Jaysen Tsao
Linear Algebra

Vector Spaces and Subspaces

Vector Spaces

A vector is an element of a vector space, which is a fundamental structure in linear algebra. In fact, the study of vector spaces is what we call linear algebra.

Definition 6: Vector Space

A vector space (𝑉,+,) over a field 𝐹, sometimes called an 𝑭-vector space, is a set 𝑉 together with a binary operation + (called vector addition) and a function :𝐹×𝑉𝑉 (called scalar multiplication) such that the following properties hold:

For all 𝑎,𝑏𝐹 and 𝐮,𝐯,𝐰𝑉:

  1. Associativity of vector addition. (𝐮+𝐯)+𝐰=𝐮+(𝐯+𝐰).
  2. Commutativity of vector addition. 𝐮+𝐯=𝐯+𝐮.
  3. Existence of vector additive identity. 𝟎𝑉 s.t. 𝐯+𝟎=𝐯.
  4. Existence of vector additive inverses. (𝐯)𝑉 s.t. 𝐯+(𝐯)=𝟎.
  5. Compatibility of field multiplicative identity. 1𝐹𝐯=𝐯.
  6. Distributivity of scalar multiplication over vector addition. 𝑎(𝐮+𝐯)=𝑎𝐮+𝑎𝐯.
  7. Distributivity of scalar multiplication over scalar addition. (𝑎+𝑏)𝐯=𝑎𝐯+𝑏𝐯.
  8. Compatibility of scalar and field multiplication. (𝑎𝑏)𝐯=𝑎(𝑏𝐯).

Together, these are called the vector space axioms. The elements of 𝑉 are called vectors. For any 𝐯𝑉, the element 𝐯 is called the additive inverse of 𝐯.

Notation

Vectors are often denoted in boldface (𝐯) or with an arrow on top (𝑣) to distinguish them from scalars.

Notation

A vector space (𝑉,+,) over a field 𝐹 is usually denoted simply as 𝑉, where the operations of vector addition and scalar multiplication are implied. When there is ambiguity, the notation +𝑉 and 𝑉 may be used to refer to the vector addition and scalar multiplication operations of 𝑉, respectively.

Important

The definition of scalar multiplication implies that scalar multiplication is closed over 𝑉, meaning that for any scalar 𝑎𝐹 and any vector 𝐯𝑉, the result of scalar multiplication 𝑎𝐯 is also an element of 𝑉.

Notation

The vector additive identity of a vector space 𝑉 may be denoted 𝟎𝑉 when there is ambiguity.

Theorem 3: Multiplication of a vector by the zero scalar

Let 𝑉 be a vector space over a field 𝐹. Then for any vector 𝐯𝑉, 0𝐹𝐯=𝟎𝑉.

Proof: Theorem 3.

Let 𝑉 be a vector space over a field 𝐹, and let 𝐯𝑉 be an arbitrary vector. Call the field additive identity 0𝐹𝐹. Then:

0𝐹𝐯=(0𝐹+0𝐹)𝐯 by the definition of additive identity 0𝐹𝐯=0𝐹𝐯+0𝐹𝐯 by distributivity of scalar multiplication over +𝐹0𝐹𝐯+(0𝐹𝐯)=0𝐹𝐯+0𝐹𝐯+(0𝐹𝐯) by left-adding 0𝐹𝐯 to both sides 𝟎𝑉=0𝐹𝐯+0𝐹𝐯+(0𝐹𝐯) by the definition of vector additive inverse 𝟎𝑉=0𝐹𝐯+(0𝐹𝐯+(0𝐹𝐯)) by associativity of vector addition 𝟎𝑉=0𝐹𝐯+𝟎𝑉 by the definition of vector additive inverse 𝟎𝑉=0𝐹𝐯 by the definition of vector additive identity 

Theorem 4: Multiplication of the zero vector by a scalar

Let 𝑉 be a vector space over a field 𝐹. Then for any scalar 𝑎𝐹, 𝑎𝟎𝑉=𝟎𝑉.

Proof: Theorem 4.

