minqi’s solutions to the exercises of sheldon axler (2014), linear algebra done ... · 2020. 10....
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Minqi’s Solutions to the Exercises of Sheldon Axler
(2014), Linear Algebra Done Right (3rd Edition)
Minqi Pan1
1Website: http://www.minqi-pan.com
Contents
Preface ix
Chapter 1. Rn and Cn 11.1. Exercise 1.A.1: Multiplicative Inverse of Complex Numbers 11.2. Exercise 1.A.2: A Cubic Root of 1 11.3. Exercise 1.A.3: Square Roots of i 21.4. Exercise 1.A.4: Commutativity of Addition of Complex Numbers 21.5. Exercise 1.A.5: Associativity of Addition of Complex Numbers 21.6. Exercise 1.A.6: Associativity of Multiplication of Complex Numbers 21.7. Exercise 1.A.7: Existence and Uniqueness of Additive Inverse 31.8. Exercise 1.A.8: Existence and Uniqueness of Multiplicative Inverse 31.9. Exercise 1.A.9: The Distributive Property of Complex Numbers 31.10. Exercise 1.A.10: Vector Addition and Scalar Multiplication in R4 41.11. Exercise 1.A.11: A Scalar Multiplication Example in C3 41.12. Exercise 1.A.12: Associativity of Vector Addition in Fn 41.13. Exercise 1.A.13: Associativity of Scalar Multiplication in Fn 51.14. Exercise 1.A.14: Scalar Multiplication by 1 in Fn 51.15. Exercise 1.A.15: Left-distributive Property of Operations in Fn 51.16. Exercise 1.A.16: Right-distributive Property of Operations in Fn 5
Chapter 2. Definition of Vector Space 72.1. Exercise 1.B.1: −(−v) = v in Vector Spaces 72.2. Exercise 1.B.2: av = 0 Implies a = 0 or v = 0 in Vector Spaces 72.3. Exercise 1.B.3: Solutions of a Vector Equation 72.4. Exercise 1.B.4: The Empty Set is NOT a Vector Space 82.5. Exercise 1.B.5: An Alternative Definition of Vector Space 82.6. Exercise 1.B.6: A Definition of R ∪ {∞} ∪ {−∞} 8
Chapter 3. Subspaces 113.1. Exercise 1.C.1: Examples and Counterexamples of Subspaces of F3 113.2. Exercise 1.C.2: Supplying Proofs for Examples of Vector Spaces 123.3. Exercise 1.C.3: A Subspace of R(−4,4) 133.4. Exercise 1.C.4: A Subspace of R[0,1] 133.5. Exercise 1.C.5: R2 is NOT a Subspace of C2 133.6. Exercise 1.C.6: Comparing Subspaces of R3 and C3 133.7. Exercise 1.C.7: A Quasi-subspace Unclosed under Scalar Multiplication 143.8. Exercise 1.C.8: A Quasi-subspace Unclosed under Vector Addition 143.9. Exercise 1.C.9: Periodic Functions is NOT a Subspace of RR 143.10. Exercise 1.C.10: The Intersection of 2 Subspaces is a Subspace 153.11. Exercise 1.C.11: The Intersection of Subspaces is a Subspace 15
v
vi CONTENTS
3.12. Exercise 1.C.12: When the Union of 2 Subspaces is a Subspace 153.13. Exercise 1.C.13: When the Union of 3 Subspaces is a Subspace 163.14. Exercise 1.C.14: An Example of Sums of Subspaces 163.15. Exercise 1.C.15: U + U = U for Any Subspace U 173.16. Exercise 1.C.16: Commutativity of the Sum of Subspaces 173.17. Exercise 1.C.17: Associativity of the Sum of Subspaces 173.18. Exercise 1.C.18: The Identity of Sum of Subspaces 183.19. Exercise 1.C.19: There is NO Cancellation Law for Sums of Subspaces 183.20. Exercise 1.C.20: A Direct Sum of 2 Subspaces that Equals to F4 183.21. Exercise 1.C.21: A Direct Sum of 2 Subspaces that Equals to F5 193.22. Exercise 1.C.22: A Direct Sum of 4 Subspaces that Equals to F5 193.23. Exercise 1.C.23: There is NO Cancellation Law for Direct Sums 203.24. Exercise 1.C.24: A Direct Sum Decomposition of RR 20
Chapter 4. Span and Linear Independence 234.1. Exercise 2.A.1: Alternative Span from an Existing Span 234.2. Exercise 2.A.2: Examples of Linearly Independent Lists 244.3. Exercise 2.A.3: a Linearly Dependent List 254.4. Exercise 2.A.4: Conditions to Linear Dependence 254.5. Exercise 2.A.5: Difference between C and R as Scalar Fields 254.6. Exercise 2.A.6: Constructing Another Independent List from a List 264.7. Exercise 2.A.7: Constructing Another Independent List from a List 274.8. Exercise 2.A.8: Linear Independent List Multiplied by Scalar 274.9. Exercise 2.A.9: Addition of Two Linearly Independent Lists 284.10. Exercise 2.A.10: Adding a Vector to a Linearly Independent List 284.11. Exercise 2.A.11: Concatenating a Vector to an Independent List 284.12. Exercise 2.A.12: Possible Linearly Independent Lists in P4(F) 294.13. Exercise 2.A.13: Possible Spans of P4(F) 294.14. Exercise 2.A.14: Another Definition of Infinite-Dimensional 294.15. Exercise 2.A.15: F∞ is Infinite-Dimensional 294.16. Exercise 2.A.16: R[0,1] is Infinite-Dimensional 304.17. Exercise 2.A.17: Polynomials that Vanish at the Same Point 30
Chapter 5. Bases 335.1. Exercise 2.B.1: Vector Spaces that Have Exactly One Basis 335.2. Exercise 2.B.2: Examples and Counterexamples of Bases 335.3. Exercise 2.B.3: Find and Extend a Basis for a Subspace of R5 345.4. Exercise 2.B.4: Find and Extend a Basis for a Subspace of C5 355.5. Exercise 2.B.5: A Basis of P3(F) without Degree-2 Polynomials 365.6. Exercise 2.B.6: Constructing Another Basis from a Basis 385.7. Exercise 2.B.7: Constructing a Counterexample 395.8. Exercise 2.B.8: Basis Obtained from a Direct Sum 40
Chapter 6. Dimension 416.1. Exercise 2.C.1: A Space with the Same Dimension of its Subspace 416.2. Exercise 2.C.2: Subspaces of R2 416.3. Exercise 2.C.3: Subspaces of R3 416.4. Exercise 2.C.4: Find and Extend a Basis for a Subspace of P4(F) 426.5. Exercise 2.C.9: dim span(v1 + w, . . . , vm + w) > m− 1 43
CONTENTS vii
Chapter 7. The Vector Space of Linear Maps 457.1. Exercise 3.A.1: Conditions for a Mapping to be Linear 457.2. Exercise 3.A.5: L(V,W ) is a vector space 467.3. Exercise 3.A.6: Algebraic Properties of Products of Linear Maps 467.4. Exercise 3.A.7: Linear Map on 1D Space is Scalar Multiplication 477.5. Exercise 3.A.8: Homogeneity Alone does NOT Imply Linearity 477.6. Exercise 3.A.9: Additivity Alone does NOT Imply Linearity 48
Chapter 8. Null Spaces and Ranges 498.1. Exercise 3.B.1: A Linear Map T with dim null T = 3,dim range T = 2 498.2. Exercise 3.B.2: range S ⊂ null T Implies (ST )2 = 0 498.3. Exercise 3.B.3: Spanning = Surjective; Independent = Injective 498.4. Exercise 3.B.4: {T ∈ L(R5,R4) : dim null T > 2} is NOT a Subspace 508.5. Exercise 3.B.5: A Linear Map T such that range T = null T 50
Chapter 9. Matrices 539.1. Exercise 3.C.1: M(T ) Has At Least dim range T Nonzero Entries 539.2. Exercise 3.C.2: Find Bases from the Matrix of D ∈ L(P3(R),P2(R)) 539.3. Exercise 3.C.3: Find Bases for V,W from M(T ) where T ∈ L(V,W ) 549.4. Exercise 3.C.4: Find Bases for W from M(T ) where T ∈ L(V,W ) 549.5. Exercise 3.C.5: Find Bases for V from M(T ) where T ∈ L(V,W ) 54
Chapter 10. Invertibility and Isomorphic Vector Spaces 5710.1. Exercise 3.D.1: (ST )−1 = T−1S−1 5710.2. Exercise 3.D.2: Noninvertible Operators on V is NOT a Subspace 5710.3. Exercise 3.D.3: Injective Subspace Mappings and Invertible Operators 5810.4. Exercise 3.D.4: null T1 = null T2 if and only if T1 = ST2 5810.5. Exercise 3.D.5: range T1 = range T2 if and only if T1 = T2S 59
Chapter 11. Products and Quotients of Vector Spaces 6111.1. Exercise 3.E.1: T is a Linear Map iff graph of T is a Subspace of V ×W 6111.2. Exercise 3.E.2: Vj is Finite-dimensional if V1 × · · · × Vm is Finite-dimensional 6111.3. Exercise 3.E.3: Example of U1 × U2 Isomorphic to Non-Direct-Sum U1 + U2 6211.4. Exercise 3.E.4: L(V1×· · ·×Vm,W ) and L(V1,W )×· · ·×L(Vm,W ) are Isomorphic 6211.5. Exercise 3.E.5: L(V,W1×· · ·×Wm) and L(V,W1)×· · ·×L(V,Wm) are Isomorphic 63
Chapter 12. Duality 6512.1. Exercise 3.F.1: Linear Functionals are either Surjective or Zero 6512.2. Exercise 3.F.2: Examples of Linear Functionals on R[0,1] 6512.3. Exercise 3.F.3: Existence of ϕ ∈ V ′ such that ϕ(v) = 1 6612.4. Exercise 3.F.10: The Dual Map Function that Takes T to T ′ is Linear 6612.5. Exercise 3.F.20: U ⊂W Implies W 0 ⊂ U0 67
Chapter 13. Polynomials 6913.1. Exercise 4.2: {0} ∪ {p ∈ P(F) : deg p = m} is NOT a Subspace 6913.2. Exercise 4.4: There Exists a n-Degree Polynomial with m Zeros 6913.3. Exercise 4.6: p has Distinct Zeros iff p and p′ have NO Common Zeros 7013.4. Exercise 4.7: An Odd-degree, Real-coefficient Polynomial has a Real Zero 7013.5. Exercise 4.8: A Linear Operator on P(R) 70
Bibliography 73
Preface
Remarkable exercises are:
(1) Exercise 1.C.9: Periodic Functions is NOT a Subspace of RR.(2) Exercise 1.C.13: When the Union of 3 Subspaces is a Subspace.(3) Exercise 2.B.7: Constructing a Counterexample.(4) Exercise 2.C.9: dim span(v1 + w, . . . , vm + w) > m− 1.(5) Exercise 3.C.5: Find Bases for V from M(T ) where T ∈ L(V,W ).(6) Exercise 3.D.4: null T1 = null T2 if and only if T1 = ST2.(7) Exercise 3.D.5: range T1 = range T2 if and only if T1 = T2S.(8) Exercise 4.A.7: An Odd-degree, Real-coefficient Polynomial has a Real Zero.
Minqi PanApril 4, 2020
ix
CHAPTER 1
Rn and Cn
1.1. Exercise 1.A.1: Multiplicative Inverse of Complex Numbers
Suppose a and b are real numbers, not both 0. Find real numbers c and d such that
1/(a+ bi) = c+ di
Answer. Since at least one of the real numbers a, b is different from 0, we have a2 + b2 6= 0.We can then let
c =a
a2 + b2,
d =−b
a2 + b2.
Then
(a+ bi)
(a
a2 + b2− b
a2 + b2i
)=
(a
a
a2 + b2− b
(− b
a2 + b2
))+
(a
(− b
a2 + b2
)+ b
a
a2 + b2
)i
= 1 + 0i
= 1.
�
1.2. Exercise 1.A.2: A Cubic Root of 1
Show that
−1 +√
3i
2
is a cube root of 1 (meaning that its cube equals 1).
Answer.
−1 +√
3i
2· −1 +
√3i
2· −1 +
√3i
2
=−2− 2
√3i
4· −1 +
√3i
2
=(−2)(−1)− (−2
√3)(√
3) + (−2√
3 + (−2√
3)(−1))i
8= 1
�
1
2 1. Rn AND Cn
1.3. Exercise 1.A.3: Square Roots of i
Find two distinct square roots of i.
Answer. Note that
i = eiπ2 .
Thus the following two numbers are two distinct square roots of i:
±eiπ4 = ±(
√2
2+
√2
2i).
�
1.4. Exercise 1.A.4: Commutativity of Addition of Complex Numbers
Show that α+ β = β + α for all α, β ∈ C.
Proof. Suppose α = a + bi and β = c + di, where a, b, c, d ∈ R. Then the definition ofaddition of complex numbers shows that
α+ β = (a+ c) + (b+ d)i
and
β + α = (c+ a) + (d+ b)i.
The equations above and the commutativity of addition of real numbers show that α + β =β + α. �
1.5. Exercise 1.A.5: Associativity of Addition of Complex Numbers
Show that (α+ β) + λ = α+ (β + λ) for all α, β, λ ∈ C.
Proof. Suppose α = a+ bi, β = c+ di and λ = e+ fi, where a, b, c, d, e, f ∈ R. Then thedefinition of addition of complex numbers shows that
(α+ β) + λ = ((a+ c) + (b+ d)i) + e+ fi
= ((a+ c) + e) + ((b+ d) + f)i
and
α+ (β + λ) = a+ bi+ ((c+ e) + (d+ f)i)
= (a+ (c+ e)) + (b+ (d+ f))i.
The equations above and the associativity of addition of real numbers show that (α+ β) + λ =α+ (β + λ). �
1.6. Exercise 1.A.6: Associativity of Multiplication of Complex Numbers
Show that (αβ)λ = α(βλ) for all α, β, λ ∈ C.
Proof. Suppose α = a+ bi, β = c+ di and λ = e+ fi, where a, b, c, d, e, f ∈ R. Then thedefinition of multiplication of complex numbers shows that
(αβ)λ = ((ac− bd) + (ad+ bc)i)(e+ fi)
= ((ac− bd)e− (ad+ bc)f) + ((ac− bd)f + (ad+ bc)e)i
= (ace− bde− adf − bcf) + (acf − bdf + ade+ bce)i
1.9. EXERCISE 1.A.9: THE DISTRIBUTIVE PROPERTY OF COMPLEX NUMBERS 3
and
α(βλ) = (a+ bi)((ce− df) + (cf + de)i)
= (a(ce− df)− b(cf + de)) + (a(cf + de) + b(ce− df))i
= (ace− adf − bcf − bde) + (acf + ade+ bce− bdf)i.
The equations above and the commutativity of addition of real numbers show that (αβ)λ =α(βλ). �
1.7. Exercise 1.A.7: Existence and Uniqueness of Additive Inverse
Show that for every α ∈ C, there exists a unique β ∈ C such that α+ β = 0.
Proof. Suppose α = a+ bi where a, b ∈ R, then β = −a− bi satisfies α+ β = 0.Suppose that there exists another β′ ∈ C such that α+ β′ = 0. Then
β = β + (α+ β′)
= (β + α) + β′
= β′.
The equations above show that β′ = β. �
1.8. Exercise 1.A.8: Existence and Uniqueness of Multiplicative Inverse
Show that for every α ∈ C with α 6= 0, there exists a unique β ∈ C such that αβ = 1.
Proof. The existence of β has been shown in Exercise 1.A.1.Suppose that there exists another β′ ∈ C such that and αβ′ = 1. Then
β = β(αβ′)
= (βα)β′
= β′.
The equations above show that β′ = β. �
1.9. Exercise 1.A.9: The Distributive Property of Complex Numbers
Show that λ(α+ β) = λα+ λβ for all λ, α, β ∈ C.
Proof. Suppose α = a+ bi, β = c+ di and λ = e+ fi, where a, b, c, d, e, f ∈ R. Then thedefinition of multiplication and addition of complex numbers shows that
λ(α+ β) = (e+ fi)((a+ c) + (b+ d)i)
= (e(a+ c)− f(b+ d)) + (e(b+ d) + f(a+ c))i
= (ea+ ec− fb− fd) + (eb+ ed+ fa+ fc)i
and
λα+ λβ = (e+ fi)(a+ bi) + (e+ fi)(c+ di)
= ((ea− fb) + (eb+ fa)i) + ((ec− fd) + (ed+ fc)i)
= (ea− fb+ ec− fd) + (eb+ fa+ ed+ fc)i.
The equations above and the commutativity of addition of real numbers show that λ(α+β) =λα+ λβ. �
4 1. Rn AND Cn
1.10. Exercise 1.A.10: Vector Addition and Scalar Multiplication in R4
Find x ∈ R4 such that
(4,−3, 1, 7) + 2x = (5, 9,−6, 8).
Answer.
x =
(1
2, 6,−7
2,
1
2
).
�
1.11. Exercise 1.A.11: A Scalar Multiplication Example in C3
Explain why there does not exist λ ∈ C such that
λ(2− 3i, 5 + 4i,−6 + 7i) = (12− 5i, 7 + 22i,−32− 9i).
Proof. If there exists such λ then the definition of scalar multiplication in C3 shows that,
λ(2− 3i) = 12− 5i.
Thus
λ =12− 5i
2− 3i= 3 + 2i.
But
λ(2− 3i, 5 + 4i,−6 + 7i) = (3 + 2i)(2− 3i, 5 + 4i,−6 + 7i)
= (12− 5i, 7 + 22i,−32 + 9i)
6= (12− 5i, 7 + 22i,−32− 9i).
Therefore λ does not exist. �
1.12. Exercise 1.A.12: Associativity of Vector Addition in Fn
Show that (x+ y) + z = x+ (y + z) for all x, y, z ∈ Fn.
