Section 11.3.
Homework #5
Masaya Sato
2. Let V the vector space of polynomials with coefficients in Q in the variable x of degree at most 5 with 1 , x , x2 , . . . , x5 as basis. Prove that the following are elements of the dual space of V and express them as linear combinations of the dual basis:
(a) E : V → Q defined by E ( p(x)) = p(3). (b) ϕ : V → Q defined by ϕ( p(x)) = (c) ϕ : V → Q defined by ϕ( p(x)) =
1
0
1
0
p(t)dt. t2 p(t)dt.
(d) ϕ : V → Q defined by ϕ( p(x)) = p(5) where p (x) denoted the usual derivative of the polynomial p(x) with respect to x. Solution: Let V ∗ denote the dual space of V . (a) For all p(x) and q (x) in V and r ∈ Q
(( p + r.q)(x)) = ( p + r.q)(3) = p(3) + rq (3) = E ( p(x)) + r.E (q (x)). E ( p(x) + r.q(x)) = E (( So E is a linear functional, i.e. an element of V ∗ . Now observe that {E 0 , E 1, E 2 , E 3 , E 4 , E 5 }, where E i : V → Q is defined by
1 if i = j 0 otherwise
E i (x j ) =
for i = 0, 1, . . . , 5, is a basis for V ∗ . Then (1)E 0 + E (x)E 1 + · · · + E (x5 )E 5 = E 0 + 3E 1 + · · · + 35 E 5 E = E (1) is a linear combination of the basis given above. (b) For all p(x) and q (x) in V and r ∈ Q 1
ϕ( p(x) + r.q(x)) = ϕ( p + r.q)(x) =
1
( p + r.q)(t)dt =
0
( p(t) + r.q(t))dt
0
1
=
1
p(t)dt +
0
r.q(t)dt
0
1
=
1
p(t)dt + r
0
q (t)dt
0
= ϕ( p(t)) + r.ϕ(q(t)) and thus ϕ ∈ V ∗ . Now consider {ϕ0 , ϕ1 , ϕ2 , ϕ3, ϕ4, ϕ5}, where each ϕi is defined by ϕi =
1 if i = j 0 otherwise
Abstract Abstract Algebra Algebra by Dummit and Foote 1
Section 11.3.
Homework #5
Masaya Sato
for i = 0, 1, . . . , 5. Then the set of linear functionals functionals given given above is a basis for the dual space. Therefore 1 1 ϕ = ϕ(1)ϕ0 + ϕ(x)ϕ1 + · · · + ϕ(x5)ϕ5 = ϕ0 + ϕ1 + · · · + ϕ5 2 6 is a linear combination of the dual basis. (c) For all p(x) and q(x) in V and r ∈ Q 1
ϕ( p(x) + r.q(x)) = ϕ( p + r.q)(x) =
1
( p + r.q)(t)dt =
0
t2 ( p(t) + r.q(t))dt
0
1
=
1
2
t p(t)dt +
0
1
2
t p(t)dt +
0
rt2 q (t)dt
0
1
=
t2 (r.q(t))dt
0
1
=
1
2
t p(t)dt + r
0
t2 q (t)dt
0
= ϕ( p(t)) + r.ϕ(q (t)). Now let {ϕ0 , ϕ1 , ϕ2 , ϕ3 , ϕ4 , ϕ5 } be a dual basis given by ϕi =
1 if i = j 0 otherwise
for each i = 0, 1, . . . , 5. Then ϕ = ϕ(1)ϕ0 + ϕ(x)ϕ1 + · · · + ϕ(x5 )ϕ5 =
1 1 1 ϕ0 + ϕ1 + · · · + ϕ5 3 4 8
is a linear combination of the dual basis given above. (d) For all p(x) and q (x) in V and r ∈ Q d ( p + r.q)(x) dx d = ( p(x) + r.q(x)) dx d d = p(x) + r.q(x) dx dx d d = p(x) + r. q(x) dx dx = p (x) + r.q (x)
ϕ( p(x) + r.q(x)) = ϕ( p + r.q)(x) = ( p + r.q) (t) =
= ϕ( p(x)) + r.ϕ(q(x)), so ϕ is a linear functional in V ∗. Then for a dual basis {ϕ0 , ϕ1 , ϕ2 , ϕ3 , ϕ4 , ϕ5 }, where ϕi =
1 if i = j 0 otherwise
Abstract Abstract Algebra Algebra by Dummit and Foote 2
Section 11.3.
Homework #5
Masaya Sato
for each i = 0, 1, . . . , 5. Then ϕ = ϕ(1)ϕ0 + ϕ(x)ϕ1 + · · · + ϕ(x5)ϕ5 = ϕ1 + 10ϕ2 + · · · + 625ϕ5
is expressed as a linear combination of the dual basis. 3. Let S be any subset V ∗ for some finite dimensional space V . Defin Definee Ann( Ann(S ) = {v ∈ V |f (v ) = 0 ∀f ∈ S }.
