Question

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

Answer 1

Do you know the conditions required to have a subspace?

Answer 2

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

Answer 3

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).

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

talking*

Answer 8

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.

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

and im assuming R^2 is the x y plane

Answer 13

and there sum is 0

Answer 14

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

Answer 15

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

Answer 16

thats about as far as i can go

Answer 17

what issue?

Answer 18

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

Answer 19

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.

Answer 20

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

Answer 21

Correct

Answer 22

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?

Answer 23

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

Answer 24

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

Answer 25

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

Answer 26

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

Answer 27

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

Answer 28

You are 100% correct!

Answer 29

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

Answer 30

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

Answer 31

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

Answer 32

It's the same thing either way.

Answer 33

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

Answer 34

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

Answer 35

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

Answer 36

thats where i run into trouble.

Answer 37

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

Answer 38

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."

Answer 39

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.

Answer 40

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

Answer 41

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

Answer 42

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

Answer 43

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.

Answer 44

Does that answer your question?

Answer 45

ok, yes it does

Answer 46

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.

Answer 47

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

Answer 48

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

Answer 49

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.

Answer 50

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 ?

Answer 51

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

Answer 52

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

Answer 53

so would the second vector consist of 0's ?

Answer 54

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.

Answer 55

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

Answer 56

is that right or wrong way to think of it?

Answer 57

ok so then again y1 = -y2

Answer 58

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

Answer 59

Yes, that is the safer assumption.

Answer 60

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

Answer 61

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

Answer 62

We need to show that its components add to 0.

Answer 63

if the vectors of x = -y ?

Answer 64

Yes, I think you could do it that way.

Answer 65

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

Answer 66

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

Answer 67

Here's what I mean.

Answer 68

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

Answer 69

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

Answer 70

Does that make sense? Any questions on that part?

Answer 71

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

Answer 72

actually i think thats wrong

Answer 73

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*)... :)

Answer 74

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

Answer 75

"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."

Answer 76

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

Answer 77

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

Answer 78

"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."

Answer 79

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

Answer 80

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

Answer 81

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

Answer 82

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

Answer 83

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.

Answer 84

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

Answer 85

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

Answer 86

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.

Answer 87

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

Answer 88

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."

Answer 89

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

Answer 90

so is there a different way to work it?

Answer 91

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

Answer 92

yes

Answer 93

OK, I'm looking at it now.

Answer 94

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

Answer 95

so is x = (1,0)

Answer 96

or x = (0,1)

Answer 97

Yep, all those are in the set so far.

Answer 98

well yes 0 *0 =0