The determinant ofAis negative if 4t -4 or t< -12

We check the results for1 1

2B

t

=

1 11

2 12

tB

t

= +

[ ] [ ]1 1 1 11 1 6

' 1 2 4 12 1 2 22 2 2

t tx B x t

t t t

+= = + = + + +

It is positive if t> -2 or t< -6

We do not observe any major difference betweenAandB.


91/306


91

ADVANCED MATHEMATICS SLIDES WEEK 3

Week 3 - Linear Algebra (II)

Klein: Chapter 5 (it excludes Cramers rule)

Determinant of matrix Section 5.1

Rank of matrix See lecture slides

Eigenvalues and eigenvectors Section 5.3

Diagonalization of a matrix Section 5.3

Cramers rule is no part of the material


92/306


92

Linear Maps

As we have seen last week, the matrix

11 1

1

n

m mn

a a

a a

can be

interpreted as a function (.): n mf , in particular:

1 11 1 1

1

( )

n

n m mn n

x a a x

f

x a a x

=

.

We now look at functions (.): n mT such that, for

, ,n c x y , ( ) ( ) ( )T T T+ = +x y x y and ( ) ( )T c cT =x x .Functions that satisfy both criteria are called linear functions (or linearmaps or linear transformations).

Example 1:

3 2

(.):T ,

1

1 2

2

1 3

3

( ) 2

xx x

T x x xx

+

= , then

1 1 1 1

1 1 2 2

2 2 2 2

1 1 3 3

3 3 3 3

1 1

1 2 1 2 1 2 1 2

2 21 3 1 3 1 3 1 3

3 3

( ) ( )2( )

( ) (2 2 ) 2

w y w yw y w y

T w y T w yw y w y

w y w y

w yw w y y w w y y

T w T yw w y y w wx y yw y

+ + + + + = + = = + + +

+ + + + +

= + = + + )

And

1 1 1

1 2 1 2

2 2 2

1 3 1 3

3 3 3

( ) ( ) ( )2 2

y cy ycy cy y y

T c y T cy c cT ycy cy y y

y cy y

+ + = = = =


93/306


93

Mapping and matrices

Theorem:Any linear map can be represented by a matrix and any matrix is a

linear map. That is, they are the same thing.Matrix representation of a linear map:

Let 1, , ne e be the unit vectors inn , then a matrix representation of

a linear map (.): n mT is:

1 2( ) ( ) ( )nA T e T e T e

=

This shows (if we proved it) that every linear map has a matrixrepresentation. The other way around (that every matrix represents alinear map) is done in the tutorial.

Example 2We represent the linear map from example 1 as a matrix:

01 0 1 0 1 1

() , ( 1 )1 2 0 1 0 2 0 00

00 0 0

( 0 )0 2 1 2

1

T T

T

+ +

= = = =

+ = =

So the map Tis represented by the matrix:

1 1 01 0 2

Lets check:

1

1 2

2

1 3

3

1 1 0

21 0 2

xx x

xx x

x

+ =

This is indeed our original map T.


94/306


94

Example 3 of linear mapping: a counter clockwise rotation of 90

degrees

We are interested in a matrixA:2 2 , which represents a counter

clockwise rotation.

The matrix0 1

1 0A

=

can be understood as follows:

First column of the matrixA.Rotating the unit vector1

0

counter clockwise we get0

1

.

Second column of the matrixA.Rotating the unit vector0

1

counterclockwise we get1

0

,

Thus, the rotation is represented by:

0 1

1 0A

=

Thus the matrix can be used to rotate any vector counter clockwise.

For instance the vector2

1

:

0 1 2 1

1 0 1 2Ax y

= = =

The rotation implies that the vector2

1x

=

is perpendicular to its

mappingy =1

2


95/306


95

Because:

1) 2 ( 1) 1 2 0x y = + = (the inner product ofxandyis zero)

2) 5x = and 5y = So that both vectorsxandyhave equal

length.

Implication 1

The matrix0 1

1 0B

=

represents a clockwise rotation: Thus

1 0

0 1B

=

and0 1

1 0B

=


96/306


96

Now we show graphically that ( ) ( ) ( )T T T+ = +x y x y . It can be seen

in the picture that it does not matter whether you first rotate yourvectors and then add them, or the other way around, i.e. first adding

them and only then rotating them.


