Clarification on Subspaces.

[Ethan3141]

Lets say I have a vector space [MATH]V[/MATH] and three subspaces [MATH]W, W_1, W_2[/MATH] with the condition that [MATH]W_1 {\subseteq} W[/MATH] then I am wondering if [MATH]W_1[/MATH] and [MATH]W{\cap}W_2[/MATH] are subspaces of [MATH]W[/MATH]. I am really convinced they are but would love confirmation in case I have made a stupid mistake in my thoughts.




Thanks for any help, Ethan.

 

[tkhunny]

We don't normally use the word "confirmation" in this context. Typically, we seek "proof".

(edited after actually reading the problem statement) :-(

 

[Ethan3141]

My main problem is normally it is that they're subspaces of the "parent space?", V. But I want to know if they're subspaces of W and I think they're. I know how to prove it I was mainly just wondering if I understand the concept correctly because if not my proof would likely be wrong. I do think it was a stupid question for me to ask but just wanted to know (I often get paranoid over very small details and like to know for definite). Anyway thanks for your input.

 

[tkhunny]

I totally misread the problem. Here's a plan, present your proof of what your question actually asks. We're pretty friendly, here. You didn't make fun of me when I was talking gibberish (now deleted, above), did you? We'll return the courtesy. Let's see what you're doing. Take a leap. Present your work.

 

[Ethan3141]

I totally misread the problem. Here's a plan, present your proof of what your question actually asks. We're pretty friendly, here. You didn't make fun of me when I was talking gibberish (now deleted, above), did you? We'll return the courtesy. Let's see what you're doing. Take a leap. Present your work.

It's fine I am certain that I am correct now, since the intersection of [MATH]W[/MATH] and [MATH]W_1[/MATH] is clearly a subset of [MATH]W[/MATH] and a subspace of [MATH]V[/MATH] (i.e it hits the subspace criterion) so it is a subspace of [MATH]W[/MATH]and the same argument goes for [MATH]W_1[/MATH] (which is defined to be a subset of [MATH]W[/MATH] anyway). I mean this is exactly what I thought in the first place but for some reason I had doubt. Thanks for the help

 

[tkhunny]

What if your intersection is Null?

 

[pka]

Lets say I have a vector space [MATH]V[/MATH] and three subspaces [MATH]W, W_1, W_2[/MATH] with the condition that [MATH]W_1 {\subseteq} W[/MATH] then I am wondering if [MATH]W_1[/MATH] and [MATH]W{\cap}W_2[/MATH] are subspaces of [MATH]W[/MATH]. I am really convinced they are but would love confirmation in case I have made a stupid mistake in my thoughts.

Note that the definition of subspace requires that the subset contain the zero therefore the intersection of two subspaces cannot be empty.
Then is the intersection necessarily closed with respect to addition & scalar multiplication?

 

[Ethan3141]

What if your intersection is Null?

I don't need to worry about that because the question I need to clarify this for is about direct sums and there is enough information for me to know that the intersection can't be null. Interesting question and I honestly am unsure since the proof I linked below seems not to mention that case. I mean I guess it still holds since the empty set is a subset of any set and it would still be closed under addition and multiplication. Ah but it doesn't contain the zero vector so I guess not. In any case I don't think that it can be null.

Note that the definition of subspace requires that the subset contain the zero therefore the intersection of two subspaces cannot be empty.
Then is the intersection necessarily closed with respect to scalar multiplication?

The intersection of two subspaces is a subspace which means it is closed under addition and scalar. Here is a proof showing this. Though we also had this assigned to us as a proof for homework so I am quite familiar with it.

 

[Steven G]

How can the intersection of two subspaces be null?

It can't but I am sure that tkhunny just wanted you to think about that. I 100% support the asking of his question. It is also good that you know that the intersection is not null, but if you didn't then you need to think about the case of the intersection being null or concluding that it can't be null.

 

[Ethan3141]

It can't but I am sure that tkhunny just wanted you to think about that. I 100% support the asking of his question. It is also good that you know that the intersection is not null, but if you didn't then you need to think about the case of the intersection being null or concluding that it can't be null.

Yeah I edited my response as you were typing yours. I agree it is a great question. I edit my responses too much because all my thoughts just come rushing out and by the time I respond I have thought about something else I would've liked to have said, sorry .

 

[Steven G]

I don't need to worry about that because the question I need to clarify this for is about direct sums and there is enough information for me to know that the intersection can't be null. Interesting question and I honestly am unsure since the proof I linked below seems not to mention that case. I mean I guess it still holds since the empty set is a subset of any set and it would still be closed under addition and multiplication. Ah but it doesn't contain the zero vector so I guess not. In any case I don't think that it can be null.


The intersection of two subspaces is a subspace which means it is closed under addition and scalar. Here is a proof showing this.

I did not read the proof. However the proof will not be valid if it does not say that the intersection is not empty. Did you understand the proof? Can you now prove it on your own? You really should be able to. It is also a great test question in my opinion.

 

[pka]

I don't need to worry about that because the question I need to clarify this for is about direct sums and there is enough information for me to know that the intersection can't be null. Interesting question and I honestly am unsure since the proof I linked below seems not to mention that case. I mean I guess it still holds since the empty set is a subset of any set and it would still be closed under addition and multiplication. Ah but it doesn't contain the zero vector so I guess not. In any case I don't think that it can be null.
The intersection of two subspaces is a subspace which means it is closed under addition and scalar. Here is a proof showing this.

If you knew all of that why did you post the question?
Any subspace of a V-space (also known as a linear space) must contain the zero.
So the intersection of two subspaces cannot be empty.

 

[Ethan3141]

If you knew all of that why did you post the question?
Any subspace of a V-space (also known as a linear space) must contain the zero.
So the intersection of two subspaces cannot be empty.

Like I said somewhere in this thread I get really caught up on minor details to the point of where I start doubting myself. So I posted here to get a conversation started which helped clear my doubt. It might seem stupid but its just what I do.