# [Linear Algebra] Show that H ∩ K is a subspace of V

• bornofflame
In summary: It is not correct to say that v is a set of vectors in V. v is a single vector in V. To say that H ∩ K = Span{v} is incorrect. H ∩ K is a set of vectors in V. It is the set of all vectors that are in both H and K. Span{v} is a set of vectors in V that are linear combinations of v. These are two different sets of vectors in V, and they are not necessarily the same.If you can show that H ∩ K is the span of some single vector v in V, then you are done. But you cannot assume that. You need to start with the assumption that u and v are arbitrary vectors in H
bornofflame

## Homework Statement

From Linear Algebra and Its Applications, 5th Edition, David Lay
Chapter 4, Section 1, Question 32
Let H and K be subspaces of a vector space V. The intersection of H and K is the set of v in V that belong to both H and K. Show that H ∩ K is a subspace of V. (See figure.) Give an example in ℝ2 to show that the union of two subspaces is not, in general, a subspace.

## Homework Equations

/theorems[/B]
Theorem 1: If v1,...vp are in vector space V, then Span{v1,...vp} is a subspace of V.

## The Attempt at a Solution

This is what I started off with:
Let u,v ∈ H; s, t ∈ K
0v ∈ H, K
u + v ∈ H
s + t ∈ K

In the middle of writing that down, I was thinking that v is a set of vectors in V, and that H ∩ K = Span{v}, therefore, per Theorem 1 in the book, H ∩ K is a subspace of V.

Written as:
v ∈ V
H ∩ K = Span{v}
∴ H ∩ K is a subspace of V per Theorem 1.

For the second part I used the following:
m, n ∈ ℝ2
m = {[x, y]: x, y ≥ 0}
n = {[x, y]: x ≤ 0, y ≥ 0}
m ∪ n = {[x, y]: x = ℝ, y ≥ 0}
let c = -1, u = [-1, 3]
cu = [1, -3]
cu ∉ m ∪ n
∴ m ∪ n is not a subspace

Is my train of thinking correct concerning this problem?

#### Attachments

• [LA]HintersectK.png
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bornofflame said:

## Homework Statement

From Linear Algebra and Its Applications, 5th Edition, David Lay
Chapter 4, Section 1, Question 32
Let H and K be subspaces of a vector space V. The intersection of H and K is the set of v in V that belong to both H and K. Show that H ∩ K is a subspace of V. (See figure.) Give an example in ℝ2 to show that the union of two subspaces is not, in general, a subspace.
View attachment 224838

## Homework Equations

/theorems[/B]
Theorem 1: If v1,...vp are in vector space V, then Span{v1,...vp} is a subspace of V.

## The Attempt at a Solution

This is what I started off with:
Let u,v ∈ H; s, t ∈ K
0v ∈ H, K
u + v ∈ H
s + t ∈ K

In the middle of writing that down, I was thinking that v is a set of vectors in V, and that H ∩ K = Span{v}, therefore, per Theorem 1 in the book, H ∩ K is a subspace of V.

Written as:
v ∈ V
H ∩ K = Span{v}
∴ H ∩ K is a subspace of V per Theorem 1.

For the second part I used the following:
m, n ∈ ℝ2
m = {[x, y]: x, y ≥ 0}
n = {[x, y]: x ≤ 0, y ≥ 0}
m ∪ n = {[x, y]: x = ℝ, y ≥ 0}
let c = -1, u = [-1, 3]
cu = [1, -3]
cu ∉ m ∪ n
∴ m ∪ n is not a subspace

Is my train of thinking correct concerning this problem?
For the first part, you would need to show that ##H \cap K## is the span of v. That is easy to do using the definition of subspace. Since I don't have your textbook at hand, I don't know what Theorem 1 says. Perhaps it covers the missing pieces.
For the second part, your example does not work because m and n are not subspaces of ℝ2 as you have defined them. (Check the definition of subspace again.)

bornofflame
bornofflame said:

## Homework Statement

From Linear Algebra and Its Applications, 5th Edition, David Lay
Chapter 4, Section 1, Question 32
Let H and K be subspaces of a vector space V. The intersection of H and K is the set of v in V that belong to both H and K. Show that H ∩ K is a subspace of V. (See figure.) Give an example in ℝ2 to show that the union of two subspaces is not, in general, a subspace.
View attachment 224838

## Homework Equations

/theorems[/B]
Theorem 1: If v1,...vp are in vector space V, then Span{v1,...vp} is a subspace of V.
I don't see that this theorem is relevant.
bornofflame said:

## The Attempt at a Solution

This is what I started off with:
Let u,v ∈ H; s, t ∈ K
0v ∈ H, K
u + v ∈ H
s + t ∈ K
This isn't a good start, since you are already given that H and K are subspaces of V. So we know that both are closed under vector addition and scalar multiplication.
A better start would be to assume that u and v are arbitrary vectors in H ∩ K. Show that their sum is also in H ∩ K, and show that any scalar multiple of either one of them (say u) is also in H ∩ K.
bornofflame said:
In the middle of writing that down, I was thinking that v is a set of vectors in V, and that H ∩ K = Span{v}, therefore, per Theorem 1 in the book, H ∩ K is a subspace of V.
But some of these vectors v don't belong to H ∩ K, so I don't think this gets you anywhere.
bornofflame said:
Written as:
v ∈ V
H ∩ K = Span{v}
∴ H ∩ K is a subspace of V per Theorem 1.