Let 𝑉 be a vector space over a field 𝐹, and let 𝑎𝐹 be an arbitrary scalar. Call the vector additive identity 𝟎𝑉𝑉. Then:

𝑎𝟎𝑉=𝑎(𝟎𝑉+𝟎𝑉) by the definition of vector additive identity 𝑎𝟎𝑉=𝑎𝟎𝑉+𝑎𝟎𝑉 by distributivity of scalar multiplication over +𝑉𝑎𝟎𝑉+(𝑎𝟎𝑉)=𝑎𝟎𝑉+𝑎𝟎𝑉+(𝑎𝟎𝑉) by left-adding 𝑎𝟎𝑉 to both sides 𝟎𝑉=𝑎𝟎𝑉+𝑎𝟎𝑉+(𝑎𝟎𝑉) by the definition of vector additive inverse 𝟎𝑉=𝑎𝟎𝑉+(𝑎𝟎𝑉+(𝑎𝟎𝑉)) by associativity of vector addition 𝟎𝑉=𝑎𝟎𝑉+𝟎𝑉 by the definition of vector additive inverse 𝟎𝑉=𝑎𝟎𝑉 by the definition of vector additive identity 

Theorem 5: Negation is Scalar Multiplication by 1

Let 𝑉 be a vector space over a field 𝐹. Then for any vector 𝐯𝑉, (1)𝐯=𝐯.

Proof: Theorem 5.

Let 𝑉 be a vector space over a field 𝐹, and let 𝐯𝑉 be an arbitrary vector. Call the field additive identity 0𝐹𝐹 and the field multiplicative identity 1𝐹. By the definition of the additive inverse, there exists a scalar 1𝐹 such that 1+(1)=0𝐹. Then:

(1+(1))𝐯=0𝐹𝐯 by right-multiplying both sides by 𝐯1𝐯+(1)𝐯=0𝐹𝐯 by distributivity of scalar multiplication over +𝐹𝐯+(1)𝐯=0𝐹𝐯 by the definition of multiplicative identity 𝐯+(1)𝐯=𝟎𝑉 by Theorem 3(𝐯)+𝐯+(1)𝐯=𝟎𝑉+(𝐯) by adding 𝐯 to both sides (𝐯)+𝐯+(1)𝐯=𝐯 by the definition of vector additive identity 𝐯+(𝐯)+(1)𝐯=𝐯 by commutativity of vector addition 𝟎𝑉+(1)𝐯=𝐯 by the definition of vector additive inverse (1)𝐯=𝐯 by the definition of vector additive identity 

The cartesian product of a field 𝐹 with itself 𝑛 times, denoted 𝐹𝑛, is the set of all 𝑛-tuples of elements of 𝐹.

It turns out that for any field 𝐹 and positive integer 𝑛, 𝐹𝑛 is a vector space over 𝐹 under componentwise addition and scalar multiplication.

Notation

In the context of introducing 𝐹𝑛, assume 𝑛 is a positive integer.

Definition 7: The set 𝐹𝑛

Let 𝐹 be a field. The set 𝐹𝑛 is the set of all 𝑛-tuples of elements of 𝐹:

𝐹𝑛={(𝑎1,𝑎2,,𝑎𝑛)|𝑎1,𝑎2,,𝑎𝑛𝐹}.

Definition 8: Operations on 𝐹𝑛

Define componentwise addition and scalar multiplication on 𝐹𝑛 as follows:

  1. Componentwise addition. For any 𝐮=(𝑢1,𝑢2,,𝑢𝑛),𝐯=(𝑣1,𝑣2,,𝑣𝑛)𝐹𝑛,

    𝐮+𝐯=(𝑢1+𝑣1,𝑢2+𝑣2,,𝑢𝑛+𝑣𝑛).
  2. Scalar multiplication. For any scalar 𝑎𝐹 and any vector 𝐯=(𝑣1,𝑣2,,𝑣𝑛)𝐹𝑛,

    𝑎𝐯=(𝑎𝑣1,𝑎𝑣2,,𝑎𝑣𝑛).

