2Linear maps

IB Linear Algebra

2.3 The first isomorphism theorem and the rank-nullity

theorem

The main theorem of this section is the rank-nullity theorem, which relates the

dimensions of the kernel and image of a linear map. This is in fact an easy

corollary of a stronger result, known as the first isomorphism theorem, which

directly relates the kernel and image themselves. This first isomorphism is an

exact analogy of that for groups, and should not be unfamiliar. We will also

provide another proof that does not involve quotients.

Theorem

(First isomorphism theorem)

.

Let

α

:

U → V

be a linear map. Then

ker α

and

im α

are subspaces of

U

and

V

respectively. Moreover,

α

induces an

isomorphism

¯α : U/ ker α → im α

(u + ker α) 7→ α(u)

Note that if we view a vector space as an abelian group, then this is exactly

the first isomorphism theorem of groups.

Proof. We know that 0 ∈ ker α and 0 ∈ im α.

Suppose u

1

, u

2

∈ ker α and λ

1

, λ

2

∈ F. Then

α(λ

1

u

1

+ λ

2

u

2

) = λ

1

α(u

1

) + λ

2

α(u

2

) = 0.

So λ

1

u

1

+ λ

2

u

2

∈ ker α. So ker α is a subspace.

Similarly, if

α

(

u

1

)

, α

(

u

2

)

∈ im α

, then

λα

(

u

1

)+

λ

2

α

(

u

2

) =

α

(

λ

1

u

1

+

λ

2

u

2

)

∈

im α. So im α is a subspace.

Now by the first isomorphism theorem of groups,

¯α

is a well-defined isomor-

phism of groups. So it remains to show that

¯α

is a linear map. Indeed, we

have

¯α(λ(u + ker α)) = α(λu) = λα(u) = λ(¯α(u + ker α)).

So ¯α is a linear map.

Definition

(Rank and nullity)

.

If

α

:

U → V

is a linear map between finite-

dimensional vector spaces over

F

(in fact we just need

U

to be finite-dimensional),

the rank of

α

is the number

r

(

α

) =

dim im α

. The nullity of

α

is the number

n(α) = dim ker α.

Corollary

(Rank-nullity theorem)

.

If

α

:

U → V

is a linear map and

U

is

finite-dimensional, then

r(α) + n(α) = dim U.

Proof.

By the first isomorphism theorem, we know that

U/ ker α

∼

=

im α

. So we

have

dim im α = dim(U/ ker α) = dim U − dim ker α.

So the result follows.

We can also prove this result without the first isomorphism theorem, and

say a bit more in the meantime.

Proposition.

If

α

:

U → V

is a linear map between finite-dimensional vector

spaces over

F

, then there are bases (

e

1

, ··· , e

m

) for

U

and (

f

1

, ··· , f

n

) for

V

such that α is represented by the matrix

I

r

0

0 0

,

where r = r(α) and I

r

is the r ×r identity matrix.

In particular, r(α) + n(α) = dim U .

Proof.

Let

e

k+1

, ··· , e

m

be a basis for the kernel of

α

. Then we can extend this

to a basis of the (e

1

, ··· , e

m

).

Let

f

i

=

α

(

e

i

) for 1

≤ i ≤ k

. We now show that (

f

1

, ··· , f

k

) is a basis for

im α

(and thus

k

=

r

). We first show that it spans. Suppose

v ∈ im α

. Then we

have

v = α

m

X

i=1

λ

i

e

i

!

for some λ

i

∈ F. By linearity, we can write this as

v =

m

X

i=1

λ

i

α(e

i

) =

k

X

i=1

λ

i

f

i

+ 0.

So v ∈ hf

1

, ··· , f

k

i.

To show linear dependence, suppose that

k

X

i=1

µ

i

f

i

= 0.

So we have

α

k

X

i=1

µ

i

e

i

!

= 0.

So

P

k

i=1

µ

i

e

i

∈ ker α. Since (e

k+1

, ··· , e

m

) is a basis for ker α, we can write

k

X

i=1

µ

i

e

i

=

m

X

i=k+1

µ

i

e

i

for some

µ

i

(

i

=

k

+ 1

, ··· , m

). Since (

e

1

, ··· , e

m

) is a basis, we must have

µ

i

= 0 for all i. So they are linearly independent.

Now we extend (f

1

, ··· , f

r

) to a basis for V , and

α(e

i

) =

(

f

i

1 ≤ i ≤ k

0 k + 1 ≤ i ≤ m

.

Example. Let

W = {x ∈ R

5

: x

1

+ x

2

+ x

3

= 0 = x

3

− x

4

− x

5

}.

What is dim W ? Well, it clearly is 3, but how can we prove it?

We can consider the map α : R

5

→ R

2

given by

x

1

.

.

.

x

5

7→

x

1

+ x

2

+ x

5

x

3

− x

4

− x

5

.

Then

ker α

=

W

. So

dim W

= 5

− r

(

α

). We know that

α

(1

,

0

,

0

,

0

,

0) = (1

,

0)

and α(0, 0, 1, 0, 0) = (0, 1). So r(α) = dim im α = 2. So dim W = 3.

More generally, the rank-nullity theorem gives that

m

linear equations in

n

have a space of solutions of dimension at least n − m.

Example.

Suppose that

U

and

W

are subspaces of

V

, all of which are finite-

dimensional vector spaces of F. We let

α : U ⊕ W → V

(u, w) 7→ u + w,

where the ⊕ is the external direct sum. Then im α = U + W and

ker α = {(u, −u) : u ∈ U ∩ W }

∼

=

dim(U ∩ W ).

Then we have

dim U + dim W = dim(U ⊕ W ) = r(α) + n(α) = dim(U + W ) + dim(U ∩ W ).

This is a result we’ve previously obtained through fiddling with basis and horrible

stuff.

Corollary.

Suppose

α

:

U → V

is a linear map between vector spaces over

F

both of dimension n < ∞. Then the following are equivalent

(i) α is injective;

(ii) α is surjective;

(iii) α is an isomorphism.

Proof.

It is clear that, (iii) implies (i) and (ii), and (i) and (ii) together implies

(iii). So it suffices to show that (i) and (ii) are equivalent.

Note that

α

is injective iff

n

(

α

) = 0, and

α

is surjective iff

r

(

α

) =

dim V

=

n

.

By the rank-nullity theorem,

n

(

α

) +

r

(

α

) =

n

. So the result follows immediately.

Lemma.

Let

A ∈ M

n,n

(

F

) =

M

n

(

F

) be a square matrix. The following are

equivalent

(i) There exists B ∈ M

n

(F) such that BA = I

n

.

(ii) There exists C ∈ M

n

(F) such that AC = I

n

.

If these hold, then

B

=

C

. We call

A

invertible or non-singular, and write

A

−1

= B = C.

Proof.

Let

α, β, γ, ι

:

F

n

→ F

n

be the linear maps represented by matrices

A, B, C, I

n

respectively with respect to the standard basis.

We note that (i) is equivalent to saying that there exists

β

such that

βα

=

ι

.

This is true iff

α

is injective, which is true iff

α

is an isomorphism, which is true

iff α has an inverse α

−1

.

Similarly, (ii) is equivalent to saying that there exists

γ

such that

αγ

=

ι

.

This is true iff

α

is injective, which is true iff

α

is isomorphism, which is true iff

α has an inverse α

−1

.

So these are the same things, and we have β = α

−1

= γ.