Proof. Suppose x = (x1, . . . , xn), y = (y1, . . . , yn) and z = (z1, . . . , zn) where x1, . . . , xn, y1, . . . , yn, z1, . . . , zn ∈F. Then the definition of addition in Fn shows that
(x+ y) + z = (x1 + y1, . . . , xn + yn) + (z1, . . . , zn)
= ((x1 + y1) + z1, . . . , (xn + yn) + zn)
and
x+ (y + z) = (x1, . . . , xn) + (y1 + z1, . . . , yn + zn)
= (x1 + (y1 + z1), . . . , xn + (yn + zn))
The equations above and the associativity of addition in F show that (x+ y) + z = x+ (y+z). �
1.16. EXERCISE 1.A.16: RIGHT-DISTRIBUTIVE PROPERTY OF OPERATIONS IN Fn 5
1.13. Exercise 1.A.13: Associativity of Scalar Multiplication in Fn
Show that (ab)x = a(bx) for all x ∈ Fn and all a, b ∈ F.
Proof. Suppose x = (x1, . . . , xn) where x1, . . . , xn ∈ F. Then the definition of scalarmultiplication in Fn shows that
(ab)x = ((ab)x1, . . . , (ab)xn)
and
a(bx) = a(bx1, . . . , bxn)
= (a(bx1), . . . , a(xn))
The equations above and the associativity of multiplication in F show that (ab)x = a(bx). �
1.14. Exercise 1.A.14: Scalar Multiplication by 1 in Fn
Show that 1x = x for all x ∈ Fn.
Proof. Suppose x = (x1, . . . , xn) where x1, . . . , xn ∈ F. Then the definition of scalarmultiplication in Fn and the fact that 1 is the multiplicative identity of F shows that
1x = (1x1, . . . , 1xn)
= (x1, . . . , xn)
= x.
�
1.15. Exercise 1.A.15: Left-distributive Property of Operations in Fn
Show that λ(x+ y) = λx+ λy for all λ ∈ F and x, y ∈ Fn.
Proof. Suppose x = (x1, . . . , xn) and y = (y1, . . . , yn) where x1, . . . , xn ∈ F and y1, . . . , yn ∈F. Then the definition of addition and scalar multiplication in Fn shows that
λ(x+ y) = λ(x1 + y1, . . . , xn + yn)
= (λ(x1 + y1), . . . , λ(xn + yn))
and
λx+ λy = (λx1, . . . , λxn) + (λy1, . . . , λyn)
= (λx1 + λy1, . . . , λxn + λyn)
The equations above and the distributive property of F show that λ(x+ y) = λx+ λy. �
1.16. Exercise 1.A.16: Right-distributive Property of Operations in Fn
Show that (a+ b)x = ax+ bx for all a, b ∈ F and all x ∈ Fn.
Proof. Suppose x = (x1, . . . , xn) where x1, . . . , xn ∈ F. Then the definition of scalarmultiplication and addition in Fn shows that
(a+ b)x = ((a+ b)x1, . . . , (a+ b)xn)
and
ax+ bx = (ax1, . . . , axn) + (bx1, . . . , bxn)
= (ax1 + bx1, . . . , axn + bxn).
The equations above and the distributive property of F show that (a+ b)x = ax+ bx. �
CHAPTER 2
Definition of Vector Space
2.1. Exercise 1.B.1: −(−v) = v in Vector Spaces
Prove that −(−v) = v for every v ∈ V .
Proof.
−(−v) = −(−v) + 0
= −(−v) + ((−v) + v)
= (−(−v) + (−v)) + v
= 0 + v
= v
�
2.2. Exercise 1.B.2: av = 0 Implies a = 0 or v = 0 in Vector Spaces
Suppose a ∈ F, v ∈ V , and av = 0. Prove that a = 0 or v = 0.
Proof. For the purpose of arriving at a contradiction, assume that both a 6= 0 and v 6= 0.Then
v = 1v = (1
aa)v =
1
a(av) = 0
which contradicts with v 6= 0.Therefore a = 0 or v = 0. �
2.3. Exercise 1.B.3: Solutions of a Vector Equation
Suppose v, w ∈ V . Explain why there exists a unique x ∈ V such that v + 3x = w.
Proof. We claim that the following x ∈ V satisfies v + 3x = w:
x =1
3(w − v).
We can verify it by showing that,
v + 3x = v + 3× 1
3(w − v)
= v + w − v= w.
Suppose that there exists another x′ ∈ V that also satisfies v + 3x′ = w, then
0 = w − w= (v + 3x)− (v + 3x′)
= v + 3x− v − 3x′
= 3x− 3x′.
7
8 2. DEFINITION OF VECTOR SPACE
Thus
x′ = x′ +1
3× 0
= x′ +1
3(3x− 3x′)
= x′ + x− x′
= x.
�
2.4. Exercise 1.B.4: The Empty Set is NOT a Vector Space
The empty set is not a vector space. The empty set fails to satisfy only one of the require-ments listed in 1.19. Which one?
Answer. The empty set fails to satisfy the “Additive Identity” requirement. Because theredoes not exist an element 0 ∈ ∅. �
2.5. Exercise 1.B.5: An Alternative Definition of Vector Space
Show that in the definition of a vector space (1.19), the additive inverse condition can bereplaced with the condition that
0v = 0,∀v ∈ V.Here the 0 on the left side is the number 0, and the 0 on the right side is the additive identity
of V .
Proof. We will show that the new condition is equal to the original condition, which statesthat for every v ∈ V , there exists w ∈ V such that v + w = 0. That the old condition impliesthe new condition has been shown in Theorem 1.29 “The number 0 times a vector”. Thereforeit suffices to show that the new condition implies the old condition.
Suppose that the new condition holds.Pick v ∈ V , and let w = (−1)v. It follows that
v + w = 1v + (−1)v
= (1− 1)v
= 0v
= 0.
Therefore the old condition holds. �
2.6. Exercise 1.B.6: A Definition of R ∪ {∞} ∪ {−∞}
Let∞ and −∞ denote two distinct objects, neither of which is in R. Define an addition andscalar multiplication on R∪{∞}∪{−∞} as you could guess from the notation. Specifically, thesum and product of two real numbers is as usual, and for t ∈ R define
t∞ =
−∞ if t < 0,
0 if t = 0,
∞ if t > 0,
t(−∞) =
∞ if t < 0,
0 if t = 0,
−∞ if t > 0,
t+∞ =∞+ t =∞, t+ (−∞) = (−∞) + t = (−∞),
∞+∞ =∞, (−∞) + (−∞) = (−∞), ∞+ (−∞) = 0.
Is R ∪ {∞} ∪ {−∞} a vector space over R? Explain.
2.6. EXERCISE 1.B.6: A DEFINITION OF R ∪ {∞} ∪ {−∞} 9
Answer. No. Because the associativity of vector addition fails:
(1024 +∞)−∞ =∞−∞ = 0
but1024 + (∞−∞) = 1024 + 0 = 1024.
�
CHAPTER 3
Subspaces
3.1. Exercise 1.C.1: Examples and Counterexamples of Subspaces of F3
For each of the following subsets of F3, determine whether it is a subspace of F3:
(a) {(x1, x2, x3) ∈ F3 : x1 + 2x2 + 3x3 = 0};(b) {(x1, x2, x3) ∈ F3 : x1 + 2x2 + 3x3 = 4};(c) {(x1, x2, x3) ∈ F3 : x1x2x3 = 0};(d) {(x1, x2, x3) ∈ F3 : x1 = 5x3}.
Answer. (a) Denote A = {(x1, x2, x3) ∈ F3 : x1 + 2x2 + 3x3 = 0}. We claim that Ais a subspace of F3 with the following verifications.(i) (0, 0, 0) ∈ A because 0 + 2 · 0 + 3 · 0 = 0.(ii) Pick u,w ∈ A, where u = (u1, u2, u3), w = (w1, w2, w3). It follows that u + w =
(u1 +w1, u2 +w2, u3 +w3). We can show that u+w ∈ A by the following equations:
(u1 + w1) + 2(u2 + w2) + 3(u3 + w3)
= (u1 + 2u2 + 3u3) + (w1 + 2w2 + 3w3)
= 0 + 0 = 0.
(iii) Pick a ∈ F and u ∈ A, where u = (u1, u2, u3). It follows that au = (au1, au2, au3).We can show that au ∈ A by the following equations:
(au1) + 2(au2) + 3(au3) = a(u1 + 2u2 + 3u3)
= a0 = 0.
(b) Denote B = {(x1, x2, x3) ∈ F3 : x1 +2x2 +3x3 = 4}. B is not a subspace of F3, becausevector (0, 0, 0) /∈ B (as 0 + 2 · 0 + 3 · 0 6= 4).
(c) Denote C = {(x1, x2, x3) ∈ F3 : x1x2x3 = 0}. C is not a subspace of F3 because bysetting u = (0, 1, 1) ∈ C and w = (1, 1, 0) ∈ C, we find out that u+ w = (1, 2, 1) /∈ C.
(d) Denote D = {(x1, x2, x3) ∈ F3 : x1 = 5x3}. We claim that D is a subspace of F3 withthe following verifications.
(i) (0, 0, 0) ∈ A because 0 = 5 ∗ 0.(ii) Pick u,w ∈ A, where u = (u1, u2, u3), w = (w1, w2, w3). It follows that u + w =
(u1 +w1, u2 +w2, u3 +w3). We can show that u+w ∈ A by the following equations:
u1 + w1 = 5u3 + 5w3
= 5(u1 + w3)
(iii) Pick a ∈ F and u ∈ A, where u = (u1, u2, u3). It follows that au = (au1, au2, au3).We can show that au ∈ A by the following equations:
au1 = a(5u3)
= 5(au3).
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11
12 3. SUBSPACES
3.2. Exercise 1.C.2: Supplying Proofs for Examples of Vector Spaces
Verify all the assertions in Example 1.35.
(a) If b ∈ F, then
{(x1, x2, x3, x4) ∈ F4 : x3 = 5x4 + b}is a subspace of F4 if and only if b = 0.
(b) The set of continuous real-values functions on the interval [0, 1] is a subspace of R[0,1].(c) The set of differentiable real-valued functions on R is a subspace of RR
(d) The set of differentiable real-valued functions f on the interval (0, 3) such that f ′(2) = bis a subspace of R(0,3) if and only if b = 0.
(e) The set of all sequences of complex numbers with limit 0 is a subspace of C∞.
Answer. (a) Denote A = {(x1, x2, x3, x4) ∈ F4 : x3 = 5x4 + b}.If b = 0, then A is a subspace by a verification similar to Exercise 1.C.2 (d).On the other hand, if A is a subspace, then 0 ∈ A. It follows that
0 = 5 ∗ 0 + b = b.
(b) Denote B = {f ∈ R[0,1] : f is continuous}.We can verify that 0 ∈ B because the constant function f = 0 is continuous.Suppose that u,w ∈ B, we can verify that (u+w) ∈ B because the addition of two
complex continuous functions is continuous by Theorem 4.9 of [Rud76].Suppose that a ∈ F and u ∈ B. Note that the constant function f = a is continuous.
We can verify that au ∈ B because the multiplication of two continuous complexfunctions is continuous by Theorem 4.9 of [Rud76].
(c) Denote C = {f ∈ RR : f is differentiable}.We can verify that 0 ∈ C because the constant function f = 0 is differentiable.Suppose that u,w ∈ C, we can verify that (u+w) ∈ C because the addition of two
differentiable real functions is differentiable by Theorem 5.3 of [Rud76].Suppose that a ∈ R and u ∈ C. Note that the constant function a(x) = a is
differentiable. We can verify that au ∈ C because the multiplication of two differentiablereal functions is differentiable by Theorem 5.3 of [Rud76].
(d) Denote D = {f ∈ R(0,3) : f is differentiable and f ′(2) = b}.Suppose that b = 0. We claim that D is a subspace of R(0,3) with the following
verifications.(i) 0 ∈ D because the constant function 0(x) = 0 is differentiable and 0′(2) = 0.(ii) Suppose that u,w ∈ D. (u+w) is also differentiable by Theorem 5.3 of [Rud76].
Also by Theorem 5.3 of [Rud76], (u + v)′(2) = u′(2) + v′(2) = 0 + 0 = 0. Thus(u+ w) ∈ D.
(iii) Suppose that a ∈ R and u ∈ D. Note that the constant function a(x) = a is differ-entiable, so au is differentiable by Theorem 5.3 of [Rud76]. Also by Theorem 5.3of [Rud76], (au)′(2) = a′(2)u(2) +a(2)u′(2) = 0×u(2) +a×0 = 0. Thus au ∈ D.
On the other hand, suppose that D is a subspace. Then it implies that 0 ∈ Dwhere 0(x) = 0, and 0′(x) = 0 implies that 0′(2) = b = 0.
(e) Denote E = {zn ∈ C∞ : limn→∞ zn = 0}.We can verify that 0n ∈ E where 0n = (0, 0, 0, . . . ) because
limn→∞
0n = 0.
Suppose that un, wn ∈ E, we can verify that (un +wn) ∈ C because limn→∞(un +wn) = limn→∞ un + limn→∞ wn = 0 + 0 = 0 by Theorem 3.3 of [Rud76].
3.6. EXERCISE 1.C.6: COMPARING SUBSPACES OF R3 AND C3 13
Suppose that a ∈ F and un ∈ E, we can verify that aun ∈ E because limn→∞ aun =a limn→∞ un = a0 = 0 by Theorem 3.3 of [Rud76].
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3.3. Exercise 1.C.3: A Subspace of R(−4,4)
Show that the set of differentiable real-valued functions f on the interval (−4, 4) such thatf ′(−1) = 3f(2) is a subspace of R(−4,4).
Proof. Denote X = {f ∈ R(−4,4) : f is differentiable and f ′(−1) = 3f(2)}(1) We claim that 0 ∈ X where 0(x) = 0. Note that 0 is differentiable and 0′(−1) =
0, 0(2) = 0, thus 0′(−1) = 3× 0(2).(2) Pick f, g ∈ X. That f + g is differentiable follows from Theorem 5.3 of [Rud76].
Also by Theorem 5.3 of [Rud76], (f + g)′(−1) = f ′(−1) + g′(−1) = 3f(2) + 3g(2) =3(f(2) + g(2)) = 3((f + g)(2)). Thus f + g ∈ X.
(3) Pick f ∈ X and a ∈ F. That af is differentiable follows from Theorem 5.3 of [Rud76]by considering a as a constant function a(x) = a. Also by Theorem 5.3 of [Rud76],(af)′(−1) = a(f ′(−1)) = a(3f(2)) = 3(af(2)) = 3((af)(2)). Thus af ∈ X.
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3.4. Exercise 1.C.4: A Subspace of R[0,1]
Suppose b ∈ R. Show that the set of continuous real-valued functions f on the interval [0, 1]
such that∫ 1
0f = b is a subspace of R[0,1] if and only if b = 0.
Proof. Denote X = {f ∈ R[0,1] :f is continuous and∫ 1
0f = b}.
Suppose that b = 0. We now verify that X is a subspace.
(1) We can verify that 0 ∈ X because 0(x) = 0 is continuous and∫ 1
00(x)dx = 0.
(2) Pick f, g ∈ X. It follows that f + g is continuous by Theorem 4.9 of [Rud76] andRiemann-integrable by Theorem 6.8 of [Rud76]. Further by Theorem 6.12 of [Rud76],
we have∫ 1
0(f(x) + g(x))dx =
∫ 1
0f(x)dx+
∫ 1
0g(x) = 0 + 0 = 0.
(3) Pick a ∈ F, f ∈ X. By Theorem 4.9 of [Rud76], af is continuous. Thus af is Riemann-integrable by Theorem 6.8 of [Rud76]. Further by Theorem 6.12 of [Rud76], we have∫ 1
0(af(x))dx = a
∫ 1
0f(x)dx = a× 0 = 0.
On the other hand, suppose that X is a subspace. Then it follows that 0 ∈ X, which implies
that∫ 1
00(x)dx = 0 = b. �
3.5. Exercise 1.C.5: R2 is NOT a Subspace of C2
Is R2 a subspace of the complex vector space C2?
Answer. No. Because multiplying the vector (1, 1) ∈ R2 by scalar i gives (i, i) which is notin R2. �
3.6. Exercise 1.C.6: Comparing Subspaces of R3 and C3
(i) Is {(a, b, c) ∈ R3 : a3 = b3} a subspace of R3?(ii) Is {(a, b, c) ∈ C3 : a3 = b3} a subspace of C3?
Answer. (i) Yes. Let’s observe the real function f(x) = x3. f(x) is a 1-1 correspon-dence between R and R. Therefore the condition a3 = b3 is equivalent to a = b.
14 3. SUBSPACES
(ii) No. Denote X = {(a, b, c) ∈ C3 : a3 = b3}. Set u = (2, 2e2π3 i, 0) and v = (e
2π3 i, 1, 0),
then u+v = (2+e2π3 i, 2e
2π3 i+1, 0). Note that u ∈ X and v ∈ X but u+v /∈ X because,(
2 + e2π3 i)3
= 3√
3i,(2e
2π3 i + 1
)3
= −3√
3i.
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3.7. Exercise 1.C.7: A Quasi-subspace Unclosed under Scalar Multiplication
Give an example of a nonempty subset U of R2 such that U is closed under addition andunder taking additive inverses (meaning −u ∈ U whenever u ∈ U), but U is not a subspace ofR2.
Answer.U = {(0, 0), (1, 1), (−1,−1)}.
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3.8. Exercise 1.C.8: A Quasi-subspace Unclosed under Vector Addition
Give an example of a nonempty subset U of R2 such that U is closed under scalar multipli-cation, but U is not a subspace of R2.
Answer.U = {(0, y) : y ∈ R} ∪ {(x, 0) : x ∈ R}.
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3.9. Exercise 1.C.9: Periodic Functions is NOT a Subspace of RR
A function f : R → R is called periodic if there exists a positive number p such thatf(x) = f(x + p) for all x ∈ R. Is the set of periodic functions from R to R a subspace of RR?Explain.
