Subspace Verification in R4: Homework Question & Solution

[iRaid]

Homework Statement

In each case below, either show that the set W is a subspace of R⁴ or give a counterexample to show it is not.
a) $W=\{(x_{1},x_{2},x_{3},x_{4})|x_{4}=x_{1}+x_{3}\}$
b) $W=\{(x_{1},x_{2},x_{3},x_{4})|x_{1}-x_{2}=1\}$

Homework Equations

The Attempt at a Solution

a) Satisfies the 3 conditions of a subspace:
Part of the 0 vector
Closed under addition. If I add the 2 same vectors together, say $(1,2,2,3)+(1,2,2,3)=(2,4,4,6)$ and this still satisfies x₄=x₁+x₃ since 2+4=6.
Closed under scalar multiplication. $cx_{4}=cx_{1}+cx_{3}\implies x_{4}=x_{1}+x_{3}$
This is a subspace of R⁴

Wondering if this makes sense. I'm also not sure if this even the right way to prove it (I know those are the conditions, I'm not sure how I would write them out).

b) First off 0 isn't in W since $x_{1}-x_{2}\not= 1$
And it isn't closed under scalar multiplication.
This is not a subspace of R⁴


Can anyone tell me if this makes sense (if not) how would I explain each of these.

Thanks.

 

[1MileCrash]

For a.), you can't prove something by picking a specific example (and why did you add the same vector to itself?). Instead, add 2 arbitrary vectors that meet the conditions of the subspace, and show that the sum meets the same conditions.

Also, you don't ever need to show that 0 is in W for W to be a subspace. It is sufficient to show that it is closed under scalar multiplication and addition, and then you are done.

 

[Mark44]

Homework Statement

In each case below, either show that the set W is a subspace of R⁴ or give a counterexample to show it is not.
a) $W=\{(x_{1},x_{2},x_{3},x_{4})|x_{4}=x_{1}+x_{3}\}$
b) $W=\{(x_{1},x_{2},x_{3},x_{4})|x_{1}-x_{2}=1\}$

Homework Equations

The Attempt at a Solution

a) Satisfies the 3 conditions of a subspace:
Part of the 0 vector

I know what you mean, but you're not saying what you mean, which is "the zero vector is in W."

Closed under addition. If I add the 2 same vectors together, say $(1,2,2,3)+(1,2,2,3)=(2,4,4,6)$ and this still satisfies x₄=x₁+x₃ since 2+4=6.

No, you can't pick a vector. You have to show that for any two arbitrary vectors in W, their sum is also in W.

Closed under scalar multiplication. $cx_{4}=cx_{1}+cx_{3}\implies x_{4}=x_{1}+x_{3}$
This is a subspace of R⁴

Wondering if this makes sense.

It's good that you were wondering, because it doesn't make sense. What you have shown is that x₄ is a linear combination of x₁ and x₃. To show that W is closed under scalar multiplication, take any arbitrary vector v in W and show that cv is also in W.

I'm also not sure if this even the right way to prove it (I know those are the conditions, I'm not sure how I would write them out).

b) First off 0 isn't in W since $x_{1}-x_{2}\not= 1$

No, you're given that x₁ - x₂ = 1.

And it isn't closed under scalar multiplication.

Show this. To show that a set isn't closed under either vector addition or scalar multiplication, you can do this by a specific example where either of these properties doesn't hold.

This is not a subspace of R⁴Can anyone tell me if this makes sense (if not) how would I explain each of these.

Thanks.

 

[iRaid]

No, you can't pick a vector. You have to show that for any two arbitrary vectors in W, their sum is also in W.

I don't understand how I would show this with arbitrary vectors. Let's say $\vec{u}+\vec{v}=\vec{w}$ How would I show that some random vectors would still work. It would be $(u_{1},u_{2},u_{3},u_{4},)+(v_{1},v_{2},v_{3},v_{4},)=\vec{w}$

It's good that you were wondering, because it doesn't make sense. What you have shown is that x₄ is a linear combination of x₁ and x₃. To show that W is closed under scalar multiplication, take any arbitrary vector v in W and show that cv is also in W.

Again, I'm not sure how I would express this without showing an example.

No, you're given that x₁ - x₂ = 0.

The problem states $x_{1}-x_{2}=1$

Show this. To show that a set isn't closed under either vector addition or scalar multiplication, you can do this by a specific example where either of these properties doesn't hold.

