IB Linear Algebra

3Duality

IB Linear Algebra

3.2 Dual maps

Since linear algebra is the study of vector spaces and linear maps between them,

after dualizing vector spaces, we should be able to dualize linear maps as well.

If we have a map

V → W

, then after dualizing, the map will go the other

direction, i.e.

∗

→ V

∗

. This is a characteristic common to most dualization

processes in mathematics.

Definition

(Dual map)

Let

V, W

be vector spaces over

and

V → W ∈

(

V, W

). The dual map to

, written

∗

→ V

∗

is given by

θ 7→ θ ◦ α

Since the composite of linear maps is linear,

∗

(

)

∈ V

∗

. So this is a genuine

map.

Proposition.

Let

α ∈ L

(

V, W

) be a linear map. Then

∗

∈ L

(

∗

, V

∗

) is a

linear map.

This is not the same as what we remarked at the end of the definition of

the dual map. What we remarked was that given any

∗

(

) is a linear map.

What we want to show here is that α

∗

itself as a map W

∗

→ V

∗

is linear.

Proof. Let λ, µ ∈ F and θ

, θ

∈ W

∗

. We want to show

∗

(λθ

+ µθ

) = λα

∗

(θ

) + µα

∗

(θ

To show this, we show that for every

v ∈ V

, the left and right give the same

result. We have

∗

(λθ

+ µθ

)(v) = (λθ

+ µθ

)(αv)

= λθ

(α(v)) + µθ

(α(v))

= (λα

∗

(θ

) + µα

∗

(θ

))(v).

So α

∗

∈ L(W

∗

, V

∗

What happens to the matrices when we take the dual map? The answer is

that we get the transpose.

Proposition.

Let

V, W

be finite-dimensional vector spaces over

and

V →

be a linear map. Let (

, ··· , e

) be a basis for

and (

, ··· , f

) be a basis

for W ; (ε

, ··· , ε

) and (η

, ··· , η

) the corresponding dual bases.

Suppose

is represented by

with respect to (

) and (

) for

and

Then α

∗

is represented by A

with respect to the corresponding dual bases.

Proof. We are given that

α(e

) =

k=1

We must compute α

∗

(η

). To do so, we evaluate it at e

. We have

∗

(η

)(e

) = η

(α(e

)) = η

k=1

= A

We can also write this as

∗

(η

)(e

) =

k=1

Since this is true for all j, we have

∗

(η

)

k=1

So done.

Note that if α : U → V and β : V → W , θ ∈ W

∗

, then

(βα)

∗

(θ) = θβα = α

∗

(θβ) = α

∗

(β

∗

(θ)).

So we have (

βα

)

∗

. This is obviously true for the finite-dimensional case,

since that’s how transposes of matrices work.

Similarly, if α, β : U → V , then (λα + µβ)

∗

= λα

∗

+ µβ

∗

What happens when we change basis? If

−1

for some invertible

and Q, then

= (Q

−1

AP )

= P

−1

)

= ((P

−1

)

−1

)

So in the dual space, we conjugate by the dual of the change-of-basis matrices.

As we said, we can use dualization to translate problems about a vector space

to its dual. The following lemma gives us some good tools to do so:

Lemma.

Let

α ∈ L

(

V, W

) with

V, W

finite dimensional vector spaces over

Then

(i) ker α

∗

= (im α)

(ii) r

(

) =

(

∗

) (which is another proof that row rank is equal to column

rank).

(iii) im α

∗

= (ker α)

At first sight, (i) and (iii) look quite similar. However, (i) is almost trivial to

prove, but (iii) is rather hard.

Proof.

(i) If θ ∈ W

∗

, then

θ ∈ ker α

∗

⇔ α

∗

(θ) = 0

⇔ (∀v ∈ V ) θα(v) = 0

⇔ (∀w ∈ im α) θ(w) = 0

⇔ θ ∈ (im α)

(ii) As im α ≤ W , we’ve seen that

dim im α + dim(im α)

= dim W.

Using (i), we see

n(α

∗

) = dim(im α)

r(α) + n(α

∗

) = dim W = dim W

∗

By the rank-nullity theorem, we have r(α) = r(α

∗

(iii)

The proof in (i) doesn’t quite work here. We can only show that one

includes the other. To draw the conclusion, we will show that the two

spaces have the dimensions, and hence must be equal.

Let θ ∈ im α

∗

. Then θ = φα for some φ ∈ W

∗

. If v ∈ ker α, then

θ(v) = φ(α(v)) = φ(0) = 0.

