ChewMA1506-14 Ch6

MA1506 Mathematics II

Chapter 6 Linear Transformation

1 Chew T S MA1506-14 Chapter 6

6.1 What is a Linear Transformation

In this chapter, transformations are mappings (rules) that send vectors to vectors

Linear transformation T further satisfies


( ) ( ) ( )T u v T u T vα β α β+ = +

Examples

Identity transformation is linear

Check



( )I u u=

( ) ( ) ( )I u v u v I u I vα β α β α β+ = + = +

Let D be the transformation ( ) 2D u u=

D is linear

( ) 2( ) ( ) ( )D u v u v D u D vα β α β α β+ = + = +

Check

Then T is a linear transformation

1 32 4

Given

Define T u1 32 4

u =

= multiplication of matrices

1 3( ) ( )

2 4T u v u vα β α β

+ = +

1 3 1 32 4 2 4

u vα β = +

1 3 1 32 4 2 4

u vα β = +

Tu Tvα β= +

Example


is 2-dim vector (2x1 matrix)


u

6.2 Linear transformations and matrices

Every vector u in 2-D space can be represented by

Let T be a linear transformation defined in 2-D space. Then


just need to know

ˆ ˆai bj+

ˆ ˆ ˆ ˆ( ) ( )T u T ai bj aTi bTj= + = +

Every vector u in 3-D space can be represented by

Let T be a linear transformation defined in 3-D space. Then



just need to know

ˆˆ ˆai bj ck+ +

ˆ ˆˆ ˆ ˆ ˆ( ) ( )T u T ai bj ck aTi bTj cTk= + + = + +

Example: Given



i

j

1ˆ ˆ ˆ4

Ti i j= + 1ˆ ˆ ˆ4

Tj i j= +

ˆ ˆ ˆ ˆ(2 3 ) 2 3T i j Ti Tj+ = +1 1ˆ ˆ ˆ ˆ2( ) 3( )4 4

i j i j= + + +11 7ˆ ˆ4 2

i j= +

1ˆ ˆ4

i j+

1 ˆ ˆ4

i j+T

We can find T(u). For example

Next we want to show that every linear Transformation T maps n-dim vectors to m-dim vectors can be represented by a Matrix A Furthermore Tu=Au=multiplication of matrices A and u



All necessary info about T

is called Matrix of T relative to

Suppose

9


a bc d

ˆ ˆ ˆ aTi ai cj

c = + =

ˆ ˆ ˆ bTj bi dj

d = + =

ˆ ˆ,i j or in coordinates(system)

ˆ ˆ( , )i j

Shall prove a bTu u

c d

=

Now we shall prove the result we just mentioned

1ˆ0

a b a b ai

c d c d c

= =

0ˆ1

a b a b bj

c d c d d

= =

a bu

c d

=

so


a bc d

α β

= +

ˆ ˆ( )a b

i jc d

α β

+


𝑢 =𝛼𝛽 = 𝛼𝚤 + 𝛽𝚥 Let

We shall compute 𝑎 𝑏𝑐 𝑑 𝑢

First compute

On the other hand, recall

a bc d

α β = +

So

Hence

Thus T can be represented by a matrix


ˆ ˆ ˆ ˆ( )ˆTu T i j Ti Tjα β α β= + = +

ˆ ˆa b

Tu uc d

= (See the above equality and previous slide)


