Chapter 1: Vector Spaces

Linear Algebra Done Right (4th Edition), by Sheldon Axler

Last updated: September 4, 2024

Contents
1A: Rn and Cn 2

1B: Definition of Vector Space 4

1C: Subspaces 6

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1A: Rn and Cn
Problem 1
Show that α + β = β + α for all α, β ∈ C.

Proof. Let α = a + bi, β = c + di. Then we have that

α + β = (a + bi) + (c + di)
= (c + di) + (a + bi)
=β+α

Problem 3
Show that (αβ)γ = α(βγ) for all α, β, γ ∈ C.

Proof. Choose arbitrary α, β, γ ∈ C. Denote α = a + bi, β = c + di, γ = e + f i.

Then we have that

(αβ)γ = ((ac − bd) + (ad + bc)i)(e + f i)

= (ace − bde − adf − bcf ) + (ade + bce + acf − bdf )i

At the same time, we have

α(βγ) = (a + bi)((ce − df ) + (cf + de)i)

= (ace − adf − bcf − bde) + (ade + acf + bce − bdf )i

Hence, (αβ)γ = α(βγ).

Problem 5
Show that for any α ∈ C, there exists a unique β ∈ C such that α + β = 0.

Proof. Denote α = a + bi. By the property of a field, we know there exists

unique c = −a and d = −b such that β = c + di and α + β = 0. Suppose for the
sake of contradiction that β is not unique, then there exists c′ + d′ i such that
(a + c′ ) + (b + d′ )i = 0 while c′ ̸= a or d′ ̸= d, contradicting the uniqueness of
additive inverse property.

Problem 8
Find two distinct squared roots of i.

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Proof. Suppose α = a + bi’s square equals one.

(a + bi)2 = a2 − b2 + 2abi = 1
Then |a| = |b|, 2ab = 1. So we get the solution to be
1 1 1 1
√ + √ , −√ − √ i
2 2 2 2

Problem 10
Show that (x + y) + z = x + (y + z) ∀x, y, z ∈ Fn .

Proof.

(x + y) + z = (x1 + y1 , . . . , xn + yn ) + (z1 , . . . , zn )
= (x1 , . . . , xn ) + (y1 + z1 , . . . , yn + zn )
= x + (y + z)

Problem 14
Show that γ(x + y) = γx + γy ∀γ ∈ F, x, y ∈ Fn .

Proof.

γ(x + y) = γ(x1 + y1 , . . . , xn + yn )
= (γ(x1 + y1 ), . . . , γ(xn + yn ))
= (γx1 + γy1 , . . . , γxn + γyn )
= γ(x1 , . . . , xn ) + γ(y1 , . . . , yn )
= γx + γy

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1B: Definition of Vector Space
Theorem 1. A vector space is a set that is closed under vector addition and
scalar multiplication. It also has the following properties:
ˆ commutativity
ˆ associativity
ˆ additive identity
ˆ additive inverse
ˆ multiplicative identity
ˆ (scalar) distributive
Notation: FS .
Explanation: If S is a set, then FS denotes the set of functions from S to F
(scalar function). e.g. f ∈ FS .
Comment: FS is a vector space; One can think of f ∈ Fn as f : {1, 2, . . . , n} →
F.

Problem 1
Prove that −(−v) = v for all v ∈ V .

Proof. We know −(−v) is the unique additive inverse of −v. At same time by
definition, v + (−v) = 0 and thus by commutativity (−v) + v = 0. This shows
that v is the unique additive inverse of (−v), and such that −(−v) = v.

Problem 2
Suppose a ∈ F, v ∈ V , and av = 0. Prove that a = 0 or v = 0.

Proof. Suppose for the sake of contradiction that a ̸= 0 and v ̸= 0 but av = 0.

v = 1v
1
v = · av
a
1
v= 0
a
This forms a contradiction.

Problem 3
Suppose v, w ∈ V . Explain why there exists a unique x ∈ V such that
v + 3x = w.

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Proof. Suppose there exists x′ which also satisfies the condition. Then we have

v + 3x = w v + 3x′ = w (1)

This gives that x = (w − v)/3 = x′ which shows x is unique.

Problem 4
The empty set is not a vector space, why?

Proof. There is no additive identity in the empty set.

Problem 7
Suppose S is a nonempty set. Let V S denotes the set of functions from
S to V . Define a natural addition and scalar multiplication on V S , and
show that V S is a vector space with these definitions.

Proof. Let f, g ∈ V S : S → V . Define the addition and multiplication to be that

f + g(x) = f (x) + g(x)

We have that
ˆ commutativity: f + g(x) = f (x) + g(x) = g(x) + f (x) = g + f (x)

ˆ associativity: (f + g) + h(x) = f (x) + g(x) + h(x) = f + (g + h)(x)

ˆ additive identity: Define 0 : S → 0 ∈ V , then f + 0(x) = 0 + f (x) = f (x)

ˆ additive inverse: for every f ∈ V S , define g(x) = −f (x) which exists by

the property of vector space and thus we have that g + f = 0 for every x
and thus that it exists.
ˆ multiplicative identity: same as above

ˆ (scalar) distributive: a(f + g)(x) = a(f (x) + g(x)) = af (x) + g(x)

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1C: Subspaces
Definition 2 (subspace). A subset U of V is called subspace of V if U is also a
vector space with the same additive identity, addition, and scalar multiplication
as on V .
Remark 3. The set {0} is the smallest subspace of V , and V itself is the largest
subspace of V .

