3Linear maps

IA Vectors and Matrices

3.3 Rank and nullity

Definition

(Rank of linear map)

.

The rank of a linear map

f

:

U → V

, denoted

by r(f), is the dimension of the image of f.

Definition

(Nullity of linear map)

.

The nullity of

f

, denoted

n

(

f

) is the

dimension of the kernel of f.

Example.

For the projection onto a plane in

R

3

, the image is the whole plane

and the rank is 2. The kernel is a line so the nullity is 1.

Theorem (Rank-nullity theorem). For a linear map f : U → V ,

r(f) + n(f) = dim(U).

Proof.

(Non-examinable) Write

dim

(

U

) =

n

and

n

(

f

) =

m

. If

m

=

n

, then

f

is

the zero map, and the proof is trivial, since

r

(

f

) = 0. Otherwise, assume

m < n

.

Suppose

{e

1

, e

2

, ··· , e

m

}

is a basis of

ker f

, Extend this to a basis of the

whole of

U

to get

{e

1

, e

2

, ··· , e

m

, e

m+1

, ··· , e

n

}

. To prove the theorem, we

need to prove that {f(e

m+1

), f(e

m+2

), ···f(e

n

)} is a basis of im(f ).

(i)

First show that it spans

im

(

f

). Take

y ∈ im

(

f

). Thus

∃x ∈ U

such that

y = f(x). Then

y = f(α

1

e

1

+ α

2

e

2

+ ··· + α

n

e

n

),

since e

1

, ···e

n

is a basis of U. Thus

y = α

1

f(e

1

) + α

2

f(e

2

) + ···+ α

m

f(e

m

) + α

m+1

f(e

m+1

) + ···+ α

n

f(e

n

).

The first

m

terms map to

0

, since

e

1

, ···e

m

is the basis of the kernel of

f

.

Thus

y = α

m+1

f(e

m+1

) + ··· + α

n

f(e

n

).

(ii) To show that they are linearly independent, suppose

α

m+1

f(e

m+1

) + ··· + α

n

f(e

n

) = 0.

Then

f(α

m+1

e

m+1

+ ··· + α

n

e

n

) = 0.

Thus

α

m+1

e

m+1

+

···

+

α

n

e

n

∈ ker

(

f

). Since

{e

1

, ··· , e

m

}

span

ker

(

f

),

there exist some α

1

, α

2

, ···α

m

such that

α

m+1

e

m+1

+ ··· + α

n

e

n

= α

1

e

1

+ ··· + α

m

e

m

.

But

e

1

···e

n

is a basis of

U

and are linearly independent. So

α

i

= 0 for all

i

.

Then the only solution to the equation

α

m+1

f

(

e

m+1

) +

···

+

α

n

f

(

e

n

) =

0

is α

i

= 0, and they are linearly independent by definition.

Example.

Calculate the kernel and image of

f

:

R

3

→ R

3

, defined by

f(x, y, z) = (x + y + z, 2x − y + 5z, x + 2z).

First find the kernel: we’ve got the system of equations:

x + y + z = 0

2x − y + 5z = 0

x + 2z = 0

Note that the first and second equation add to give 3

x

+6

z

= 0, which is identical

to the third. Then using the first and third equation, we have

y

=

−x − z

=

z

.

So the kernel is any vector in the form (−2z, z, z) and is the span of (−2, 1, 1).

To find the image, extend the basis of

ker

(

f

) to a basis of the whole of

R

3

:

{

(

−

2

,

1

,

1)

,

(0

,

1

,

0)

,

(0

,

0

,

1)

}

. Apply

f

to this basis to obtain (0

,

0

,

0)

,

(1

, −

1

,

0)

and (1

,

5

,

2). From the proof of the rank-nullity theorem, we know that

f

(0

,

1

,

0)

and f (0, 0, 1) is a basis of the image.

To get the standard form of the image, we know that the normal to the plane

is parallel to (1

, −

1

,

0)

×

(1

,

5

,

2)

k

(1

,

1

, −

3). Since

0 ∈ im

(

f

), the equation of

the plane is x + y − 3z = 0.