a bc d

ˆ ˆ ˆ aTi ai cj

c = + =

ˆ ˆ ˆ bTj bi dj

d = + =

in coordinates(system) ˆ ˆ( , )i j

Example

Identity transformation

Hence



Iu u=

1 00 1

I =

1ˆ ˆ0

Ii i = =

0ˆ ˆ1

Ij j = =

Example

Linear transformation T



114

1 14

T

=

1ˆ ˆ ˆ4

Ti i j= +1ˆ ˆ ˆ4

Tj i j= +

Example

1 2 34 5 67 8 9

T =



ˆˆ ˆ ˆ4 7Ti i j k= + +ˆˆ ˆ ˆ2 5 8Tj i j k= + +

ˆ ˆˆ ˆ3 6 9Tk i j k= + +

Example

T : 2-D vectors 3-D vectors



ˆˆ ˆ ˆ 2Ti i j k= + +

ˆˆ ˆ 3Tj i k= −

1 11 02 3

T =

−

Example




ˆ ˆ2Ti i=ˆ ˆ ˆTj i j= +

ˆ ˆ ˆTk i j= −

2 1 10 1 1

T = −

Dimension of linear transformation


T is 2 dimensional if


T is 3 dimensional if



Recall if T maps 2-dim vectors to 2 dim vectors then T can be represented by 2x2 matrix We define det(T)=det( )

6.3 Determinant of linear transformation

a bT

c d

=

a bc d

Determinant of 2-D transformation


Determinant of 3-D Transformation

Recall if T maps 3-dim vectors to 3-dim vectors then T can be represented by 3x3 matrix A

We define det(T)=det(A)


6.3 Determinant of linear transformation

6.4 Volume and Determinant

Area of parallelogram = base x height

normal multiplication

cross product 20 Chew T S MA1506-14 Chapter 6

v

uθ

sinu v θ= ×

u v= ×

= |𝑎𝑑 − 𝑏𝑐| = |detA|

𝑢 = 𝑎𝑏 𝑣 = 𝑐

𝑑

A= 𝑎 𝑐𝑏 𝑑

Let

Let which is the matrix induced by vectors u and v

sinu v u v nθ× =

n is unit vector n u⊥

n v⊥

00

( )

i j ku v a b

c d

ad bc k

× =

= −

vol of parallelepiped = base area x height


6.4 Volume and Determinant

Let A=𝑎11 𝑎12 𝑎13𝑎21 𝑎22 𝑎23𝑎31 𝑎32 𝑎33

𝑢 =𝑎11𝑎21𝑎31

𝑣 =𝑎12𝑎22𝑎32

𝑤 =𝑎13𝑎23𝑎33

= |detA| Proof omitted

One to one (non–singular ) linear transformation

6.5 Properties of linear transformation

T(0)=T(0+0)=T(0)+T(0)

Hence T(0)=0 (Property 1)

Suppose transformation T is linear. Then

A transformation T is said to be one to one if

i.e., T(u)=T(v), then u=v

0=zero vector


it never maps distinct vectors to the same vector

Property 2: Let T be a linear transformation. Then the following properties are equivalent: (1) T is 1-1 (2) If T(u)=0, then u=0=zero vector, i.e., only one vector maps to zero vector, that vector is zero Proof: (1) implies (2) Suppose T(u)=0. We know that T(0)=0 By (1), u=0 (2) implies (1)

Suppose T(u)=T(v). Then T(u-v)=0. By (2), u-v=0 23 Chew T S MA1506-14 Chapter 6


From now onwards, we only consider linear transformations T from n-dim space to n-dim space. We know that T can be represented by a square matrix A and Tu=Au. Here Au is the multiplication of two matrices. So properties of T can be induced from properties of the square matrix For example T(0)=0 (Property 1). It is clear since A0=0



Suppose T=nxn matrix A. Then properties of T can be induced from properties of A From Chapter 5,Section 5.5, we know that System of linear equations AX=B has unique solution X , for any given n-dim vector B, iff det(A) not zero

Hence if det(A) not zero, then T is 1-1(unique soln) and onto (has soln)



Example: Non Singular (1-1)Transformation

So is 1-1 linear transformation

from 2-dim space onto 2-dim space



0 1det 1

1 0 −

=

0 11 0

T−

=

Suppose linear transformation T=nxn matrix A and det(A)=0

T is not 1-1 ( is singular)

Then from Chapter 5, Section 5.5, we know that system of linear equations AX=0 has infinitely many solutions X In other words, TX=0 has infinitely many solutions X Hence T maps many nonzero vectors to zero vector So T is not 1-1(is singular)



Rank

A transformation T from 3-D to 3-D has ONLY three cases

Case1: det(T) not zero T maps 3-d space ONTO 3-d space

Case 2: det(T)=0

For this case, we say T is of rank 3

T maps 3-d space ONTO a plane For this case, we say T is of rank 2

Subcase 1

Example



2 1 11 2 10 0 0

Why? see slide 33

Subcase 2

Rank (cont.)

