Compliments in Linear Algebra 80146
by Amit Novick
First compiled: 08/01/2017
Examples of Dual Spaces
• Let V be some vector − space.
Suppose B = (e1 , ..., en ) is some basis of V (not necessarily the standard
basis)
Pn
As such, ∀v ∈ V there is a unique representation v = i=0 xi ei .
Then, B defines an isomorphism []B : V → Fncol :
 
x1
 . 
 
n

∀v ∈ V : [v]B = 
 .  ∈ Fcol .
 . 
xn
Essentially, f i (v) = xi , and as such, f i : V → F is a linear transf ormation,
specifically, f i ∈ V ∗ , i.e. f i is a f unctional.
Indeed, every basis B of V defines f unctionals f 1 , ..., f n ∈ V ∗ .
• Lemma: Suppose B = (e1 , ..., en ) ⊂ V is a basis.
If ∀i : 1 ≤ i ≤ n , f i : V → F is a linear transf ormation such that
f i (v) = xi ,
then (f 1 , ..., f n ) ⊂ V ∗ is a basis.
Proof: Let u ∈ V ∗ . We want to show that there is a unique representation
for u as a linear combination of (f 1 , ..., f n ).
Pn
– proposition: u = i=1 u(ei )f i
proof: Let v ∈ V . Since B is a basis of V , then there is a unique
Pn
representation v = i=1 xi ei .
Pn
Note that u(v) = i=1 xi u(ei )
Pn
Pn
In fact, v = i=1 u(ei )f i = i=1 xi u(ei ) (f i (v) = xi per f ’s definition)
Thus, Span(f 1 , ..., f n ) = V ∗ .
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– proposition: (f 1 , ..., f n ) is linearly independent.
proof: (negative proof) Suppose ∃a1 , a2 , ..., an ∈ F
Pn
such that i=1 ai f i = 0, and ∃ai0 6= 0.
Pn
Then, 0 = 0(ei0 ) = i=1 ai f i (ei0 ) = ai0 6= 0, in contradiction.
With the last proposition, we completed our proof. • Notes: (i) The basis (f 1 , ..., f n ) ⊂ V ∗ is called dual in relation
to the basis (e1 , ..., en ) ⊂ V , and satisfies:
(
1 i=j
fi (ej ) =
0 i 6= j
In fact, every f unctional in f 1 , ..., f n is dependent upon
the choice of the respective vector in the basis (e1 , ..., en ).
(ii) Exercise: Suppose (e1 , ..., en ) ⊂ V is a basis,
and suppose (f 1 , ..., f n ) ⊂ V ∗ is its dual basis,
Pn
Pn
Let v ∈ V , and let u ∈ V ∗ , suppose v = i=1 ei ej , and u = i=1 ai f i .
 1
b
.


Pn

Then u(v) = i=1 ai bi = (a1 , ..., an ) · 
.
.
bn
(iii) Corollary: dimV = dimV ∗ = dimV ∗ ∗
Reminder on Linear T ransf ormations and M atrices
• Theorem:
Let V, U each be vector − spaces on some f ield F.
Suppose B = (u1 , ..., un ) ⊂ U is some basis,
And suppose C = (v1 , ..., vm ) ⊂ V is some set of vectors.
Pm
Define T : U → V such that Tui = j=1 aji vj
 1
 1
y
x
 . 
 . 
 
 

 
If ∀u ∈ U , [Tu ]C = 
 . , [u]B =  . 
 . 
 . 
ym
xn
Pn
Then, y j = i=1 aji xi
Examples:
(
Suppose
y 1 = a11 x1 + a12 x2 + a13 x3 ∈ U3
y 2 = a21 x1 + a22 x2 + a23 x3 ∈ U3
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Define T : U3 → V2 such that:
T (u1 ) = a11 v1 + a21 v2
T (u2 ) = a12 v1 + a22 v2
T (u3 ) = a13 v1 + a23 v2
 1
1
a1
1
1
a a2 a3
t

a12
→ A = 21
,
and
A
=
a1 a22 a23
a13

a21
a22 .
a23
• Definition: Suppose T : U → V is a linear transf ormation,
The transf ormation T ∗ : V ∗ → U ∗ such that ∀ϕ ∈ V ∗ :T ∗ (ϕ)(u) =
ϕ(T (u)),
is a linear transf ormation, and may also be called T ’s dual linear
transf ormation.
An illustration:
• Theorem: Let U, V be two vector − spaces each over some f ield F.
Suppoes B ⊂ U is a basis,
and suppose C ⊂ V is also a basis.
Suppose T : U → V is a linear transf ormation, such that A = [T ]B
C.
If B ∗ , C ∗ are the dual bases of bases B, C respectively,
∗
t
Then, [T ∗ ]C
B∗ = A
and, suppose dimU = n, dimV = m then:
T : U → V in matrix form is in M at(m, n)
T ∗ : V ∗ → U ∗ in matrix form is in M at(n, m).
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