Proving Subspace

  • #1
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.
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
41,833
964
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
162
0
Sorry, I always do that, I have attached it now.
 

Attachments

  • Untitled.png
    Untitled.png
    4 KB · Views: 309
  • #4
162
0
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
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
15,045
1,629
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
162
0
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
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
15,045
1,629
I'm saying your notation doesn't make sense. Is it supposed to be an order paired of vectors?
 
  • #8
162
0
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
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
15,045
1,629
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
162
0
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
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
15,045
1,629
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.
 

Related Threads on Proving Subspace

  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
7
Views
1K
  • Last Post
Replies
22
Views
3K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
1
Views
941
Replies
1
Views
1K
Replies
0
Views
859
Replies
6
Views
1K
  • Last Post
Replies
4
Views
872
Top