not sure where to start for this proof.
Determine whether the following sets form subspaces of R^2
{(x1,x2)^T | x1+x2=0}

- anonymous

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- jtvatsim

Do you know the conditions required to have a subspace?

- anonymous

it has to be nonempty
alpha x is in the subset for any alpha
x+y is in the subset
that is what my book says. but not sure how to apply that to this

- jtvatsim

OK, there are three properties as your book suggests. Another equivalent version says:
1) the zero vector must be in the subset (this ensures nonemptiness, as well as other useful things)
2) constant multiples must be in the subset (alpha x, where alpha is a scalar and x is a vector)
3) vector sums must be in the subset (x + y, where x and y are both vectors).

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## More answers

- jtvatsim

The trick, as you mention, is figuring out how to apply it. :)

- jtvatsim

We are dealing with the subset {(x1,x2)^T | x1 + x2 = 0} correct? Are you clear on what all the notation means here?

- anonymous

i think one of the problems is that calculus 1 was the only pre req for linear but trying to read the book i think its assumed you have had discrete or some other math proof based class. at least thats what i get from talling to other students

- anonymous

talking*

- jtvatsim

Yes, you are probably correct... I imagine it is quite a bit more challenging in that case... :) It is easy enough once you get the basics. I'll try to walk you through this one if you think that will be most helpful.

- anonymous

anything would be. im lost and tried googling stuff but its just more proofs and symbols that i then have to try and google what those symboles mean

- jtvatsim

OK. Let's start with step one. Identify what all the important concepts, symbols, and definitions are.

- anonymous

ok. i know that x1 and x2 are vectors in R^2

- anonymous

and im assuming R^2 is the x y plane

- anonymous

and there sum is 0

- jtvatsim

Good, so far. There is a minor issue however.

- freckles

well I think (x1,x2)^T is a vector in R^2
not x1 and x2

- anonymous

thats about as far as i can go

- anonymous

what issue?

- jtvatsim

That's a good start. As @freckles mentioned, we need to be careful to distinguish that x1 and x2 are not vectors separately.

- jtvatsim

The whole expression is (x1,x2) which is a vector. x1 is the x-component, and x2 is the y-component. These individual components sum to 0.

- anonymous

so its a single vector like x = (x1
x2)

- jtvatsim

Correct

- jtvatsim

It is useful to think of an actual example just so we have something concrete in mind. So some examples of vectors that satisfy this are (1,-1) since 1 + (-1) = 0, (-2, 2) since (-2) + 2 = 0. and so on. Make sense so far?

- anonymous

yes. i guess "abstractly" x1 = negative x2 so that they sum to 0

- jtvatsim

Exactly! In fact, that is what the expression x1 + x2 = 0 implies. Solving for x1 gives as you say, x1 = -x2.

- jtvatsim

Alright, so we still need to decide whether this set of vectors is in fact a subspace.

- jtvatsim

Let's start with the first condition: 1) Does the zero vector live in this set?

- anonymous

i would say yes because 0 + (-0) is still 0. is that the right way to think of it?

- jtvatsim

You are 100% correct!

- jtvatsim

(0,0) satisfies the condition that x1 + x2 = 0, since 0 + 0 = 0. Good, our set passes condition 1.

- jtvatsim

Now, to check condition 2: If c is some multiple and v is in the set, is cv in the set?

- jtvatsim

Sorry, I guess I should say If alpha is some multiple and x is in the set, is alpha x in the set?

- jtvatsim

It's the same thing either way.

- jtvatsim

So, this is a little weird, but here goes.

- anonymous

so alphax1 + alphax2 = 0
so alpha is any number
alpha = 100
x1 = 5 so x2 would = -5 so someting like
100(5) + 100(-5) = 0
because 500 + -500 = 0

- jtvatsim

Great, that's the spirit of the proof! Now we just make the argument as general as possible.

- anonymous

thats where i run into trouble.

- anonymous

becasue i asked him if i can give an example on a test that is coming up and he said no because one example might be right but there is posibly infinite examples and one could be false and some other stuff i didnt quite catch

- jtvatsim

Here is how to phrase the general proof. Look it over and if you have any questions, we will spend some time unpacking it.
"Suppose we have x = (x1,x2) in the set. So, x1 + x2 = 0. Now, look at alpha x.
alpha x = (alpha x1, alpha x2). Notice that alpha x1 + alpha x2 = alpha (x1 + x2). But we know that x1 + x2 = 0! So, alpha x1 + alpha x2 = alpha (x1 + x2) = 0 as we wanted."

- jtvatsim

Notice that we need to start with x. We do not know if alpha x1 + alpha x2 actually equals 0. This is what we are trying to show... Instead we start with x and show that alpha x works too. With a couple of clever algebraic tricks we pull it off.

- anonymous

ok i think i get it, one question though, might be dumb but where you (and my book) say constant multiples must be in the subset. lets say the subset is 1 2 3 4 5 does that mean alpha can only be 1 2 3 4 5 and not 6 or anything higher or am i completely wrong on that

- anonymous

but what you did is only for the second condition, right?

