Is Every Set in ℝ3 a Subspace?

If any of the properties are violated, the set is not a subspace. You don't need to go back to the original 10 axioms and check them all again. You have already satisfied them by being a subset of ##\mathbb{R^3}##.So you have to go back to the original question and check each of the sets to see if they satisfy these three properties. That is a much simpler task than checking the full set of 10 axioms for each set.
  • #1
MarcL
170
2

Homework Statement



A. {(x,y,z)ι x<y<z }
B.{(x,y,z)ι -4x+2y=0, -5x-7z=0 }
C.{(x,y,z)ι -9x-3y+8z=7}
D.{-7x-8y,9x+6y,3x-6y ι x,y arbitrary numbers}
E.{(x,y,z)ι x+y+z=0 }
F.{(x,x+4,x-2 }

Homework Equations



So it is a subspace therefore I need the additive axiom: u+v = v+u, the one where u + o = u and the additive inverse where u + (-u) = (-u)+u=0

The Attempt at a Solution



I thought all of them would be a subspace of ℝ3 because they all have 3 components and ℝ3 includes all numbers in the subspace, or am I seeing this wrong?

sorry for the picture, edit done!
 
Last edited:
Physics news on Phys.org
  • #2
I'm not seeing it at all. Posting unreadable images doesn't help anyone. Type your questions.
 
  • #4
I am sorry , I couldn't write them down, was on my chromebook, was really hard to do, ill edit right now.
 
  • #5
Edit done, sorry about that!
 
  • #6
MarcL said:

Homework Statement



A. {(x,y,z)ι x<y<z }
B.{(x,y,z)ι -4x+2y=0, -5x-7z=0 }
C.{(x,y,z)ι -9x-3y+8z=7}
D.{-7x-8y,9x+6y,3x-6y ι x,y arbitrary numbers}
E.{(x,y,z)ι x+y+z=0 }
F.{(x,x+4,x-2 }

Homework Equations



So it is a subspace therefore I need the additive axiom: u+v = v+u, the one where u + o = u and the additive inverse where u + (-u) = (-u)+u=0

The Attempt at a Solution



I thought all of them would be a subspace of ℝ3 because they all have 3 components and ℝ3 includes all numbers in the subspace, or am I seeing this wrong?

sorry for the picture, edit done!

Yes you are seeing it wrong. You have to check the axioms. Just to get you started, in part A is (0,0,0) in the described set?
 
  • #7
forgot the statement. Which of the following sets are subspaces of
f6d400a46c9c7f3c6c5770295a75701.png
was the problem statement. However because of the restriction it wouldn't be right?
 
  • #8
MarcL said:
forgot the statement. Which of the following sets are subspaces of
f6d400a46c9c7f3c6c5770295a75701.png
was the problem statement. However because of the restriction it wouldn't be right?

You need to quote what you are replying to, otherwise it becomes impossible to follow the thread. Because of what restriction what wouldn't be right and why? Explanation required. Maybe even some math statements.
 
  • #9
LCKurtz said:
You need to quote what you are replying to, otherwise it becomes impossible to follow the thread. Because of what restriction what wouldn't be right and why? Explanation required. Maybe even some math statements.

This would mean that my 4th axiom would not hold because u + o = u, my "0" vector would be (0,0,0) to satisfy the axiom, but it would not satisfy my restriction where x<y<z, right? ( if that is what you were trying to get to?)
 
  • #10
MarcL said:
This would mean that my 4th axiom would not hold because u + o = u, my "0" vector would be (0,0,0) to satisfy the axiom, but it would not satisfy my restriction where x<y<z, right? ( if that is what you were trying to get to?)

Yes, that is what I was getting at. So your first example is not a subspace and that is the reason. It is easy to show something isn't a subspace when it isn't because you only have to find one axiom that doesn't work. There may be some others in your problem that aren't subspaces and you have to show why. What's worse is that some of them are subspaces and you can't just say that I see they are. You have to check that all the axioms work, which can be lots of busy work.

Also, even though your problem image is now readable, the reason posting problems in images is frowned on is that we can't copy, for example, part B to talk about it when we are discussing that problem.
 
  • Like
Likes MarcL
  • #11
Thank you, I'll give a shot to the rest of them and post if I have another question ,and I understand about the picture. won't happen again :)
 
  • #12
You don't need to confirm that all 10 (or 11 or 12, whatever) vector space axioms are satisfied, because all of your sets already belong to a vector space, ##\mathbb{R^3}##.

All you need to check are these properties:
1. The 0 element belongs to the set.
2. The set is closed under vector addition. IOW, if u and v are in the set, then u + v is also in the set.
3. The set is closed under scalar multiplication. IOW, if u is in the set and k is a scalar, the ku is also in the set.

If a particular set satisfies all three of these properties, the set is a subspace of the larger vector space.
 

FAQ: Is Every Set in ℝ3 a Subspace?

What is a vector subspace problem?

A vector subspace problem is a mathematical problem that involves determining whether a given set of vectors form a subspace of a larger vector space. A subspace is a subset of a vector space that is closed under addition and scalar multiplication.

How do you check if a set of vectors is a subspace?

To check if a set of vectors is a subspace, you need to make sure that the vectors satisfy the three conditions for a subspace: closure under addition, closure under scalar multiplication, and containing the zero vector. This can be done by performing operations on the given vectors and checking if the resulting vector is also in the set.

What is the significance of vector subspaces?

Vector subspaces are important because they allow us to break down a larger vector space into smaller, more manageable subspaces. This can make solving problems in linear algebra and other fields of mathematics easier and more efficient.

What happens if a set of vectors does not form a subspace?

If a set of vectors does not form a subspace, it means that it does not satisfy one or more of the conditions for a subspace. This could be due to the vectors not being closed under addition or scalar multiplication, or not containing the zero vector. In this case, the set is not considered to be a subspace.

Can a vector subspace problem have multiple solutions?

Yes, a vector subspace problem can have multiple solutions. This is because there can be multiple sets of vectors that satisfy the conditions for a subspace. However, the solutions may not always be unique, as there can be multiple ways to represent a subspace using different sets of vectors.

Back
Top