# Homework Help: Subspace math question

1. Dec 8, 2013

### iRaid

1. The problem statement, all variables and given/known data
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) $W=\{(x_{1},x_{2},x_{3},x_{4})|x_{1}-x_{2}=1\}$

2. Relevant equations

3. 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.

2. Dec 9, 2013

### 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.

3. Dec 9, 2013

### Staff: Mentor

I know what you mean, but you're not saying what you mean, which is "the zero vector is in W."
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.
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.
No, you're given that x1 - x2 = 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.

Last edited: Dec 9, 2013
4. Dec 9, 2013

### iRaid

I don't understand how I would show this with arbitrary vectors. Lets say $\vec{u}+\vec{v}=\vec{w}$ How would I show that some random vectors would still work. It would be $(u_{1},u_{2},u_{3},u_{4},)+(v_{1},v_{2},v_{3},v_{4},)=\vec{w}$
Again, I'm not sure how I would express this without showing an example.
The problem states $x_{1}-x_{2}=1$
Again unsure on how to show this.

Thanks for the help Mark.

5. Dec 9, 2013

### iRaid

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

6. Dec 9, 2013

### R136a1

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

7. Dec 9, 2013

### iRaid

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

8. Dec 9, 2013

### Staff: Mentor

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.
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.
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.

9. Dec 9, 2013

### Staff: Mentor

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).
Correct.

10. Dec 9, 2013

### iRaid

Ok lets 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, $w_{4}=u_{4}+v_{4}$

11. Dec 9, 2013

### vela

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

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

This follows from the usual definition of vector addition.

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

12. Dec 9, 2013

### 1MileCrash

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.