Theorem 6: The set 𝐹𝑛 is a vector space over 𝐹

Let 𝐹 be a field. Then 𝐹𝑛 is a vector space over 𝐹 under the operations defined in Definition 8.

Proof: Theorem 6.

Let 𝐹 be a field, and let 𝐹𝑛 be the set as defined in Definition 7. Define vector addition and scalar multiplication on 𝐹𝑛 as in Definition 8. We will verify that 𝐹𝑛 satisfies all vector space axioms under these operations.

Let 𝑎,𝑏𝐹 be arbitrary scalars, and let 𝐮=(𝑢1,𝑢2,,𝑢𝑛),𝐯=(𝑣1,𝑣2,,𝑣𝑛),𝐰=(𝑤1,𝑤2,,𝑤𝑛)𝐹𝑛 be arbitrary vectors. Then:

  1. Associativity of vector addition.

    (𝐮+𝐯)+𝐰=((𝑢1,𝑢2,,𝑢𝑛)+(𝑣1,𝑣2,,𝑣𝑛))+(𝑤1,𝑤2,,𝑤𝑛)=((𝑢1+𝑣1)+𝑤1,(𝑢2+𝑣2)+𝑤2,,(𝑢𝑛+𝑣𝑛)+𝑤𝑛)=(𝑢1+(𝑣1+𝑤1),𝑢2+(𝑣2+𝑤2),,𝑢𝑛+(𝑣𝑛+𝑤𝑛))=(𝑢1,𝑢2,,𝑢𝑛)+(𝑣1+𝑤1,𝑣2+𝑤2,,𝑣𝑛+𝑤𝑛)=𝐮+(𝐯+𝐰).
  2. Commutativity of vector addition.

    𝐮+𝐯=(𝑢1,𝑢2,,𝑢𝑛)+(𝑣1,𝑣2,,𝑣𝑛)=(𝑢1+𝑣1,𝑢2+𝑣2,,𝑢𝑛+𝑣𝑛)=(𝑣1+𝑢1,𝑣2+𝑢2,,𝑣𝑛+𝑢𝑛)=(𝑣1,𝑣2,,𝑣𝑛)+(𝑢1,𝑢2,,𝑢𝑛)=𝐯+𝐮.
  3. Existence of vector additive identity. Choose 𝟎𝑉=(0𝐹,0𝐹,,0𝐹), where 0𝐹 is the additive identity of 𝐹. For any vector 𝐯=(𝑣1,𝑣2,,𝑣𝑛)𝐹𝑛, we have:

    𝐯+𝟎𝑉=(𝑣1+0𝐹,𝑣2+0𝐹,,𝑣𝑛+0𝐹)=(𝑣1,𝑣2,,𝑣𝑛)=𝐯.
  4. Existence of vector additive inverses. For any vector 𝐯=(𝑣1,𝑣2,,𝑣𝑛)𝐹𝑛, choose 𝐯=(𝑣1,𝑣2,,𝑣𝑛), where 𝑣𝑖 is the additive inverse of 𝑣𝑖 in 𝐹. Then:

    𝐯+(𝐯)=(𝑣1+(𝑣1),𝑣2+(𝑣2),,𝑣𝑛+(𝑣𝑛))=(0𝐹,0𝐹,,0𝐹)=𝟎𝑉.
  5. Compatibility of field multiplicative identity. For any vector 𝐯=(𝑣1,𝑣2,,𝑣𝑛)𝐹𝑛:

    1𝐹𝐯=(1𝐹𝑣1,1𝐹𝑣2,,1𝐹𝑣𝑛)=(𝑣1,𝑣2,,𝑣𝑛)=𝐯.
  6. Distributivity of scalar multiplication over vector addition. For any scalar 𝑎𝐹 and any vectors 𝐮=(𝑢1,𝑢2,,𝑢𝑛),𝐯=(𝑣1,𝑣2,,𝑣𝑛)𝐹𝑛:

    𝑎(𝐮+𝐯)=𝑎((𝑢1,𝑢2,,𝑢𝑛)+(𝑣1,𝑣2,,𝑣𝑛))=𝑎(𝑢1+𝑣1,𝑢2+𝑣2,,𝑢𝑛+𝑣𝑛)=(𝑎(𝑢1+𝑣1),𝑎(𝑢2+𝑣2),,𝑎(𝑢𝑛+𝑣𝑛))=(𝑎𝑢1+𝑎𝑣1,𝑎𝑢2+𝑎𝑣2,,𝑎𝑢𝑛+𝑎𝑣𝑛)=(𝑎𝑢1,𝑎𝑢2,,𝑎𝑢𝑛)+(𝑎𝑣1,𝑎𝑣2,,𝑎𝑣𝑛)=𝑎(𝑢1,𝑢2,,𝑢𝑛)+𝑎(𝑣1,𝑣2,,𝑣𝑛)=𝑎𝐮+𝑎𝐯.
  7. Distributivity of scalar multiplication over scalar addition. For any scalars 𝑎,𝑏𝐹 and any vector 𝐯=(𝑣1,𝑣2,,𝑣𝑛)𝐹𝑛:

    (𝑎+𝑏)𝐯=(𝑎+𝑏)(𝑣1,𝑣2,,𝑣𝑛)=((𝑎+𝑏)𝑣1,(𝑎+𝑏)𝑣2,,(𝑎+𝑏)𝑣𝑛)=(𝑎𝑣1+𝑏𝑣1,𝑎𝑣2+𝑏𝑣2,,𝑎𝑣𝑛+𝑏𝑣𝑛)=(𝑎𝑣1,𝑎𝑣2,,𝑎𝑣𝑛)+(𝑏𝑣1,𝑏𝑣2,,𝑏𝑣𝑛)=𝑎(𝑣1,𝑣2,,𝑣𝑛)+𝑏(𝑣1,𝑣2,,𝑣𝑛)=𝑎𝐯+𝑏𝐯.
  8. Compatibility of scalar and field multiplication. For any scalars 𝑎,𝑏𝐹 and any vector 𝐯=(𝑣1,𝑣2,,𝑣𝑛)𝐹𝑛:

    (𝑎𝑏)𝐯=(𝑎𝑏)(𝑣1,𝑣2,,𝑣𝑛)=((𝑎𝑏)𝑣1,(𝑎𝑏)𝑣2,,(𝑎𝑏)𝑣𝑛)=(𝑎(𝑏𝑣1),𝑎(𝑏𝑣2),,𝑎(𝑏𝑣𝑛))=𝑎(𝑏𝑣1,𝑏𝑣2,,𝑏𝑣𝑛)=𝑎(𝑏(𝑣1,𝑣2,,𝑣𝑛))=𝑎(𝑏𝐯).
Example.
An element of 𝐹3 is a triple (𝑎,𝑏,𝑐) where 𝑎,𝑏,𝑐𝐹.
Example.
Elements of 2, a vector space over , can represent points in the 2D Cartesian plane.

Terminology: Real and Complex Vector Spaces

A vector space over is called a real vector space, and a vector space over is called a complex vector space.

Corollary 1

𝑛 is a real vector space, and 𝑛 is a complex vector space.

Subspaces

A subspace of an 𝐹-vector space 𝑉 is a subset of 𝑉 which is itself a vector space under the same vector addition and scalar multiplication as 𝑉.

Definition 9: Subspace

Let 𝑉 be a vector space over a field 𝐹, and let 𝐻 be a subset of 𝑉. Let +𝑉 denote the vector addition operation on 𝑉, and let 𝑉 denote the scalar multiplication function on 𝑉 using scalars from 𝐹.

Then 𝐻 is a subspace of 𝑉 iff 𝐻 is itself a vector space under the vector addition +𝑉 and scalar multiplication 𝑉.