Again unsure on how to show this.

Thanks for the help Mark.

 

[iRaid]

For a.), you can't prove something by picking a specific example (and why did you add the same vector to itself?). Instead, add 2 arbitrary vectors that meet the conditions of the subspace, and show that the sum meets the same conditions.

Also, you don't ever need to show that 0 is in W for W to be a subspace. It is sufficient to show that it is closed under scalar multiplication and addition, and then you are done.

So showing 0 is in W doesn't work? Let's say 0 isn't in W, would that automatically mean it is not a subspace?

 

[R136a1]

Also, you don't ever need to show that 0 is in W for W to be a subspace. It is sufficient to show that it is closed under scalar multiplication and addition, and then you are done.

If that were true, then $\emptyset$ would be a subspace...

 

[iRaid]

If that were true, then $\emptyset$ would be a subspace...

Well would it be better to say W contains $\vec{0}$?

 

[Mark44]

I don't understand how I would show this with arbitrary vectors. Let's say $\vec{u}+\vec{v}=\vec{w}$ How would I show that some random vectors would still work. It would be $(u_{1},u_{2},u_{3},u_{4},)+(v_{1},v_{2},v_{3},v_{4},)=\vec{w}$

You're given the condition for a vector to be in W; namely, that x₄ = x₁ + x₃.

So two arbitrary vectors might be u = <u₁, u₂, u₃, u₄>. Since u is in W, it must be true that u₄ = u₁ + u₃. Now, come up with another arbitrary vector, add the two vectors together, and show that their sum is in W.

Again, I'm not sure how I would express this without showing an example.

The problem states $x_{1}-x_{2}=1$

That's what I meant (and have changed in my earlier post). When you wrote that it wasn't equal to 1, that threw me off.

Again unsure on how to show this.

Thanks for the help Mark.

If you're proving that a set IS a subspace, you have to show that any vectors in the set satisfy closure under vector addition and scalar multiplication. If you are showing that a set IS NOT a subspace, specific examples will work, because you're showing that not all vectors in the set satisfy the closure and/or other conditions.

 

[Mark44]

So showing 0 is in W doesn't work?

Sure, that works (assuming you're talking about the 1st problem). If you're out to show that a certain set is a subspace of some vector space, the set can't be empty (it has to have at least the zero vector in it).

Let's say 0 isn't in W, would that automatically mean it is not a subspace?

Correct.

 

[iRaid]

You're given the condition for a vector to be in W; namely, that x₄ = x₁ + x₃.

So two arbitrary vectors might be u = <u₁, u₂, u₃, u₄>. Since u is in W, it must be true that u₄ = u₁ + u₃. Now, come up with another arbitrary vector, add the two vectors together, and show that their sum is in W.

Ok let's see... So I would say something like: Let $\vec{u}=(u_{1},u_{2},u_{3},u_{4})$ and $\vec{v}=(v_{1},v_{2},v_{3},v_{4})$. You can then say: $\vec{w}=\vec{u}+\vec{v}=(w_{1},w_{2},w_{3},w_{4})$. Then, $u_{4}=u_{1}+u_{3}$ and $v_{4}=v_{1}+v_{3}$. So finally, $w_{4}=u_{4}+v_{4}$

 

[vela]

Ok let's see... So I would say something like: Let $\vec{u}=(u_{1},u_{2},u_{3},u_{4})$ and $\vec{v}=(v_{1},v_{2},v_{3},v_{4})$. You can then say: $\vec{w}=\vec{u}+\vec{v}=(w_{1},w_{2},w_{3},w_{4})$.

You need to say $\vec{u}$ and $\vec{v}$ are elements of W.

Then, $u_{4}=u_{1}+u_{3}$ and $v_{4}=v_{1}+v_{3}$.

This follows from the fact $\vec{u}$ and $\vec{v}$ are in W.

So finally, $w_{4}=u_{4}+v_{4}$

This follows from the usual definition of vector addition.

Now you need to show that $\vec{w}$ is in W. That is, show that $w_4 = w_1 + w_3$.

 

[1MileCrash]

If that were true, then $\emptyset$ would be a subspace...

Fair enough.

If you know W is not empty (which is clear for the OPs subspaces), then you do not need to show that 0 is in W.