Remark 4. The subspace of R2 are precisely {0}, all lines in R2 containing the
origin, R2 .
Definition 5 (Sum of subspace). Suppose V1 , · · · , Vm are subspaces. The sum
of them, denoted by V1 + · · · + Vm , is the set of all possible sums of element of
V1 , · · · , Vm . Specifically,

V1 + · · · + Vm = {v1 + · · · + vm : v1 ∈ V1 , · · · , vm ∈ Vm }
Lemma 6. Suppose V1 , · · · , Vm are subspaces of V . Then V1 + · · · + Vm is the
smallest subspace of V containing V1 , · · · , Vm .
Definition 7 (Direct Sum). Suppose V1 , · · · , Vm are subspaces of V .

ˆ The sum V1 +· · ·+Vm is called a direct sum if each element of V1 +· · ·+Vm

can be written only as a sum of v1 , · · · , vm , where each vk ∈ Vk .
ˆ If V1 + · · · + Vm is a direct sum, then V1 ⊕ · · · ⊕ Vm denotes V1 + · · · + Vm ,
with ⊕ serving as the indication that this is a direct sum.

Example. Suppose Vk is a subspace of Fn of those vectors whose coordinates are

all zero but k-th coordinate. Then we have

Fn = V1 ⊕ · · · + Vm

Example (Sum that is not a direct sum). Suppose

V1 = {(x, y, 0) ∈ F3 : x, y ∈ F}
V2 = {(0, 0, z) ∈ F3 : z ∈ F}
V3 = {(0, y, y) ∈ F3 : y ∈ F}

Then F 3 = V1 + V2 + V3 because for every (x, y, z) ∈ F3 ,

(x, y, z) = (x, y, 0) + (0, 0, z) + (0, 0, 0)

However, F 3 ̸= V1 ⊕ V2 ⊕ V3 since

(0, 0, 0) = (0, −1, 0) + (0, 0, −1) + (0, 1, 1)

= (0, 0, 0) + (0, 0, 0) + (0, 0, 0)

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Theorem 8. Suppose V1 , . . . , Vm are subspaces of V . Then V1 + · · · + Vm is a
direct sum if and only if the only way to write 0 as a sum of v1 + · · · + vm , where
vk ∈ Vk , is by taking each vk to equal 0.
Theorem 9. Suppose that U and W are subspaces of V . Then

U + W is a direct sum ⇐⇒ U ∩ W = {0}

Problem 1
Verify the following examples to be valid subspaces:
1. If b ∈ F, then

{(x1 , x2 , x3 , x4 ) ∈ F4 : x3 = 5x4 + b}

is a subspace of F4 if and only if b = 0.

2. The set of continuous real-valued functions on the interval [0, 1] is
a subspace of R[0,1] .
3. The set of differential real-valued functions on R is a subspace of
RR .
4. The set of differentiable real-valued functions f on the interval (0, 3)
such that f ′ (2) = b is a subspace of R(0,3) if and only if b = 0.

5. The set of all sequences of complex numbers with limit 0 is a

subspace of C∞ .

Proof. 1. ⇒ (0, 0, 0, 0) is an element and thus 0 = 0 + b, b = 0 ⇐ Easy to

verify.
2. 0 function is cts; cts functions are closed under addition and scalar multi-
plication.
3. 0 function is differentiable; differentiable functions are closed under addition
and scalar multiplication.
4. For this to be closed under addition, one needs to restrict that f ′ (2)+g ′ (2) =
b + b = b and thus b = 0.
5. limn→∞ a(S1 + S2 ) = limn→∞ aS1 + limn→∞ aS2 = 0 + 0 = 0. At the same
time, the 0 sequence has limit 0.

Problem 4
Suppose b ∈ R. Show that theRset of continuous real-valued functions f
1
on the interval [0, 1] such that 0 f = b is a subspace of R[0,1] if and only
if b = 0.

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 R1 R1 R1
Proof. ⇒ 0 f + g = 0 f + 0 g = 2b = b so b = 0.
⇐ 0 is in the set; closed under addition/multiplication.

Problem 5
Prove that R2 is not a subspace of C2 over the field C.

Proof. This does not hold for scalar multiplication since we’ve defined scalar to
be complex numbers. To see this, Let a = (x + yi) ∈ C and z = (z1 , z2 ) ∈ R2 .
/ R2 .
We have that az = (z1 (x + yi), z2 (x + yi)) ∈

Problem 7
Prove or disprove: If U is a nonempty subset of R2 such that U is closed
under addition and under taking additive inverse (−u ∈ U), then U is a
subspace in R2 .

