Department

Mathematics

Course Code

MAT224H1

Professor

Sean Uppal

MAT224H1b.doc

Page 1 of 15

Linear Transformations

EXAMPLES AND ELEMENTARY PROPERTIES

Definition

If V and W are two vector spaces, a function WVT

→

: is called a linear transformation if it satisfies the

following axioms:

T1:

(

)

(

)

(

)

(

)

(

)

WvTVTVvvvTVTvvTWV

∈

∈

+

=

+

1111 ,,,, .

T2:

(

)

(

)

VvrvTrvrTWV

∈

∈

⋅

=

⋅

,, R.

Example

Define 222

:MPT

→

, and

( )

+−++−

++−−+

=→∈++= cbacba

cbacba

vTPcxbxav2

1

2

2. Show that T is linear.

• T1:

•

(

)

(

)

(

)

(

)

(

)

( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( )

+++−++++++−

+++++−+−+++

=

+++++=+

111111

111111

2

1111

2

1

ccbbaaccbbaa

ccbbaaccbbaa

xccxbbaaTvvT

.

•

(

)

(

)

(

)

(

)

( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( )

+++−++++++−

+++++−+−+++

=

+−++−

++−−+

+

+−++−

++−−+

=

+++++=+

111111

111111

111111

111111

2

111

2

1

2

1

2

1

2

1

ccbbaaccbbaa

ccbbaaccbbaa

cbacba

cbacba

cbacba

cbacba

xcxbaTcxbxaTvTvT

.

• T2:

•

( )

(

)

( )

vTr

cbacba

cbacba

r

rcrbrarcrbra

rcrbrarcrbra

rcxrbxraTvrT

⋅=

+−++−

++−−+

=

+−++−

++−−+

=++=⋅

2

1

2

1

2

.

• Therefore, T is linear.

Example

The following are linear transformations:

• 1

:−

→

nn PPD where

( )( ) ( )( )

′

=xpxpDnn – ex:

(

)

323

2+=+ xxxD.

• 1

;+

→

nn PPI where

( )( ) ( )

=x

ann dyypxpI.

Theorem

Let WVT

→

: be a linear transformation.

1)

(

)

WV

T00

=

.

2)

(

)

(

)

VvvTvT

∈

∀

−

=

−

, .

3)

( )

==

⋅=

⋅

n

i

ii

n

i

ii vTavaT

11

.

www.notesolution.com

MAT224H1b.doc

Page 2 of 15

Theorem

Let WVT

→

: and WVS

→

: be two linear transformations. Suppose that

{

}

n

vvV,,span1

=

. If

(

)

(

)

ivSvTii

∀

=

, , then ST

=

.

Proof: Let Vvav

n

i

ii ∈⋅=

=1

. So

( ) ( )

==

⋅=

⋅=

n

i

ii

n

i

ii vTavaTvT

11

, and

( ) ( )

==

⋅=

⋅=

n

i

ii

n

i

ii vSavaSvS

11

. Thus,

(

)

(

)

vSvT

=

.

Theorem

Let V and W be vector spaces, and

{

}

n

ee ,,

1 a basis of V. Given any vector Www n

∈

,,

1, there exits a

unique linear transformation WVT

→

: satisfying

(

)

iweTii

∀

=

, . In fact, the action of T is as follows:

Given Vvav

n

i

ii ∈⋅=

=1

, then

( ) ( )

=

⋅=

n

i

ii vTavT

1

.

Example

Find a linear transformation 222

:MPT

→

such that

( )

=+ 00

01

1xT,

(

)

=+ 01

10

2

xxT, and

(

)

=+ 10

00

12

xT.

•

(

)

(

)

(

)

{

}

22 1,,1xxxx +++ is a basis of P2.

•

(

)

(

)

(

)

( ) ( ) ( )

+−

=

++−

=

−+

=

=−−

=−−

=−−

=−−+−−+−−

+++++=++

2

2

2

0

0

0

0

11

3

2

1

32

21

31

2

322131

2

3

2

21

2

cba

c

cba

c

cba

c

ccc

ccb

cca

xcccxccbcca

xcxxcxccxbxa

.

•

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

( )

+−++−

++−−+

=

+−

+

++−

+

−+

=

+

+

=

+⋅++⋅++⋅=+++++=

22

22

10

00

2

01

10

2

00

01

2

10

00

01

10

00

01

1111

321

2

3

2

21

2

3

2

21

cbacba

cbacba

vT

cbacbacba

ccc

xTcxxTcxTcxcxxcxcTvT

.

KERNEL AND IMAGE OF A LINEAR TRANSFORMATION

Definition

Let WVT

→

: be a linear transformation. Then:

•

(

)

{

}

0|ker

=

∈

=

vTVvT.

www.notesolution.com

MAT224H1b.doc

Page 3 of 15

•

(

)

{

}

VvvTT

∈

=

|im.

Theorem

If WVT

→

: is a linear transformation, then Tker is a subspace of V, and

T

im

is a subspace of W.

Definition

(

)

(

)

TT kerdimnullity

=

.

(

)

(

)

TT imdimrank

=

.

Example

Given an m×n matrix A, show that ATAcolim

=

(so ATArankrank

=

), where

(

)

AXXTT mn

A=→|:RR .

• Write

[

]

n

CCA

1

=

where Ci are columns, and

[

]

R∈=i

T

nxxxX,

1.

• Then

[

]

[

]

ACxCxxxCCAXA nn

T

nn colim1111 =+=== .

ONE-TO-ONE AND ONTO TRANSFORMATION

Definition

Let WVT

→

: be a linear transformation. Then:

• T is said to be onto if WT

=

im.

• T is said to be one-to-one if

(

)

(

)

11 vvvTvT

=

=

(each vector in W corresponds to only one

element in V).

Theorem

If WVT

→

: is a linear transformation, then T is one-to-one if and only if 0ker

=

T.

Proof:

• Want: T is one-to-one 0ker

=

T.

• Let Tvker

∈

.

•

(

)

(

)

000

=

=

=

vTvT because T is one-to-one. So 0ker

=

T.

• Want: 0ker

=

T T is one-to-one.

• Let

(

)

(

)

(

)

0

11

=

−

=

vvTvTvT.

• But since 0ker

=

T, 11 0vvvv

=

=

−

. So T is one-to-one.

Example

Given

(

)

(

)

xyxyxyxTT ,,,|:32 −+=→RR , show that T is one-to-one but not onto.

• Want: T is one-to-one.

•

(

)

(

)

{

}

0,|,ker

=

=

yxTyxT.

•

( ) ( )

=

=

=

=−

=+

=−+= 0

0

0

0

0

0,,, y

x

x

yx

yx

xyxyxyxT. So 0ker

=

T.

• Since 0ker

=

T, T is one-to-one.

www.notesolution.com

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