97/306


97

Example 4: Interpretation of matrix multiplication (I)

We have two matricesAandB:

0 1

1 0A

=

3 0

0 3B

=

Question: what is the meaning of ABx?

1)

First multiplicationBx: Implies a multiplication of both elementsof the vectorxby a factor 3.

3 0

0 3Bx x

=

3 0 1 31

0 3 0 0Be

= =

and

3 0 0 02

0 3 1 3Be

= =

2) Second multiplicationABx:Counter clockwise rotation

Thus :

0 1 3 0 0 3

1 0 0 3 3 0C AB

= = =

Mapping C:

Step 1:Bx: three times larger length of the vector

Step 2:ABx: Counter clockwise rotation by 90 degrees


98/306


98

We apply the mapping on

2

1x

=

0 3 2 3

3 0 1 6

=

2

1x

=

is perpendicular toy =

3

6

Because:1) 2 ( 3) 1 6 0x y = + =

2) 5x = and 45 3 5y = =

so that 3y x=


99/306


99

Finally, we show that ( ) ( )T c cT =x x .


100/306


100

Also here, it does not matter whether we first multiply our vector witha number and then rotate it or the other way around.


101/306


101

Multiplication by a common factor

Example 5:

3 3

Ax= 3x

3 0 0 2 6

0 3 0 1 3

0 0 3 3 9

=

Example 6:

3 3

Ax x= ,

0 0 2 2 20 0 1 1

0 0 3 3 3

= =


102/306


102

Interpretation of matrix multiplication (II)

Example 7:

To make the length of vector in 3 three times larger:

3 0 0

0 3 0

0 0 3

A

=

The product ofAAmakes the length of a vector 9 times larger:

2

2

2

3 0 0 3 0 0 3 0 0

0 3 0 0 3 0 0 3 0

0 0 3 0 0 3 0 0 3

=

The product ofAAAmakes the length of a vector 27 times larger:

3

3

3

3 0 0 3 0 0 3 0 0 3 0 0

0 3 0 0 3 0 0 3 0 0 3 0

0 0 3 0 0 3 0 0 3 0 0 3

=

Et cetera


103/306


103

Space and subspace

3 has a dimension of 3.

It means that it can be spanned by three independent vectors atmaximum.

Example 8:

In 3 the line spanned by

2

1

3

has a dimension of 1. The line is

referred to as a subspace of 3 . It means that

2

1

3

, , is part of

this subspace.

Example 9:

In 3 the sub-space spanned by the vectors

2

1

3

and

1

0

2

has a

dimension of 2. A vector in this subspace can be characterized as

2 2 1

1 1 0

3 2 3 2

+ = + +

,


104/306


104

Example 10:

2 1 0

1 0 0

3 2 0

A

=

will be a mapping into the subspace spanned by the

vectors

2

1

3

and

1

0

2

. Thus, the image of each vector will be part of

this subspace. The matrixAis non-invertible. It has a rank of 2.


105/306


105

Interpretation of the determinant of a matrix

The determinant measures how much a linear map blows up theimage.

Consider the map in the figure. It maps the blue square into thegreen one.

The determinant is the ratio between the area of the green squareand the blue one.

Since the blue one is the unit square, while the green has area 8, thedeterminant of the matrix associated with this map is 8.

Negative determinants occur when positive vectors are mapped intonegative ones.

Note that for any shape to which we apply a linear map, thisrelation between area before and after the map is the same. Inhigher dimensions we do not talk of the area but of the volume.


106/306


106


107/306


107

Thus interpretation of determinant for 2 (no proof):

Lets consider the mapping

a c

b d

We consider the size of the area spanned by:

1)0 0

0 0

a c

b d

=

2)1

0

a c a

b d b

=

3)1

1

a c a c

b d b d

+ = +

4)0

1

a c c

b d d

=

It can be shown that the area spanned by0

0

,a

b

,a c

b d

+ +

,c

d

is ad bc,which is the determinant of the matrix

a c

b d


108/306


108

Example 11:The determinant of

3 0

0 3

A =

equals nine. Thus det(A)=9. (The area of the mapping spanned by the

four vectors0

0

,3

0

,3

3

,0

3

equals 9).