- jtvatsim

This is only the second condition, yes. And your question is a good one. Let me think how to answer it.

- jtvatsim

alpha can be as large or as small as it wants. It must work for any alpha. So, in your example, you give a very small subset. {1, 2, 3, 4, 5}. This is actually NOT a subspace since alpha = 10 breaks it. In fact 10*1 = 10 which is not in the set.

- jtvatsim

Does that answer your question?

- anonymous

ok, yes it does

- jtvatsim

Great. One last note on condition 2. We have to start with x1 + x2 = 0 NOT alpha x1 + alpha x2 = 0 since the set only talks about x1 + x2 = 0 in the definition.

- jtvatsim

That was a minor difference in your proof versus my general proof.

- jtvatsim

Just wanted to be sure you picked up on that. :)

- jtvatsim

So where are we so far? We have shown that condition 1 and 2 both work. All that remains is condition 3. If it works, then we have a subspace. If it fails, we do not have a subspace.

- anonymous

i scrolled up and read what you said the three conditinos are and number 3 says someting about 2 vectors but dont be only have 1 vector x ?

- jtvatsim

Good point. We are allowed to assume we have a second vector y for condition 3's proof.

- jtvatsim

After all, with condition 2 we just showed that this set probably has an infinite number of vectors (every multiple works).

- anonymous

so would the second vector consist of 0's ?

- jtvatsim

Well, we don't exactly know. It is safer to say that the second vector y = (y1,y2) will just act like the other vectors in this set. That is, y1 + y2 = 0.

- anonymous

because anything added to 0 is still whatever that number was. so 5 + 0 = 5?

- anonymous

is that right or wrong way to think of it?

- anonymous

ok so then again y1 = -y2

- jtvatsim

I see your point. However, your example is dealing with numbers rather than vectors.

- jtvatsim

Yes, that is the safer assumption.

- jtvatsim

But hold your thought, you might need just a slight change in thinking to see what we are getting at.

- jtvatsim

So, how will we know if x + y is in the set?

- jtvatsim

We need to show that its components add to 0.

- anonymous

if the vectors of x = -y ?

- jtvatsim

Yes, I think you could do it that way.

- anonymous

is that not the right way? or was there something else?

- jtvatsim

Ah... wait. We aren't saying that x + y = 0. We want to say that (first components) + (second components) = 0.

- jtvatsim

Here's what I mean.

- jtvatsim

First, we need to know what x + y looks like.
x + y = (x1,x2) + (y1,y2) = (x1+y1,x2+y2)

- jtvatsim

Then, we need to find a way to show that x1+y1 + x2+y2 = 0. That will be the trick.

- jtvatsim

Does that make sense? Any questions on that part?

- anonymous

i think so. wouldt you just move x2 and y2 over? so
x1 + x2 = -x2 - y2 ??

- anonymous

actually i think thats wrong

- jtvatsim

You could do that. But that won't be a proof. We need to start with the most basic assumption and build up to what we want. So here we go (*long inhale*)... :)

- jtvatsim

"Suppose we have two vectors x and y in the set."

- jtvatsim

"Then, x = (x1,x2) so x1 + x2 = 0 and y = (y1,y2) so y1 + y2 = 0. This is true since they are in the set."

- jtvatsim

"Now, I'd like to think about x + y. x + y looks like this:
x + y = (x1,x2) + (y1,y2) = (x1+y1,x2+y2)."

- jtvatsim

"Now, do these components sum to 0? I notice that
(x1 + y1) + (x2 + y2) = x1 + y1 + x2 + y2 = (x1 + x2) + (y1 + y2)"

- jtvatsim

"But, wait, I just said before that x1 + x2 = 0 and y1 + y2 = 0. So, the components do sum to 0 after all! In fact, (x1 + y1) + (x2 + y2) = (x1 + x2) + (y1 + y2) = 0 + 0 = 0 as I wanted."

- jtvatsim

Do you follow? There are little nuances here and there, but this is really the same format you follow for any "subspace proof". :)

- jtvatsim

I'll make a summary of the highlights so you can bring all these scattered ideas together. :)

- anonymous

ok. im still reading everything you just typed and comprehending, might take a minute

- jtvatsim

No worries, take your time, these things don't happen in a flash. It took me a lot of practice to get comfortable. :)

- jtvatsim

This is what a finished proof might look like (you might find it helpful to see the big picture all at once):
{(x1,x2)^T | x1 + x2 = 0} is a subspace of R^2.
Proof:
A set is a subspace if it is 1) nonempty, 2) cx is in the set for any scalar c, and 3) if x + y is in the set.
1) (0,0) is in the set since 0 + 0 = 0. So, the set is nonempty.
2) Suppose x is in the set and c is some scalar. Then x = (x1,x2) and x1 + x2 = 0.
Think about cx. cx = (cx1, cx2). Now cx1 + cx2 = c(x1 + x2) = c(0) = 0. So cx is in the set.
3) Suppose x and y are in the set. Then x = (x1,x2) and x1 + x2 = 0, also y = (y1,y2) and y1 + y2 = 0.
Think about x + y. x + y = (x1,x2) + (y1,y2) = (x1+y1,x2+y2). Now (x1 + y1) + (x2 + y2) = (x1 + x2) + (y1 + y2) = 0 + 0 = 0. So x + y is in the set.
Since all three conditions are met, the set is a subspace.

