Linear Functionals

Vectors and Linear Maps

Definition 2.3.2 A linear map (or transformation) from the complex vec-

2.5 Linear Functionals

Theorem 2.4.7 The vectors{|e_i,J|e_i}^mi=1withN=2mform an orthonor- mal basis for the real vector space V with inner product | ^R= | . In particular,Vmust be even-dimensional for it to have a complex structureJ.

Definition 2.4.8 IfVis a real vector space, thenC⊗V, together with the complex multiplication rule

α

β⊗ |a

=(αβ)⊗ |a, α, β∈C,

is a complex vector space called thecomplexification of V and denoted complexification byV^C. In particular,(Rⁿ)^C≡C⊗Rⁿ∼=Cⁿ.

Note that dimCV^C = dimRV and dimRV^C =2 dimRV. In fact, if {|a_k}^N_k=1 is a basis of V, then it is also a basis of V^C as a complexvec- tor space, while{|a_k, i|a_k}^N_k=1is a basis ofV^Cas arealvector space.

After complexifying a real vector spaceVwith inner product | ^R= | , we can define an inner product on it which is sesquilinear (or hermitian) as follows

α⊗a|β⊗b ≡ ¯αβa|b.

It is left to the reader to show that this inner product satisfies all the proper- ties given in Definition2.2.1.

To complexify a real vector spaceV, we have to “multiply” it by the set of complex numbers:V^C=C⊗V. As a result, we get arealvector space of twice the original dimension. Is there a reverse process, a “division” of a (necessarily even-dimensional) real vector space? That is, is there a way of getting a complex vector space of half complex dimension, starting with an even-dimensional real vector space?

LetVbe a 2m-dimensional real vector space. LetJbe a complex structure onV, and{|e_i,J|e_i}^mi=1a basis ofV. On the subspaceV₁≡Span{|e_i}^mi=1, define the multiplication by a complex number by

(α+iβ)⊗ |v1 ≡(α1+βJ)|v1, α, β∈R, |v1 ∈V₁. (2.22) It is straightforward to show that this process turns the 2m-dimensional real vector spaceVinto them-dimensional complex vector spaceV^C

1.

(a) Let|a =(α₁, α₂, . . . , α_n)be inCⁿ. Defineφ:Cⁿ→Cby φ

|a

= n k=1

α_k.

Then it is easy to show thatφis a linear functional.

(b) Let μ_ij denote the elements of an m×n matrix M. Define ω: M^m×n→Cby

ω(M)= m i=1

n j=1

μ_ij.

Then it is easy to show thatωis a linear functional.

(c) Letμijdenote the elements of ann×nmatrixM. Defineθ:M^n×n→ Cby

θ(M)= n j=1

μ_jj,

the sum of the diagonal elements ofM. Then it is routine to show that θis a linear functional.

(d) Define the operatorint:C⁰(a, b)→Rby

integration is a linear functional on the space of continuous functions int(f )=

_b

a

f (t ) dt.

Thenintis a linear functional on the vector spaceC⁰(a, b).

(e) LetVbe a complex inner product space. Fix|a ∈V, and letγ_a:V→ Cbe defined by

γ_a

|b

= a|b. Then one can show thatγ_ais a linear functional.

(f) Let{|a1,|a2, . . . ,|am}be an arbitrary finite set of vectors inV, and {φ₁,φ₂, . . . ,φ_m}an arbitrary set of linear functionals onV. Let

A≡ m k=1

|akφ_k∈End(V)

be defined by A|x =

m k=1

|akφ_k

|x

= m k=1

φ_k

|x

|ak.

ThenAis a linear operator onV.