T maps 3-d space ONTO a line

For this case, we say T is of rank 1

Example



1 1 11 1 10 0 0

Why? See slide 34

Rank 3: det(T) non zero

no direction destroyed

Image still 3-d space

T maps 3-d space ONTO 3-d space



Rank 2: det(T) = 0

1 direction destroyed No k direction, image is a plane

T maps 3-d space ONTO a plane



2 1 11 2 10 0 0

Example

210

Ti =

120

Tj =

110

Tk =

1 2 1 2 11 11 1 2 1 23 3

0 0 0 0 0α β

= + = +

is on the plane induced by and Tk

Ti Tj

Rank 1: det(T) = 0

2 directions destroyed

Image is a line

T maps 3-d space ONTO a line



1 1 11 1 10 0 0

Example

110

=



1 2 34 5 67 8 9

The det of this matrix is zero

It maps 3-d space ONTO a plane. why? Since the image is either a plane or a line, see two subcases. The image is not a line, so it should be a plane If it is a line then 1

47

= 𝛼258

for some α

It is not possible

Example

Rank 2: det(T) = 0

has no solution

has solution in fact infinitely

many solutions

Image is a plane

(1,2,4) not on the plane

(1,2,3) on the plane



1 2 3 14 5 6 27 8 9 4

xyz

=

1 2 3 14 5 6 27 8 9 3

xyz

=

T = 1 2 34 5 67 8 9

Det T=0

Example

recall (1) if det(A)=0, then AX=0 has infinitely many solutions (2) If det(A)=0, then AX=B has two cases: AX=B has NO solution or AX=B has infinitely many solutions (Why infinitely many solutions, see next slide)


To understand the previous example, 6.5 Properties of linear transformation

Chew T S MA1506-14 Chapter 6 36


If det(A)=0, then AX=0 has nonzero solution say . hX Then 0 0h hA X AXα α α= = =So if det(A)=0, then AX=0 has infinitely many nonzero solutions.

Now suppose AX=B has a solution, say pXThen pAX B=Hence

( ) 0p h p hA X X AX AX B Bα α+ = + = + =



The above holds for any real number and is nonzero. So if det(A)=0 and AX=B has a solution, then it has infinitely many solutions

αhX

( ) 0p h p hA X X AX AX B Bα α+ = + = + =

Why infinitely many solutions?


1 2 3 1 04 5 6 2 07 8 9 1 0

α − =

α any real number

6.5 Properties of linear transformation 1 2 3 14 5 6 27 8 9 3

xyz

=

1 2 3 1/ 3 14 5 6 2 / 3 27 8 9 0 3

− =

1 2 3 1 04 5 6 2 07 8 9 1 0

− =

1 2 3 1/ 3 1 14 5 6 2 / 3 2 27 8 9 0 1 3

α − + − =

We shall discuss this matrix again in T9 Q 6

How to check that the vector [1,2,3] is on the plane, vector [1,2,4] is not on the plane ?


1 2 34 5 67 8 9

End of Chapter 6

Recall if the plane is generated by two vectors u and v ,then (uxv).w=0 iff vector w is on the plane

u

v w uxv


𝑢 =𝑎11𝑎21𝑎31

𝑣 =𝑎12𝑎22𝑎32

𝑤 =𝑎13𝑎23𝑎33

𝑇 =𝑎11 𝑎12 𝑎13𝑎21 𝑎22 𝑎23𝑎31 𝑎32 𝑎33

(uxv).w=0 iff detT=0

ChewMA1506-14 Ch6

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