Lemma 3.1. f(x) = cos(πx) + cos(x) is not periodic.
Proof. Suppose that f(x) = cos(πx) + cos(x) is periodic, then there exists θ ∈ R such thatθ > 0 and f(x+ θ) = f(x) for all x ∈ R. Pick x = 0, we have
(3.1) f(θ + 0) = cos(πθ) + cos(θ) = f(0) = cos(0) + cos(0) = 2.
Note that −1 6 cos(πθ) 6 1 and −1 6 cos(θ) 6 1. However, if cos(πθ) < 1 and cos(θ) < 1,then cos(πθ) + cos(θ) < 1. Thus (3.1) implies that{
cos(πθ) = 1
cos(θ) = 1.
This further implies that there exists some m,n ∈ Z such that,{πθ = 2mπ
θ = 2nπ.
Since θ 6= 0, dividing the above two equations yields
π =m
n.
This contradicts with the fact that π is irrational. �
3.12. EXERCISE 1.C.12: WHEN THE UNION OF 2 SUBSPACES IS A SUBSPACE 15
Main Answer. No, because Lemma 3.1 gives an example of two periodic functions wheretheir addition is not periodic. �
3.10. Exercise 1.C.10: The Intersection of 2 Subspaces is a Subspace
Suppose U1 and U2 are subspaces of V . Prove that the intersection U1 ∩U2 is a subspace ofV .
Proof. We prove by verifying the three conditions for U1 ∩ U2 being a subspace.
(1) The premise that U1 is a subspace implies 0 ∈ U1. The premise that U2 is a subspaceimplies 0 ∈ U2. Therefore 0 ∈ U1 ∩ U2.
(2) Pick u,w ∈ U1 ∩ U2. The premise that U1 is a subspace implies u + w ∈ U1. Thepremise that U2 is a subspace implies u+ w ∈ U2. Therefore u+ w ∈ U1 ∩ U2.
(3) Pick a ∈ F and u ∈ U1 ∩ U2. The premise that U1 is a subspace implies that au ∈ U1.The premise that U2 is a subspace implies au ∈ U2. Therefore au ∈ U1 ∩ U2.
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3.11. Exercise 1.C.11: The Intersection of Subspaces is a Subspace
Prove that the intersection of every collection of subspaces of V is a subspace of V .
Proof. Pick set I and (Si)i∈I , a family of sets indexed over set I. Suppose that Si isa subspace of V for all i ∈ I. We prove by verifying the three conditions for ∩i∈ISi being asubspace.
(1) For all i ∈ I, the premise that Si is a subspace implies that 0 ∈ Si. Therefore 0 ∈ ∩i∈ISi.(2) Pick u,w ∈ ∩i∈ISi. For all i ∈ I, the premise that Si is a subspace implies u+w ∈ Si.
Therefore u+ w ∈ ∩i∈ISi.(3) Pick a ∈ F and u ∈ ∩i∈ISi. For all i ∈ I, the premise that Ui is a subspace implies that
au ∈ Ui. Therefore au ∈ ∩i∈ISi.
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3.12. Exercise 1.C.12: When the Union of 2 Subspaces is a Subspace
Prove that the union of two subspaces of V is a subspace of V if and only if one of thesubspaces is contained in the other.
Proof. Pick two subspaces V1, V2 of V .
(1) If V1 ⊂ V2, then V1 ∪ V2 = V2. Since V2 is a subspace, it follows that V1 ∪ V2 is asubspace.
(2) Suppose that V1 ∪ V2 is a subspace. To prove by contradiction, suppose that V1 6⊂ V2
and V2 6⊂ V1. Thus there exists v1 ∈ V1 such that v1 /∈ V2, and there exists v2 ∈ V2
such that v2 /∈ V1.We claim that v1 + v2 /∈ V1. Because if v1 + v2 was in V1, the premise that V1 is a
subspace would imply that v2 = (v1 + v2)− v1 is also in V1.We claim that v1 + v2 /∈ V2. Because if v1 + v2 was in V2, the premise that V2 is a
subspace would imply that v1 = (v1 + v2)− v2 is also in V2.Thus v1 + v2 /∈ V1 ∪ V2, contradicting with the premise that V1 ∪ V2 is a subspace.
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16 3. SUBSPACES
3.13. Exercise 1.C.13: When the Union of 3 Subspaces is a Subspace
Prove that the union of three subspaces of V is a subspace of V if and only if one of thesubspaces contains the other two.
Proof. Pick three subspaces V1, V2, V3 of V .
(1) If V1 ⊂ V3 and V2 ⊂ V3, then V1 ∪ V2 ∪ V3 = V3. Since V3 is a subspace, it follows thatV1 ∪ V2 ∪ V3 is a subspace.
(2) Suppose that V1 ∪ V2 ∪ V3 is a subspace of V .If V1 ⊂ V2, then V1 ∪ V2 ∪ V3 becomes V2 ∪ V3. By Exercise 1.C.12, it follows that
either V2 ⊂ V3 or V3 ⊂ V2. Thus either V3 or V2 is the subspace that contains the othertwo.
If V2 ⊂ V1, then V1 ∪ V2 ∪ V3 becomes V1 ∪ V3. By Exercise 1.C.12, it follows thateither V1 ⊂ V3 or V3 ⊂ V1. Thus either V3 or V1 is the subspace that contains the othertwo.
If V1 6⊂ V2 and V2 6⊂ V1, then there exists u ∈ V1 and w ∈ V2 such that u /∈ V2 andw /∈ V1.(i) We claim that V1 \ (V1 ∩V2) ⊂ V3. Pick x ∈ V1 \ (V1 ∩V2) ⊂ V3. The premise that
V1∪V2∪V3 is a subspace implies that x+w ∈ V1∪V2∪V3 and 2x+w ∈ V1∪V2∪V3.If x+w was in V1, then w = (x+w)−x would be in V1 which is a contradiction. If2x+w was in V1, then w = (2x+w)− 2x would be in V1 which is a contradiction.If x+w was in V2, then x = (x+w)−w would be in V2 which is a contradiction. If2x+w was in V2, then 2x = (2x+w)−w would be in V2 which is a contradiction.Therefore x+w ∈ V3 and 2x+w ∈ V3. And it follows that x = (2x+w)− (x+w)is in V3.
(ii) We claim that V2 \ (V1 ∩ V2) ⊂ V3. Pick y ∈ V2 \ (V1 ∩ V2) ⊂ V3. The premise thatV1∪V2∪V3 is a subspace implies that u+y ∈ V1∪V2∪V3 and u+2y ∈ V1∪V2∪V3.If u+ y was in V1, then y = (u+ y)−u would be in V1 which is a contradiction. Ifu+ 2y was in V1, then 2y = (u+ 2y)− u would be in V1 which is a contradiction.If u+ y was in V2, then u = (u+ y)− y would be in V2 which is a contradiction. Ifu+ 2y was in V2, then u = (u+ 2y)− 2y would be in V2 which is a contradiction.Therefore u+ y ∈ V3 and u+ 2y ∈ V3. And it follows that y = (u+ 2y)− (u+ y)is in V3.
(iii) We claim that V1 ∩ V2 ⊂ V3. Pick z ∈ V1 ∩ V2. The premise that V1 is a subspaceimplies that u + z ∈ V1. If u + z ∈ V1 ∩ V2, then u = (u + z) − z ∈ V2 which is acontradiction. Thus u+ z ∈ V1 \ (V1 ∩ V2). By (i) we have u+ z ∈ V3. Also by (i)we have u ∈ V3. The premise that V3 is a subspace implies that z = (u+ z)− u isin V3.
The above (i) and (iii) implies that V1 ⊂ V3. The above (ii) and (iii) implies thatV2 ⊂ V3.
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3.14. Exercise 1.C.14: An Example of Sums of Subspaces
Verify the assertion in Example 1.38: Suppose that U = {(x, x, y, y) ∈ F4 : x, y ∈ F} andW={(x, x, x, y) ∈ F4 : x, y ∈ F}. Then
U +W = {(x, x, y, z) ∈ F4 : x, y, z ∈ F}.
Proof. (1) Pick x ∈ U + W . Then there exists u = (xu, xu, yu, yu) ∈ U and w =(xw, xw, xw, yw) ∈ W such that xu, yu, xw, yw ∈ F and x = u + w. Thus x = (xu +
3.17. EXERCISE 1.C.17: ASSOCIATIVITY OF THE SUM OF SUBSPACES 17
xw, xu + xw, yu + xw, yu + yw), which shows that x ∈ {(x, x, y, z) ∈ F4 : x, y, z ∈ F}.Therefore U +W ⊂ {(x, x, y, z) ∈ F4 : x, y, z ∈ F}.
(2) Pick v ∈ {(x, x, y, z) ∈ F4 : x, y, z ∈ F}. Then there exists x, y, z ∈ F such thatv = (x, x, y, z). By choosing u = (x, x, y, y) ∈ U and w = (0, 0, 0, z − y) ∈ W , we haveu+ w = v. Thus v ∈ U +W . Therefore {(x, x, y, z) ∈ F4 : x, y, z ∈ F} ⊂ U +W .
The above (1) and (2) imply that U +W = {(x, x, y, z) ∈ F4 : x, y, z ∈ F}. �
3.15. Exercise 1.C.15: U + U = U for Any Subspace U
Suppose U is a subspace of V . What is U + U?
Proof. We claim that U + U = U .
(1) Pick z ∈ U +U , then there exists x, y ∈ U such that z = x+ y. Since U is closed underaddition, we have x+ y ∈ U , i.e. z ∈ U . Thus U + U ⊂ U .
(2) Pick u ∈ U , then u = 1u = ( 12 + 1
2 )u = 12u+ 1
2u, i.e. u ∈ U + U . Thus U ⊂ U + U .
The above (1) and (2) imply that U + U = U . �
3.16. Exercise 1.C.16: Commutativity of the Sum of Subspaces
Is the operation of addition on the subspaces of V commutative? In other words, if U andW are subspaces of V , is U +W = W + U?
Proof. We claim that U +W = W + U .
(1) Pick v ∈ U + W , then there exists u ∈ U and w ∈ W such that v = u + w. Since theaddition on V is commutative, u + w = w + u, and this shows that v = w + u is inW + U . Thus U +W ⊂W + U .
(2) Pick v ∈ W + U , then there exists w ∈ W and u ∈ U such that v = w + u. Since theaddition on V is commutative, w + u = u + w, and this shows that v = u + w is inU +W . Thus W + U ⊂ U +W .
The above (1) and (2) imply that U +W = W + U . �
3.17. Exercise 1.C.17: Associativity of the Sum of Subspaces
Is the operation of addition on the subspaces of V associative? In other words, if U1, U2, U3
are subspaces of V , is
(U1 + U2) + U3 = U1 + (U2 + U3)?
Proof. We claim that (U1 + U2) + U3 = U1 + (U2 + U3).
(1) Pick v ∈ (U1 + U2) + U3, then there exists u1, u2, u3 ∈ U such that v = (u1 + u2) + u3.Since the addition on V is associative, (u1 + u2) + u3 = u1 + (u2 + u3), and this showsthat v = u1 + (u2 + u3) is in U1 + (U2 + U3). Thus (U1 + U2) + U3 ⊂ U1 + (U2 + U3).
(2) Pick v ∈ U1 + (U2 + U3), then there exists u1, u2, u3 ∈ U such that v = u1 + (u2 + u3).Since the addition on V is associative, u1 + (u2 + u3) = (u1 + u2) + u3, and this showsthat v = (u1 + u2) + u3 is in (U1 + U2) + U3. Thus U1 + (U2 + U3) ⊂ (U1 + U2) + U3.
The above (1) and (2) imply that (U1 + U2) + U3 = U1 + (U2 + U3). �
18 3. SUBSPACES
3.18. Exercise 1.C.18: The Identity of Sum of Subspaces
Does the operation of addition on the subspaces of V have an additive identity? Whichsubspaces have additive inverses?
Proof. We claim that the subspace {0} is the additive identity, i.e. U + {0} = U for allsubspaces U .
(1) Pick v ∈ U + {0}, then there exists u ∈ U such that v = u + 0, and this shows thatv = u is in U . Thus U + {0} ⊂ U .
(2) Pick u ∈ U , then u = u+ 0 shows that u ∈ U + {0}. Thus U ⊂ U + {0}.The above (1) and (2) imply that U + {0} = U .
We claim that for all subspaces U and W , U +W = {0} implies U = W = {0}, which meansthat the subspaces that have additive inverses are only {0} itself.
Pick subspaces U and W . Assume U + W = {0}. 1.39 of [Axl14] shows that U ⊂ U + Wand W ⊂ U + W . Thus U ⊂ {0} and W ⊂ {0}. Note that the only two subsets of {0} are∅ and {0}. Also, the premise that U and W are subspaces implies 0 ∈ U and 0 ∈ W , thusU = W = {0}. �
3.19. Exercise 1.C.19: There is NO Cancellation Law for Sums of Subspaces
Prove or give a counterexample: if U1, U2,W are subspaces of V such that
U1 +W = U2 +W,
then U1 = U2.
Counterexample. Set V = R2, U1 = R2, U2 = {(x, 0) : x ∈ R},W = {(0, x) : x ∈ R}.Then U1, U2,W are subspaces of V , U1 +W = R2, U2 +W = R2 but U1 6= U2. �
3.20. Exercise 1.C.20: A Direct Sum of 2 Subspaces that Equals to F4
Suppose
U = {(x, x, y, y) ∈ F4 : x, y ∈ F}.Find a subspace W of F4 such that F4 = U ⊕W .
Proof. We claim that the following W is a subspace of F4 such that F4 = U ⊕W :
W = {(0, x, 0, y) ∈ F4 : x, y ∈ F}.We first verify that W is a subspace of F4 by 1.34 of [Axl14].
(1) We can show that 0 ∈W by choosing x = 0, y = 0 in (0, x, 0, y).(2) Suppose that u,w ∈ W where u = (0, xu, 0, yu) and w=(0, xw, 0, yw). Then u + w =
(0, xu + xw, 0, yu + yw) shows that u+ w is in W .(3) Pick a ∈ F and u ∈ W where u = (0, xu, 0, yu). Then au = (0, axu, 0, ayu) shows that
au is in W .
We then verify that U +W is a direct sum by 1.45 of [Axl14].Pick v ∈ U ∩W . Then there exists x, y, s, t ∈ F such that v = (x, x, y, y) = (0, s, 0, t), which
implies that x = 0, s = 0, y = 0, t = 0, i.e. v = 0. Therefore U +W is a direct sum.Finally, we verify that F4 = U ⊕W .
(1) 1.39 of [Axl14] implies that U ⊕W is a subspace of F4. Therefore U ⊕W ⊂ F4.(2) Pick v ∈ F4 where v = (a, b, c, d) and a, b, c, d ∈ F. Define
u = (a, a, c, c), w = (0, b− a, 0, d− c).Then u ∈ U,w ∈W and v = u+ w, which shows that v ∈ U ⊕W . Thus F4 ⊂ U ⊕W .
3.22. EXERCISE 1.C.22: A DIRECT SUM OF 4 SUBSPACES THAT EQUALS TO F5 19
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3.21. Exercise 1.C.21: A Direct Sum of 2 Subspaces that Equals to F5
Suppose
U = {(x, y, x+ y, x− y, 2x) ∈ F5 : x, y ∈ F}.Find a subspace W of F5 such that F = U ⊕W .
Proof. We claim that the following W is a subspace of F5 such that F4 = U ⊕W :
W = {(0, 0, x, y, z) ∈ F5 : x, y, z ∈ F}.
We first verify that W is a subspace of F5 by 1.34 of [Axl14].
(1) We can show that 0 ∈W by choosing x = 0, y = 0, z = 0 in (0, 0, x, y, z).(2) Suppose that u,w ∈W where
u = (0, 0, xu, yu, zu), w = (0, 0, xw, yw, zw).
Then u+ w = (0, 0, xu + xw, yu + yw, zu + zw) shows that u+ w is in W .(3) Pick a ∈ F and u ∈W where u = (0, 0, xu, yu, zu). Then
au = (0, 0, axu, ayu, azu)
shows that au is in W .
We then verify that U +W is a direct sum by 1.45 of [Axl14].Pick v ∈ U ∩W . Then there exists x, y, a, b, c ∈ F such that v = (x, y, x + y, x − y, 2x) =
(0, 0, a, b, c), which implies that x = 0, y = 0, a = 0, b = 0, c = 0, i.e. v = 0. Therefore U +W isa direct sum.
Finally, we verify that F5 = U ⊕W .
(1) 1.39 of [Axl14] implies that U ⊕W is a subspace of F5. Therefore U +W ⊂ F5.(2) Pick v ∈ F5 where v = (a, b, c, d, e) and a, b, c, d, e ∈ F. Define u = (a, b, a + b, a −
b, 2a), w = (0, 0, c− a− b, d− a+ b, e− 2a). Then u ∈ U,w ∈W and v = u+w, whichshows that v ∈ U +W . Thus F5 ⊂ U ⊕W .
�
3.22. Exercise 1.C.22: A Direct Sum of 4 Subspaces that Equals to F5
Suppose
U = {(x, y, x+ y, x− y, 2x) ∈ F5 : x, y ∈ F}.Find three subspaces W1,W2,W3 of F5, none of which equals {0}, such that F5 = U ⊕W1 ⊕W2 ⊕W3.
Proof. We claim that the following W1,W2,W3 are subspaces of F5 such that F5 = U ⊕W1 ⊕W2 ⊕W3:
W1 = {(0, 0, x, 0, 0) ∈ F5 : x ∈ F},W2 = {(0, 0, 0, x, 0) ∈ F5 : x ∈ F},W3 = {(0, 0, 0, 0, x) ∈ F5 : x ∈ F}.
