In the article An Introduction to Vector Spaces, the meaning of vector and vector space are presented . Since a vector space is a set (with a bundle of properties), a question that can be raised is: “Are subsets of a vector space V also vector spaces under the addition and scalar multiplication defined on V?” Stated another way, if V is a vector space, W ⊆ V, is W a vector space under the addition and scalar multiplication defined on V? Not necessarily! Let’s look at Example 1 below.

 

Example 1

Let V = \mathbb{R} \times \mathbb{R}. The addition in V is defined by (x,y) + (x’,y’) = (x+x’,y+y’) and the scalar multiplication is defined by k(x,y) = (kx,ky). It can be shown that V is a vector space over the field \mathbb{R}. Now, let W = \mathbb{R} \times \{ 1 \}. Suppose that the addition and scalar multiplication in W are as defined on V, i.e. (x,y) + (x’,y’) = (x+x’,y+y’) and k(x,y) = (kx,ky), respectively. W is a subset of V, but W is not a vector space. As a counterexample, note that (3,1) ∈ W and (10,1) ∈ W. By the definition of addition in V, (3,1) + (10,1) = (13,2), however (13,2) ∉ W. So, W is not a vector space.

 

Example 2

In Example 1, let W = \{ (x,y) \in \matbb{R} \times \mathbb{R} \vert x = y \}. Suppose that the addition and scalar multiplication operations in W are as defined on V above, namely (x,y) + (x’,y’) = (x+x’,y+y’) and k(x, y) = (kx,ky). Note that W ⊆ V. In this example, W is a vector space over the field \mathbb{R}.  Prove that W is a vector space over the field!

Answer

To prove that W is closed with respect to addition [i.e. u, v ∈ W  ⇒ (u + v) ∈ W], let u and v be any member of W. Let u = (u0,u0) and v = (v0,v0). By the definition of the addition in W, u + v = (u0+v0,u0+v0) ∈ W. So, W is closed with respect to addition.

To prove that W is closed with respect to scalar multiplication [i.e. (k \in \mathbb{R} and u ∈W) ⇒ ku ∈ W], let k be any real number and let u = (u0,u0) be any member of W. By the definition of the scalar multiplication in W, ku = (ku0,ku0) ∈ W. Consequently, W is closed with respect to scalar multiplication.

To prove the associative property of addition, let u, v, and w are any members of W. Since W ⊆ V, it holds that u, v, w ∈ V. Since V is a vector space, then u + (v + w) = (u + v) + w. (Proven). [Note: As the proof has shown, the associative property of addition in W is inherited from the vector space V.]

To prove the commutative property of addition, let u and v be any member of W. Since W ⊆ V, it holds that u, v  ∈ V. Since V is a vector space, then u + v = v + u. (Proven) [As before, the commutative property of addition in W is inherited from the vector space V.]

To prove the existence of zero, note that (0,0) ∈ W. We claim that (0,0) is the zero in W, i.e. 0 = (0,0). To prove it, let v be any member of W and let v = (v0,v0). As a consequence, 0 + v = (0,0) + (v0,v0) = (0+v0,0+v0) = (v0,v0) = v. Hence, 0 + v = v. In addition, v + 0 = (v0,v0) + (0,0) = (v0+0,v0+0) = (v0,v0) = v. Therefore, v + 0 = v. To sum up, W has the zero, i.e. 0 = (0,0), which proves the existence of zero.

To prove the existence of additive inverses, let u be any member of W and u = (u0,u0). Select –u = (-u0,-u0) ∈ W. As a consequence, u + (-u) = (u0,u0) + (-u0,-u0) = (u0+(-u0),u0+(-u0)) = (u0-u0,u0-u0) = (0,0) = 0. Also, (-u) + u = (-u0,-u0) + (u0,u0) = (-u0+u0,-u0+ u0) = (0,0) = 0. (Proven)

Next, we will prove the first property of scalar multiplication: for every r \in \mathbb{R} and every u, v ∈ W, r(u+v) = ru + rv. To prove this, let r be any member of the field \mathbb{R} and u, v be any members of W. Since W ⊆ V, it follows that u, v ∈ V. As V is a vector space, then r(u+v) = ru + rv. (Proven)