For the matrix:

1/ 3 0

0 1/ 3

B =

we have det(B)=1/9

Properties of determinants:1)It can be shown that in general det(AB)=det(A)det(B)

2)It can be show that 1 1

det( )det( )

AA

=

Note that for the particular example 11AB=I

3 0 1/ 3 0 1 0

0 3 0 1/3 0 1

=

And that the determinant of the unit matrix equals 1.


109/306


109

Determinants in 3

For 3 the determinant ofA

11 12 13

21 22 23

31 32 33

a a aA a a a

a a a

=

can be calculated as follows:

The determinant can be calculated for each of the three rows or eachof the three columns.

1) For the first row of the matrix:

22 23 21 23 21 22

11 12 13

32 33 31 33 31 32

| |a a a a a a

A a a aa a a a a a

= +

For which22 23

32 33

a a

a ais the minor of 11a (the determinant of the sub-

matrix of 11a )

21 23

31 33

a a

a ais the minor of

12a

21 22

31 32

a a

a ais the minor of

13a

2)One can also take the second row of the matrix A:

12 13 21 23 11 12

21 22 23

32 33 31 33 31 32

| |a a a a a a

A a a aa a a a a a

= +

etc


110/306


110

Thus for each of the minors it is multiplied by:

11 12 13

21 22 23

31 32 33

a a a

A a a aa a a

=

Conclusion:The determinant of an upper-diagonal matrix

11 12 13

22 23

33

0

0 0

a a a

A a a

a

=

equals 11 22 33| |A a a a=

And for a diagonal matrix:

11

22

33

0 0

0 00 0

a

A aa

=

Hence: the determinant ofAis equal to zero if one of the diagonalelements equals zero.

11 22 33| |A a a a=


111/306


111

Example 12

The determinant of the matrix

2 1 1

1 4 4

1 0 2

A

=

equals 6.

For a 3x3 matrix, there are six possibilities to calculate thedeterminant of the matrix:

First row ofA:

4 4 1 4 1 4

2 1 1 60 2 1 2 1 0

+ =

or (second row ofA):

1 1 2 1 2 11 4 4 6

0 2 1 2 1 0 + + =

or (third row ofA):

1 1 2 11 2 6

4 4 1 4+ =

or(first column ofA):

4 4 1 4 1 42 1 1 6

0 2 1 2 1 0

+ =

or(second column ofA):

1 4 2 11 4 6

1 2 1 2

+ =

or(third column ofA):

1 4 2 1 2 11 4 2 6

1 0 1 0 1 4+ + =


112/306


112

Example 13

The determinant of the matrix

1 0

2 2

0 1

a

A a

a

=

equals 3 4a a .


113/306


113

Eigenvalues and eigenvectors

Compute the eigenvalues of the nx nsquare matrixAas follows:

Ax x=

( ) 0A I x =

For which x is a non-zero vector, andIis an nx nidentity matrix. Inorder for a non-zero vectorxto satisfy this equation, ( )A I must

not be invertible. If ( )A I has an inverse:

1 1( ) ( ) ( ) 0A I A I x A I = or

1( ) 0x A I =

thusx= 0.

The matrix is non-invertible if its determinant is zero:

| | 0A I =

IfAis a 2 x 2 matrix:11 12

21 22

a aA

a a

=

Then | | 0A I = becomes

11 12

21 22

0a a

a a

=

The characteristic equation is a second-order polynomial:

11 11 21 12( )( ) 0a a a a =

or2

11 22 11 22 21 12( ) ( ) 0a a a a a a + + =

For which: the trace of the matrixAis defined as the sum of the


114/306


114

diagonal elements: 11 22tr( )A a a= +

11 22 21 12| |A a a a a=

1,2

tr (tr )2 4 | |

2

A A A

=

Thus two solutions: two real eigenvalues at maximum


115/306


115

A matrix is diagonalizable:

The square matrix can be diagonalized as follows:

AP P=

For which is a diagonal matrix with the eigenvalues ofAon themain diagonal. For a 2 x 2 matrixA, the diagonal matrix is

1

2

0

0

=

P is a matrix that is spanned by the eigenvectors

[ 1, 2]P p p= .