Suppose that
(3.2) 0 = u+ w1 + w2 + w3
20 3. SUBSPACES
for some u = (x, y, x+ y, x− y, 2x), w1 = (0, 0, a, 0, 0), w2 = (0, 0, 0, b, 0), w3 = (0, 0, 0, 0, c) wherex, y, a, b, c ∈ F. Then (3.2) is equivalent to the following system of linear equations.
1 0 0 0 00 1 0 0 01 1 1 0 01 −1 0 1 02 0 0 0 1
xyabc
=
00000
.It is now clear that the coefficient matrix is nonsingular, thus x = y = a = b = c = 0 is the
only solution. By 1.44 of [Axl14], we conclude that F5 = U ⊕W1 ⊕W2 ⊕W3. �
3.23. Exercise 1.C.23: There is NO Cancellation Law for Direct Sums
Prove or give a counterexample: if U1, U2,W are subspaces of V such that
V = U1 ⊕W and V = U2 ⊕W,
then U1 = U2.
Counterexample. Set V = R2, U1 = {(x, x) : x ∈ R}, U2 = {(−x, x) : x ∈ R},W ={(0, x) : x ∈ R}. Then U1, U2,W are subspaces of V , U1 ⊕ W = R2, U2 ⊕ W = R2 butU1 6= U2. �
3.24. Exercise 1.C.24: A Direct Sum Decomposition of RR
A function f : R→ R is called even if
f(−x) = f(x)
for all x ∈ R. A function f : R→ R is called odd if
f(−x) = −f(x)
for all x ∈ R. Let Ue denote the set of real-valued even functions on R and let Uo denote the setof real-valued odd functions on R. Show that RR = Ue ⊕ Uo.
Proof. We first verify that Ue and Uo are subspaces of RR by 1.34 of [Axl14].
(1) We can show that 0 ∈ Ue and 0 ∈ Uo because the 0(−x) = 0(x) = 0 and 0(−x) =−0(x) = 0 for all x ∈ R.
(2) Suppose that ue, we ∈ Ue, uo, wo ∈ Uo, then
(ue + we)(−x) = ue(−x) + we(−x)
= ue(x) + we(x)
= (ue + we)(x).
Thus ue + we ∈ Ue. Also,
(uo + wo)(−x) = uo(−x) + wo(−x)
= −uo(x)− wo(x)
= −(uo(x) + wo(x))
= −(uo + wo)(x).
Thus uo + wo ∈ Uo.
3.24. EXERCISE 1.C.24: A DIRECT SUM DECOMPOSITION OF RR 21
(3) Pick a ∈ F and ue ∈ Ue, uo ∈ Uo, then
(aue)(−x) = a · ue(−x)
= a · ue(x)
= (aue)(x).
Thus aue ∈ Ue. Also,
(auo)(−x) = a · uo(−x)
= −a · uo(x)
= −(auo)(x).
Thus auo ∈ Uo.
We then verify that Ue + Uo is a direct sum by 1.45 of [Axl14].Pick u ∈ Ue ∩ Uo, then u(−x) = u(x) = −u(x) for all x ∈ R. Thus u(x) = 0 for all x ∈ R,
which means that u is the 0 function.Finally, we prove that RR = Ue ⊕ Uo.
(1) 1.39 of [Axl14] implies that Ue ⊕ Uo is a subspace of RR. Therefore Ue ⊕ Uo ⊂ RR.(2) Pick f(x) ∈ RR. Define
fe(x) =f(x) + f(−x)
2, fo(x) =
f(x)− f(−x)
2.
Then
fe(−x) =f(−x) + f(x)
2
=f(x) + f(−x)
2= fe(x),
which implies that fe(x) ∈ Ue. Also,
fo(−x) =f(−x)− f(x)
2
= −f(x)− f(−x)
2= −fo(x),
which implies that fo(x) ∈ Uo. And,
fe(x) + fo(x) =f(x) + f(−x)
2+f(x)− f(−x)
2
= −f(x) + f(−x) + f(x)− f(−x)
2= f(x),
which shows that f(x) ∈ Ue ⊕ Uo. Therefore RR ⊂ Ue ⊕ Uo.
�
CHAPTER 4
Span and Linear Independence
4.1. Exercise 2.A.1: Alternative Span from an Existing Span
Suppose v1, v2, v3, v4 spans V . Prove that the list
v1 − v2, v2 − v3, v3 − v4, v4
also spans V .
Proof. Denote w1 = v1 − v2, w2 = v2 − v3, w3 = v3 − v4, w4 = v4. Let
A =
1 −1 0 00 1 −1 00 0 1 −10 0 0 1
,then A is nonsingular and
A−1 =
1 1 1 10 1 1 10 0 1 10 0 0 1
.Moreover,
A
v1
v2
v3
v4
=
w1
w2
w3
w4
,v1
v2
v3
v4
= A−1
w1
w2
w3
w4
.(1) Pick x ∈ span(v1, v2, v3, v4), then there exists x1, x2, x3, x4 ∈ F such that
v =[x1 x2 x3 x4
] v1
v2
v3
v4
=[x1 x2 x3 x4
] 1 1 1 10 1 1 10 0 1 10 0 0 1
w1
w2
w3
w4
=[x1 x1 + x2 x1 + x2 + x3 x1 + x2 + x3 + x4
] w1
w2
w3
w4
.Therefore v ∈ span(w1, w2, w3, w4).
23
24 4. SPAN AND LINEAR INDEPENDENCE
(2) Pick w ∈ span(w1, w2, w3, w4), then there exists y1, y2, y3, y4 ∈ F such that
w =[y1 y2 y3 y4
] w1
w2
w3
w4
=[y1 y2 y3 y4
] 1 −1 0 00 1 −1 00 0 1 −10 0 0 1
v1
v2
v3
v4
=[y1 y2 − y1 y3 − y2 y4 − y3
] v1
v2
v3
v4
.Therefore w ∈ span(v1, v2, v3, v4).
The above (1) and (2) imply that span(w1, w2, w3, w4) = span(v1, v2, v3, v4). �
4.2. Exercise 2.A.2: Examples of Linearly Independent Lists
Verify the assertions in Example 2.18.
(a) A list v of one vector v ∈ V is linearly independent if and only if v 6= 0.(b) A list of two vectors in V is linearly independent if and only if neither vector is a scalar
multiple of the other.(c) (1, 0, 0, 0), (0, 1, 0, 0), (0, 0, 1, 0) is linearly independent in F4.(d) The list 1, z, . . . , zm is linearly independent in P(F) for each nonnegative integer m.
Proof. (a) If the list v is linearly independent but v = 0, then 1v = 0 is a nontrivialrepresentation of 0 using v which is a contradiction. If v 6= 0 but the list v is not linearlyindependent, then there exists a ∈ F such that av = 0 and a 6= 0 which contradicts 1.30of [Axl14].
(b) If a list of two vectors v1, v2 in V is linearly independent and assume that, without lossof generality, v1 = av2 for some a ∈ F, then 1v1− av2 = 0 is a nontrivial representationof 0 using v1, v2 which is a contradiction. If a list of two vectors v1, v2 in V is notlinearly independent, then there exist some a1, a2 ∈ F such that a1v1 + a2v2 = 0 andnot both of a1 and a2 are zero. If a1 6= 0, then v1 = −a2
a1v2; if a1 = 0, then v2 = −a1
a2v1.
(c) Consider the equation x(1, 0, 0, 0) + y(0, 1, 0, 0) + z(0, 0, 1, 0) = 0 where x, y, z ∈ F. Inmatrix form, 1 0 0
0 1 00 0 1
xyz
=
000
Since the coefficient matrix is nonsingular, the only solution is x = y = z = 0.
(d) Consider the function f(z) = x0 + x1z + · · · + xN−1zN−1 where N − 1 ≥ 0 and
x0, x1, . . . , xN−1 ∈ F. Suppose f(z) = 0 for all z ∈ F. Then, in particular, f(1) =
4.5. EXERCISE 2.A.5: DIFFERENCE BETWEEN C AND R AS SCALAR FIELDS 25
0, f(2) = 0, . . . , f(N) = 0. In matrix form,
1 1 1 1 · · · 11 2 22 23 · · · 2N−1
1 3 32 33 · · · 3N−1
1 4 42 43 · · · 4N−1
......
......
. . ....
1 N N2 N3 · · · NN−1
x0
x1
...xN−1
=
00...0
.
Note that the above coefficient matrix is a Vandermonde matrix. Since all the num-bers 1, 2, . . . , N are distinct, its determinant is non-zero. Therefore the only solution isx0 = x1 = · · · = xN−1 = 0.
�
4.3. Exercise 2.A.3: a Linearly Dependent List
Find a number t such that
(3, 1, 4), (2,−3, 5), (5, 9, t)
is not linearly independent in R3.
Proof. We claim that t = 2 makes the list not linearly independent in R3. This can beshown by
− 3 · (3, 1, 4) + 2 · (2,−3, 5) + 1 · (5, 9, 2)
= ((−9) + 4 + 5, (−3) + (−6) + 9, (−12) + 10 + 2)
= 0.
�
4.4. Exercise 2.A.4: Conditions to Linear Dependence
Verify the assertion in the second bullet point in Example 2.20: the list
(2, 3, 1), (1,−1, 2), (7, 3, c)
is linearly dependent in F3 if and only if c = 8.
Proof. Let
A =
2 1 73 −1 31 2 c
.Then the list (2, 3, 1), (1,−1, 2), (7, 3, c) is linearly dependent if and only if det(A) = 0. Since
det(A) = 40− 5c, it follows that det(A) = 0 if and only if c = 8. �
4.5. Exercise 2.A.5: Difference between C and R as Scalar Fields
(a) Show that if we think of C as a vector space over R, then the list (1+ i, 1− i) is linearlyindependent.
(b) Show that if we think of C as a vector space over C, then the list (1+ i, 1− i) is linearlydependent.
26 4. SPAN AND LINEAR INDEPENDENCE
Proof. Suppose that there exists a+ bi, c+ di ∈ C such that a, b, c, d ∈ R and (a+ bi)(1 +i) + (c+ di)(1− i) = 0, then
(a− b+ c+ d) + (a+ b− c+ d)i = 0,
Rewrite the above linear system in the matrix form,
[1 −1 1 11 1 −1 1
]abcd
=
[00
],
then its solutions are −yxxy
: x, y ∈ R
.
(a) If we think of C as a vector space over R, then the list is linearly independent because−yxxy
: x, y ∈ R
∩x0y0
: x, y ∈ R
=
0000
.
(b) If we think of C as a vector space over C, then the list is linearly dependent because−1111
∈−yxxy
: x, y ∈ R
.
�
4.6. Exercise 2.A.6: Constructing Another Independent List from a List
Suppose v1, v2, v3, v4 is linearly independent in V . Prove that the list
v1 − v2, v2 − v3, v3 − v4, v4
is also linearly independent.
Proof. Use the same notation as the proof of Exercise 2.A.1. If the list
v1 − v2, v2 − v3, v3 − v4, v4
is linearly dependent, then there exists a1, a2, a3, a4 ∈ F, such that
[a1 a2 a3 a4
] w1
w2
w3
w4
= 0
and [a1 a2 a3 a4
]6=[0 0 0 0
].
Therefore,
[a1 a2 a3 a4
]A
v1
v2
v3
v4
= 0.
4.8. EXERCISE 2.A.8: LINEAR INDEPENDENT LIST MULTIPLIED BY SCALAR 27
Since A is nonsingular, [a1 a2 a3 a4
]A 6=
[0 0 0 0
].
Therefore it follows that the list v1, v2, v3, v4 is linearly dependent. �
4.7. Exercise 2.A.7: Constructing Another Independent List from a List
Prove or give a counterexample: If v1, v2, . . . , vm is a linearly independent list of vectors inV , then
5v1 − 4v2, v2, v3, . . . , vm
is linearly independent.
Proof. We claim that the proposition is true. Define
A =
5 −4 0 . . . 00 1 0 . . . 00 0 1 . . . 0...
......
. . ....
0 0 0 . . . 1
.Suppose that
5v1 − 4v2, v2, v3, . . . , vm
is linearly dependent, then there exists a1, a2, a3, . . . , am ∈ F, such that
[a1 a2 a3 . . . am
]A
v1
v2
v3
. . .vm
= 0
and [a1 a2 a3 . . . am
]6=[0 0 0 . . . 0
].
Since det(A) = 5, the matrix A is nonsingular. Thus[a1 a2 a3 . . . am
]A 6=
[0 0 0 . . . 0
].
Therefore it follows that the list v1, v2, . . . , vm is linearly dependent. �
4.8. Exercise 2.A.8: Linear Independent List Multiplied by Scalar
Prove or give a counterexample: If v1, v2, . . . , vm is a linearly independent list of vectors inV and λ ∈ F with λ 6= 0, then λv1, λv2, . . . , λvm is linearly independent.
Proof. We claim that the proposition is true. Suppose that λv1, λv2, . . . , λvm is linearlydependent, then there exists a1, a2, . . . , am, not all 0, such that
a1λv1 + a2λv2 + · · ·+ amλvm = 0.
Since λ 6= 0, λ−1 exists in F. Multiplying the above equation by λ−1 yields
a1v1 + a2v2 + · · ·+ amvm = 0.
Note that not all of a1, a2, . . . , am are zero. Therefore it follows that v1, v2, . . . , vm is alinearly dependent list. �
28 4. SPAN AND LINEAR INDEPENDENCE
4.9. Exercise 2.A.9: Addition of Two Linearly Independent Lists
Prove or give a counterexample: If v1, . . . , vm and w1, . . . , wm are linearly independent listsof vectors in V , then v1 + w1, . . . , vm + wm is linearly independent.
Counterexample. Set V = R2,m = 2, v1 = (0, 1), v2 = (1, 0), w1 = (0,−1), w2 = (−1, 0).Then v1, v2 is a linearly independent list, and w1, w2 is also a linearly independent list. But
v1 + w1 = (0, 0)
and
v2 + w2 = (0, 0)
is not a linearly independent list. �
4.10. Exercise 2.A.10: Adding a Vector to a Linearly Independent List
Suppose v1, . . . , vm is linearly independent in V and w ∈ V . Prove that if v1 +w, . . . , vm +wis linearly dependent, then w ∈ span(v1, . . . , vm).
Proof. If v1 + w, . . . , vm + w is linearly dependent, then there exists a1, . . . , am, not all 0,such that
a1(v1 + w) + · · ·+ am(vm + w) = 0.
The above equation implies that
w = − a1
a1 + · · ·+ amv1 − · · · −
ama1 + · · ·+ am
vm,
thus w ∈ span(v1, . . . , vm). �
4.11. Exercise 2.A.11: Concatenating a Vector to an Independent List
Suppose v1, . . . , vm is linearly independent in V and w ∈ V . Show that v1, . . . , vm, w islinearly independent if and only if
w /∈ span(v1, . . . , vm).
Proof. (1) Suppose that v1, . . . , vm, w is linearly dependent, then there exists a1, . . . , am, aw ∈F, not all 0, such that
a1v1 + · · ·+ amvm + aww = 0.
If aw = 0, then the above equation would imply that v1, . . . , vm is linearly de-pendent, which is a contradiction. Therefore aw 6= 0 and the above equation impliesthat
w = − a1
awv1 − · · · −
amaw
vm.
Thus w ∈ span(v1, . . . , vm).(2) Suppose that w ∈ span(v1, . . . , vm), then there exists a1, . . . , am ∈ F such that
w = a1v1 + · · ·+ amvm.
The above equation implies that
a1v1 + · · ·+ amvm − w = 0.
Thus v1, . . . , vm, w is linearly dependent.�
4.15. EXERCISE 2.A.15: F∞ IS INFINITE-DIMENSIONAL 29
4.12. Exercise 2.A.12: Possible Linearly Independent Lists in P4(F)
Explain why there does not exist a list of six polynomials that is linearly independent inP4(F).
Proof. It is obvious that the list 1, z, z2, z3, z4 spans P4(F), which has length 5. By 2.23 of[Axl14], the length of every linearly independent list of vectors in P4(F) is therefore less thanor equal to 5. Thus there does not exist a list of 6 polynomials that is linearly independent inP4(F). �
4.13. Exercise 2.A.13: Possible Spans of P4(F)
Explain why no list of four polynomials spans P4(F).
Proof. Exercise 2.A.2 (d) has already proved that the list 1, z, z2, z3, z4 is linearly inde-pendent in P4(F), which has length 5. By 2.23 of [Axl14], the length of every spanning list ofP4(F) is greater than or equal to 5. Thus no list of 4 polynomials spans P4(F). �
4.14. Exercise 2.A.14: Another Definition of Infinite-Dimensional
Prove that V is infinite-dimensional if and only if there is a sequence v1, v2, . . . of vectors inV such that v1, . . . , vm is linearly independent for every positive integer m.
Proof. (1) Suppose that V is finite-dimensional, then there exists a list u1, . . . , ukin V that spans V . By 2.23 of [Axl14], the length of every linearly independent listof vectors in V is less than or equal to k. Therefore, if we pick any vector sequencev1, v2, . . . in V , v1, . . . , vk+1 must be linearly dependent.
(2) Suppose that V is infinite-dimensional. We can construct, by the following steps, asequence v1, v2, . . . of vectors in V such that v1, . . . , vm is linearly independent forevery positive integer m:
Step (1): Pick any v ∈ V such that v 6= 0, and let v1 = v. Note that the list v1 is linearlyindependent.
Step (n): Suppose that Step (n− 1) has already created a vector list v1, . . . , vn−1 such thatv1, . . . , vm is linearly independent for every positive integer m ∈ {1, . . . , n − 1}.Since V is infinite-dimensional, there exists w ∈ V such that w /∈ span(v1, . . . , vn−1).Set vn = w, and we claim that v1, . . . , vn is linearly independent. To prove this,suppose that there exists a1, . . . , an ∈ F, not all zero, such that
a1v1 + · · ·+ anvn = 0.