Proof. No. U = {(x1 , x2 ) : x1 , x2 ∈ Z}. Then 12 (1, 1) ∈

/ U.

Problem 9
A function f : R → R is called periodic if there exists a positive number
p s.t. f (x) = f (x + p) for all x ∈ R. Is the set of periodic functions from
R to R a subspace of RR ?

Proof. No, problem occurs with the addition. Suppose we have f (x) = f (x + p)
and g(x) = g(x + q). Then (f + g)(x) = f (x) + g(x) = f (x + p) + g(x + q) ̸=
(f + g)(x + l) for some fixed l for all p, q.

Problem 11
Prove that the intersection of every collection of subspaces of V is a
subspace of V .

T
Proof. Let i Vi denote the collection
T of subspaces of V . ThenTwe know 0 ∈
bigcupi Vi . Let a ∈ F, x, y ∈ i Vi . We have that a(x + y) ∈ i Vi and thus
finish the proof.

Problem 12
Prove that the union of two subspaces of V is a subspace of V if and only
if one of the subspaces is contained in the other.

Proof. Let V1 , V2 be two subspaces of V .

⇒ Suppose for the sake of contradiction that there exists v1 ∈ V1 s.t. v1 ∈/ V2
and v2 ∈ V2 s.t. v2 ∈
/ V1 . Then by assumption we have that v1 +v2 ∈ V1 ∪V2 . Here
we can also show that v1 + v2 ∈/ V1 because if it does, v1 + v2 + (−v1 ) = v2 ∈ V1 .
Similarly, v1 + v2 ∈
/ V2 . Thus we’ve reached a contradiction.

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 ⇐ This direction is trivial.

Problem 14
Suppose

U = {(x, −x, 2x) ∈ F3 : x ∈ F} and W = {(x, x, 2x) ∈ F3 : x ∈ F}

Describe U + W.

Proof.
(x, −x, 2x) + (y, y, 2y) = (x + y, −x + y, 2(x + y))
One can think of this as U + W = {(a, b, 2a) : a, b ∈ F}.

Problem 15
Suppose U is a subspace of V , what is U + U?

Proof. U + U = U.
Take u1 + u2 ∈ U + U, then u1 + u2 ∈ U. Conversely, take u ∈ U, then
u = u + 0 ∈ U + U.

Problem 16
Is the operation of addition on the subspaces of V commutative (U + W =
W + U)?

Proof. Take u ∈ U, w ∈ W, then u + w = w + u, implying the conclusion.

Problem 18
Does the operation of addition on the subspaces of V have an additive
identity? Which subspaces have additive inverses?

Proof. Yes, the zero subspace i.e. {0} is the additive identity. The subspace
that have additive inverses is only {0}.

Problem 19
Prove or disprove: If V1 , V2 , U are subspaces of V such that

V1 + U = V2 + U

then V1 = V2 .

F
Proof. Counterexample: Consider when U = V1 V2 , then the relation holds
while V1 ̸= V2 .

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 Problem 20
Suppose
U = {(x, x, y, y) ∈ F4 : x, y ∈ F}
Find a subspace W ∈ F4 s.t. F4 = U ⊕ W.

Proof. Define W = {(0, a, b, 0) ∈ F4 : a, b ∈ F} to be a subspace of F4 .

Then first W + U ⊆ F4 . Take (q, w, e, r) ∈ F4 , we can have (q, w, e, r) =
(q, q, r, r) + (0, w − q, e − r, r) ∈ U + W. We have F4 = U + W. Furthermore,
take (x, x, y, y) ∈ U, (0, a, b, 0) ∈ W. For the element to be in the intersection,
we need to have (x, x, y, y) = (0, a, b, 0) which implies that x = y = a = b = 0
and thus W ∩ U = {0}.

Problem 21
Suppose
U = {x, y, x + y, x − y, 2x} ∈ F5 : x, y ∈ F
Find a subspace W ∈ F5 s.t. F5 = U ⊕ W.

Proof. Define
W = {(0, 0, m, n, z) : m, n, z ∈ F}
Then (a, b, c, d, e) = (a, b, a + b, a − b, 2a) + (0, 0, c − (a + b), d − (a − b), e − 2a).
The rest follows exactly as in P20.

Problem 23
Prove or disprove: If V1 , V2 , U are subspaces of V s.t.

V = V1 ⊕ U and V = V2 ⊕ U,

then V1 = V2 .

Proof. Counterexample: Let U = {(x, x) : x ∈ F}, V1 = {(x, 0) : x ∈ F}, V2 =

{(0, x) : x ∈ F}.

Problem 24
A function f : R → R is called even if f (−x) = f (x) and odd if f (−x) =
−f (x) for all x ∈ R. Let Ve denote the set of real-valued even functions
on R and let Vo denote the set of real-valued odd functions on R. Show
that RR = Ve ⊕ Vo .

Proof. ⇐ Trivial direction.

⇒ f (x) = f (x)+f
2
(−x)
+ f (x)−f
2
(−x)

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