Verifying Subspace Properties of S+T

  • Thread starter charlies1902
  • Start date
  • Tags
    Subspace
In summary, the conversation discusses verifying the 0 vector and the closure properties for subspaces. The problem involves using words to answer and proving that if S and T are subspaces, then S+T is also a subspace. The conversation also clarifies the use of vectors and subspaces in the problem.
  • #1
charlies1902
162
0
The problem has been attached. I am having difficulty expressing myself. The professor said for this problem, it would be best if I use words to answer it.

1. I must verify the 0 vector is in S+T. Since S and T are subspaces, the 0 vector must exist in both S and T. Thus 0+0=0 and 0 vector is in S+T



2. I must verify that if X and Y are subspaces in S+T, then I need to check if X+Y is still in the subspace S+T.
Note: X and Y are of the form S+T, which yields the following:
So I define X as U1+U2, where U1 is in S and U2 is in T.
I define Y as V1+V2, where V1 is in S and V2 is in T.
X+Y=[U1+V1, U2+V2] Since U1 and V1 are both in S, then U1+V1 is in S. Since U2 and V2 are both in T, then U2+V2 are both in T. Thus X+Y is in S.



3. Lastly I must verify that if Y is in S+T, then I need to check if cY is in S+T, where c is some scalar. Using the same Y=V1+V2 where V1 is in S and V2 is in T. I get cY=c(V1+V2)=cV1+cV2, this satisfies that cV1 is in S and cV2 is in T. Thus cY is in S+T.


Is this correct?
Also, if a certain part is incorrect, can you refer clearly to the part or quote just that part? I get confused easily.

Thanks.
 
Physics news on Phys.org
  • #2
charlies1902 said:
The problem has been attached.
Where? There is no attachment. You seem to be trying to prove "If S and T are vector spaces then S+ T is a vector space but what does the "+" mean here? A S and T assumed to be subspaces of some larger space? If so then S+ T is the "direct sum" (more commonly written S⊕ T), the set of all vectors of the form v= s+ t where s is in S and t is in T.

I am having difficulty expressing myself. The professor said for this problem, it would be best if I use words to answer it.

1. I must verify the 0 vector is in S+T. Since S and T are subspaces, the 0 vector must exist in both S and T. Thus 0+0=0 and 0 vector is in S+T.
Okay, that is good.



2. I must verify that if X and Y are subspaces in S+T, then I need to check if X+Y is still in the subspace S+T.
NO, you don't! You must verify that if x and y are vectors in S+ T, then x+ y is still a vector in S+ T.

Note: X and Y are of the form S+T, which yields the following:
So I define X as U1+U2, where U1 is in S and U2 is in T.
I define Y as V1+V2, where V1 is in S and V2 is in T.
This makes sense if X, Y, U1, and V1 are vectors not subspaces.

X+Y=[U1+V1, U2+V2] Since U1 and V1 are both in S, then U1+V1 is in S. Since U2 and V2 are both in T, then U2+V2 are both in T. Thus X+Y is in S.



3. Lastly I must verify that if Y is in S+T, then I need to check if cY is in S+T, where c is some scalar. Using the same Y=V1+V2 where V1 is in S and V2 is in T. I get cY=c(V1+V2)=cV1+cV2, this satisfies that cV1 is in S and cV2 is in T. Thus cY is in S+T.


Is this correct?
Also, if a certain part is incorrect, can you refer clearly to the part or quote just that part? I get confused easily.

Thanks.
As long as you understand that all of your X, Y, U1, and U2 are vectors, not subspaces, then you proof will work.
 
  • #3
Sorry, I always do that, I have attached it now.
 

Attachments

  • Untitled.png
    Untitled.png
    3.6 KB · Views: 380
  • #4
HallsofIvy said:
NO, you don't! You must verify that if x and y are vectors in S+ T, then x+ y is still a vector in S+ T.

.

Oops, I meant to say "vectors," not subspaces.
 
  • #5
charlies1902 said:
So I define X as U1+U2, where U1 is in S and U2 is in T.
I define Y as V1+V2, where V1 is in S and V2 is in T.
X+Y=[U1+V1, U2+V2]
I'm not sure what the notation in the last line is supposed to mean.
 