If an = 0, then a1, . . . , an−1 would be linearly dependent, which is a contradiction.If an 6= 0, then
vn = − a1
anv1 − · · · −
an−1
anvn−1,
which implies that vn ∈ span(v1, . . . , un−1), which is also a contradiction.�
4.15. Exercise 2.A.15: F∞ is Infinite-Dimensional
Prove that F∞ is infinite-dimensional.
30 4. SPAN AND LINEAR INDEPENDENCE
Proof. Suppose that F∞ is finite-dimensional, then there exists a list v1, . . . , vk ∈ F∞ suchthat F∞ = span(v1, . . . , vk). But we may construct the following independent list of vectors inF∞ which has length k + 1:
v1 = (1, 0, 0, . . . , 0, 0, 0, 0, . . . )
v2 = (0, 1, 0, . . . , 0, 0, 0, 0, . . . )
v3 = (0, 0, 1, . . . , 0, 0, 0, 0, . . . )
...
vk−1 = (0, 0, 0, . . . , 1, 0, 0, 0, . . . )
vk = (0, 0, 0, . . . , 0, 1, 0, 0, . . . )
vk+1 = (0, 0, 0, . . . , 0, 0, 1, 0, . . . )
This contradicts with 2.23 of [Axl14]. �
4.16. Exercise 2.A.16: R[0,1] is Infinite-Dimensional
Prove that the real vector space of all continuous real-valued functions on the interval [0, 1]is infinite-dimensional.
Proof. Suppose that R[0,1] is finite-dimensional, then there exists a list
v1, . . . , vk ∈ R[0,1]
such that R[0,1] = span(v1, . . . , vk). But we may construct the following independent list ofvectors in R[0,1] which has length k + 1:
v1(x) = 1 if x =1
k + 2; otherwise 0
v2(x) = 1 if x =2
k + 2; otherwise 0
v3(x) = 1 if x =3
k + 2; otherwise 0
...
vk−1(x) = 1 if x =k − 1
k + 2; otherwise 0
vk(x) = 1 if x =k
k + 2; otherwise 0
vk+1(x) = 1 if x =k + 1
k + 2; otherwise 0.
This contradicts with 2.23 of [Axl14]. �
4.17. Exercise 2.A.17: Polynomials that Vanish at the Same Point
Suppose p0, p1, . . . , pm are polynomials in Pm(F) such that pj(2) = 0 for each j. Prove thatp0, p1, . . . , pm is not linearly independent in Pm(F).
4.17. EXERCISE 2.A.17: POLYNOMIALS THAT VANISH AT THE SAME POINT 31
Proof. Denote pj(z) = aj0 + aj1z + aj2z2 + · · ·+ ajmz
m for each j and let
A =
a00 a01 a02 . . . a0m
a00 a01 a02 . . . a0m
......
.... . .
...am0 am1 am2 . . . amm
.That pj(2) = 0 for each j implies
A
20
21
22
...2m
=
000...0
.Thus A is a singular matrix, and so is AT .Suppose there exists s0, s1, . . . , sm ∈ F such that s0p0 + s1p1 + · · ·+ smpm = 0, then
a00 a10 . . . am0
a01 a11 . . . am1
......
. . ....
a0m a1m . . . amm
s0
s1
...sm
=
00...0
.Note that the coefficient matrix of the above linear system is AT , which we have shown is
a singular matrix. So the linear system has a nonzero solution. We therefore conclude that thelist p0, p1, . . . , pm is linearly dependent. �
CHAPTER 5
Bases
5.1. Exercise 2.B.1: Vector Spaces that Have Exactly One Basis
Find all vector spaces that have exactly one basis.
Proof. We claim that no vector spaces over F (which is either R or C) have exactly onebasis.
Given a vector space V . Suppose that v1, . . . , vk is the only basis of V . Denote
v =
v1
...vk
.Note that Av is also a basis of V for any nonsingular matrix A, then
Av = v, for all nonsingular A.
In particular, take A = 2I. Then
(2I − I)v = 0,
which implies that v = 0. But the list 0 is not linearly dependent, contradicting with thepresumption that v1, . . . , vk is a basis. �
5.2. Exercise 2.B.2: Examples and Counterexamples of Bases
Verify all the assertions in Example 2.28.
(a) The list (1, 0, . . . , 0), (0, 1, 0, . . . , 0), . . . , (0, . . . , 0, 1) is a basis of Fn, called the standardbasis of Fn.
(b) The list (1, 2), (3, 5) is a basis of F2.(c) The list (1, 2,−4), (7,−5, 6) is linearly independent in F3 but is not a basis of F3 because
it does not span F3.(d) The list (1, 2), (3, 5), (4, 13) spans F2 but is not a basis of F2 because it is not linearly
independent.(e) The list (1, 1, 0), (0, 0, 1) is a basis of {(x, x, y) ∈ F3 : x, y ∈ F}.(f) The list (1,−1, 0), (1, 0,−1) is a basis of
{(x, y, z) ∈ F3 : x+ y + z = 0}.
(g) The list 1, z, . . . , zm is a basis of Pm(F).
Proof. (a) Pick v = (v1, v2, . . . , vn) ∈ Fn where v1, v2, . . . , vn ∈ F. Then it is obviousthat v can be written uniquely as
v1(1, 0, . . . , 0) + v2(0, 1, 0, . . . , 0) + · · ·+ vn(0, . . . , 0, 1).
33
34 5. BASES
(b) Pick v = (v1, v2) ∈ F2. Let v = x(1, 2) + y(3, 5) where x, y ∈ F, which corresponds tothe following linear system [
1 32 5
] [xy
]=
[v1
v2
].
Since the coefficient of the above linear system is nonsingular, there is exactly one
solution[x y
]Tfor every
[v1 v2
]T. Therefore (1, 2), (3, 5) is a basis of F2.
(c) As shown in (1), there exists a linearly independent list of length 3 in F3, thus no listsof length 2 can span F3 by 2.23 of [Axl14].
(d) As shown in (1), there exists a list of length 2 that spans F2, thus no lists of length 3can be linearly independent in F2 by 2.23 of [Axl14].
(e) Pick v = (x, x, y) ∈ F3 where x, y ∈ F. Then it is obvious that v can be written uniquelyas
x(1, 1, 0) + y(0, 0, 1).
(f) Denote V = {(x, y, z) ∈ F3 : x+ y+ z = 0}. Pick v = (a, b, c) ∈ V then c = −a− b. Letv = x(1,−1, 0) + y(1, 0,−1) where x, y ∈ F, which corresponds to the following linearsystem 1 1
−1 00 −1
[xy
]=
ab
−a− b
.Note that the above linear system is an overdetermined system, so it has either no
solutions or a unique solution. Moreover, it can be shown that[−ba+ b
]is one possible solution. Therefore every v ∈ V can be written uniquely using (1,−1, 0), (1, 0,−1).
(g) Pick v(z) = a0 + a1z + a2z2 + · · ·+ amz
m ∈ Pm(F) where a0, a1, a2, . . . , am ∈ F. Then4.7 of [Axl14] implies that v can be written uniquely as
a0 · 1 + a1 · z + a2 · z2 + · · ·+ am · zm.
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5.3. Exercise 2.B.3: Find and Extend a Basis for a Subspace of R5
(a) Let U be the subspace of R5 defined by
U = {(x1, x2, x3, x4, x5) ∈ R5 : x1 = 3x2 and x3 = 7x4}.
Find a basis of U .(b) Extend the basis in part (a) to a basis of R5.(c) Find a subspace W of R5 such that R5 = U ⊕W .
Proof. (a) Let
u1 = (3, 1, 0, 0, 0)
u2 = (0, 0, 7, 1, 0)
u3 = (0, 0, 0, 0, 1).
We claim that the list u1, u2, u3 is a basis of U . To prove this, pick u = (x1, x2, x3, x4, x5) ∈U , then it is obvious that u can be uniquely represented as
x2u1 + x4u2 + x5u3.
5.4. EXERCISE 2.B.4: FIND AND EXTEND A BASIS FOR A SUBSPACE OF C5 35
(b) Let
v1 = (1, 0, 0, 0, 0)
v2 = (0, 1, 0, 0, 0)
v3 = (0, 0, 1, 0, 0)
v4 = (0, 0, 0, 1, 0)
v5 = (0, 0, 0, 0, 1).
Then v1, v2, v3, v4, v5 is a basis of R5. Thus the list
u1, u2, u3, v1, v2, v3, v4, v5
spans R5. The procedure of the proof of 2.31 of [Axl14] deletes v2, v4, v5 and leavesv1, v3 unchanged, resulting in the following basis of R5 extended from the basis in part(a):
u1, u2, u3, v1, v3.
(c) Let W = span(v1, v3). We claim that R5 = U ⊕W . To prove this, by 1.45 of [Axl14],we only need to show that
R5 = U +W and U ∩W = {0}.
To prove the 1st equation above, suppose v ∈ R5. Then because the list u1, u2, u3, v1, v3
spans R5, there exists a1, a2, a3, b1, b3 ∈ F such that
v = a1u1 + a2u2 + a3u3 + b1v1 + b3v3.
In other words, we have v = u + w, where u = (a1u1 + a2u2 + a3u3) ∈ U andw = (b1v1 + b3v3) ∈W . Thus v ∈ U +W , completing the proof that R5 = U +W .
To show that U ∩ W = {0}, suppose v ∈ U ∩ W . Then there exist scalarsa1, a2, a3, b1, b3 ∈ F such that
v = a1u1 + a2u2 + a3u3 = b1v1 + b3v3.
Thus
a1u1 + a2u2 + a3u3 − b1v1 − b3v3 = 0.
Because u1, u2, u3, v1, v3 is linearly independent, this implies that a1 = a2 = a3 =b1 = b3 = 0. Thus v = 0, completing the proof that U ∩W = {0}.
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5.4. Exercise 2.B.4: Find and Extend a Basis for a Subspace of C5
(a) Let U be the subspace of C5 defined by
U = {(z1, z2, z3, z4, z5) ∈ C5 : 6z1 = x2 and z3 + 2z4 + 3z5 = 0}.
Find a basis of U .(b) Extend the basis in part (a) to a basis of C5.(c) Find a subspace W of C5 such that C5 = U ⊕W .
Proof. (a) Let
u1 = (1, 6, 0, 0, 0)
u2 = (0, 0,−2, 1, 0)
u3 = (0, 0,−3, 0, 1).
36 5. BASES
We claim that the list u1, u2, u3 is a basis of U . To prove this, pick u = (z1, z2, z3, z4, z5) ∈U , then it is obvious that u can be uniquely represented as
x1u1 + x4u2 + x5u3.
(b) Let
v1 = (1, 0, 0, 0, 0)
v2 = (0, 1, 0, 0, 0)
v3 = (0, 0, 1, 0, 0)
v4 = (0, 0, 0, 1, 0)
v5 = (0, 0, 0, 0, 1).
Then v1, v2, v3, v4, v5 is a basis of C5. Thus the list
u1, u2, u3, v1, v2, v3, v4, v5
spans C5. The procedure of the proof of 2.31 of [Axl14] deletes v2, v4, v5 and leavesv1, v3 unchanged, resulting in the following basis of C5 extended from the basis in part(a):
u1, u2, u3, v1, v3.
(c) Let W = span(v1, v3). We claim that C5 = U ⊕W . To prove this, by 1.45 of [Axl14],we only need to show that
C5 = U +W and U ∩W = {0}.To prove the 1st equation above, suppose v ∈ C5. Then because the list u1, u2, u3, v1, v3
spans C5, there exists a1, a2, a3, b1, b3 ∈ F such that
v = a1u1 + a2u2 + a3u3 + b1v1 + b3v3.
In other words, we have v = u + w, where u = (a1u1 + a2u2 + a3u3) ∈ U andw = (b1v1 + b3v3) ∈W . Thus v ∈ U +W , completing the proof that C5 = U +W .
To show that U ∩ W = {0}, suppose v ∈ U ∩ W . Then there exist scalarsa1, a2, a3, b1, b3 ∈ F such that
v = a1u1 + a2u2 + a3u3 = b1v1 + b3v3.
Thus
a1u1 + a2u2 + a3u3 − b1v1 − b3v3 = 0.
Because u1, u2, u3, v1, v3 is linearly independent, this implies that a1 = a2 = a3 =b1 = b3 = 0. Thus v = 0, completing the proof that U ∩W = {0}.
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5.5. Exercise 2.B.5: A Basis of P3(F) without Degree-2 Polynomials
Prove or disprove: there exists a basis p0, p1, p2, p3 of P3(F) such that none of the polynomialsp0, p1, p2, p3 has degree 2.
Proof. We claim that the following list is a possible choice of such p0, p1, p2, p3:
p0(z) = 1 + z3
p1(z) = z + z3
p2(z) = z2 + z3
p3(z) = z3.
5.5. EXERCISE 2.B.5: A BASIS OF P3(F) WITHOUT DEGREE-2 POLYNOMIALS 37
To prove this, note that p0, p1, p2, p3 each has degree 3. Thus it only suffices to show thatp0, p1, p2, p3 is a basis of P3(F).
Pick p(z) ∈ P3(F). As shown in Exercise 2.B.2, 1, z, z2, z3 is a basis P3(F). Thus there existsa unique list of scalars a0, a1, a2, a3 ∈ F such that
p(z) = a0 + a1z + a2z2 + a3z
3.
Let us denote
A =
1 0 0 10 1 0 10 0 1 10 0 0 1
,then A is a nonsingular matrix and
A−1 =
1 0 0 −10 1 0 −10 0 1 −10 0 0 1
.Using this notation, we have
p0
p1
p2
p3
= A
1zz2
z3
,and
A−1
p0
p1
p2
p3
=
1zz2
z3
.Therefore
p(z) =[a0 a1 a2 a3
] 1zz2
z3
=([a0 a1 a2 a3
]A−1
)p0
p1
p2
p3
.The above equation implies that p0, p1, p2, p3 spans P3(F).To show that p0, p1, p2, p3 is linearly independent, suppose b0, b1, b2, b3 ∈ F are such that
0 =[b0 b1 b2 b3
] p0
p1
p2
p3
.Then the above equation implies that
0 =([b0 b1 b2 b3
]A)
1zz2
z3
.
38 5. BASES
Since 1, z, z2, z3 is linearly independent, this implies that
0 =[b0 b1 b2 b3
]A.
Since A is nonsingular, this further implies that
0 =[b0 b1 b2 b3
].
Thus p0, p1, p2, p3 is linearly independent and hence is a basis of P3(F). �
5.6. Exercise 2.B.6: Constructing Another Basis from a Basis
Suppose v1, v2, v3, v4 is a basis of V . Prove that
v1 + v2, v2 + v3, v3 + v4, v4
is also a basis of V .
Proof. Denote
u1 = v1 + v2
u2 = v2 + v3
u3 = v3 + v4
u4 = v4.
Pick v ∈ V , then there exists a unique list of scalars a1, a2, a3, a4 ∈ F such that
v = a1v1 + a2v2 + a3v3 + a4v4.
Let us denote
A =
1 1 0 00 1 1 00 0 1 10 0 0 1
,then A is a nonsingular matrix and
A−1 =
1 −1 1 −10 1 −1 10 0 1 −10 0 0 1
.Using this notation, we have
u1
u2
u3
u4
= A
v1
v2
v3
v4
,and
A−1
u1
u2
u3
u4
=
v1
v2
v3
v4
.
5.7. EXERCISE 2.B.7: CONSTRUCTING A COUNTEREXAMPLE 39
Therefore
v =[a0 a1 a2 a3
] v1
v2
v3
v4
=([a0 a1 a2 a3
]A−1
)u1
u2
u3
u4
.The above equation implies that u1, u2, u3, u4 spans V .To show that u1, u2, u3, u4 is linearly independent, suppose b1, b2, b3, b4 ∈ F are such that
0 =[b1 b2 b3 b4
] u1
u2
u3
u4
.Then the above equation implies that
0 =([b1 b2 b3 b4
]A)v1
v2
v3
v4
.Since v1, v2, v3, v4 is linearly independent, this implies that
0 =[b1 b2 b3 b4
]A.
Since A is nonsingular, this further implies that
0 =[b1 b2 b3 b4
].
Thus u1, u2, u3, u4 is linearly independent and hence is a basis of V . �
5.7. Exercise 2.B.7: Constructing a Counterexample
Prove or give a counterexample: If v1, v2, v3, v4 is a basis of V and U is a subspace of V suchthat v1, v2 ∈ U and v3 /∈ U and v4 /∈ U , then v1, v2 is a basis of U .
Counterexample. Take V = R4 and define
v1 = (2, 0, 1, 0),
v2 = (0, 0, 0, 1),
v3 = (0, 3,1
2, 1),
v4 = (3, 0, 1,1
2).
We claim that v1, v2, v3, v4 is a basis of V , which can be shown by the following determinantcalculation:
det
2 0 1 00 0 0 10 3 1
2 13 0 1 1
2
= −3.
DefineU = {(2x, y, x, z) ∈ R4 : x, y, z ∈ R}.
40 5. BASES
We claim that U is a subspace of V . To show this, first 0 = (2 · 0, 0, 0, 0) is in U . Supposeu = (2x, y, x, z) ∈ U and w = (2a, b, a, c) ∈ U , we can check that u + w = (2x + 2a, y + b, x +a, z + c) = (2(x+ a), y + b, x+ a, z + c) is in U . Suppose f ∈ F and u = (2x, y, x, z) ∈ U , we cancheck that fu = (2fx, y, fx, z) = (2(fx), y, fx, z) is also in U .
Further, it is clear that v1, v2 ∈ U, v3 /∈ U, v4 /∈ U .Finally, since (0, 1, 0, 0) ∈ U cannot be written as a linear combination of v1 and v2, we
conclude that v1, v2 is not a basis of U . �
5.8. Exercise 2.B.8: Basis Obtained from a Direct Sum
Suppose U and W are subspaces of V such that V = U ⊕W . Suppose also that u1, . . . , umis a basis of U and w1, . . . , wn is a basis of W . Prove that
u1, . . . , um, w1, . . . , wn
is a basis of V .