An example of linear isomorphism is that between a vector space and its dual space, which we discuss now. Consider anN-dimensional vector space with a basisB= {|a1,|a2, . . . ,|a_N}. For any given set ofNscalars,

{α₁, α₂, . . . , α_N}, define the linear functionalφ_α byφ_α|a_i =α_i. Whenφ_α acts on any arbitrary vector|b =N

i=1βi|aiinV, the result is φ_α|b =φ_α

_N

i=1

β_i|a_i

= N i=1

β_iφ_α|a_i = N

i=1

β_iα_i. (2.23) This expression suggests that|bcan be represented as a column vector with entriesβ₁, β₂, . . . , β_N andφ_α as a row vector with entriesα₁, α₂, . . . , α_N. Thenφ_α|bis merely the matrix product¹⁴ of the row vector (on the left) and the column vector (on the right).

φ_α is uniquely determined by the set{α₁, α₂, . . . , α_N}. In other words, corresponding to every set ofN scalars there exists a unique linear func- tional. This leads us to a particular set of functionals,φ₁,φ₂, . . . ,φ_N corre- sponding, respectively, to the sets of scalars{1,0,0, . . . ,0},{0,1,0, . . . ,0}, . . . ,{0,0,0, . . . ,1}. This means that

Every set ofNscalars defines a linear functional.

φ₁|a1 =1 and φ₁|aj =0 forj=1, φ₂|a2 =1 and φ₂|aj =0 forj=2,

... ... ...

φ_N|a_N =1 and φ_N|a_j =0 forj=N, or that

φ_i|aj =δij, (2.24)

whereδij is the Kronecker delta.

The functionals of Eq. (2.24) form a basis of the dual spaceV^∗. To show this, consider an arbitraryγ∈V^∗, which is uniquely determined by its action on the vectors in a basis B= {|a₁,|a₂, . . . ,|a_N}. Let γ|a_i =γ_i ∈C.

Then we claim thatγ =N

i=1γ_iφ_i. In fact, consider an arbitrary vector|a inVwith components(α₁, α₂, . . . , α_N)with respect toB. Then, on the one hand,

γ|a =γ _N

i=1

αi|ai

= N i=1

αiγ|ai = N i=1

αiγi.

On the other hand, _N

i=1

γ_iφ_i

|a = _N

i=1

γ_iφ_{i N}

j=1

α_j|a_j

= N i=1

γ_i N j=1

α_jφ_i|a_j = N i=1

γ_i N j=1

α_jδ_ij= N

i=1

γ_iα_i.

14Matrices will be taken up in Chap.5. Here, we assume only a nodding familiarity with elementary matrix operations.

Since the actions ofγ andN

i=1γ_iφ_i yield equal results for arbitrary|a, we conclude thatγ=N

i=1γ_iφ_i, i.e.,{φ_i}^N_i=1spanV^∗. Thus, we have the following result.

Theorem 2.5.2 For every basis B= {|a_j}^N_j=1 inV,there corresponds a unique basisB^∗= {φ_i}^Ni=1inV^∗with the property thatφ_i|a_j =δ_ij.

By this theorem the dual space of anN-dimensional vector space is also N-dimensional, and thus isomorphic to it. The basisB^∗is called thedual

dual basis basisofB. A corollary to Theorem2.5.2is that to every vector inVthere

corresponds auniquelinear functional inV^∗. This can be seen by noting that every vector|ais uniquely determined by its components(α₁, α₂, . . . , α_N) in a basisB. The unique linear functional φ_a corresponding to |a, also called the dual of|a, is simplyN

i=1αiφ_i, withφ_i∈B^∗.

Definition 2.5.3 Anannihilator of |a ∈Vis a linear functional φ∈V∗

such thatφ|a =0. LetWbe a subspace ofV. The set of linear functionals inV^∗that annihilate all vectors inWis denoted byW⁰.

The reader may check that W⁰ is a subspace ofV∗. Moreover, if we extend a basis{|ai}^ki=1of Wto a basis B= {|ai}^Ni=1of V, then we can

annihilator of a vector and a subspace

show that the functionals{φ_j}^Nj=k+1, chosen from the basisB^∗= {φ_j}^Nj=1

dual toB, spanW⁰. It then follows that

dimV=dimW+dimW⁰. (2.25)

We shall have occasions to use annihilators later on when we discuss sym- plectic geometry.

