# When is a subset a subspace

• I
Let ##\mathbb{V}## be a vector space and ##\mathbb{W}## be a subset of ##\mathbb{V}##, with the same operations.

Claim:
If ##\mathbb{W}## is non-empty, closed under addition and scalar multiplication, then ##\mathbb{W}## is a subspace of ##\mathbb{V}##.

A set is a vector space if it satisfies 10 properties:

1. Closure under addition
2. Closure under scalar multiplication
3. Commutativity under addition
4. Associativity under addition
5. Existence of additive identity
6. Existence of additive inverse
7. Distributivity for scalar multiplication over addition in scalars
8. Distributivity for scalar multiplication over addition in vectors
9. Associativity under scalar multiplication
10. Identity for scalar multiplication

The properties 1 and 2 are given. 3,4,7,8,9,10 are easily verified as ##\mathbb{W}## and ##\mathbb{V}## share same operations. For example 3:

Let ##\vec u , \vec v \epsilon \mathbb{W}##
$$\vec u + \vec v (in~ \mathbb{W}) \\ = \vec u + \vec v (in~ \mathbb{V}) \text{(same operation)} \\ = \vec v + \vec u (in~ \mathbb{V}) (\mathbb{V} \text{is a vector space.)} \\ = \vec v + \vec u (in~ \mathbb{W}) ~\text{(same operation)} \\$$

The problem is 5 and 6.
The book I'm using ( S. Andrilli and D. Hecker) says:-

Let ##\vec u~ \epsilon ~\mathbb{W}##
$$\Rightarrow~~ 0\vec u ~ \epsilon ~\mathbb{W} ~\text{(closed under scalar multiplication.)}\\ \Rightarrow ~~ 0\vec u ~ \epsilon ~ \mathbb{V} ~\text{(property of subset)}\\ \Rightarrow ~~ \vec 0 ~\epsilon ~ \mathbb{V} ~\text{(property of vector space. Which we proved earlier.)}\\ \Rightarrow ~~ \vec 0 ~ \epsilon ~ \mathbb{W}~ \text{(as they share the same operations)}$$

what??

How can same operations imply that if ##\vec 0## is in ##\mathbb{V}## it also must be in ##\mathbb{W}## ??

## Answers and Replies

PeroK
Science Advisor
Homework Helper
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Let ##\mathbb{V}## be a vector space and ##\mathbb{W}## be a subset of ##\mathbb{V}##, with the same operations.

Claim:
If ##\mathbb{W}## is non-empty, closed under addition and scalar multiplication, then ##\mathbb{W}## is a subspace of ##\mathbb{V}##.

A set is a vector space if it satisfies 10 properties:

1. Closure under addition
2. Closure under scalar multiplication
3. Commutativity under addition
4. Associativity under addition
5. Existence of additive identity
6. Existence of additive inverse
7. Distributivity for scalar multiplication over addition in scalars
8. Distributivity for scalar multiplication over addition in vectors
9. Associativity under scalar multiplication
10. Identity for scalar multiplication

The properties 1 and 2 are given. 3,4,7,8,9,10 are easily verified as ##\mathbb{W}## and ##\mathbb{V}## share same operations. For example 3:

Let ##\vec u , \vec v \epsilon \mathbb{W}##
$$\vec u + \vec v (in~ \mathbb{W}) \\ = \vec u + \vec v (in~ \mathbb{V}) \text{(same operation)} \\ = \vec v + \vec u (in~ \mathbb{V}) (\mathbb{V} \text{is a vector space.)} \\ = \vec v + \vec u (in~ \mathbb{W}) ~\text{(same operation)} \\$$

The problem is 5 and 6.
The book I'm using ( S. Andrilli and D. Hecker) says:-

Let ##\vec u~ \epsilon ~\mathbb{W}##
$$\Rightarrow~~ 0\vec u ~ \epsilon ~\mathbb{W} ~\text{(closed under scalar multiplication.)}\\ \Rightarrow ~~ 0\vec u ~ \epsilon ~ \mathbb{V} ~\text{(property of subset)}\\ \Rightarrow ~~ \vec 0 ~\epsilon ~ \mathbb{V} ~\text{(property of vector space. Which we proved earlier.)}\\ \Rightarrow ~~ \vec 0 ~ \epsilon ~ \mathbb{W}~ \text{(as they share the same operations)}$$

what??

How can same operations imply that if ##\vec 0## is in ##\mathbb{V}## it also must be in ##\mathbb{W}## ??

If ##W## is closed under scalar multiplication and it is non-empty then ##\exists \ \vec{w} \in W## and ##0 \vec{w} = \vec{0} \in W##.

An important subtlety is that you know that ##0 \vec{w} = \vec{0}## because ##W \subset V## and this property holds for all vectors in ##V##, hence all vectors in ##W##.

You can show that ##-\vec{w} \in W## by a similar argument with the scalar ##-1##.

If WW is closed under scalar multiplication, and it is non-empty then ∃ →w∈W\exists \ \vec{w} \in W and 0→w=→0∈W0 \vec{w} = \vec{0} \in W.

That's the problem, I don't know that 0##\vec w = \vec 0## in W because it is not a Vector Space yet.

fresh_42
Mentor
The problem is 5 and 6.
The book I'm using ( S. Andrilli and D. Hecker) says:-

Let ##\vec u~ \epsilon ~\mathbb{W}##
$$\Rightarrow~~ 0\vec u ~ \epsilon ~\mathbb{W} ~\text{(closed under scalar multiplication.)}\\ \Rightarrow ~~ 0\vec u ~ \epsilon ~ \mathbb{V} ~\text{(property of subset)}\\ \Rightarrow ~~ \vec 0 ~\epsilon ~ \mathbb{V} ~\text{(property of vector space. Which we proved earlier.)}\\ \Rightarrow ~~ \vec 0 ~ \epsilon ~ \mathbb{W}~ \text{(as they share the same operations)}$$

what??

How can same operations imply that if ##\vec 0## is in ##\mathbb{V}## it also must be in ##\mathbb{W}## ??
What is said here is, that ##0_\mathbb{V} = 0 \cdot w =: 0_\mathbb{W}## is the same vector.

PeroK
PeroK
Science Advisor
Homework Helper
Gold Member
2020 Award
That's the problem, I don't know that 0##\vec w = \vec 0## in W because it is not a Vector Space yet.

An important subtlety is that you know that ##0 \vec{w} = \vec{0}## because ##W \subset V## and this property holds for all vectors in ##V##, hence all vectors in ##W##.

Kaguro
Oh! Of course..
Now when you put it this way, it seems clear!

Thank you both!

PeroK
Science Advisor
Homework Helper
Gold Member
2020 Award
Let ##\vec u~ \epsilon ~\mathbb{W}##
$$\Rightarrow~~ 0\vec u ~ \epsilon ~\mathbb{W} ~\text{(closed under scalar multiplication.)}\\ \Rightarrow ~~ 0\vec u ~ \epsilon ~ \mathbb{V} ~\text{(property of subset)}\\ \Rightarrow ~~ \vec 0 ~\epsilon ~ \mathbb{V} ~\text{(property of vector space. Which we proved earlier. ***)}\\ \Rightarrow ~~ 0\vec u = \vec{0} ~\text{(property of vector space. Which we proved earlier.+++)}\\\Rightarrow ~~ \vec 0 ~ \epsilon ~ \mathbb{W}~ \text{(as they share the same operations)}$$

Actually, I would say this is not quite right. I would replace the statement marked *** with the one marked +++.

Actually, I would say this is not quite right. I would replace the statement marked *** with the one marked +++.
Yes, the for all elements is the important thing.