Section 11.1 and 11.2.
Homework #4
Masaya Sato
Section 11.1 4. Prove that the space of real-valued functions on the closed interval [ a, b] is an infinite
dimensional vector space over
R,
where a < b.
Proof. First observe that the space F ([a, ([a, b], R) of real-valued function on [a, [ a, b] is an abelian group under the pointwise pointwise addition, i.e.
(f + g)(x )(x) = f ( ([a, b], R) ∀x ∈ [a, b]. f (x) + g (x) ∀f, g ∈ F ([a, Then for all f, g ∈ F ([a, ([a, b], R) and r, s ∈ R 1. (rs) rs)f = r(sf ) 2. 1f = f 3. (r + s)f = rf + sf 4. r(f + g) = rf + rg , where the ring action is given by the pointwise multiplication rf ( ([a, b], R) rf (x) for all f ∈ F ([a, and all r ∈ R. Thus F ([a, ([a, b], R) is a vector space over R. Now we want to claim that F ([a, ([a, b], R) is infinite dimensional. So suppose by contradiction that F ([a, ([a, b], R) is finite dimensional with a basis {f i , . . . , fn | f i ∈ F ([a, ([a, b], R) ∀i = 1, . . . , n}. Since each f i is chosen to be a basis element, f i is not a zero zero function function.. So each each compact compact support C i of f i is not empty empty. Therefor Thereforee for every every f ∈ F ([a, ([a, b], R) there are a1 , . . . an ∈ R such that f = a1 f 1 + · · · + an f n , and the compact support of f is given by C i ∪ ·· · ∪ C i , where 1 ≤ i1 ≤ · · · ≤ ik ≤ n. However F ([a, ([a, b], R) contains a real-valued function g whose compact support is distinct from any of C of C i ∪ · · · ∪ C i . This contradicts that the space F ([a, ([a, b], R) is finite dimensional. dimensional. Hence F ([a, ([a, b], R) is infinite dimensional. 1
1
k
k
5. Prove that the space of continuous real-valued functions on the closed interval [ a, b] is an
infinite dimensional vector space over
R,
where a < b. b.
Proof. The same argument above applies to this problem, and the space C ([a, ([a, b], R) of con-
tinuous real-valued functions is a vector space over R. Now Now consi conside derr the space space R[x] of polynomials of one variable x defined over the closed interval [a, [ a, b]. [x] is a subspace of ([a, b], R), and a basis is given by C ([a,
{1, x , x2 , . . . }. Thus
R[x]
is infinite dimensional and hence C ([a, ([a, b], R) is infinite dimensional as well.
6. Let V be a vector space of finite dimension. If ϕ is any linear transformation from V to
V prove there is an integer m such that the intersection of the image of ϕm and the kernel of ϕm is {0}. Abstract Abstract Algebra Algebra by Dummit and Foote 1
Section 11.1 and 11.2.
Homework #4
Proof. Observe first that ϕ(V ) V ) ⊆ V . V . Then there exists some integer m ∈
Masaya Sato Z>0
such that
· · · = ϕm+1 (V ) V ) = ϕm (V ) V ) ⊆ ϕm 1 (V ) V ) ⊆ ··· ⊆ ϕ(V ) V ) ⊆ V −
because V is finite dimension dimensional. al. Now let let v ∈ im ϕm ∩ ker ϕm . Then Then there there is some w ∈ V such that v = ϕm (w) since v ∈ im ϕ. Moreover ϕm (v ) = 0 since v ∈ ker ϕ. Therefore v = ϕm (w) = ϕ2m (w) = ϕm ◦ ϕm (w) = ϕm (v ) = 0 and hence im ϕm ∩ ker ϕm = ∅. Sec 11.2 9. If W is a subspace of the vector space V stable under the linear transformation ϕ (i.e.