Similar to subfields, if we already know that some set 𝐻 is a subset of a vector space 𝑉, then only three criteria need to be checked (rather than rechecking all axioms):

Theorem 7: Subspace Criteria

Let 𝑉 be a vector space over a field 𝐹, and let 𝐻 be a subset of 𝑉. Let 𝟎𝑉 denote the vector additive identity of 𝑉. Then 𝐻 is a subspace of 𝑉 if and only if the following criteria are met:

  1. Existence of additive identity. 𝟎𝑉𝐻.
  2. Closure under vector addition. 𝐮,𝐯𝐻,𝐮+𝐯𝐻.
  3. Closure under scalar multiplication. 𝑎𝐹,𝐯𝐻,𝑎𝐯𝐻.
Proof: Subspace Criteria.

Suppose 𝐻 is a subset of 𝑉. Let 𝑝 be the property that 𝐻 is a subspace of 𝑉, and let 𝑞 be the property that 𝐻 satisfies the three conditions listed in Theorem 7.

𝒑𝒒. Assume 𝐻 is a subspace of 𝑉. The vector additive identity of 𝑉 is unique. Since 𝐻 is a subspace of 𝑉, its vector additive identity must be the same as that of 𝑉, so 𝟎𝑉𝐻. 𝐻 is a vector space under the same operations as 𝑉, so 𝐻 is closed under vector addition and scalar multiplication. Therefore, 𝑞 is true.

𝒒𝒑. Assume that 𝟎𝑉𝐻, 𝐻 is closed under vector addition, and 𝐻 is closed under scalar multiplication. Then, axiom (3) is satisfied, as well as the requirement that +𝑉:𝐻×𝐻𝐻 and 𝑉:𝐹×𝐻𝐻. Since 𝐻𝑉, the axioms of vector spaces that involve only elements of 𝐻 and the operations +𝑉 and 𝑉 (namely axioms 1, 2, 5, 6, 7, and 8) must also hold for 𝐻 since they hold for all elements of 𝑉. Finally, axiom (4) is satisfied since for any 𝐯𝐻, we know that (1𝐹)𝐯𝐻 by closure of 𝐻 under scalar multiplication, and (1𝐹)𝐯=𝐯, so 𝐯𝐻 by Theorem 5. Thus, 𝐻 satisfies all vector space axioms under the same operations as 𝑉, and 𝐻 is a subset of 𝑉, so 𝐻 is a subspace of 𝑉. Therefore, 𝑝 is true.

Since 𝑝𝑞 and 𝑞𝑝, we have 𝑝𝑞. ∎

Property: {𝟎} is a subspace of every vector space

If 𝟎𝑉 is the vector additive identity of a vector space 𝑉, then {𝟎𝑉} is a subspace of 𝑉. (This is called the trivial subspace of 𝑉.)

Proof: {𝟎} is a subspace of every vector space.

Suppose 𝑉 is a vector space over 𝐹, and 𝟎𝑉 is the vector additive identity of 𝑉. Since 𝟎𝑉𝑉, it follows that {𝟎𝑉}𝑉.

  1. Existence of additive identity. 𝟎𝑉{𝟎𝑉}.
  2. Closure under vector addition. For any 𝐮,𝐯{𝟎𝑉}, we have 𝐮=𝐯=𝟎𝑉, so 𝐮+𝐯=𝟎𝑉+𝟎𝑉=𝟎𝑉{𝟎𝑉}.
  3. Closure under scalar multiplication. For any scalar 𝑎𝐹 and any vector 𝐯{𝟎𝑉}, we have 𝐯=𝟎𝑉, so 𝑎𝐯=𝑎𝟎𝑉=𝟎𝑉{𝟎𝑉}.

By Theorem 7, {𝟎𝑉} is a subspace of 𝑉. ∎

Property: Every vector space is a subspace of itself

If 𝑉 is a vector space, then 𝑉 is a subspace of itself, 𝑉.

Sums of Vector Spaces

Definition 10: Sum of Vector Spaces

Let 𝐻1,𝐻2,𝐻𝑛 be vector spaces. The sum 𝐻1+𝐻2++𝐻𝑛 is the set of all vectors that can be written as the sum of vectors from each of the vector spaces:

𝑖=1𝑛𝐻𝑖=𝐻1+𝐻2++𝐻𝑛={𝐯1+𝐯2++𝐯𝑛|𝐯1𝐻1,𝐯2𝐻2,,𝐯𝑛𝐻𝑛}.

