anonymous
  • anonymous
not sure where to start for this proof. Determine whether the following sets form subspaces of R^2 {(x1,x2)^T | x1+x2=0}
Linear Algebra
chestercat
  • chestercat
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

jtvatsim
  • jtvatsim
Do you know the conditions required to have a subspace?
anonymous
  • 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
  • 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).

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

jtvatsim
  • jtvatsim
The trick, as you mention, is figuring out how to apply it. :)
jtvatsim
  • 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
  • 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
  • anonymous
talking*
jtvatsim
  • 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
  • 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
  • jtvatsim
OK. Let's start with step one. Identify what all the important concepts, symbols, and definitions are.
anonymous
  • anonymous
ok. i know that x1 and x2 are vectors in R^2
anonymous
  • anonymous
and im assuming R^2 is the x y plane
anonymous
  • anonymous
and there sum is 0
jtvatsim
  • jtvatsim
Good, so far. There is a minor issue however.
freckles
  • freckles
well I think (x1,x2)^T is a vector in R^2 not x1 and x2
anonymous
  • anonymous
thats about as far as i can go
anonymous
  • anonymous
what issue?
jtvatsim
  • 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
  • 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
  • anonymous
so its a single vector like x = (x1 x2)
jtvatsim
  • jtvatsim
Correct
jtvatsim
  • 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
  • anonymous
yes. i guess "abstractly" x1 = negative x2 so that they sum to 0
jtvatsim
  • jtvatsim
Exactly! In fact, that is what the expression x1 + x2 = 0 implies. Solving for x1 gives as you say, x1 = -x2.
jtvatsim
  • jtvatsim
Alright, so we still need to decide whether this set of vectors is in fact a subspace.
jtvatsim
  • jtvatsim
Let's start with the first condition: 1) Does the zero vector live in this set?
anonymous
  • anonymous
i would say yes because 0 + (-0) is still 0. is that the right way to think of it?
jtvatsim
  • jtvatsim
You are 100% correct!
jtvatsim
  • jtvatsim
(0,0) satisfies the condition that x1 + x2 = 0, since 0 + 0 = 0. Good, our set passes condition 1.
jtvatsim
  • jtvatsim
Now, to check condition 2: If c is some multiple and v is in the set, is cv in the set?
jtvatsim
  • 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
  • jtvatsim
It's the same thing either way.
jtvatsim
  • jtvatsim
So, this is a little weird, but here goes.
anonymous
  • 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
  • jtvatsim
Great, that's the spirit of the proof! Now we just make the argument as general as possible.
anonymous
  • anonymous
thats where i run into trouble.
anonymous
  • 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
  • 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
  • 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
  • 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
  • anonymous
but what you did is only for the second condition, right?
jtvatsim
  • jtvatsim
This is only the second condition, yes. And your question is a good one. Let me think how to answer it.
jtvatsim
  • 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
  • jtvatsim
Does that answer your question?
anonymous
  • anonymous
ok, yes it does
jtvatsim
  • 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
  • jtvatsim
That was a minor difference in your proof versus my general proof.
jtvatsim
  • jtvatsim
Just wanted to be sure you picked up on that. :)
jtvatsim
  • 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
  • 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
  • jtvatsim
Good point. We are allowed to assume we have a second vector y for condition 3's proof.
jtvatsim
  • jtvatsim
After all, with condition 2 we just showed that this set probably has an infinite number of vectors (every multiple works).
anonymous
  • anonymous
so would the second vector consist of 0's ?
jtvatsim
  • 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
  • anonymous
because anything added to 0 is still whatever that number was. so 5 + 0 = 5?
anonymous
  • anonymous
is that right or wrong way to think of it?
anonymous
  • anonymous
ok so then again y1 = -y2
jtvatsim
  • jtvatsim
I see your point. However, your example is dealing with numbers rather than vectors.
jtvatsim
  • jtvatsim
Yes, that is the safer assumption.
jtvatsim
  • jtvatsim
But hold your thought, you might need just a slight change in thinking to see what we are getting at.
jtvatsim
  • jtvatsim
So, how will we know if x + y is in the set?
jtvatsim
  • jtvatsim
We need to show that its components add to 0.
anonymous
  • anonymous
if the vectors of x = -y ?
jtvatsim
  • jtvatsim
Yes, I think you could do it that way.
anonymous
  • anonymous
is that not the right way? or was there something else?
jtvatsim
  • jtvatsim
Ah... wait. We aren't saying that x + y = 0. We want to say that (first components) + (second components) = 0.
jtvatsim
  • jtvatsim
Here's what I mean.
jtvatsim
  • jtvatsim
First, we need to know what x + y looks like. x + y = (x1,x2) + (y1,y2) = (x1+y1,x2+y2)
jtvatsim
  • jtvatsim
Then, we need to find a way to show that x1+y1 + x2+y2 = 0. That will be the trick.
jtvatsim
  • jtvatsim
Does that make sense? Any questions on that part?
anonymous
  • anonymous
i think so. wouldt you just move x2 and y2 over? so x1 + x2 = -x2 - y2 ??
anonymous
  • anonymous
actually i think thats wrong
jtvatsim
  • 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
  • jtvatsim
"Suppose we have two vectors x and y in the set."
jtvatsim
  • 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
  • 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
  • jtvatsim
"Now, do these components sum to 0? I notice that (x1 + y1) + (x2 + y2) = x1 + y1 + x2 + y2 = (x1 + x2) + (y1 + y2)"
jtvatsim
  • 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
  • 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
  • jtvatsim
I'll make a summary of the highlights so you can bring all these scattered ideas together. :)
anonymous
  • anonymous
ok. im still reading everything you just typed and comprehending, might take a minute
jtvatsim
  • 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
  • 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
  • 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
  • jtvatsim
Yes, proofs are quite picky, but the format is almost always the same.
jtvatsim
  • 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
  • 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
  • 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
  • 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
  • anonymous
so is there a different way to work it?
jtvatsim
  • 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
  • anonymous
yes
jtvatsim
  • jtvatsim
OK, I'm looking at it now.
jtvatsim
  • jtvatsim
Well, let's just go through the list. Does (0,0) work?
anonymous
  • anonymous
so is x = (1,0)
anonymous
  • anonymous
or x = (0,1)
jtvatsim
  • jtvatsim
Yep, all those are in the set so far.
anonymous
  • anonymous
well yes 0 *0 =0
jtvatsim
  • jtvatsim
Good, so no way are we going to break rule 1. Pass.
anonymous
  • anonymous
ok so as long as one of them is 0 them it passes
jtvatsim
  • jtvatsim
Yes, I like that summary. Much simpler than what I was going to say. :)
jtvatsim
  • 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
  • jtvatsim
Do you see the problem here?
anonymous
  • anonymous
yes. it would be 1 and not 0
jtvatsim
  • 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
  • anonymous
ok. before that, is there a reason you went straight to 3 and skipped 2??
jtvatsim
  • 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
  • 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
  • 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
  • jtvatsim
Does that make sense? :)
anonymous
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • anonymous
but thanks for all your help

Looking for something else?

Not the answer you are looking for? Search for more explanations.