  • #6
vela said:
I'm not sure what the notation in the last line is supposed to mean.

U1, U2, V1, V2 are vectors.

When I add X+Y, this yields U1+V1, U2+V2


Is that what you were asking?
 
  • #7
I'm saying your notation doesn't make sense. Is it supposed to be an order paired of vectors?
 
  • #8
vela said:
I'm saying your notation doesn't make sense. Is it supposed to be an order paired of vectors?
Uh, I defined X and Y in the same form as S+T. I included a picture in the 3rd post on this thread. I'm not sure what you mean by an ordered pair?
What I was trying to show with X+Y=[U1+V1, U2+V2], is that since U1 and V1 were in S, then U1+V1 must be in S, and since U2 and V2 are in T, then U2+V2 must be in T, thus X+Y is in S+T.
 
  • #9
I know what you mean. I'm saying you shouldn't write "X+Y=[U1+V1, U2+V2]". What you wrote in words
Since U1 and V1 are both in S, then U1+V1 is in S. Since U2 and V2 are both in T, then U2+V2 are both in T. Thus X+Y is in S.
is sufficient. If you want to be a bit more explicit, you can say "Thus X+Y = (U1+V1)+(U2+V2) is in S."

In other words, I know what mathematical object (U1+V1)+(U2+V2) is. On the other hand, what exactly is [U1+V1, U2+V2] supposed to be? You didn't define what [ , ] means.
 
  • #10
vela said:
I know what you mean. I'm saying you shouldn't write "X+Y=[U1+V1, U2+V2]". What you wrote in words
is sufficient. If you want to be a bit more explicit, you can say "Thus X+Y = (U1+V1)+(U2+V2) is in S."

In other words, I know what mathematical object (U1+V1)+(U2+V2) is. On the other hand, what exactly is [U1+V1, U2+V2] supposed to be? You didn't define what [ , ] means.

Oh I see, I think I'll stick with the words rather than using that. So my statement would be sufficient to show that X+Y is in S+T?

I'm also trying to explain the 3rd condition in words. I repasted the part below:
3. Lastly I must verify that if Y is in S+T, then I need to check if cY is in S+T, where c is some scalar. Using the same Y=V1+V2 where V1 is in S and V2 is in T. I get cY=c(V1+V2)=cV1+cV2, this satisfies that cV1 is in S and cV2 is in T. Thus cY is in S+T.
Is this worded correctly?
 
  • #11
Instead of "this satisfies that cV1 is in S and cV2 is in T", I'd say, "Because S and T are subspaces, cV1 and cV2 are elements of S and T, respectively." It's a bit clearer and states explicitly what your logic is. But, yeah, your argument is fine.
 

FAQ: Verifying Subspace Properties of S+T

1. What is a subspace property?

A subspace property is a characteristic or rule that applies to all elements within a subspace. For example, a common subspace property is closure under addition and multiplication, meaning that if two elements are in the subspace, their sum and product will also be in the subspace.

2. How do you verify subspace properties of S+T?

To verify subspace properties of S+T, you need to check if the sum of two elements from the subspaces S and T also belongs to their sum, S+T. This can be done by checking if the sum satisfies all the subspace properties of S and T.

3. Can you give an example of verifying subspace properties of S+T?

Say we have two subspaces, S = {(x,y)| x,y are real numbers} and T = {(a,b)| a,b are real numbers}. To verify the subspace property of closure under addition, we can choose two elements from S and T, (x1,y1) and (a1,b1), and add them together: (x1+a1, y1+b1). If this new element is also in S+T, then the subspace property is satisfied.

4. What happens if one of the subspaces is not a subspace?

If one of the subspaces is not a subspace, then the subspace property of S+T may not hold. For example, if one of the subspaces does not contain the zero vector, then the subspace property of closure under addition will not hold for S+T.

5. Why is it important to verify subspace properties of S+T?

Verifying subspace properties of S+T is important because it ensures that S+T is also a subspace. This is necessary for performing operations and calculations within the subspace, and it also helps to understand the properties and structure of the subspace.

Similar threads

Replies
1
Views
1K
Replies
43
Views
3K
Replies
4
Views
1K
Replies
17
Views
5K
Replies
3
Views
2K
Replies
6
Views
1K
Back
Top