Next, we will prove the second property of scalar multiplication: for every r,s \in \mathbb{R} and every u ∈ W, it holds that (r+s)u = ru + su. To prove this, let r and s be any members of the field \mathbb{R} and u be any member of W. Since W ⊆ V, it holds that u ∈ V. Since V is a vector space, then (r+s)u = ru + su. (Proven)

Let’s turn to the proof of the third property of scalar multiplication: for every r, s \in \mathbb{R} and every u ∈ W, it holds that (rs)u = r(su). To prove this, let r and s be any members of the field \mathbb{R} and u be any member of W. Because W ⊆ V, it follows that u ∈ V. Since V is a vector space, then (rs)u = r(su).

Next, we will prove the fourth property of scalar multiplication: for every u ∈ W it holds that 1u = u. To prove this, let u be any member of W. Because W ⊆ V, it follows that u ∈ V. Since V is a vector space, then 1u = u. (Proven)

 

Compare Example 1 with Example 2. There are two similarities among the examples. First, in both examples, W ⊆ V. Second, in both examples, the addition and scalar multiplication in W are as defined in the vector space V. In other words, W “inherits” the addition and scalar multiplication defined on V. Despite the similarities, it turns out that W in Example 1 is not a vector space while W in Example 2 is a vector space. In other words, we say that W in Example 2 is a vector subspace of V.

 

Definition of A Vector Subspace

Let V be a vector space over the field F and W ⊆ V. W is a vector subspace of V if W is a vector space over the field F with the addition and scalar multiplication operations defined on V.

 

Now, how to prove that a subset of a vector space V is also a vector space under the operations defined on the vector space V? In other words, if V is a vector space, W ⊆ V, and In W, the addition and scalar multiplication are defined as in V., how to prove that W is a vector subspace of V? The first method is to prove that W satisfies all the properties of a vector space. It has been demonstrated in Example 2. As Example 2 shows, some properties are easily proven because they are “inherited” from the vector space V such as the commutative and associative properties of addition, as well as the properties of scalar multiplication. However, there is another method to prove that W is a vector subspace of V. The following theorem can be applied.

 

Theorem

Let V be a vector space over the field F and W ⊆ V. Then, W is a subspace of V if and only if both of the following conditions are satisfied: a) If u, v ∈ W then (u + v) ∈ W, and b) If k ∈ F and u ∈ W then ku ∈ W.

 

By the theorem, to prove that W is a vector subspace of V, it is sufficient to prove that W is closed with respect to addition and scalar multiplication operations defined on V.

 

Example 3

Let V = {\mathbb{R}}^3 and in V, the addition and scalar multiplication are defined by (u1,u2,u3) + (v1,v2,v3) = (u1+v1,u2+v2,u3+v3) and k(u1,u2,u3) = (ku1,ku2,ku3), where k is any member of the field \mathbb{R}. Let W = \{ (x,y,z) \in {\mathbb{R}}^3 \vert y = x+z \}. Is W a vector subspace of V?

Answer

Let u and v be any members of W, u = (u1,u2,u3), and v = (v1,v2,v3). Consequently, u + v = (u1+v1,u2+v2,u3+v3). Since u, v ∈ W, it holds that u2 = u1+u3 and v2 = v1+v3. Moreover, u + v = (u1+v1,(u1+u3)+(v1+v3),u3+v3) = (u1+v1, (u1+v1)+(u3+v3),u3+v3) ∈ W. Accordingly, W is closed with respect to the addition defined in V.

Let k be any member of the field \mathbb{R} and u is any member of W. Let u = (u1,u2,u3). Therefore, ku = (ku1,ku2,ku3). Since u ∈ W, it holds that u2 = u1 + u3, which results in ku = (ku1,k(u1+u3),ku3) = (ku1,ku1+ku3,ku3) ∈ W. Hence, W is closed with respect to the scalar multiplication defined in V.

We have proved that W is closed with respect to addition and scalar multiplication defined on V. So, by the theorem, W is a vector subspace of V.

 

 

 

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