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

Subspace proof

  • Thread starter gavin1989
  • Start date
  • #1
8
0
Prove that if L and P are three dimentional subspaces of R5, then L and P must have a nonzero constant in common.

i am just stuck now on how to get this proof started... any thoughts on how to start?
 

Answers and Replies

  • #2
matt grime
Science Advisor
Homework Helper
9,395
3
L is thee dimensional: let a,b,c be a basis. P is three dimensional: let x,y,z be a basis. What can you say about a,b,c,x,y,z given that they live in a 5 dimensional space?
 
  • #3
8
0
okay, if so, they will form a six dimentinal subspace, which dont span the vector spcce.


thanks a lot!
 
  • #4
matt grime
Science Advisor
Homework Helper
9,395
3
How do you know that these 6 vectors do not span the space? They may or may not span the space, but that is immaterial: you have 6 vectors in a 5 dimensional space, therefore they are .....?
 
  • #5
8
0
they are not linearly independent? and thus not a basis for such space
 
  • #6
Matterwave
Science Advisor
Gold Member
3,965
326
The part that they are "not linearly independent" is what is pertinent to the question. You don't care whether they make a basis for a 6 dimensional space.

If your 6 basis vectors are not linearly independent, then they share...

What does it mean if the vectors are not linearly independent?
 
  • #7
8
0
then they dont share common subspaces...
 
  • #8
matt grime
Science Advisor
Homework Helper
9,395
3
I think you're just guessing and hoping now. I don't know what Matterwave means, but all I want you to do is look at the definition of linearly dependent (or the negation of linear independence).
 
  • #9
HallsofIvy
Science Advisor
Homework Helper
41,793
922
L is thee dimensional: let a,b,c be a basis. P is three dimensional: let x,y,z be a basis. What can you say about a,b,c,x,y,z given that they live in a 5 dimensional space?
okay, if so, they will form a six dimentinal subspace, which dont span the vector spcce.


thanks a lot!
How do you know that these 6 vectors do not span the space? They may or may not span the space, but that is immaterial: you have 6 vectors in a 5 dimensional space, therefore they are .....?
Notice that you did not use the fact that the two subspaces have no non-zero vector (you said "constant" but I presume you meant vector) in common. That's the important thing.
 

Related Threads for: Subspace proof

  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
8
Views
2K
  • Last Post
Replies
4
Views
3K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
2
Views
3K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
9
Views
2K
  • Last Post
Replies
15
Views
2K
Top