Basic Analysis – Gently Done Topological Vector Spaces

44 Basic Analysis
Finally, suppose that X/V is one-dimensional. For any x ∈ X, we have [x] =
t[z] in X/V, for some t ∈ K and some z /∈ V. Hence x = v +tz for some v ∈ V.
In particular, any linear subspace containing V as a proper subset must contain
z. The only such subspace is X itself, and we deduce that V is a maximal proper
subspace.
Definition 5.11 Let X and Y be vector spaces over K, i.e., both over R or both
over C. A map T : X → Y is said to be a linear mapping if
T(αx ′ +x ′′ ) = αTx ′ +Tx ′′ , for α ∈ K and x ′ ,x ′′ ∈ X.
A map T : X → Y, with K = C, is called conjugate linear if
T(αx ′ +x ′′ ) = αTx ′ +Tx ′′ , for α ∈ C and x ′ ,x ′′ ∈ X.
A linear map λ : X → K is called a linear functional or linear form. If K = R,
then λ is called a real linear functional, whereas if K = C, λ is called a complex
linear functional. The algebraic dual of X (often denoted X ′ ) is the vector space
(over K) of all linear functionals on X equipped with the obvious operations of
addition and scalar multiplication.
Example 5.12 Let X = C n . For any given u = (u 1 ,...,u n ) ∈ C n , the map
z = (z 1 ,...,z n ) ↦→ � n
i=1 u i z i
z ↦→ � n
i=1 u i z i
is a (complex) linear functional, whereas the map
is a conjugate linear functional. In fact, every linear (respectively,
conjugate linear) functional on C n has this form. We can see this by induction.
Indeed, if λ : C → C is a linear functional on C, then λ(z 1 ) = u 1 z 1 for any z 1 ∈ C,
where u 1 = λ(1), and so the statement is true for n = 1.
Suppose now that is true for n = k and let λ : C k+1 → C be a linear functional.
Let e i , i = 1,...,k + 1 be the standard basis vectors for C k+1 , that is, e 1 =
(1,0,...,0), e 2 = (0,1,0,...,0) etc. The map (z 1 ,...,z k ) ↦→ λ((z 1 ,...,z k ,0)) is
a linear functional on C k and so, by the induction hypothesis, there are elements
u 1 ,...,u k ∈ C such that
λ((z 1 ,...,z k ,0)) =
Hence, for any (z 1 ,...,z k+1 ) ∈ C k+1 , we have
k�
i=1
u i z i .
λ((z 1 ,...,z k+1 )) = λ((z 1 ,...,z k ,0))+λ((0,...,0,z k+1 ))
=
k�
i=1
k+1 �
=
i=1
u i z i +z k+1 λ((0,...,0,1))
u i z i
Department of Mathematics King’s College, London