Corollary 2

If 𝐻1,𝐻2,,𝐻𝑛 are subspaces of a vector space 𝑉, then 𝐻1+𝐻2++𝐻𝑛 is a subspace of 𝑉.

Example.

Let 𝑋={(𝑥,0,0)|𝑥} and 𝑌={(0,𝑦,0)|𝑦} be subspaces of 3. Then:

𝑋+𝑌={(𝑥,𝑦,0)|𝑥,𝑦}.

Theorem 8: Sum of Subspaces is the Smallest Containing Subspace

Let 𝑉 be a vector space, and suppose 𝐻1,𝐻2,,𝐻𝑛 be subspaces of 𝑉. Then 𝐻1+𝐻2++𝐻𝑛 is the smallest subspace of 𝑉 containing 𝐻1,𝐻2,,𝐻𝑛, meaning that if 𝐻 is any subspace of 𝑉 such that 𝐻𝑖𝐻 for all 𝑖, then 𝐻1+𝐻2++𝐻𝑛𝐻.

Definition 11: Direct Sum of Subspaces

Let 𝐻1,𝐻2,,𝐻𝑛 be subspaces of a vector space 𝑉.

  • The sum 𝐻1+𝐻2++𝐻𝑛 is a direct sum iff each vector in 𝐻1+𝐻2++𝐻𝑛 can be written as a unique sum of vectors from each subspace.
  • If 𝐻1+𝐻2++𝐻𝑛 is a direct sum, we denote it as 𝐻1𝐻2𝐻𝑛 or 𝑖=1𝑛𝐻𝑖.
Example.

Let 𝑋={(𝑥,0,0)|𝑥} and 𝑌={(0,𝑦,0)|𝑦} be subspaces of 3. Then:

𝑋𝑌={(𝑥,𝑦,0)|𝑥,𝑦}.
Nonexample.

Let 𝑋={(𝑥,0,0)|𝑥} and 𝑍={(𝑧,𝑧,0)|𝑧} be subspaces of 3. Then:

𝑋+𝑍={(𝑥+𝑧,𝑧,0)|𝑥,𝑧}.

However, 𝑋+𝑍 is not a direct sum because the vector (1,1,0) can be written as both (1,0,0)+(0,1,0) and (0,0,0)+(1,1,0).

Corollary 3: Direct Sum of Two Subspaces

Let 𝑉 be a vector space, and let 𝐻1,𝐻2 be subspaces of 𝑉. Then 𝐻1+𝐻2 is a direct sum if and only if for some 𝐮1,𝐮2𝐻1, 𝐯1,𝐯2𝐻2:

𝐮1+𝐯1=𝐮2+𝐯2(𝐮1=𝐮2 and 𝐯1=𝐯2).

Theorem 9: Formalism for Direct Sum

Let 𝑉 be a vector space, and let 𝐻1,𝐻2,,𝐻𝑛 be subspaces of 𝑉. Then 𝐻1+𝐻2++𝐻𝑛 is a direct sum if and only if the only way to write 𝟎𝑉 as a sum of vectors from each subspace is the trivial combination 𝟎𝑉=𝟎𝐻1+𝟎𝐻2++𝟎𝐻𝑛.

Theorem 10: Direct Sum of Two Subspaces

Let 𝑉 be a vector space, and let 𝐻1,𝐻2 be subspaces of 𝑉. 𝐻1+𝐻2 is a direct sum if and only if 𝐻1𝐻2={𝟎𝑉}.

Definition 12: Complement of a Subspace

Let 𝑉 be a vector space, and let 𝐻 be a subspace of 𝑉. A complement of 𝐻 in the ambient space 𝑉 is a subspace 𝐾 of 𝑉 such that 𝐻𝐾=𝑉.

