3Linear maps

IA Vectors and Matrices

3.2 Linear Maps

Definition

(Domain, codomain and image of map)

.

Consider sets

A

and

B

and mapping

T

:

A → B

such that each

x ∈ A

is mapped into a unique

x

0

=

T

(

x

)

∈ B

.

A

is the domain of

T

and

B

is the co-domain of

T

. Typically,

we have T : R

n

→ R

m

or T : C

n

→ C

m

.

Definition

(Linear map)

.

Let

V, W

be real (or complex) vector spaces, and

T : V → W . Then T is a linear map if

(i) T (a + b) = T (a) + T (b) for all a, b ∈ V .

(ii) T (λa) = λT (a) for all λ ∈ R (or C).

Equivalently, we have T (λa + µb) = λT (a) + µT (b).

Example.

(i)

Consider a translation

T

:

R

3

→ R

3

with

T

(

x

) =

x + a

for some fixed,

given

a

. This is not a linear map since

T

(

λx

+

µy

)

6

=

λx

+

µy

+ (

λ

+

µ

)

a

.

(ii) Rotation, reflection and projection are linear transformations.

Definition

(Image and kernel of map)

.

The image of a map

f

:

U → V

is the

subset of V {f(u) : u ∈ U}. The kernel is the subset of U {u ∈ U : f (u) = 0}.

Example.

(i)

Consider

S

:

R

3

→ R

2

with

S

(

x, y, z

) = (

x

+

y,

2

x − z

). Simple yet

tedious algebra shows that this is linear. Now consider the effect of

S

on

the standard basis.

S

(1

,

0

,

0) = (1

,

2),

S

(0

,

1

,

0) = (1

,

0) and

S

(0

,

0

,

1) =

(0

, −

1). Clearly these are linearly dependent, but they do span the whole

of R

2

. We can say S(R

3

) = R

2

. So the image is R

2

.

Now solve

S

(

x, y, z

) =

0

. We need

x

+

y

= 0 and 2

x − z

= 0. Thus

x

=

(

x, −x,

2

x

), i.e. it is parallel to (1

, −

1

,

2). So the set

{λ

(1

, −

1

,

2) :

λ ∈ R}

is the kernel of S.

(ii)

Consider a rotation in

R

3

. The kernel is the zero vector and the image is

R

3

.

(iii)

Consider a projection of

x

onto a plane with normal

ˆn

. The image is the

plane itself, and the kernel is any vector parallel to ˆn

Theorem.

Consider a linear map

f

:

U → V

, where

U, V

are vector spaces.

Then im(f) is a subspace of V , and ker(f) is a subspace of U .

Proof. Both are non-empty since f(0) = 0.

If

x, y ∈ im

(

f

), then

∃a, b ∈ U

such that

x

=

f

(

a

)

, y

=

f

(

b

). Then

λx

+

µy

=

λf

(

a

) +

µf

(

b

) =

f

(

λa

+

µb

). Now

λa

+

µb ∈ U

since

U

is a vector

space, so there is an element in

U

that maps to

λx

+

µy

. So

λx

+

µy ∈ im

(

f

)

and im(f) is a subspace of V .

Suppose

x, y ∈ ker

(

f

), i.e.

f

(

x

) =

f

(

y

) =

0

. Then

f

(

λx

+

µy

) =

λf

(

x

) +

µf(y) = λ0 + µ0 = 0. Therefore λx + µy ∈ ker(f).