Proof. (1) We first prove that u1, . . . , um, w1, . . . , wn spans V . Pick v ∈ V . By thedefinition of direct sum, v can be written in the only one way as a sum u + w, whereu ∈ U and w ∈W . By the definition of basis, u can be written as u = a1u1 +· · ·+amumand w can be written as w = b1w1 + · · ·+ bnwn where a1, . . . , am, b1, . . . , bn ∈ F. Takentogether,
v = a1u1 + · · ·+ amum + b1w1 + · · ·+ bnwn.
(2) We then prove that u1, . . . , um, w1, . . . , wn is linearly independent. Suppose there existx1, . . . , xm, y1, . . . , yn ∈ F such that
x1u1 + · · ·+ xmum + y1w1 + · · ·+ ynwn = 0.
Thenx1u1 + · · ·+ xmum = −y1w1 − · · · − ynwn.
Denote p = x1u1 + · · · + xmum and q = −y1w1 − · · · − ynwn, then p ∈ U ∩W andq ∈ U ∩W . By 1.45 of [Axl14], p = q = 0.
Further, since u1, . . . , um is linearly independent, p = 0 implies that all of x1, . . . , xmare 0. Similarly, since w1, . . . , wn is linearly independent, q = 0 implies that all ofy1, . . . , yn are 0.
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CHAPTER 6
Dimension
6.1. Exercise 2.C.1: A Space with the Same Dimension of its Subspace
Suppose V is finite-dimensional and U is a subspace of V such that dimU = dimV . Provethat U = V .
Proof. Suppose dimU = dimV = n and u1, . . . , un is a basis of U . Since u1, . . . , un is alinearly independent list in V with length dimV , 2.39 of [Axl14] implies that u1, . . . , un is abasis of V . Hence
U = span(u1, . . . , un) = V.
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6.2. Exercise 2.C.2: Subspaces of R2
Show that the subspaces of R2 are precisely {0}, R2, and all lines in R2 through the origin.
Proof. Pick a subspace U of R2, then dimU ∈ {0, 1, 2} by 2.39 of [Axl14].
(1) If dimU = 0, then U = {0}.(2) If dimU = 1, then there exists a vector v = (x, y) ∈ R2 such that
U = span(v).
Hence, geometrically, U can be viewed as a line in R2 passing through points (0, 0) and(x, y).
(3) If dimU = 2, then Exercise 2.C.1 implies that U = R2.
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6.3. Exercise 2.C.3: Subspaces of R3
Show that the subspaces of R3 are precisely {0}, R3, all lines in R3 through the origin, andall planes in R3 through the origin.
Proof. Pick a subspace U of R3, then dimU ∈ {0, 1, 2, 3} by 2.39 of [Axl14].
(1) If dimU = 0, then U = {0}.(2) If dimU = 1, then there exists a vector v = (x, y, z) ∈ R3 such that
U = span(v).
Hence, geometrically, U can be viewed as a line in R3 passing through points (0, 0, 0)and (x, y, z).
(3) If dimU = 2, then there exists a vector v = (v1, v2, v3), w = (w1, w2, w3) in R3 suchthat
U = span(v, w).
Geometrically, since the list v, w is linearly independent, v and w are not on thesame line in R3. Hence U is a plane in R3 that passes the following three points:(v1, v2, v3), (w1, w2, w3) and (0, 0, 0).
41
42 6. DIMENSION
(4) If dimU = 3, then Exercise 2.C.1 implies that U = R3.
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6.4. Exercise 2.C.4: Find and Extend a Basis for a Subspace of P4(F)
(a) Let U = {p ∈ P4(F) : p(6) = 0}. Find a basis of U .(b) Extend the basis in part (a) to a basis of P4(F).(c) Find a subspace W of P4(F) such that P4(F) = U ⊕W .
Proof. (a) We first verify that U is a subspace of P4. First 0 ∈ U since 0(6) = 0. Ifp, q ∈ U , then p+ q ∈ U because (p+ q)(6) = p(6) + q(6) = 0. If a ∈ F and p ∈ U , thenap ∈ U because (ap)(6) = ap(6) = 0.
We claim that the following list is a basis of U :
z − 6, z2 − 62, z3 − 63, z4 − 64.
Clearly each of the above polynomials is in U . Suppose a, b, c, d ∈ F and
a(z − 6) + b(z2 − 62) + c(z3 − 63) + d(z4 − 64) = 0
for every z ∈ F. Without explicitly expanding the left side of the equation above, wecan see that the left side has a dz4 term. Because the right side has no z4 term, thisimplies that d = 0. Because d = 0, we see that the left side has a cz3 term, whichimplies that c = 0. Similarly, b = a = 0.
Thus the equation above implies that a = b = c = d = 0. Hence the list is linearlyindependent in U . Thus dimU > 4. Considering U as a subspace of P4(F), which hasdimension 5, we conclude that
4 6 dimU 6 5.
But Exercise 2.A.17 has shown that any list of vectors in U that has length 5 isnot linearly independent, which means that dimU 6= 5. Thus dimU = 4 and 2.39 of[Axl14] implies that the list is a basis of U .
(b) We claim that the following list is a basis of P4(F):
1, z − 6, z2 − 62, z3 − 63, z4 − 64.
Clearly each of the above polynomials is in P4(F). Suppose
a0, a1, a2, a3, a4 ∈ F
and
a0 + a1(z − 6) + a2(z2 − 62) + a3(z3 − 63) + a4(z4 − 64) = 0
for every z ∈ F. Without explicitly expanding the left side of the equation above, wecan see that the left side has a a4z
4 term. Because the right side has no z4 term, thisimplies that a4 = 0. Because a4 = 0, we see that the left side has a a3z
3 term, whichimplies that a3 = 0. Similarly, a2 = a1 = 0. Because a4 = a3 = a2 = a1 = 0, we canalso conclude that a0 = 0.
Hence the list is linearly independent in P4(F). Since dimP4(F) = 5, 2.39 of[Axl14] implies that the list is a basis of P4.
(c) Define
W ≡ {a0 : a0 ∈ F}and we claim that P4(F) = U ⊕W .
To prove this, by 1.45 of [Axl14], we only need to show that
P4(F) = U +W and U ∩W = {0}.
6.5. EXERCISE 2.C.9: dim span(v1 + w, . . . , vm + w) > m− 1 43
To prove the first equation above, suppose v ∈ P4(F). Then, because the list1, z − 6, z2 − 62, z3 − 63, z4 − 64 spans P4, there exists a0, a1, a2, a3, a4 ∈ F such that
v = a0︸︷︷︸w
+ a1(z − 6) + a2(z2 − 62) + a3(z3 − 63) + a4(z4 − 64)︸ ︷︷ ︸u
.
In other words, we have v = u+w, where u ∈ U and w ∈W are defined as above.Thus v ∈ U +W , completing the proof that P4(F) = U +W .
To show that U ∩ W = {0}, suppose v ∈ U ∩ W . Then there exist scalarsa0, a1, a2, a3, a4 ∈ F such that
v = a0 = a1(z − 6) + a2(z2 − 62) + a3(z3 − 63) + a4(z4 − 64).
Thus
a0 − a1(z − 6)− a2(z2 − 62)− a3(z3 − 63)− a4(z4 − 64) = 0.
Because because the list 1, z − 6, z2 − 62, z3 − 63, z4 − 64 is linearly independent,this implies that a0 = a1 = a2 = a3 = a4 = 0. Thus v = 0, completing the proof thatU ∩W = {0}.
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6.5. Exercise 2.C.9: dim span(v1 + w, . . . , vm + w) > m− 1
Suppose v1, . . . , vm is linearly independent in V and w ∈ V . Prove that
dim span(v1 + w, . . . , vm + w) > m− 1.
Proof. Define
U ≡ span(v1 + w, . . . , vm + w),
w1 ≡ (v1 + w)− (v2 + w) = v1 − v2,
w2 ≡ (v2 + w)− (v3 + w) = v2 − v3,
...
wm−1 ≡ (vm−1 + w)− (vm + w) = vm−1 − vm,W ≡ span(w1, . . . , wm−1).
Then W is a subspace of U .Suppose a1, . . . , am−1 ∈ F and a1w1 + · · ·+ am−1wm−1 = 0, then
[a1 a2 . . . am−1
]
1 −1 0 . . . 0 00 1 −1 . . . 0 0...
......
. . ....
...0 0 0 . . . 1 −1
v1
v2
...vm−1
vm
= 0.
Since v1, . . . , vm is linearly independent, the above equation implies
[a1 a2 . . . am−1
]
1 −1 0 . . . 0 00 1 −1 . . . 0 0...
......
. . ....
...0 0 0 . . . 1 −1
= 0.
44 6. DIMENSION
Transposing the above equation gives the following linear system:
1 0 . . . 0−1 1 . . . 00 −1 . . . 0...
.... . .
...0 0 . . . 10 0 . . . −1
a1
a2
...am−1
= 0.
Denote Ei,j as the elementary matrix produced by adding row i of the m×m identity matrixto row j, then left multiplication by Em−1,mEm−2,m−1 . . . E2,3E1,2 gives
1 0 . . . 00 1 . . . 00 0 . . . 0...
.... . .
...0 0 . . . 10 0 . . . 0
a1
a2
...am−1
= 0.
Hence a1 = a2 = · · · = am−1 = 0 and w1, . . . , wm−1 is linearly independent. It is clear thatw1, . . . , wm−1 spans W . Hence w1, . . . , wm−1 is a basis of W . Therefore
dimW = m− 1.
Finally, 2.38 of [Axl14] implies
dimU > dimW = m− 1.
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CHAPTER 7
The Vector Space of Linear Maps
7.1. Exercise 3.A.1: Conditions for a Mapping to be Linear
Suppose b, c ∈ R. Define T : R3 → R2 by
T (x, y, z) = (2x− 4y + 3z + b, 6x+ cxyz).
Show that T is linear if and only if b = c = 0.
Proof. (1) Suppose b = c = 0, then
T (x, y, z) = (2x− 4y + 3z, 6x).
Pick u = (u1, u2, u3), v = (v1, v2, v3), then
T (u+ v) = (2(u1 + v1)− 4(u2 + v2) + 3(u3 + v3), 6(u1 + v1))
= (2u1 + 2v1 − 4u2 − 4v2 + 3u3 + 3v3, 6u1 + 6v1)
and
Tu+ Tv = (2u1 − 4u2 + 3u3, 6u1) + (2v1 − 4v2 + 3v3, 6v1)
= (2u1 − 4u2 + 3u3 + 2v1 − 4v2 + 3v3, 6u1 + 6v1).
Hence T (u+ v) = Tu+ Tv.Pick λ ∈ F, then
T (λv) = T (λu1, λu2, λu3)
= (2λu1 − 4λu2 + 3λu3, 6λu1)
and
λT (v) = λ(2u1 − 4u2 + 3u3, 6u1)
= (2λu1 − 4λu2 + 3λu3, 6λu1).
Hence T (λv) = λT (v).(2) By definition, if T is linear, then T (u + v) = Tu + Tv for all u, v ∈ R3. In particular,
take u = (1, 1, 1), v = (1, 1, 1). Then
T (u+ v) = (2 · 2− 4 · 2 + 3 · 2 + b, 6 · 2 + c · 2 · 2 · 2)
= (2 + b, 12 + 8c)
= Tu+ Tv
= (2− 4 + 3 + b+ 2− 4 + 3 + b, 6 + c+ 6 + c)
= (2 + 2b, 12 + 2c)
Hence b = 2b and 8c = 2c, we thus conclude that b = c = 0.�
45
46 7. THE VECTOR SPACE OF LINEAR MAPS
7.2. Exercise 3.A.5: L(V,W ) is a vector space
Prove the assertion in 3.7 of [Axl14]: L(V,W ) is a vector space.
Proof. Define 0 as the linear map in L(V,W ) such that
0(v) = 0 ∀v ∈ V.
Pick R,S, T ∈ L(V,W ), v ∈ V and a, b ∈ F.
(1) Commutativity: The commutativity of vector space W implies Sv + Tv = Tv + Sv.Hence S + T = T + S.
(2) Associativity: The associativity of vector space W implies
(Rv + Sv) + Tv = Rv + (Sv + Tv)
and
(ab)Sv = a(bSv).
Hence (R+ S) + T = R+ (S + T ) and (ab)S = a(bS).(3) Additive identity: It is clear that T + 0 = T .(4) Additive inverse: Define −T as the linear map in L(V,W ) such that
−T (v) = −(Tv) ∀v ∈ V.
Then T + (−T ) = 0.(5) Multiplicative identity: The multiplicative identity of vector space W implies (1T )(v) =
1(Tv) = Tv. Hence 1T = T .(6) Distributive properties: The distributive properties of W implies
a(Sv + Tv) = aSv + aTv
and
(a+ b)Sv = aSv + bSv.
Hence a(S + T ) = aS + aT and (a+ b)S = aS + bS.
�
7.3. Exercise 3.A.6: Algebraic Properties of Products of Linear Maps
Prove the assertions in 3.9 of [Axl14]: The associativity, identity and distributive propertiesof Products of Linear Maps.
Proof. (1) Associativity: pick linear maps T1, T2 and T3 such that T3 maps into thedomain of T2, and T2 maps into the domain of T1. Then for any v in the domain of T3,
((T1T2)T3)(v) = (T1T2)(T3(v)) = T1(T2(T3(v))) = T1(T2T3)(v).
Hence
(T1T2)T3 = T1(T2T3).
(2) Identity: Pick T ∈ L(V,W ). Define IV as the identity map on V and IW as the identitymap on W . Then for any v ∈ V ,
(TIV )(v) = T (IV (v)) = T (v) = IW (T (v)) = (IWT )(v).
Hence
TIV = IWT = T.
7.5. EXERCISE 3.A.8: HOMOGENEITY ALONE DOES NOT IMPLY LINEARITY 47
(3) Distributive properties. Pick T, T1, T2 ∈ L(U, V ) and S, S1, S2 ∈ L(V,W ). Then forany u ∈ U ,
((S1 + S2)T )(u) = S1(T (u)) + S2(T (u))
= (S1T )(u) + (S2T )(u)
= (S1T + S2T )(u)
and
(S(T1 + T2))(u) = S(T1(u) + T2(u))
= S(T1(u)) + S(T2(u))
= (ST1)(u) + (ST2)(u)
= (ST1 + ST2)(u).
Hence
(S1 + S2)T = S1T + S2T
and
S(T1 + T2) = ST1 + ST2.
�
7.4. Exercise 3.A.7: Linear Map on 1D Space is Scalar Multiplication
Show that every linear map from a 1-dimensional vector space to itself is multiplication bysome scalar. More precisely, prove that if dimV = 1 and T ∈ L(V, V ), then there exists λ ∈ Fsuch that Tv = λv for all v ∈ V .
Proof. Since dimV = 1, there exists a vector b which is a basis of V . Pick any T ∈ L(V, V ),then, by the definition of basis, there exists λ ∈ F such that
T (b) = λb.
We claim that Tv = λv for all v ∈ V . To show this, pick any v ∈ V , then there exists α ∈ Fsuch that
v = αb.
Hence
T (v) = T (αb) = αT (b) = αλb = λαb = λv.
�
7.5. Exercise 3.A.8: Homogeneity Alone does NOT Imply Linearity
Give an example of a function ϕ : R2 → R such that
ϕ(av) = aϕ(v)
for all a ∈ R and all v ∈ R2 but ϕ is not linear.
Proof. Let ϕ(x, y) ≡ (x3 + y3)13 where (x, y) ∈ R2.
(1) Pick any a ∈ R, then
ϕ(ax, ay) = (a3x3 + a3y3)13
= (a3(x3 + y3))13
= a(x3 + y3)13
= aϕ(x, y).
48 7. THE VECTOR SPACE OF LINEAR MAPS
(2) ϕ is not linear because
ϕ((1, 2) + (3, 4)) = (43 + 63)13
= 28013
6= 913 + 91
13
= ϕ(1, 2) + ϕ(3, 4)
�
7.6. Exercise 3.A.9: Additivity Alone does NOT Imply Linearity
Give an example of a function ϕ : C→ C such that
ϕ(w + z) = ϕ(w) + ϕ(z)
for all w, z ∈ C but ϕ is not linear.
Proof. Let ϕ(z) = <(z) where z ∈ C.
(1) Pick any w, z ∈ C, then
ϕ(w + z) = <(w + z)
= <(w) + <(z)
= ϕ(w) + ϕ(z).
(2) ϕ is not linear because
iϕ(i) = i<(i)
= 0
6= −1
= <(−1)
= ϕ(i · i).�
CHAPTER 8
Null Spaces and Ranges
8.1. Exercise 3.B.1: A Linear Map T with dim null T = 3,dim range T = 2
Give an example of a linear map T such that
dim null T = 3 and dim range T = 2.
Proof. Define T as the linear map in L(R5,R5) such that
T (a, b, c, d, e) = (a, b, 0, 0, 0)
for all (a, b, c, d, e) ∈ R5.It is clear that range T = 2 and dimV = 5. Hence, by the fundamental theorem of linear
maps,
dim null T = dimV − dim range T = 3.
�
8.2. Exercise 3.B.2: range S ⊂ null T Implies (ST )2 = 0
Suppose V is a vector space and S, T ∈ L(V, V ) are such that
range S ⊂ null T.
Prove that (ST )2 = 0.
Proof. Pick any v ∈ V , then
v ∈ V⇓
T (v) ∈ V⇓
S(T (v)) ∈ range S⇓
S(T (v)) ∈ null T⇓
T (S(T (v))) = 0⇓
S(T (S(T (v)))) = 0.
Hence (ST )2 = 0. �
8.3. Exercise 3.B.3: Spanning = Surjective; Independent = Injective
Suppose v1, . . . , vm is a list of vectors in V . Define T ∈ L(Fm, V ) by
T (z1, . . . , zm) = z1v1 + · · ·+ zmvm.