ϕ(W ) W ) ⊆ W ), W ), show that ϕ induces linear transformations ϕ|W on W and ϕ on the quotient vector space V /W . If ϕ|W and ϕ are nonsingular prove ϕ is nonsingular. Prove the converse /W . If ϕ holds if V has finite dimension and give a counterexample with V infinite dimensional. Proof. Since ϕ(W ) W ) ⊆ W , W , ϕ|W is naturally induced by
ϕ|W ( W (w) = ϕ(w) ∀w ∈ W . W . It is immediate to show that ϕ|W is well-defined since ϕ is well-define well-defined. d. Also ϕ is induced by ϕ(v ) = ϕ(v ) + W . W . Then for all v1 = v2 ∈ V /W there exists some w ∈ W such that v1 = v2 + w, and ϕ(v1 ) = ϕ(v2 + w) = ϕ(v2 + w) + W = ϕ(v2 ) + ϕ(w) + W = ϕ(v2) + W = ϕ(v2 ) since W is invariant under ϕ. So ϕ is well-defined. Now suppose that both ϕ|W and ϕ are nonsingular. nonsingular. Then for all v ∈ ker ϕ if v ∈ W ⊆ V , V , then 0 = ϕ(v ) = ϕ|W ( W (v ). So v ∈ ker ϕ|W = 0 and thus v = 0. Otherwise if v ∈ (V − W ) W ) ∪ {0} for v ∈ ker ϕ, then ϕ(v ) = 0 ⇒ ϕ(v ) + W = 0 + W ⇒ ϕ(v) = 0. Because ϕ is nonsingular, v = 0. Therefore v ∈ W implies that v = 0 since v ∈ (V − W ) W ) ∪{ 0}. Thus ϕ is nonsingular. Abstract Abstract Algebra Algebra by Dummit and Foote 2
Section 11.1 and 11.2.
Homework #4
Masaya Sato
Next suppose that V is finite dimensional and ϕ is nonsingular. Since ϕ is an isomorphism with a subspace W stable under ϕ, its restriction ϕ|W is also an isomorphism on W . W . So ker ϕ|W = 0 and ϕ is nonsingular. nonsingular. Similarly Similarly for a basis {b1 , . . . , bk } for W it extends to a basis {b1 , . . . , bk , bk+1 , . . . , bn } for V . V . Then a basis for V /W is given by
{bk+1 + W , . . . , bn + W }. This basis is mapped to another basis
{ϕ(b1 ) + W , . . . , ϕ( ϕ(bn ) + W } via the map ϕ since ϕ is nonsingular. nonsingular. Therefor Thereforee ϕ is an isomorphism and hence ϕ is nonsingular. However for an infinite dimensional vector space V = C (R), which is the set of all continuous functions defined over R, consider a linear map Ψ : V → V defined by Ψ(f Ψ(f ) = 2f . f . Observe that Ψ is injective, i.e. kerΨ = o, where o : R → R denotes the zero function. So Ψ is nonsingular. Now let W ⊆ V be the set of all continuous functions g defined over R, whose support is [0, [0, 1]. 1]. Th Then en let let W be the set of all continuous functions in W ∪ {o} restricted to [0, [0, 1]. 1]. Th Then en W is a subspace of V and moreover it is stable under Ψ |W . Ther eref efor oree W . Th kerΨ|W = W and Ψ|W is singular. singular.
11. Let ϕ be a linear transformation form the finite dimensional vector space V to itself
such that ϕ2 = ϕ. (a) Prove that im ϕ ∩ ker ϕ = 0. (b) Prove Prove that that V = im ϕ ⊕ ker ϕ. (c) Prove that there is a basis of V such that the matrix of ϕ with respect to this basis is a diagonal matrix whose entries are all 0 or 1. Proof. (a) Let v ∈ im ϕ ∩ ker ϕ be taken arbitrarily. Since v ∈ im ϕ, there exists some w ∈ V
so that v = ϕ(w). Moreover ϕ(v) = 0 since v ∈ ker ϕ. Then ϕ2 = ϕ implies that v = ϕ(w) = ϕ2 (w) = ϕ(v ) = 0 and thus v = 0. (b) It is sufficient to show that V ⊆ im ϕ ⊕ ker ϕ. For every v ∈ V ( v − ϕ(v )) v = ϕ(v) + (v and observe that ϕ(v ) ∈ im ϕ and ϕ(v − ϕ(v )) = ϕ(v ) − ϕ2 (v ) = ϕ(v ) − ϕ(v ) = 0. Abstract Abstract Algebra Algebra by Dummit and Foote 3
Section 11.1 and 11.2.