Exercises

Exercise 7.
Exercise 8.
Let 𝐹 be a field. 𝐹 is the set of all infinite sequences of elements of 𝐹, that is, 𝐹={(𝑎1,𝑎2,𝑎3,)|𝑎1,𝑎2,𝑎3,𝐹}. Show that 𝐹 is a vector space over 𝐹 under componentwise addition and scalar multiplication.
Exercise 9.
Show that the set of all polynomials with real coefficients, (), is a vector space over under polynomial addition and scalar multiplication.
Exercise 10.
Let 𝐶[0,1]() be the set of all real-valued functions that are continuous on the closed interval [0,1]. Show that 𝐶[0,1]() is a vector space over .
Exercise 11.

Let 𝑉=+ be the set of all positive real numbers. Define 𝑉 to be a vector space over the field under the following operations:

𝑥+𝑉𝑦=𝑥𝑦 for all 𝑥,𝑦𝑉,𝑎𝑉𝑥=𝑥𝑎 for all 𝑎,𝑥𝑉.

Show that (𝑉,+𝑉,𝑉) is in fact a vector space.

Exercise 12.
Show that is a vector space over , the field of rational numbers.
Exercise 13.

Let 𝑆 be a nonempty set and let 𝑉=𝒫(𝑆) be the power set of 𝑆, i.e. the set of all subsets of 𝑆. Let 𝑉 be a vector space over the finite field {0,1} under the operations +𝑉 and 𝑉 defined as follows:

𝐴+𝑉𝐵=(𝐴\𝐵)(𝐵\𝐴) for all 𝐴,𝐵𝒫(𝑆),0𝑉𝐴= for all 𝐴𝒫(𝑆),1𝑉𝐴=𝐴 for all 𝐴𝒫(𝑆).

Show that (𝑉,+𝑉,𝑉) is in fact a vector space.

Exercise 14.

Recall 𝐶[0,1]() from Exercise 10. Let 𝐻 be a subset of 𝐶[0,1]() defined by:

𝐻={𝑓𝐶[0,1]()|01𝑓(𝑥)d𝑥=0}.

Show that 𝐻 is a subspace of 𝐶[0,1]().

Exercise 15.
Let 𝑉 be a vector space, and let 𝐻1,𝐻2,,𝐻𝑝 be subspaces of 𝑉. Show that the intersection 𝐻1𝐻2𝐻𝑝 is also a subspace of 𝑉.
Exercise 16.
Let 𝐻1,𝐻2 be subspaces of a vector space 𝑉. Show that 𝐻1𝐻2 is a subspace if and only if 𝐻1𝐻2 or 𝐻2𝐻1.
Exercise 17.
Let 𝑉 be a vector space, and let 𝑋,𝑌,𝑍 be subspaces of 𝑉 such that 𝑋𝑍. Show that 𝑋+(𝑌𝑍)=(𝑋+𝑌)𝑍.
Exercise 18.
Show that if 𝐻1,𝐻2,,𝐻𝑛 are subspaces of a vector space 𝑉, then 𝐻1+𝐻2++𝐻𝑛𝐻1𝐻2𝐻𝑛.
Exercise 19.
Let 𝑋={(𝑥,0,0)|𝑥} and 𝑌={(0,𝑦,0)|𝑦} be subspaces of 3. Show that 𝑋+𝑌 is a direct sum.
Exercise 20.
Show that for any vector space 𝑉, 𝑉{𝟎}=𝑉.
Exercise 21.
Let 𝑋={(𝑥,0,0)|𝑥} and 𝑌={(𝑦,𝑦,0)|𝑦} be subspaces of 3. Show that 𝑋𝑌=3.
Exercise 22.

Let 𝐹 be the set of all real-valued functions from . Let 𝑈 be the subset of 𝐹 consisting of all even functions, and let 𝑉 be the subset of 𝐹 consisting of all odd functions:

𝑈={𝑓𝐹|𝑥,𝑓(𝑥)=𝑓(𝑥)},𝑉={𝑓𝐹|𝑥,𝑓(𝑥)=𝑓(𝑥)}.

Show that 𝑈 and 𝑉 are subspaces of 𝐹, and that 𝑈𝑉=𝐹.