- anonymous

that was both kind of morerinvolved than i was thinking but not that bad once you explained it. im still reading. but it is making sense

- jtvatsim

Yes, proofs are quite picky, but the format is almost always the same.

- jtvatsim

The most frustrating thing is not being able to just start with the goal.
We cannot say: "Suppose alpha x1 + alpha x2 = 0..." and go from there.
We must say: "Suppose x is in the set. Then x1 + x2 = 0..." and somehow manipulate this to get to "alpha x1 + alpha x2 = 0". It's a subtle but crucial difference in direction.

- anonymous

ok. i have some similar problens to work on. like x1*x2 = 0 and a few others but im going to try and apply what you explained to them. i might hit you up if i get stuck but im going to try them first

- jtvatsim

Sounds good. Remember that not all sets will be subspaces. If you an find a way to break any of the three rules you are allowed to say "No way, this is not a subspace."

- jtvatsim

Also a "Pro Tip," subspaces are linear, so if you have multiplication of variables (like x1*x2) you will probably not have a subspace.

- anonymous

so is there a different way to work it?

- jtvatsim

If you don't have a subspace, just write down the example that breaks it. Are you working on the x1*x2 = 0 one?

- anonymous

yes

- jtvatsim

OK, I'm looking at it now.

- jtvatsim

Well, let's just go through the list. Does (0,0) work?

- anonymous

so is x = (1,0)

- anonymous

or x = (0,1)

- jtvatsim

Yep, all those are in the set so far.

- anonymous

well yes 0 *0 =0

- jtvatsim

Good, so no way are we going to break rule 1. Pass.

- anonymous

ok so as long as one of them is 0 them it passes

- jtvatsim

Yes, I like that summary. Much simpler than what I was going to say. :)

- jtvatsim

Alright, but look at rule 3 for a second. x + y.
Well (1,0) is in the set, and (0,1) is in the set. But what about if we add these together?
(1,0) + (0,1) = (1,1)... Uh oh...

- jtvatsim

Do you see the problem here?

- anonymous

yes. it would be 1 and not 0

- jtvatsim

And that is a BIG problem. So, how would we write this as a proof? Many ways to do it, but here's one.

- anonymous

ok. before that, is there a reason you went straight to 3 and skipped 2??

- jtvatsim

Well, I happened to realize that rule 3 would break immediately and didn't want to waste brain energy on rule 2. After all, if only one breaks, we know instantly that it is not a subspace.

- jtvatsim

Actually, if you think about it, rule 2 works since we have at least one 0 in the component, and multiplying 0 by any number won't do anything.
c(x1,0) = (cx1,0)
or
c(0,x2) = (0,cx2)

- jtvatsim

So Rule 1 and Rule 2 pass. But Rule 3 fails. Once we know that a single rule fails, we are allowed to stop and declare the result.

- jtvatsim

Does that make sense? :)

- anonymous

yes. im still working on writing a "proof" but yeah step 3 fails so we would say it is not in the subspace of R^2 because it failed step 3

- jtvatsim

Here's an example proof that I came up with to help:
{(x1,x2) : x1*x2 = 0} is not a subspace.
Proof:
If the set were a subspace x + y should be in the set whenever x and y are in the set.
But, look at (1,0) and (0,1). These are both in the set since 1*0 = 0 and 0*1 = 0. But, (1,0) + (0,1) = (1,1). 1*1 is not 0, so the sum is not in the set.
Therefore, the set is not a subspace.

- jtvatsim

A proof is really just the "short version" of the long conversations we are having here. There are very few people who can think like a proof. Most of us have to stumble through examples and making sense of the question before we then go back through our thought process and figure out what was important.

- anonymous

ok. so the proof is using variables like x1, x2, y1, y2 etc for the "abstract" big picture and we can include numerical examples if they would fail one of the 3 steps

- jtvatsim

Yes. It is much easier to prove something false, than to prove something true.
When proving something is true, we must use the abstract big picture to account for every single possible, imaginable, thing.
When proving something is false, we can use simple numerical examples since if a statement fails even once it is false.

- jtvatsim

This is why it is truly amazing that anything could be a subspace, because the requirements are very tight and narrow. It is mind blowing that those three rules would hold forever no matter how many examples you tried.

- anonymous

is kind of cool. someone a lot smarter than me must have had to do a ton of problems to figure that out. i have more problems with variations of x1 and x2 but im going to close this question and try and work them. and if i get stuck ill @ you or something. might even do it just to check me if you dont mind

- anonymous

but thanks for all your help

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