We have “dualed” a vector, a basis, and a complete vector space. The only object remaining is a linear transformation.

Definition 2.5.4 LetT:V→Ube a linear map. DefineT^∗:U^∗→V^∗by¹⁵

dual, or pull back, of a linear transformation

#T^∗(γ)$

|a =γ T|a

∀|a ∈V, γ∈U^∗, T^∗is called thedualorpullback, ofT.

One can readily verify thatT^∗∈L(U∗,V∗), i.e., thatT^∗is alinearoper- ator onU^∗. Some of the mapping properties ofT^∗are tied to those ofT. To see this we first consider the kernel ofT^∗. Clearly,γ is in the kernel ofT^∗if and only ifγannihilates all vectors of the formT|a, i.e., all vectors inT(V).

It follows thatγ is inT(V)⁰. In particular, ifTis surjective,T(V)=U, andγ annihilates all vectors inU, i.e., it is the zero linear functional. We conclude that kerT^∗=0, and therefore,T^∗ is injective. Similarly, one can show that ifTis injective, thenT^∗is surjective. We summarize the discussion above:

15Do not confuse this “*” with complex conjugation.

Proposition 2.5.5 Let Tbe a linear transformation and T^∗ its pull back.

ThenkerT^∗=T(V)⁰.IfTis surjective(injective),thenT^∗is injective(sur- jective).In particular,T^∗is an isomorphism ifTis.

It is useful to make a connection between the inner product and linear functionals. To do this, consider a basis{|a₁,|a₂, . . . ,|a_N}and letα_i = a|a_i. As noted earlier, the set of scalars{α_i}^Ni=1 defines a unique linear functionalγ_a (see Example 2.5.1) such thatγ_a|ai =αi. Sincea|ai is also equal toαi, it is natural to identifyγ_a with the symbola|, and write γ_a→ a|.

duals and inner products

It is also convenient to introduce the notation¹⁶ |a†

≡ a|, (2.26)

where the symbol † means “dual, or dagger of”. Now we ask: How does this dagger operation act on a linear combination of vectors? Let |c = dagger of a linear

combination of vectors

α|a +β|b and take the inner product of|c with an arbitrary vector|x using linearity in the second factor:x|c =αx|a +βx|b. Now complex conjugate both sides and use the (sesqui)symmetry of the inner product:

(LHS)^∗= x|c^∗= c|x,

(RHS)^∗=α^∗x|a^∗+β^∗x|b^∗=α^∗a|x +β^∗b|x

=

α^∗a| +β^∗b|

|x.

Since this is true for all|x, we must have (|c)^†≡ c| =α^∗a| +β^∗b|. Therefore, in a duality “operation” the complex scalars must be conjugated.

So, we have

α|a +β|b†

=α^∗a| +β^∗b|. (2.27) Thus, unlike the association|a →γ_awhich is linear, the associationγ_a→ a|is not linear, but sesquilinear:

γ_αa+βb→α^∗a| +β^∗b|. It is convenient to represent|a ∈Cⁿas a column vector

|a =

⎛

⎜⎜

⎜⎝ α₁ α₂ ... αn

⎞

⎟⎟

⎟⎠.

Then the definition of the complex inner product suggests that the dual of

|amust be represented as a row vector with complex conjugate entries:

a| =

α^∗₁ α₂^∗ . . . α_n^∗

, (2.28)

16The significance of this notation will become clear in Sect.4.3.

and the inner product can be written as the (matrix) product

a|b =

α^∗₁ α₂^∗ · · · α^∗_n

⎛

⎜⎜

⎜⎝ β₁ β2

... β_n

⎞

⎟⎟

⎟⎠= n i=1

α_i^∗βi.

Compare (2.28) with the comments after (2.23).

The complex

conjugation in (2.28) is the result of the sesquilinearity of the association|a ↔ a|.