So im α

∗

⊆ (ker α)

But we know

dim(ker α)

+ dim ker α = dim V,

So we have

dim(ker α)

= dim V − n(α) = r(α) = r(α

∗

) = dim im α

∗

Hence we must have im α

∗

= (ker α)

Not only do we want to get from

∗

, we want to get back from

∗

We can take the dual of

∗

to get a

∗∗

. We already know that

∗∗

is isomorphic

, since

∗

is isomorphic to

already. However, the isomorphism between

∗

and

are not “natural”. To define such an isomorphism, we needed to pick

a basis for

and consider a dual basis. If we picked a different basis, we would

get a different isomorphism. There is no natural, canonical, uniquely-defined

isomorphism between V and V

∗

However, this is not the case when we want to construct an isomorphism

V → V

∗∗

. The construction of this isomorphism is obvious once we think hard

what

∗∗

actually means. Unwrapping the definition, we know

∗∗

(

∗

, F

Our isomorphism has to produce something in

∗∗

given any

v ∈ V

. This is

equivalent to saying given any

v ∈ V

and a function

θ ∈ V

∗

, produce something

in F.

This is easy, by definition

θ ∈ V

∗

is just a linear map

V → F

. So given

and θ, we just return θ(v). We now just have to show that this is linear and is

bijective.

Lemma.

Let

be a vector space over

. Then there is a linear map

V →

∗

)

∗

given by

ev(v)(θ) = θ(v).

We call this the evaluation map.

We call this a “canonical” map since this does not require picking a particular

basis of the vector spaces. It is in some sense a “natural” map.

Proof.

We first show that

(

)

∈ V

∗∗

for all

v ∈ V

, i.e.

(

) is linear for any

v. For any λ, µ ∈ F, θ

, θ

∈ V

∗

, then for v ∈ V , we have

ev(v)(λθ

+ µθ

) = (λθ

+ µθ

)(v)

= λθ

(v) + µθ

(v)

= λ ev(v)(θ

) + µ ev(v)(θ

So done. Now we show that

itself is linear. Let

λ, µ ∈ F

, v

∈ V

. We

want to show

ev(λv

+ µv

) = λ ev(v

) + µ ev(v

To show these are equal, pick θ ∈ V

∗

. Then

ev(λv

+ µv

)(θ) = θ(λv

+ µv

)

= λθ(v

) + µθ(v

)

= λ ev(v

)(θ) + µ ev(v

)(θ)

= (λ ev(v

) + µ ev(v

))(θ).

So done.

In the special case where V is finite-dimensional, this is an isomorphism.

Lemma. If V is finite-dimensional, then ev : V → V

∗∗

is an isomorphism.

This is very false for infinite dimensional spaces. In fact, this is true only

for finite-dimensional vector spaces (assuming the axiom of choice), and some

(weird) people use this as the definition of finite-dimensional vector spaces.

Proof.

We first show it is injective. Suppose

(

) =

for some

v ∈ V

. Then

(

) =

(

)(

) = 0 for all

θ ∈ V

∗

. So

dimhvi

dim V

∗

dim V

. So

dimhvi

= 0. So

is injective. Since

and

∗∗

have the same

dimension, this is also surjective. So done.

From now on, we will just pretend that

and

∗∗

are the same thing, at

least when V is finite dimensional.

Note that this lemma does not just say that

is isomorphic to

∗∗

(we

already know that since they have the same dimension). This says there is a

completely canonical way to choose the isomorphism.

In general, if

is infinite dimensional, then

is injective, but not surjective.

So we can think of V as a subspace of V

∗∗

in a canonical way.

Lemma.

Let

V, W

be finite-dimensional vector spaces over

after identifying

(V and V

∗∗

) and (W and W

∗∗

) by the evaluation map. Then we have

(i) If U ≤ V , then U

= U.

(ii) If α ∈ L(V, W ), then α

∗∗

= α.

Proof.

(i)

Let

u ∈ U

. Then

(

) =

(

) = 0 for all

θ ∈ U

. So

annihilates

everything in U

. So u ∈ U

. So U ⊆ U

. We also know that

dim U = dim V − dim U

= dim V − (dim V − dim U

) = dim U

So we must have U = U

(ii)

The proof of this is basically — the transpose of the transpose is the

original matrix. The only work we have to do is to show that the dual of

the dual basis is the original basis.

Let (

, ··· , e

) be a basis for

and (

, ··· , f

) be a basis for

, and let

(

, ··· , ε

) and (

, ··· , η

) be the corresponding dual basis. We know

that

(ε

) = δ

= ε

), f

(η

) = δ

= η

So (e

, ··· , e

) is dual to (ε

, ··· , ε

), and similarly for f and η.

is represented by

, then

∗

is represented by

. So

∗∗

is represented

by (A

)

= A. So done.

Proposition.

Let

be a finite-dimensional vector space

and

are

subspaces of V . Then we have

(i) (U

+ U

)

= U

∩ U

(ii) (U

∩ U

)

= U

+ U

Proof.

(i) Suppose θ ∈ V

∗

. Then

θ ∈ (U

+ U

)

⇔ θ(u

+ u

) = 0 for all u

∈ U

⇔ θ(u) = 0 for all u ∈ U

∪ U

⇔ θ ∈ U

∩ U

(ii) We have

∩ U

)

= ((U

)

∩ (U

)

= (U

+ U

)

= U

+ U

So done.