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 . 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 . Now, let . 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 . 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 . 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** = (u_{0},u_{0}) and **v** = (v_{0},v_{0}). By the definition of the addition in W, **u** + **v** = (u_{0}+v_{0},u_{0}+v_{0}) ∈ W. So, W is closed with respect to addition.

To prove that W is closed with respect to scalar multiplication [i.e. ( and **u** ∈W) ⇒ k**u** ∈ W], let k be any real number and let **u** = (u_{0},u_{0}) be any member of W. By the definition of the scalar multiplication in W, k**u** = (ku_{0},ku_{0}) ∈ 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** = (v_{0},v_{0}). As a consequence, **0** + **v** = (0,0) + (v_{0},v_{0}) = (0+v_{0},0+v_{0}) = (v_{0},v_{0}) = **v**. Hence, **0** + **v** = **v**. In addition, **v** + **0** = (v_{0},v_{0}) + (0,0) = (v_{0}+0,v_{0}+0) = (v_{0},v_{0}) = **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 = (u_{0},u_{0}). Select –**u** = (-u_{0},-u_{0}) ∈ W. As a consequence, **u** + (-**u**) = (u_{0},u_{0}) + (-u_{0},-u_{0}) = (u_{0}+(-u_{0}),u_{0}+(-u_{0})) = (u_{0}-u_{0},u_{0}-u_{0}) = (0,0) = **0**. Also, (-**u**) + **u** = (-u_{0},-u_{0}) + (u_{0},u_{0}) = (-u_{0}+u_{0},-u_{0}+ u_{0}) = (0,0) = **0**. (Proven)

Next, we will prove the first property of scalar multiplication: for every and every **u**, **v** ∈ W, r(**u**+**v**) = r**u** + r**v**. To prove this, let r be any member of the field 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**) = r**u** + r**v**. (Proven)

Next, we will prove the second property of scalar multiplication: for every and every **u** ∈ W, it holds that (r+s)**u** = r**u** + s**u**. To prove this, let r and s be any members of the field 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** = r**u** + s**u**. (Proven)

Let’s turn to the proof of the third property of scalar multiplication: for every and every **u** ∈ W, it holds that (rs)**u** = r(s**u**). To prove this, let r and s be any members of the field 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(s**u**).

Next, we will prove the fourth property of scalar multiplication: for every **u** ∈ W it holds that 1**u** = **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 1**u** = **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 k**u** ∈ 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 and in V, the addition and scalar multiplication are defined by (u_{1},u_{2},u_{3}) + (v_{1},v_{2},v_{3}) = (u_{1}+v_{1},u_{2}+v_{2},u_{3}+v_{3}) and k(u_{1},u_{2},u_{3}) = (ku_{1},ku_{2},ku_{3}), where k is any member of the field . Let . Is W a vector subspace of V?

**Answer**

Let **u **and **v** be any members of W, **u** = (u_{1},u_{2},u_{3}), and **v** = (v_{1},v_{2},v_{3}). Consequently, **u** + **v** = (u_{1}+v_{1},u_{2}+v_{2},u_{3}+v_{3}). Since **u**, **v** ∈ W, it holds that u_{2} = u_{1}+u_{3} and v_{2} = v_{1}+v_{3}. Moreover, **u** + **v** = (u_{1}+v_{1},(u_{1}+u_{3})+(v_{1}+v_{3}),u_{3}+v_{3}) = (u_{1}+v_{1}, (u_{1}+v_{1})+(u_{3}+v_{3}),u_{3}+v_{3}) ∈ W. Accordingly, W is closed with respect to the addition defined in V.

Let k be any member of the field and **u** is any member of W. Let **u** = (u_{1},u_{2},u_{3}). Therefore, k**u** = (ku_{1},ku_{2},ku_{3}). Since **u** ∈ W, it holds that u_{2} = u_{1} + u_{3}, which results in k**u** = (ku_{1},k(u_{1}+u_{3}),ku_{3}) = (ku_{1},ku_{1}+ku_{3},ku_{3}) ∈ 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.