(a) What property of T corresponds to v1, . . . , vm spanning V ?(b) What property of T corresponds to v1, . . . , vm being linearly independent?
49
50 8. NULL SPACES AND RANGES
Proof. (a) We claim that T being surjective corresponds to v1, . . . , vm spanning V .If T is surjective, then range T = V . Since range T = span(v1, . . . , vm), we con-
clude that v1, . . . , vm spans V .If v1, . . . , vm spans V , then range T = V . Hence T is surjective.
(b) We claim that T being injective corresponds to v1, . . . , vm being linearly independent.If T is injective, then 3.16 of [Axl14] implies that null T = {0}. Hence T (z1, . . . , zm) =
z1v1 + · · · + zmvm = 0 implies z1 = · · · = zm = 0. Therefore v1, . . . , vm is linearly in-dependent.
If v1, . . . , vm is linearly independent, then z1v1 + · · ·+ zmvm = 0 implies z1 = · · · =zm = 0. Hence T (z1, . . . , zm) = 0 implies z1 = · · · = zm = 0. By 3.16 of [Axl14], T istherefore injective.
�
8.4. Exercise 3.B.4: {T ∈ L(R5,R4) : dim null T > 2} is NOT a Subspace
Show that
{T ∈ L(R5,R4) : dim null T > 2}is not a subspace of L(R5,R4).
Proof. Let S, T be the linear maps in L(R5,R4) such that
S(x1, x2, x3, x4, x5) = (x1, x2, 0, 0)
and
T (x1, x2, x3, x4, x5) = (x1, 0, x3, 0)
for all (x1, x2, x3, x4, x5) ∈ R5. Then
(S + T )(x1, x2, x3, x4, x5) = (2x1, x2, x3, 0)
for all (x1, x2, x3, x4, x5) ∈ R5.Clearly, dim range S = dim range T = 2 and dim range (S+T ) = 3. Hence the fundamental
theorem of linear maps implies
dim null S = dim null T = 5− 2 = 3 > 2
and
dim null (S + T ) = 5− 3 = 2 6> 2.
We have thus shown that
{T ∈ L(R5,R4) : dim null T > 2}is not closed under addition, hence it is not a subspace of L(R5,R4). �
8.5. Exercise 3.B.5: A Linear Map T such that range T = null T
Give an example of a linear map T : R4 → R4 such that
range T = null T.
Proof. Let T be the linear map in L(R4,R4) such that
T (x1, x2, x3, x4) = (0, 0, x1, x2)
for all (x1, x2, x3, x4) ∈ R4. Then
range T = {(0, 0, x1, x2) : x1, x2 ∈ R}and
null T = {(0, 0, x3, x4) : x3, x4 ∈ R}.
8.5. EXERCISE 3.B.5: A LINEAR MAP T SUCH THAT range T = null T 51
Hencerange T = null T.
�
CHAPTER 9
Matrices
9.1. Exercise 3.C.1: M(T ) Has At Least dim range T Nonzero Entries
Suppose V and W are finite-dimensional and T ∈ L(V,W ). Show that with respect to eachchoice of bases of V and W , the matrix of T has at least dim range T nonzero entries.
Proof. Choose any basis v1, . . . , vn of V and basis w1, . . . , wm of W , suppose the matrixof T with respect to those bases has k nonzero entries where
(9.1) k < dim range T.
Then the matrix has at most k nonzero columns. Consequently, by the definition of matrixof a linear map, at most k vectors of Tv1, . . . , T vn are nonzero. Hence
dim range T 6 k,
which contradicts with (9.1). �
9.2. Exercise 3.C.2: Find Bases from the Matrix of D ∈ L(P3(R),P2(R))
Suppose D ∈ L(P3(R),P2(R)) is the differentiation map defined by Dp = p′. Find a basisof P3(R) and a basis of P2(R)) such that the matrix of D with respect to these bases is1 0 0 0
0 1 0 00 0 1 0
.Proof. Define
w1 ≡ 1, w2 ≡ x,w3 ≡ x2
and
v1 ≡ x, v2 ≡1
2x2, v3 ≡
1
3x3, v4 ≡ 1.
Clearly, w1, w2, w3 is a basis of P2(R) and v1, v2, v3, v4 is a basis of P3(R). Moreover,
Dv1 = w1, Dv2 = w2, Dv3 = w3, Dv4 = 0.
Therefore the matrix of D under these bases is1 0 0 00 1 0 00 0 1 0
.�
53
54 9. MATRICES
9.3. Exercise 3.C.3: Find Bases for V,W from M(T ) where T ∈ L(V,W )
Suppose V and W are finite-dimensional and T ∈ L(V,W ). Prove that there exist a basis ofV and a basis of W such that with respect to these bases, all entries of M(T ) are 0 except thatthe entries in row j, column j, equal 1 for 1 6 j 6 dim range T .
Proof. Let u1, . . . , um be a basis of null T ; thus dim null T = m. By 2.33 of [Axl14], thelinearly independent list u1, . . . , um can be extended to a basis
(9.2) v1, . . . , vn, u1, . . . , um
of V . Thus dimV = m+ n. It is easy to verify that Tv1, . . . , T vn is a basis of range T . Again,by 2.33 of [Axl14], the linearly independent list Tv1, . . . , T vn can be extended to a basis
(9.3) Tv1, . . . , T vn, s1, . . . , sl
of W . Thus dimW = n + l. The matrix of T under bases (9.2) and (9.3) is then of size(n + l) × (n + m), and all of its entries equal to 0 except that the entries in row j, column j,equal 1 for 1 6 j 6 n. �
9.4. Exercise 3.C.4: Find Bases for W from M(T ) where T ∈ L(V,W )
Suppose v1, . . . , vm is a basis of V and W is finite-dimensional. Suppose T ∈ L(V,W ). Provethat there exists a basis w1, . . . , wn of W such that all the entries in the first column of M(T )(with respect to the bases v1, . . . , vm and w1, . . . , wn) are 0 except for possibly a 1 in the firstrow, first column.
Proof. If Tv1 = 0, then any basis of W would make all the entries in the first column ofM(T ) equal to 0.
If Tv1 6= 0, then let w1 ≡ Tv1. By 2.33 of [Axl14], the linearly independent list w1 canbe extended to a basis w1, . . . , wn of W . Under this basis, all the entries in the first column ofM(T ) are 0 except for a 1 in the first row, first column. �
9.5. Exercise 3.C.5: Find Bases for V from M(T ) where T ∈ L(V,W )
Suppose w1, . . . , wn is a basis of W and V is finite-dimensional. Suppose T ∈ L(V,W ).Prove that there exists a basis v1, . . . , vm of V such that all the entries in the first row of M(T )(with respect to the bases v1, . . . , vm and w1, . . . , wn) are 0 except for possibly a 1 in the firstrow, first column.
Proof. Pick any basis
v1, . . . , vm
of V . Denote M to be the matrix of T under this basis. If all the entries in the first row of Mare 0, then the proof is done.
Otherwise, suppose column j of the first row of M contains a nonzero entry. Let
v′1 ≡1
M1,jvj
and
{v′2, . . . , v′m} ≡ {v1, . . . , vj−1, vj+1, . . . , vm}.Then v′1, . . . , v
′m is also a basis of V .
Denote M ′ as the matrix of T under the basis
v′1, . . . , v′m.
9.5. EXERCISE 3.C.5: FIND BASES FOR V FROM M(T ) WHERE T ∈ L(V,W ) 55
Further, let
v′′1 ≡ v′1,v′′2 ≡ v′2 −M ′1,2v′1,
...
v′′m ≡ v′m −M ′1,mv′1.Then v′′1 , . . . , v
′′m is also a basis of V . Moreover,
T (v′′1 ) = T (v′1) = T (1
M1,jvj)
=1
M1,j(M1,jw1 +M2,jw2 + · · ·+Mn,jwn)
= w1 +M2,j
M1,jw2 + · · ·+ Mn,j
M1,jwn.
And for any i = 2, . . . ,m,
T (v′′i ) = T (v′i −M ′1,iv′1)
= T (v′i)−M ′1,iT (v′1)
= M ′1,iw1 + · · ·+M ′n,iwn −M ′1,i(w1 +M2,j
M1,jw2 + · · ·+ Mn,j
M1,jwn)
= (M ′2,i −M ′1,iM2,j
M1,j)w2 + · · ·+ (M ′n,i −
M ′1,iMn,j
M1,j)wn.
Hence all the entries in the first row ofM(T ) with respect to the bases v′′1 , . . . , v′′m and w1, . . . , wn
are 0 except for a 1 in the first row, first column. �
CHAPTER 10
Invertibility and Isomorphic Vector Spaces
10.1. Exercise 3.D.1: (ST )−1 = T−1S−1
Suppose T ∈ L(U, V ) and S ∈ L(V,W ) are both invertible linear maps. Prove that ST ∈L(U,W ) is invertible and that (ST )−1 = T−1S−1.
Proof. Since T and S are invertible, 3.56 of [Axl14] implies that T and S are injectiveand surjective.
Suppose S(T (u1)) = S(T (u2)) where u1, u2 ∈ U , then that S is injective implies T (u1) =T (u2), and that T is injective implies u1 = u2. Hence ST is injective.
Pick any w ∈ W . That S is surjective implies the existence of a v ∈ V such that S(v) = w.Further, that T is surjective implies the existence of a u ∈ U such that T (u) = v. HenceS(T (u)) = w. Thus ST is surjective.
Thus 3.56 of [Axl14] implies that ST is invertible.Finally, since
(T−1S−1)(ST ) = T−1(S−1S)T = T−1T = I
and
(ST )(T−1S−1) = S(TT−1)S−1 = SS−1 = I,
we conclude that
(ST )−1 = T−1S−1.
�
10.2. Exercise 3.D.2: Noninvertible Operators on V is NOT a Subspace
Suppose V is finite-dimensional and dimV > 1. Prove that the set of noninvertible operatorson V is not a subspace of L(V ).
Proof. Let v1, v2, . . . , vn be a basis of V where n > 2. Let S be the operator on V definedas
S(a1v1 + a2v2 + · · ·+ anvn) = a1v1,
and let T be the operator on V defined as
T (a1v1 + a2v2 + · · ·+ anvn) = a2v2 + a3v3 + · · ·+ anvn.
Then S is noninvertible because v2 ∈ null S, and T is noninvertible because v1 ∈ null T .However, S + T is invertible because S + T equals the identity operator.
We have shown that the set of noninvertible operators on V is not closed under addition,hence it is not a subspace of L(V ). �
57
58 10. INVERTIBILITY AND ISOMORPHIC VECTOR SPACES
10.3. Exercise 3.D.3: Injective Subspace Mappings and Invertible Operators
Suppose V is finite-dimensional, U is a subspace of V , and S ∈ L(U, V ). Prove there existsan invertible operator T ∈ L(V ) such that Tu = Su for every u ∈ U if and only if S is injective.
Proof. Suppose there exists an invertible operator T ∈ L(V ) such that Tu = Su for everyu ∈ U . It follows that
(10.1) T−1(Su) = T−1(Tu) = u
for every u ∈ U . Pick any u1, u2 ∈ U such that Su1 = Su2. Then T−1(Su1) = T−1(Su2), and(10.1) implies that u1 = u2. Hence S is injective.
Suppose S is injective. Pick a basis u1, . . . , uk of U and extend it to a basis
u1, . . . , uk, vk+1, . . . , vn
of V ; thus dimU = k,dimV = n. Since S is injective, S is an isomorphism between U andrange S. Hence
Su1, . . . , Suk
is a basis of range T and can be extended to a basis
Su1, . . . , Suk, tk+1, . . . , tn
of V . Define T as the operator on V such that
T (ui) = Sui where i = 1, . . . , k
and
T (vi) = ti where i = k + 1, . . . , n.
Clearly, T is invertible and Tu = Su for every u ∈ U . �
10.4. Exercise 3.D.4: null T1 = null T2 if and only if T1 = ST2
Suppose W is finite-dimensional and T1, T2 ∈ L(V,W ). Prove that null T1 = null T2 if andonly if there exists an invertible operator S ∈ L(W ) such that T1 = ST2.
Proof. Suppose there exists an invertible operator S ∈ L(W ) such that T1 = ST2, then
S−1T1 = T2.
Pick any v1 ∈ null T1, then T1v1 = 0 and
T2v1 = S−1T1(v1) = 0.
Hence null T1 ⊂ null T2. Also, pick any v2 ∈ null T2, then
T1v2 = ST2(v2) = 0.
Hence null T2 ⊂ null T1. Therefore
null T1 = null T2.
On the other hand, suppose null T1 = null T2. Denote
N ≡ null T1 = null T2
and
m ≡ dimW.
Pick a basis w1, . . . , wk of range T2 and extend it to a basis of W :
w1, . . . , wk, wk+1, . . . , wm.
10.5. EXERCISE 3.D.5: range T1 = range T2 IF AND ONLY IF T1 = T2S 59
Then there exists nonzero vectors v1, . . . , vk ∈ V such that
T2vi = wi, ∀i ∈ {1, . . . , k}.Then it is easy to verify that
V = N ⊕ span(v1, . . . , vk).
and
T1v1, . . . , T1vk
is a linearly independent list, which can be extended to a basis of W :
T1v1, . . . , T1vk, w′k+1, . . . , w
′m.
Let S ∈ L(W ) be defined by
S(wi) ≡ T1vi, ∀i ∈ {1, . . . , k},and
S(wi) ≡ w′i, ∀i ∈ {k + 1, . . . ,m}.We claim that T1 = ST2. To show this, pick any v ∈ V . Then there exists v0 ∈ N anda1, . . . , ak ∈ F such that
v = v0 + a1v1 + · · ·+ akvk
and
T1(v) = T1(v0 + a1v1 + · · ·+ akvk)
= a1T1v1 + · · ·+ akT1vk
= a1Sw1 + · · ·+ akSwk
= S(a1w1 + · · ·+ akwk)
= S(T2(a1v1 + · · ·+ akvk))
= S(T2(v0 + a1v1 + · · ·+ akvk))
= ST2(v).
�
10.5. Exercise 3.D.5: range T1 = range T2 if and only if T1 = T2S
Suppose V is finite-dimensional and T1, T2 ∈ L(V,W ). Prove that range T1 = range T2 ifand only if there exists an invertible operator S ∈ L(V ) such that T1 = T2S.
Proof. Suppose there exists an invertible operator S ∈ L(V ) such that T1 = T2S. Pick anyw ∈ range T1, then there exists v1 ∈ V such that T1v1 = w. Hence T2Sv1 = w and w ∈ range T2.Also, pick any w ∈ range T2, then there exists v2 ∈ V such that T2v2 = w. Hence T1S
−1v2 = wand w ∈ range T1. Therefore range T1 = range T2.
On the other hand, suppose range T1 = range T2. Denote
R ≡ range T1 = range T2
and
n ≡ dimV.
We claim that R is finite-dimensional. To show this, pick a basis u1, . . . , uk of null T1 and extendit to a basis of V :
u1, . . . , uk, uk+1, . . . , un.
Then it is easy to verify that
T1uk+1, . . . , T1un
60 10. INVERTIBILITY AND ISOMORPHIC VECTOR SPACES
is a basis of R; thusdimR = n− k.
The fundamental theorem of linear maps then implies
k = dim null T1 = dimV − dimR = dim null T2.
Moreover, there exist vk+1, . . . , vn ∈ V such that
T2vi = T1ui, ∀i ∈ {k + 1, . . . , n}.Pick a basis v1, . . . , vk of null T2. We claim that
v1, . . . , vk, vk+1, . . . , vn
is a basis of V . By 2.39 of [Axl14], it suffices to show its linear independence. Suppose thereexists a1, . . . , an ∈ F such that
a1v1 + · · ·+ anvn = 0.
Then
T2(a1v1 + · · ·+ anvn) = 0⇓
ak+1T2vk+1 + · · ·+ anT2vn = 0⇓
ak+1T1uk+1 + · · ·+ anT1un = 0⇓
ak+1 = · · · = an = 0⇓
a1v1 + · · ·+ akvk = 0⇓
a1 = · · · = ak = 0.
Let S ∈ L(V ) be defined byS(ui) ≡ vi, ∀i ∈ {1, . . . , n}.
We claim that T1 = T2S. To show this, pick any v ∈ V . Then there exists a1, . . . , an ∈ F suchthat
v = a1u1 + · · ·+ anun.
Then
T1(v) = T1(a1u1 + · · ·+ anun)
= ak+1T1uk+1 + · · ·+ anT1un
= ak+1T2vk+1 + · · ·+ anT2vn
= T2(ak+1vk+1 + · · ·+ anvn)
= T2(ak+1Suk+1 + · · ·+ anSun)
= T2(S(ak+1uk+1 + · · ·+ anun))
= T2(S(v)).
�
CHAPTER 11
Products and Quotients of Vector Spaces
11.1. Exercise 3.E.1: T is a Linear Map iff graph of T is a Subspace of V ×W
Suppose T is a function from V to W . The graph of T is the subset of V ×W defined by
graph of T = {(v, Tv) ∈ V ×W : v ∈ V }.
Prove that T is a linear map if and only if the graph of T is a subspace of V ×W .
Proof. Suppose T is a linear map. By 3.11 of [Axl14], T (0) = 0. Hence
(0, 0) ∈ graph of T.
Suppose (u, Tu), (w, Tw) ∈ graph of T . Then T (u+ w) = Tu+ Tw ∈W implies
(u+ w, Tu+ Tw) ∈ graph of T.
Suppose a ∈ F and (u, Tu) ∈ graph of T . Then T (au) = aTu implies
(au, aTu) ∈ graph of T.
By 13.4 of [Axl14], we therefore conclude that the graph of T is a subspace of V ×W .On the other hand, suppose the graph of T is a subspace of V ×W . Pick any u, v ∈ V such
that (u, Tu), (v, Tv) ∈ graph of T . Then (u+ v, Tu+ Tv) ∈ graph of T . Hence
T (u+ v) = Tu+ Tv.
