• Support PF! Buy your school textbooks, materials and every day products via PF Here!

Linear Algebra: Subspace Proof

  • Thread starter *melinda*
  • Start date
86
0
Prove that the intersection of any collection of subspaces of V is a subspace of V.

Ok, I know I need to show that:

1. For all u and v in the intersection, it must imply that u+v is in the intersection, and

2. For all u in the intersection and c in some field, cu must be in the intersection.

I can show both 1. and 2. for the trivial case where the intersection is zero, but I'm not sure what I need to do for the arbitrary case.

Any suggestions?
 

Hurkyl

Staff Emeritus
Science Advisor
Gold Member
14,845
17
Well, I'm not sure where you're stuck, so I'll throw out two random hints.


(a) Let W be one of the spaces in the intersection...

(b) What is the definition of "x is in the intersection"?
 
86
0
Since u and v are elements of the intersection, u and v will also be elements of any subspace W that is in the intersection. And since u and v are in W and W is a subspace, this guarantees that u+v will also be in W. This same argument would apply to scalar multiplication.

Is that the right idea?
 

Hurkyl

Staff Emeritus
Science Advisor
Gold Member
14,845
17
Yes, that is one of the essential points of the proof.
 
86
0
Thanks!
 

Related Threads for: Linear Algebra: Subspace Proof

  • Posted
Replies
2
Views
2K
  • Posted
Replies
8
Views
2K
Replies
7
Views
7K
  • Posted
Replies
4
Views
3K
  • Posted
Replies
1
Views
3K
  • Posted
Replies
6
Views
1K
Replies
0
Views
1K
  • Posted
Replies
4
Views
3K

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top