(a) Prove Prove that Ann( Ann(S ) is a subspace of V . (b) Let W 1 and W 2 be subspaces of V ∗ . Prove that Ann( W 1 + W 2) = Ann(W 1 ) ∩ Ann(W 2 ) and Ann(W 1 ∩ W 2 ) = Ann(W 1 ) + Ann(W 2 ). (c) Let W 1 and W 2 be subspaces of V ∗ . Prove Prove that that W 1 = W 2 if and only if Ann( W 1) = Ann(W 2 ). (d) Prove that the annihilator of S is the same as the annihilator of the subspace of V ∗ spanned by S . (e) Assume that V is finite dimensional with basis v1, . . . , vn . Prove that if S = {v1∗ , . . . , vk∗ } for some k ≤ n, then Ann(S ) is the subspace spanned by {vk+1, . . . , vn }. (f) Assume V is finite dimensional. Prove that if W ∗ is any subspace of V ∗ then dim Ann Ann((W ∗ ) = dim V − dim W ∗ . Proof. Let V be an n-dimensional vector space over some field F with a basis {v1, . . . , vn }. Let {v1∗, . . . , vn∗ } be a basis for V ∗ dual to {v1, . . . , vn }. (a) For all v, w ∈ Ann(S ) and r ∈ F , v + rw ∈ V and for all f ∈ S ⊆ V ∗ f (v + rw ) = f (v ) + f (rw ) = f (v ) + rf (w) = 0 + 0 = 0.
So Ann(S ) is a subspace of V . (c) (⇒) It is obvious that Ann(W 1 ) = Ann(W 2 ) if W 1 = W 2 . (⇐) Suppose that Ann( W 1 ) = Ann(W 2 ), where W 1 and W 2 are both subspaces of V ∗ . Then let {vi∗ , . . . , vi∗ } denote a basis for W 1 with 1 ≤ i1 ≤ ·· ···· ≤ ik ≤ n. For every f ∈ W 1 1
k
f (v ) = 0.
So the subspace Ann( W 1) has the basis {v j , . . . , v j }, where every index in { j1 , . . . , jl } is distinct from any of {i1 , . . . , ik }. More Moreov over er Ann( Ann(W 2) also has the basis {v j , . . . , v j } by assumption Ann(W 2 ) = Ann(W 1). Therefore a basis for W 2 contains {vi∗ , . . . , vi∗ } and thus f ∈ W 2 . Similarly if f ∈ W 2 then f ∈ W 1 as well. Hence W 1 = W 2 as desired. (d) Let W ∗ denote the subspace generated by S . Th Then en suppos supposee that v ∈ Ann(S ). ) . Th Then en for every f ∈ S f (v ) = 0. Now Now observe observe that that S ⊆ Span(S ) = W ∗. So f ∈ W ∗ . Theref Therefore ore ∗ ∗ ∗ Conver versely sely suppose that that v ∈ Ann(W ). Recall Recall that that W is the subspace of v ∈ Ann(W ). Con 1
l
1
1
Abstract Abstract Algebra Algebra by Dummit and Foote 3
l
k
Section 11.3.
Homework #5
Masaya Sato
Therefore v ∈ Ann(S ) as V ∗ generated by S . So f (v ) = 0 for all f ∈ S ⊆ Span(S ) = W ∗ . Therefore desired and hence Ann(S ) = Ann(W ∗ ). (e) Recall from part (a) that Ann(S ) = {v ∈ V |vi∗ (v ) = 0 ∀i = 1, . . . , k} is a subspace of V , where S = {v1∗ , . . . , vk∗ }. Observe that Ann(S ) does not contain any of v1, . . . , vk
since vi∗ (vi ) = 1 for i = 1, . . . , k. Therefore the subspace Ann( S ) is generated by the set
{vk+1, . . . , vn }. (f) Since W ∗ is a subspace of a finite dimensional (dual) vector space V ∗ with basis {v1∗ , . . . , vn∗ }, W ∗ has a basis S = {vi∗ , . . . , vi∗ }. By results from part (d) and (e), 1
k
Ann(W ∗ ) = Ann(S ) is a subspace of V , and Ann(W ∗ ) is generated by n − k elements {v1∗, . . . , vn∗ } − {vi∗ , . . . , vi∗ }. Moreover Moreover those n − k elements are linearly independent since {v1∗ , . . . , vn∗ } is a dual basis for V ∗. Therefore dim Ann Ann((W ∗) = n − k = dim V − dim W ∗ 1
k
as desired. 4. If V is infinite dimensional with basis A, prove that A∗ = {v ∗ |v ∈ A} does not span V ∗ .
Proof. Let V be an infinite dimensional vector space over F = Z2 = {0, 1} with a basis A = {vα }α∈Λ , where Λ is an index set and the cardinality of Λ is infinite. Since the scalars are 0 or 1, every v ∈ V expressed as a finite linear combination is finite sum of elements of
A. So |A| ≥ |V |. Now consider its dual space V ∗ . For every linear functional functional f ∈ V ∗ is a linear combination of some subset of the dual basis {f α }α∈Λ , where f α(vβ ) =
1 if α = β 0 otherwise .
Therefore there is a one-to-one correspondence between linear functionals in V ∗ and all subsets subsets of A. Hence the cardinality |V ∗ | is equal to |P (A)|, the cardinality of the power set of A. Thus |V ∗ | = |P (A)| > |A| ≥ |V | and then V ∗ cannot be spanned by A∗ . Abstract Abstract Algebra Algebra by Dummit and Foote 4