For which 1 corresponds to the vector 1p and 2 corresponds to

the vector 2p

Notation:The diagonalization AP P= can also written as

a) 1A P P= b)

1P AP =


116/306


116

Example 14

The matrix2 2

2 5A

=

has the eigenvalues 1 1 = and 2 6 =

1 00 6

=

The eigenvectors belonging to

1 1 = are

21

1p

=

and1

22

p =

Note that the eigenvectors are orthogonal (because the matrixAissymmetric, what we will not proof here)

Example 15

The matrix2 4

1 1A

=

has the eigenvalues 1 3 = and 2 2 =

1 0

0 6

=

The eigenvectors belonging to

1 1 = are 41

1p =

and 12

1p =

for

2 2 =


117/306


117

Example 16 (it is hard example because of the third order

polynomial characteristic equation)

The matrix

2 1 1

2 3 4

1 1 2

A

=

has the characteristic equation

2 1 1

| | 2 3 4 ( 1)( 1)( 3) 0

1 1 2

A I

= = + =

The eigenvalues are 1, -1 and 3.

The associated eigenvectors are:

1 1 = :

1

1 1

0

p

=

2 1 = :

0

2 1

1

p

=

2 3 = :

2

3 3

1

p

=


118/306


118

Wrapping up: under which conditions is a matrixAinvertible?

Think of matrices as maps, the they are invertible if they are both one-to-one (injective) and onto (surjective).

This map is not injective.

This map is not surjective.


119/306


119

This map is both injective and surjective, and is therefore invertible.

Thus: A square matrixA n n is invertible if :

The matrixAis of full rank n;

The columns of the matrixAare linearly independent.

None of the real eigenvalues is equal to zero.

The determinant ofAis nonzero.


120/306


120


Exercises with an asterisk (*) are meant to deepen the knowledge of the material of that

week. Hence, these exercises are less likely to be asked on the exam.

Technical tutorial of week 3 - SolutionsExercise1.

Let A =

1 0

2 1

0 1 1

t

t

For what value of tdoesAhave an inverse?

Solution:The matrix no inverse if

1 0

2 1 0

0 1 1

t

t =

Which gives

1 2 11 0

1 1 0 1

tt+ =

So that(1 ) 2 0t t + =

Thus the matrixAhas an inverse if 1t

* Exercise2.

Find the determinant of

11 1

1

n

m mn

a a

a a

.

Solution:

We explained the procedure in class.

* Exercise3.Given a parallelepiped C of a certain volume and a linear map T with associated matrix A,find Vol(T(C)).

Solution:

( ( )) det( ) ( )Vol T C A Vol C =

* Exercise4.Consider the following maps and show that they are linear, without deriving their matrix

representation. Also derive and show their eigenvectors (if any).a) Blow-up of a vector along the x-axis by 100%, while the y-axis remains unchanged.


121/306


121

b)

Projection of a vector on the y-axis.

c) A counter-clockwise rotation of a vector by 90 .

Solution:a)

In the figure we drew the transformation for a specific vector. Note that algebraically, this

transformation amounts to 1 2 1 2: ( , ) (2 , )T a a a a . In green and yellow are two eigenvectors

for this transformation, we return to them shortly. First we show that the map T is linear. For

that we have to show that ( ) ( ) ( )T T T+ = +a b a b and ( ) ( )T r rT =a a . We do this both

graphically and algebraically.

In the figure we constructed ( )T +a b and it can be seen to be equal to ( ) ( )T T+a b (what does


122/306


122

this mean? Try to do the transformation T on +a b and see that it gives the same result asadding ( ) ( )T T+a b .) Algebraically, we have:

1 2 1 2 1 1 2 2 1 1 2 2

1 2 1 2 1 2 1 2 1 1 2 2

( ) (( , ) ( , )) ( , ) (2( ), )

( ) ( ) ( , ) ( , ) (2 , ) (2 , ) (2( ), )

T T a a b b T a b a b a b a b

T T T a a T b b a a b b a b a b

+ = + = + + = + +

+ = + = + = + +

a b

a b

So they are equal.

Again, we see in the figure that ( ) ( )T r rT =a a . Algebraically we have:

1 2 1 2 1 2

1 2 1 2 1 2

( ) ( ( , )) ( , ) (2 , )

( ) ( , ) (2 , ) (2 , )

T r T r a a T ra ra ra ra

rT rT a a r a a ra ra

= = =

= = =

a

a

We see again that they are equal.For the eigenvectors, we return to the first figure. Two examples of them are drawn in greenand yellow. First consider a vector along the y-axis. What would happen to it under thistransformation? Absolutely nothing. So it is an eigenvector with eigenvalue 1.