Pick any a ∈ F, then (au, aTu) ∈ graph of T . Hence
T (au) = aTu.
Therefore, by definition, we conclude that T is a linear map. �
11.2. Exercise 3.E.2: Vj is Finite-dimensional if V1 × · · · × Vm is Finite-dimensional
Suppose V1, . . . , Vm are vector spaces such that V1 × · · · × Vm is finite-dimensional. Provethat Vj is finite-dimensional for each j = 1, . . . ,m.
Proof. Suppose there exists j ∈ {1, . . . ,m} such that Vj is infinite-dimensional. Then,by 4.14, there is a sequence vj1, vj2, . . . of vectors in Vj such that vj1, vj2, . . . , vjk is linearlyindependent for every positive integer k.
Let the sequence v1, v2, . . . of vectors in V1×· · ·×Vm be defined by: for each positive integeri, the vector vi has its jth slot equal to vji and other slots equal to 0. Then v1, v2, . . . , vk islinearly independent for every positive integer k. Therefore, again by 4.14, V1 × · · · × Vm isinfinite-dimensional. �
61
62 11. PRODUCTS AND QUOTIENTS OF VECTOR SPACES
11.3. Exercise 3.E.3: Example of U1 × U2 Isomorphic to Non-Direct-Sum U1 + U2
Give an example of a vector space V and subspaces U1, U2 of V such that U1×U2 is isomorphicto U1 + U2 but U1 + U2 is not a direct sum.
Proof. Let V ≡ U1 ≡ R∞. Let
U2 ≡ {(x, 0, 0, . . . ) : x ∈ R}.Then V is a vector space and U1, U2 are two subspaces of V , and
U1 + U2 = R∞.Since U1 ∩ U2 = U2 6= {0}, U1 + U2 is not a direct sum. Let function T : U1 × U2 → U1 + U2 bedefined by
T ((x, 0, 0, . . . ), (y1, y2, . . . )) = (x, y1, y2, . . . ).
We claim that T is a linear map. Suppose
u = ((ux, 0, 0, . . . ), (uy1 , uy2 , . . . ))
andv = ((vx, 0, 0, . . . ), (vy1 , vy2 , . . . )).
Then
T (u+ v) = (ux + vx, uy1 + vy1 , uy2 + vy2 , . . . )
= (ux, uy1, uy2
, . . . ) + (vx, vy1, vy2
, . . . )
= T (u) + T (v).
Pick any a ∈ F,
T (au) = (aux, auy1, auy2
, . . . )
= a(ux, uy1 , uy2 , . . . )
= aT (u).
Clearly T is both injective and surjective. Therefore, U1 × U2 is isomorphic to U1 + U2. �
11.4. Exercise 3.E.4: L(V1× · · · × Vm,W ) and L(V1,W )× · · · ×L(Vm,W ) are Isomorphic
Suppose V1, . . . , Vm are vector spaces. Prove that L(V1× · · ·×Vm,W ) and L(V1,W )× · · ·×L(Vm,W ) are isomorphic vector spaces.
Proof. Let function T : L(V1 × · · · × Vm,W )→ L(V1,W )× · · · × L(Vm,W ) be defined by
T (P ) = (P1, . . . , Pm)
where
P1(v1) = P (v1, 0, . . . , 0) ∀v1 ∈ V1,
P2(v2) = P (0, v2, . . . , 0) ∀v2 ∈ V2,
...
Pm(vm) = P (0, 0, . . . , vm) ∀vm ∈ Vm.We claim that T is a linear map. Pick any P1, P2 ∈ L(V1 × · · · × Vm,W ), then
T (P1 + P2) = ((P1 + P2)1, . . . , (P1 + P2)m)
= ((P1)1 + (P2)1, . . . , (P1)m + (P2)m)
= ((P1)1, . . . , (P1)m) + ((P2)1, . . . , (P2)m)
= T (P1) + T (P2).
11.5. EXERCISE 3.E.5: L(V,W1 × · · · ×Wm) AND L(V,W1)× · · · × L(V,Wm) ARE ISOMORPHIC 63
Pick any λ ∈ F, P ∈ L(V1 × · · · × Vm,W ), then
T (λP ) = ((λP )1, . . . , (λP )m)
= λ(P1, . . . , Pm)
= λT (P ).
We claim that T is injective. Suppose T (P ) = 0. Then,
P (v1, 0, . . . , 0) = 0 ∀v1 ∈ V1,
P (0, v2, . . . , 0) = 0 ∀v2 ∈ V2,
...
P (0, 0, . . . , vm) = 0 ∀vm ∈ Vm.
Hence, for all v1 ∈ V1, . . . , vm ∈ Vm,
P (v1, . . . , vm) = P (v1, . . . , 0) + · · ·+ P (0, . . . , vm) = 0.
Thus
P = 0.
We claim that T is surjective. Pick any (P1, . . . , Pm) ∈ L(V1,W ) × · · · × L(Vm,W ), letP ∈ L(V1 × · · · × Vm,W ) be defined by
P (v1, . . . , vm) = P1(v1) + · · ·+ Pm(vm) ∀v1 ∈ V1, . . . , vm ∈ Vm.
Then
T (P ) = (P1, . . . , Pm).
Therefore T is an isomorphism from L(V1× · · ·×Vm,W ) to L(V1,W )× · · ·×L(Vm,W ). �
11.5. Exercise 3.E.5: L(V,W1× · · ·×Wm) and L(V,W1)× · · ·×L(V,Wm) are Isomorphic
Suppose W1, . . . ,Wm are vector spaces. Prove that L(V,W1 × · · · ×Wm) and L(V,W1) ×· · · × L(V,Wm) are isomorphic vector spaces.
Proof. Let
Π1 ∈ L(W1 × · · · ×Wm,W1),
Π2 ∈ L(W1 × · · · ×Wm,W2),
...
Πm ∈ L(W1 × · · · ×Wm,Wm)
be defined by
Π1(w1, . . . , wm) = w1,
Π2(w1, . . . , wm) = w2,
...
Πm(w1, . . . , wm) = wm
for all (w1, . . . , wm) ∈W1 × · · · ×Wm.Let function T : L(V,W1 × · · · ×Wm)→ L(V,W1)× · · · × L(V,Wm) be defined by
T (P ) = (PΠ1, . . . , PΠm)
64 11. PRODUCTS AND QUOTIENTS OF VECTOR SPACES
where
PΠ1(v) = Π1(P (v)),
PΠ2(v) = Π2(P (v)),
...
PΠm(v) = Πm(P (v)).
for all v ∈ V .We claim that T is a linear map. Pick any P1, P2 ∈ L(V,W1 × · · · ×Wm), then
T (P1 + P2) = ((P1 + P2)Π1, . . . , (P1 + P2)Πm)
= ((P1)Π1 + (P2)Π1 , . . . , (P1)Πm + (P2)Πm)
= ((P1)Π1 , . . . , (P1)Πm) + ((P2)Π1 , . . . , (P2)Πm)
= T (P1) + T (P2).
Pick any λ ∈ F, P ∈ L(V1 × · · · × Vm,W ), then
T (λP ) = ((λP )Π1 , . . . , (λP )Πm)
= λ(PΠ1 , . . . , PΠm)
= λT (P ).
We claim that T is injective. Suppose T (P ) = 0. Then,
Π1(P (v)) = 0,
Π2(P (v)) = 0,
...
Πm(P (v)) = 0.
for all v ∈ V . Hence, P (v) = 0 for all v ∈ V . Thus
P = 0.
We claim that T is surjective. Pick any (P1, . . . , Pm) ∈ L(V,W1) × · · · × L(V,Wm), letP ∈ L(V,W1 × · · · ×Wm) be defined by
P (v) = (P1(v), . . . , Pm(vm)) ∀v ∈ V.Then
T (P ) = (P1, . . . , Pm).
Therefore T is an isomorphism from L(V,W1×· · ·×Wm) to L(V,W1)×· · ·×L(V,Wm). �
CHAPTER 12
Duality
12.1. Exercise 3.F.1: Linear Functionals are either Surjective or Zero
Explain why every linear functional is either surjective or the zero map.
Proof. Suppose ϕ is a linear functional from V to F and ϕ is neither surjective nor thezero map. Then there exists y0 ∈ F such that
ϕ(v) 6= y0, ∀v ∈ V.
Also, there exist x0 ∈ F, v0 ∈ V such that
x0 6= 0 and ϕ(v0) = x0.
Since ϕ is a linear map,
ϕ(y0
x0· v0) =
y0
x0· ϕ(v0)
=y0
x0· x0
= y0,
which contradicts with
ϕ(v) 6= y0, ∀v ∈ V.
�
12.2. Exercise 3.F.2: Examples of Linear Functionals on R[0,1]
Give three distinct examples of linear functionals on R[0,1].
Proof. Pick any c ∈ R. Let ϕ : R[0,1] → R be defined by
ϕ(f) = f(c), ∀f(x) ∈ R[0,1].
Then we claim that ϕ is a linear map. To show this, pick any g(x), f(x) ∈ R[0,1]. Then
ϕ(f + g) = f(c) + g(c) = ϕ(f) + ϕ(g).
Also, pick any λ ∈ R. Then
ϕ(λf) = (λf)(c) = λ(f(c)) = λϕ(f).
Let c ≡ 1, 2, 3 respectively, we then obtained three distinct linear functionals on R[0,1]. �
65
66 12. DUALITY
12.3. Exercise 3.F.3: Existence of ϕ ∈ V ′ such that ϕ(v) = 1
Suppose V is finite-dimensional and v ∈ V with v 6= 0. Prove that there exists ϕ ∈ V ′ suchthat ϕ(v) = 1.
Proof. Pick any basis v1, . . . , vn of V and its dual basis ϕ1, . . . , ϕn of V ′, then there existsa1, . . . , an ∈ F such that
v = a1v1 + · · ·+ anvn.
Since v 6= 0, there exists k ∈ {1, . . . , n} such that
ak 6= 0.
Let
ϕ ≡ 1
akϕk,
then ϕ ∈ V ′ and
ϕ(v) =1
akϕk(a1v1 + · · ·+ anvn)
=1
aka1ϕk(v1) + · · ·+ 1
akanϕk(vn)
=1
akakϕk(vk)
=1
akak
= 1.
�
12.4. Exercise 3.F.10: The Dual Map Function that Takes T to T ′ is Linear
Prove the first two bullet points in 3.101.
(1) (S + T )′ = S′ + T ′ for all S, T ∈ L(V,W ).(2) (λT )′ = λT ′ for all λ ∈ F and all T ∈ L(V,W ).
Proof. (1) Pick any S, T ∈ L(V,W ), ϕ ∈ W ′, v ∈ V . By definition, (S + T )′(ϕ) =ϕ ◦ (S + T ). Since
ϕ ◦ (S + T )(v) = ϕ(Sv + Tv) = ϕ(Sv) + ϕ(Tv),
we conclude that
ϕ ◦ (S + T ) = ϕ ◦ S + ϕ ◦ T = S′(ϕ) + T ′(ϕ).
Therefore (S + T )′ = S′ + T ′.(2) Pick any λ ∈ F, any T ∈ L(V,W ), any ϕ ∈ W ′ and any v ∈ V . By definiton,
(λT )′(ϕ) = ϕ ◦ (λT ). Since
ϕ ◦ (λT )(v) = ϕ(λTv) = λϕ(Tv) = λ(ϕ ◦ T )(v),
we conclude that
ϕ ◦ (λT ) = λ(ϕ ◦ T ) = λ(T ′(ϕ)).
Therefore (λT )′ = λT ′.�
12.5. EXERCISE 3.F.20: U ⊂ W IMPLIES W 0 ⊂ U0 67
12.5. Exercise 3.F.20: U ⊂W Implies W 0 ⊂ U0
Suppose U and W are subsets of V with U ⊂W . Prove that W 0 ⊂ U0.
Proof. Pick any ϕ ∈ W 0, w ∈ W,u ∈ U . Since U ⊂ W , it follows that u ∈ W . Henceϕ(u) = 0 and ϕ ∈ U0. Therefore W 0 ⊂ U0. �
CHAPTER 13
Polynomials
13.1. Exercise 4.2: {0} ∪ {p ∈ P(F) : deg p = m} is NOT a Subspace
Suppose m is a positive integer. Is the set
{0} ∪ {p ∈ P(F) : deg p = m}
a subspace of P(F)?
Proof. Denote
Am ≡ {0} ∪ {p ∈ P(F) : deg p = m}.Let m ≡ 2, and we claim that A2 is not a subspace of P(F). To show this, let p(z) ≡
z2 + z + 1, q(z) ≡ −z2 + z + 1. Then both p(z) and q(z) are in A2. However,
deg (p(z) + q(z)) = deg (2z + 2) = 1 6= 2.
Hence p(z) + q(z) /∈ A2. �
13.2. Exercise 4.4: There Exists a n-Degree Polynomial with m Zeros
Suppose m and n are positive integers with m 6 n, and suppose λ1, . . . , λm ∈ F. Prove thatthere exists a polynomial p ∈ P(F ) with deg p = n such that 0 = p(λ1) = · · · = p(λm) and suchthat p has no other zeros.
Proof. Let
p(z) ≡ (z − λ1) . . . (z − λm)n−m+1.
Then, by definition, deg p = m − 1 + (n −m + 1) = n and 0 = p(λ1) = · · · = p(λm). Hence itonly suffices to show that p has no other zeros.
Suppose p has another zero τ such that
τ 6= λi, ∀i ∈ {1, . . . ,m}.
Then
0 6= τ − λi, ∀i ∈ {1, . . . ,m}and
0 6= (τ − λ1) . . . (τ − λm)n−m+1.
However,
p(τ) = (τ − λ1) . . . (τ − λm)n−m+1,
which contradicts with
p(τ) = 0.
�
69
70 13. POLYNOMIALS
13.3. Exercise 4.6: p has Distinct Zeros iff p and p′ have NO Common Zeros
Suppose p ∈ P(C) has degree m. Prove that p has m distinct zeros if and only if p and itsderivative p′ have no zeros in common.
Proof. By 4.14 of [Axl14], p has the following factorization
p(z) = c(z − λ1) . . . (z − λm),
where c, λ1, . . . , λm ∈ C. By the product rule,
p′(z) = c
m∑i=1
∏j 6=i
(z − λj)
.
If p has m distinct zeros, then
p′(λi) = c∏j 6=i
(λi − λj) 6= 0, ∀i ∈ {1, . . . ,m}.
Hence p and p′ have no zeros in common.On the other hand, if λa = λb where a, b ∈ {1, . . . ,m} and a 6= b, then
p′(λa) = c(λa − λb)∏
j 6=a and j 6=b
(λa − λj) = 0.
Hence p and p′ share the zero λa. �
13.4. Exercise 4.7: An Odd-degree, Real-coefficient Polynomial has a Real Zero
Prove that every polynomial of odd degree with real coefficients has a real zero.
Proof. Suppose p ∈ P(R), and deg p = 2n+ 1 where n ∈ {0, 1, 2, . . . }.We prove by induction on n.Suppose n = 0, then p(z) = az + b for some a, b ∈ R and a 6= 0. Obviously p has a real zero
−ba.
Suppose p ∈ P(R) has a real zero when deg p = 2n+1. Pick any p ∈ P(R) with deg p = 2n+3.By the Fundamental Theorem of Algebra, p has a zero λ in C. If λ is real then the proof is done;otherwise 4.15 of [Axl14] implies that λ̄ is also a zero of p and 4.11 of [Axl14] then implies
p(z) = (z − λ)(z − λ̄)q(z),
where deg q = 2n+ 1 and the coefficients of q are real because
(z − λ)(z − λ̄) = z2 − 2(<(λ))z + |λ|2,
in which case q(z) has a real zero and consequently p(z) has a real zero. �
13.5. Exercise 4.8: A Linear Operator on P(R)
Define T : P(R)→ RR by
Tp =
{p−p(3)x−3 if x 6= 3,
p′(3) if x = 3.
Show that Tp ∈ P(R) for every polynomial p ∈ P(R) and that T is a linear map.
13.5. EXERCISE 4.8: A LINEAR OPERATOR ON P(R) 71
Proof. Letp(x) = a0 + a1x+ · · ·+ amx
m,
where a0, . . . , am ∈ R. By Theorem 8.4 of [Rud76],
p(x) =
m∑n=0
p(n)(3)
n!(x− 3)n.
Hence, when x 6= 3,
p− p(3)
x− 3=
m∑n=1
p(n)(3)
n!(x− 3)n−1.
When x = 3,m∑
n=1
p(n)(3)
n!(3− 3)n−1 =
p(1)(3)
1!· 00 = p′(3) = Tp(3).
Hence
Tp =
m∑n=1
p(n)(3)
n!(x− 3)n−1.
Since p(n)(3)n! ∈ R for all n ∈ {1, . . . ,m}, we conclude that Tp ∈ P(R).
Pick anyq(x) = b0 + b1x+ · · ·+ blx
l.
Then
L(p+ q) =
max(m,l)∑n=1
(p+ q)(n)(3)
n!(x− 3)n−1
=
max(m,l)∑n=1
p(n)(3)
n!(x− 3)n−1 +
max(m,l)∑n=1
q(n)(3)
n!(x− 3)n−1
=
m)∑n=1
p(n)(3)
n!(x− 3)n−1 +
l)∑n=1
q(n)(3)
n!(x− 3)n−1
= Lp+ Lq.
Pick any λ ∈ R, then
L(λp) =
m∑n=1
(λp)(n)(3)
n!(x− 3)n−1
=
m∑n=1
λp(n)(3)
n!(x− 3)n−1
= λ
m∑n=1
p(n)(3)
n!(x− 3)n−1
= λLp.
Therefore L is a linear map. �
Bibliography
[Axl14] S. Axler, Linear algebra done right, Undergraduate Texts in Mathematics, Springer International Pub-
lishing, 2014.
[Rud76] W. Rudin, Principles of mathematical analysis, International series in pure and applied mathematics,McGraw-Hill, 1976.
73