Homework #4
Masaya Sato
Therefore v − ϕ(v ) ∈ ker ϕ and hence v ∈ im ϕ + ker ϕ. Thus V ⊂ im ϕ ⊕ ker ϕ since im ϕ ∩ ker ϕ = 0 by (a). (c) Let V be n dimensional and {b1 , . . . , bn } a basis for V . Also let let A = [a1 , . . . , an ] be a V . Also representation matrix of ϕ with respect to {b1 , . . . , bn }, where each ai is an n-dimensional column vector for i = 1, . . . , n. n. Since ϕ2 = ϕ, A2 = A. If ϕ is nonsingular, then A = I with respect to the standard basis {e1 , . . . , en }, where I is the identity matrix. Otherwise if ϕ is singular, say ker ϕ is k dimensional, then
1 0 A = .. .
0 ··· 0 1 ··· 0 .. . . .. , . . . 0 0 ··· 0
where a j = (0, (0, . . . , 0)T for j = k, k + 1, 1, . . . , n. n. Then
1 0 A = AA = .. .
1 0 ...
0 ··· 0 1 ··· 0 .. . . .. . . . 0 0 ··· 0
2
0 ··· 0 1 1 ··· 0 0 = .. . . .. .. . . . . 0 0 ··· 0 0
0 ··· 0 1 ··· 0 .. . . .. = A . . . 0 ··· 0
with respect to the standard basis. 12. Let V =
2
, v1 = (1, (1, 0), v2 = (0, (0, 1), so that v1 , v2 are a basis for V . V . Let ϕ be the linear 2 1 transformation of V to itself whose matrix respect to this basis is . Prove that if W 0 2 is the subspace generated by v1 then W is stable under the action of ϕ. Prove that there is no subspace subspace W invariant invariant under ϕ so that V = W ⊕ W . R
Proof. Let M =
2 1 0 2
. Since W = {v ∈ V |v = kv1 ∀k ∈ R}, for all v ∈ W M v = M (kv1 ) = kM v1 = 2kv1 ∈ W . W .
So W is stable under the action of ϕ. Now suppose by contradic contradiction tion that there exists W such that V = W ⊕ W
and W is invariant under ϕ. Since
dim V + dim W ∩ W = dim W + dim W , where V and W are 2- and 1-dimensional space, respectively,
dim W = dim V + dim W ∩ W − dim W = 2 + 0 − 1 = 1.
Abstract Abstract Algebra Algebra by Dummit and Foote 4
Section 11.1 and 11.2.
Homework #4
Masaya Sato
For a basis element v3 = (a, b) ∈ V with b = 0
W = {v ∈ V |v = kv 3 ∀k ∈ R} = {(x, y) ∈ R2 |bx − ay = 0}. So (2a + 1, 1, 2b)T . W M v3 = (2a
However (2a + b) − a(2b (2b) = b2 = 0 b(2a and this contradicts that (2a (2a + 1, 1, 2b)T is in W . Therefore W is not invariant under ϕ and hence there is no subspace of V that is stable under ϕ.
38. Let A and B be square matrices. matrices. Prove Prove that the trace of their Kroneck Kronecker er product is the
product of their traces: tr( A ⊗ B ) = tr(A tr(A)tr(B )tr(B ). (Recall the the trace of a square matrix is the sum of its diagonal entries.) Proof. Let A and B be an n × n and m × m matrices, respectively. Then
a B A ⊗ B = ...
···
a1n B 11 .. . . an1 B · · · ann B
Therefore tr(A tr(A ⊗ B ) = a11 tr(B tr(B ) + · · · + ann tr(B tr(B ) = (a11 + · · · + ann )tr(B )tr( B ) = tr(A tr(A)tr(B )tr(B ).
Abstract Abstract Algebra Algebra by Dummit and Foote 5