(0, ) 1 (0, )T y y=

Now consider a vector along the x-axis. What will happen to it under T? It will get doubled.So it is an eigenvector with eigenvalue 2.

( ,0) (2 , 0) 2 ( ,0)T x x x= =


123/306


123

b)

To make my life easier, all in one picture this time. A projection simple takes any vector and

only keeps the y-part of it: : ( , ) (0, )T x y y . Form the picture it is again clear that the map

is a linear one. Note that this time we checked ( ) ( )T r rT =

a a for r


124/306


124

c)

The figure shows that the first condition for linearity holds.


125/306


125

This figure shows that the second condition for linearity also holds. We drew it here for r


126/306


126

2

2 0 0

( ) 1 3 5 0 (2 )((3 )( 1 ) 5) (2 )( 2 8)

1 1 1

(2 )(4 )( 2 )

P

= = = = =

So we find 1 2 32, 4, 2 = = = . (You have to be lucky to be able to solve a cubic equation

this way. Dont worry; on an exam you will always be lucky.)

Now we find the associated eigenvectors 1 2 3, ,v v v by solving the equation:

1 1( ) 0A I =v

We solve by sweeping:

2 2 0 0 0 0 0 0 0 1 1 5 0

1 3 2 5 0 1 1 5 0 ~ 0 0 1 0

1 1 1 2 0 1 1 3 0 0 0 0 0

=

So we find 1

0

pp

=

v for any p, with associated eigenvalue 1 2 = .

We check our result:

2 0 0 2

1 3 5 3 2

1 1 1 0 0

p p p

p p p p

p p

= =

Such a relief!

We move on to 2v with associated eigenvalue 2 4 =

2 4 0 0 0 2 0 0 0 1 0 0 0 1 0 0 0

1 3 4 5 0 1 1 5 0 0 1 5 0 0 1 5 0

1 1 1 4 0 1 1 5 0 0 1 5 0 0 0 0 0

=

So we find 2

0

5q

q

=

v for any q, with associated eigenvalue2 4 =


2 0 0 0 0 0

1 3 5 5 20 4 5

1 1 1 4

q q q

q q q

= =

Hurrah!

We move on to 3v and 3 2 = .

2 2 0 0 0 4 0 0 0 4 0 0 0 1 0 0 0

1 3 2 5 0 1 5 5 0 0 5 5 0 0 1 1 0

1 1 1 2 0 1 1 1 0 0 1 1 0 0 0 0 0

=

So we find 3

0

rr

= v for any r, with associated eigenvalue 3 2 = .


127/306


127


2 0 0 0 0 0

1 3 5 2 2

1 1 1 2

r r r

r r r

= =

Again it works out and we have found all our eigenvectors.


128/306


128

Broad tutorial (Friday)

* Exercise 1.

Consider the Markov matrix

20 0

3

1 3 1

6 4 3

1 1 2

6 4 3

A

=

. Diagonalize it and use your result to determine

the long term state of the population (no matter what the starting state was) by calculating

1

2

3

lim kk

x

A x

x

for general population

1

2

3

x

x

x

.

Solution:

BackgroundAt the tutorial there was more explanation about the background of Markov transitionmatrices. It describes transition in the labour market, for which there are three states (e.g. state1: employment; state 2: unemployment; state 3: non-participation).

The matrix describes the probabilities in the transitions across the three states between periodtand period t+1.

Note that the numbers in the matrix should be read as conditional probabilities.2/3 = Pr(employed in period t+1 | someone was employed in period t)1/6 = Pr(unemployed in period t+1 | someone was employed in period t)1/6 = Pr(non-participant in period t+1 | someone was employed in period t)These probabilities add up to one exactly.

3/4 = Pr(unemployed in period t+1 | someone was unemployed in period t)1/4 = Pr(non-participant in period t+1 | someone was unemployed in period t)These probabilities add up to one exactly

2/3 = Pr(non participant in period t+1 | someone was non participant in period t)1/3 = Pr(unemployed in period t+1 | someone was non participant in period t)These probabilities add up to one exactly

Note that 1 2 3 1x x x+ + =

Thus

1

2

3

20 0

3

1 3 1

6 4 3

1 1 2

6 4 3

x

Ax x

x

=

is informative about the states in period t+1

To diagonalize the transition-matrixA, we have to start by finding the eigenvectors and


129/306


129

eigenvalues. We look at the characteristic equation (we showed in the lecture why thisequation matters):

2

20 0

3

1 3 1 2 3 2 1 2 1 17 1

( ) ( )(( )( ) ) ( )( ) 06 4 3 3 4 3 12 3 2 12 12

1 1 2

6 4 3

P

= = = + =

22 20 ( )(5 17 12 ) ( )(1 )(5 12 )3 3

= + =

This gives us three eigenvalues:1 2 3

2 51, ,

3 12 = = = . (You have to be lucky to be able to

solve a cubic equation this way. Dont worry; on an exam you will always be lucky.)