For the second part I used the following:
m, n ∈ ℝ2
m = {[x, y]: x, y ≥ 0}
n = {[x, y]: x ≤ 0, y ≥ 0}
m ∪ n = {[x, y]: x = ℝ, y ≥ 0}
let c = -1, u = [-1, 3]
cu = [1, -3]
cu ∉ m ∪ n
∴ m ∪ n is not a subspace

Is my train of thinking correct concerning this problem?

bornofflame
tnich said:
For the second part, your example does not work because m and n are not subspaces of ℝ2 as you have defined them. (Check the definition of subspace again.)

Oops. I either misread that or wasn't thinking. Either way. Ugh. I'll go check that now to see fix my mistake.

Mark44 said:
I don't see that this theorem is relevant.
This isn't a good start, since you are already given that H and K are subspaces of V. So we know that both are closed under vector addition and scalar multiplication.
I see what you mean. That didn't occur to me.

Mark44 said:
A better start would be to assume that u and v are arbitrary vectors in H ∩ K. Show that their sum is also in H ∩ K, and show that any scalar multiple of either one of them (say u) is also in H ∩ K.
But some of these vectors v don't belong to H ∩ K, so I don't think this gets you anywhere.

Second stab:
H, K are subspaces in V
H ∩ K = {v} in V
To be a subspace of V, H ∩ K must comply with the following:
1. Contain the 0 vector:
0v ∈ H ∩ K b/c H, K are subspaces and therefore contain the 0v

2. Closed under vector addition:
Let u, v ∈ H ∩ K, then u, v ∈ H, K which means that u + v ∈ H, K which means that u + v ∈ H ∩ K

3. Closed under scalar multiplication:
Let c ∈ ℝ, u ∈ H ∩ K, then u ∈ H, K and cu ∈ H, K which means that cu ∈ H ∩ K

It took me a second b/c I don't think I really understood what was being asked. Once that clicked, the work became much easier.

I'm going to work on part 2 now.

For part deux:
m, n are subspaces of V
To show that m ∪ n is NOT a subspace of V, it must fail one of the following:
1. Contains the 0v: yes, move on
2. Closed under vector addition:
Let m = {[x, y]: x = 0, y = ℝ}, n = {[x, y]: x = ℝ, y = 0}
Let u = [0, 1], v = [1, 0]
u + v ∉ m
u + v ∉ n
∴ u + v ∉ m ∪ n

bornofflame said:
For part deux:
m, n are subspaces of V
To show that m ∪ n is NOT a subspace of V, it must fail one of the following:
1. Contains the 0v: yes, move on
2. Closed under vector addition:
Let m = {[x, y]: x = 0, y = ℝ}, n = {[x, y]: x = ℝ, y = 0}
Let u = [0, 1], v = [1, 0]
u + v ∉ m
u + v ∉ n
∴ u + v ∉ m ∪ n
This is a good counterexample, but your notation could use a couple of tweaks.
Let M = {u ∈ R2 : x = 0, y ∈ R}, and let N = {v ∈ R2 : x ∈ R, y = 0 }
Let u = <0, 1>, v = <1, 0>
u + v = <1, 1> ∉ M ∪ N
So M ∪ N is not closed under vector addition, and isn't a subspace of R2

bornofflame

## 1. What is the definition of a subspace?

A subspace is a subset of a vector space that satisfies the three properties of closure under addition, closure under scalar multiplication, and contains the zero vector.

## 2. What is the definition of intersection?

The intersection of two sets is the set of all elements that are common to both sets. In this case, the intersection of two subspaces H and K is the set of all vectors that belong to both H and K.

## 3. How can we show that H ∩ K is a subspace of V?

In order to show that H ∩ K is a subspace of V, we need to prove that it satisfies the three properties of a subspace: closure under addition, closure under scalar multiplication, and contains the zero vector. This can be done by showing that any linear combination of vectors in H ∩ K is also in H ∩ K, and that the zero vector is in H ∩ K.

## 4. Why is it important to prove that H ∩ K is a subspace of V?

Proving that H ∩ K is a subspace of V is important because it allows us to use the properties of subspaces in our calculations. This can make solving problems in linear algebra much simpler and more efficient.

## 5. Can H ∩ K be empty?

Yes, H ∩ K can be empty if there are no common elements between the two subspaces. In this case, H ∩ K would still be considered a subspace of V, as it satisfies the three properties of a subspace, but it would only contain the zero vector.

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