To get the associated eigenvectors1 2 3, ,v v v , we use the equation:

1 1( ) 0A I =v W sweep:

2 1 11 0 0 0 0 0 0 0 0 0

3 3 3

1 3 1 1 1 1 41 0 0 0 1 0

6 4 3 6 4 3 3

1 1 2 1 1 1 0 0 0 01 0 0

6 4 3 6 4 3

=

So 1

0

4

3

r

r

=

v for any r. Of course, if 1v is to represent a population, then 17

r= .

We check if 1v is indeed an eigenvector with eigenvalue 1:

20 0

3 0 0 01 3 1

4 3 1 46 4 3

3 2 31 1 2

6 4 3

r r r r

r r r r

= + = +

So it is as we wanted.

The eigenvector for2

2

3 = :

2 20 0 0 1

0 0 0 0 1 2 03 32

1 3 2 1 1 1 10 0 0 1 2 0

6 4 3 3 6 12 30 0 0 0

1 1 2 2 1 10 0 0

6 4 3 3 6 4

=


130/306


130

So 2

3

2

p

p

p

=

v for any p. We check if 2v is indeed an eigenvector with eigenvalue2

3.

2

0 0 2 23 3 31 3 1 1 3 1 4 2

2 26 4 3 2 2 3 3 3

1 1 2 1 1 2 2

6 4 3 2 2 3 3

p pp p

p p p p p p

p p

p p p p

= + + = =

+ +

So again we made no error in calculation.

Finally we solve for the eigenvector associated with3

5

12 =

2 5 10 0 0 0 0 0

1 0 0 03 12 4 1 0 0 01 3 5 1 1 1 1 1 1

0 0 0 0 0 1 1 06 4 12 3 6 3 3 3 3

0 0 0 01 1 2 5 1 1 1 1 1

0 0 0 06 4 3 12 6 4 4 4 4

=

So 3

0

q

q

=

v for any q. We check if 3v is indeed an eigenvector with eigenvalue5

12.

2

0 0 0 03 0 01 3 1 3 1 5 5

6 4 3 4 3 12 12

51 1 2 1 2

126 4 3 4 3

q q q q q

q q

qq q

= = =

Yippee.Now were almost ready to diagonalize our matrix. Recall that we want to write

1A CDC

= , where Cis the matrix of eigenvectors, so

0 3 0

4 2 1

3 1 1

C

=

, where we picked

easy values for r,p and q, andDis the diagonal matrix of eigenvalues, so

1 0 0

20 0

3

50 0

12

D

=

.

We still need to find 1C . Because we do not show how to find an inverse of a matrix in thiscourse its not hard, but we can only do so much we simply postulate that


131/306


131

1

1 1 1

7 7 7

10 0

3

2 3 421 7 7

C

=

and check if this is indeed true:

1

1 1 1

7 7 70 3 0 1 0 01

4 2 1 0 0 0 1 03

3 1 1 0 0 12 3 4

21 7 7

C C

= =

How good of us.

Finally we have diagonalizedA: 1A CDC= .Now we want to use this to make it easy to raise A to a certain power. Notice that:

1 1 1 1 1 1( ) ( ) ( ) ( ) ( )k k k

k

A CDC CDC CDC CDC CDC CD C

= = =((((((((((((((((

So we only have to raise D to the power k, and D is a diagonal matrix, so:

1 1

1 0 0

20 0

3

50 0

12

k

k

k k

k

A CD C C C

= =

What we wanted was to take the limit of this for to infinity, to see what would happen afterinfinitely many periods, i.e. in the long run. But, because two of our eigenvalues are smallerthan one, they tend to zero as k tends to infinity. So:

1 1

1 1 11 0 0

7 7 71 0 0 0 3 0 1 0 02 1

lim lim 0 0 0 0 0 4 2 1 0 0 0 0 03 3

0 0 0 3 1 1 0 0 02 3 45

0 021 7 712

1 1 1

0 3 0 7 7 7

4 2 1 0 0 0

3 1 1 0

k

k

k

k k

k

A C C C C

= = = =

0 0 0

4 4 4

7 7 70 0

3 3 3

7 7 7

=

So now we can calculate:


132/306


132

1

2 1 2 3

3

1 2 3

0 0 0 0 0

4 4 4 4 4( )

7 7 7 7 7

3 3 3 3 3( )7 7 7 7 7

x

x x x x

x

x x x

= + + =

+ +

The last equality follows that a population vector has 1 2 3( ) 1x x x+ + = . So in the long run the

population will be divided over states 2 and 3 in the proportions 4:3, while nobody will be instate 1. (Can you understand just by looking at matrixAwhy that might be the case?)Finally, not that the long run population vector that we found is also an eigenvector witheigenvalue 1. That is no coincidence: it is almost always the case with Markov chains, in factalways if the long-run state is well defined. The reason is as follows: In the long run, weexpect a steady state, so nothing changes anymore. So we want a vector such that, ifAworkson it, we get our vector back. But that is just an eigenvector with eigenvalue 1.Dont get angry, we did not sweat for nothing. Although it is true that it is much easier to findthe long run state by looking for an eigenvector with eigenvalue 1, our method is the only way

( I know of) to fairly easily find kA for any large k.

*Exercise 2.We show that the determinant-volume formula holds in a special case and discuss the general

proof.

Solution:

We start with a unit square C, characterized by the vectors (1,0), (0,1) and investigate what

happens under transformation

2 2

:T with associated matrix

a c

b d

.If we let Twork

on our two vectors, we get

1

0

0

1

a c a

b d b

a c c

b d d

=

=

So our transformation on the unit square looks like this:


133/306


133

Now we know the area of the unit square is 1, so to calculate the determinant of the matrix, allwe have to do is calculate the area of the resulting parallelogram. To calculate this, we needone geometric fact, which we now illustrate.

The parallelogram with the blue sides has the same area as the parallelogram with the redsides. In general, if you keep one side of a parallelogram fixed and you move the opposingside along a parallel line, the area of the resulting parallelogram is the same as that of theoriginal.We now use this fact to transform our parallelogram given by (a,b) and (c,d) into a moremanageable one with the same area. We actually use it twice, to transform it into a rectangle:


134/306


134

So we see that the rectangle given by (p,0) and (0,q) has the same area. Clearly that area equalpq. So what we have still to do is calculate p and q. We start withp. What did we do in thefirst step, the first shift of parallelograms? We took our point (a,b) and went in the direction of

(c,d) until we reached the x-axis, so we have ( ) ( ) ( )0a b r c d p = for some r to be

determined. We know 0 b

b rd r d

= = , so bc

p a rc ad

= = .

Thats one out of the way. Actually, finding q is easier. We see in the figure that going from(c,d) to (0,q) is a horizontal shift, so the y-coordinate does not change: q=d.

So the area of our rectangle and therefore also our original parallelogram is( )

bcpq a d ad bc

d= =

This is indeed the determinant formula for the two-dimensional case.Of course, this is no proof of the formula in general. For that we would have to show that itholds for all shapes we could start with, not just the unit square. The way to do that is not byextending the argument we gave above (just imagine doing this for general parallelograms inhigher-dimensional spaces). Instead what mathematicians do is very different: they look atvolume as a function of a shape and show that is must have certain properties (for instance, ifyou translate a shape, its volume does not change). Then they show that there can be only onesuch function. And then they show that the determinant also has these properties. Then they

can conclude that determinant indeed gives the volume of a transformation. We wont tracetheir steps here, as that would take as much too far afield, but you might be interested to seehow you can handle such a seemingly awesome problem.

* Exercise 3.

Prove that 1 1

det( )det( )

AA

= (if 1A exists of course).

Solution:

If 1A exists, then we can say 1A A I = , so 1det( ) det( ) 1A A I = = . Now recall that


135/306


135

det( ) det( )det( )C B C B = , so 1 1 1 1

1 det( ) det( ) det( ) det( )det( )

A A A A AA

= = = .

* Exercise 4.

Prove that det( )ii

A =

, ifAis diagonalizable, where

i

are the eigenvalues ofA.

Solution:

We first establish the intuition. SupposeAis 2x2. BecauseAis diagonalizable, we know it has

2 eigenvectors ,v w with associated eigenvalues , . Now consider the parallelogram given

by ,v w and consider what would happen to it when multiplied byA. We call the original

parallelogram P and the new one Q.

In the figure it is quite clear that ( ) ( )Vol Q Vol P = . Therefore we should find that indeed

det( ) ii

A = . We now proceed to prove this.

From class we know that for a diagonalizable matrixAthe following holds:

AC CD= , where Cis the matrix with for every column an eigenvector ofA, andDis adiagonal matrix with the associated eigenvalues on the diagonal. Now we know:

det( ) det( ) det( ) det( ) det( ) det( ) det( ) det( )AC CD A C C D A D= = =

ButDis a diagonal matrix, so

1 0 0

det( ) det( ) 0 0

0 0

i i

i

n

A D

= = =

.

We have quite a powerful apparatus by now. This was not such an easy theorem tounderstand, but the proof is just a few lines.

Consequence: if one of the eigenvalues ofAequals zero, the determinant of the matrixAwillbe zero. If one of the eigenvalues ofAequals zero, the inverse of the matrixAdoes not exist.


136/306


136

ADVANCED MATHEMATICS TAKE HOME ASSIGNMENT

MATERIAL OF WEEK 3

Take home assignments of week 4 (on material of week 3)

Question 1

i)

Compute the matrix decomposition 1P AP =

for1 2

3 0A

=

ii)

Using the decomposition, compute 4A

Solution

Compute the eigenvalues ofA:

1 2

(1 ) 6 03A I

= = =

So that the characteristic equation is ( 3)( 2) 0 + =

For 3= , the eigenvector is

1 2

1 2

2 2 0

3 3 0

x x

x x

+ =

= so that 1

1

1v

is an eigenvector

For 2=

, the eigenvector is

1 2

1 2

3 2 0

3 2 0

x x

x x

+ =

+ = so that 2

2

3v

is an eigenvector

Thus:

3 0

0 2

=

1 2

1 3P

=

1

3 2

1 2 3 21 5 5

1 3 1 1 1 15

5 5

P

= = =

Check: 1

3 2

1 2 1 05 5

1 3 1 1 0 1

5 5

PP

= =


137/306


137

1P AP =

Thus:

3 2 9 6

1 2 1 2 1 2 3 05 5 5 5

1 1 3 0 1 3 2 2 1 3 0 2

5 5 5 5

= =

Using the decomposition, compute 4A

1A P P=

4 1 1 1 1 4 1A P P P P P P P P P P = =

4

44

3 2 3 2 3 2

1 2 1 2 81 0 81 323 0 5 5 5 5 5 5

1 3 1 1 1 3 0 16 1 1 81 48 1 10 2

5 5 5 5 5 5

275 130

55 265 5

195 210 39 42

5 5

A

= = = =

= =


138/306

ADVANCED MATHEMATICS ADDITIONAL EXERCISES

WEEK 3

138

One additional exercise of week 3

Exercise 1.

Determine the dimension of the span of the following vectors:

11 1

, 10 2

01 1

u v w

= = =

.

Solution:The dimension of the span of a set of vectors is equal to the number of linearly independentvectors in the set. Vectors are linearly independent if the following equation has only the

solution 1 2 3 0 = = = . 1 2 3 1 2 3

1 01 1

1 00 2

0 01 1

u v w

+ + = + + =

.

We can rewrite this equation as:

1

2

3

1 1 1 0

0 2 1 0

1 1 0 0

=

. We sweep the matrix:

1 1 1 0 1 1 1 0 1 1 1 0

0 2 1 0 ~ 0 2 1 0 ~ 0 2 1 0

1 1 0 0 0 2 1 0 0 0 0 0

We can stop here, since we are not interested in the explicit solution and it is clear that wehave all the zero-rows that we will get. The number of unknowns (the three lambdas) minus

the number of zero rows is the number of free variables that we have, i.e. the number oflambdas that we can pick non-zero. This means that there is one linear dependent vector inthe three and two linearly independent. So the dimension of the span of u,vand wis 2.


139/306


139

Documents

Advanced Mathematics 2012 2013 Week 1 - Week 8 - 21 Jan 2013