Proof That L and P Have a Common Nonzero Constant in R5

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Homework Help Overview

The discussion revolves around proving that two three-dimensional subspaces, L and P, of R5 must share a common nonzero vector. Participants are exploring the implications of linear dependence and the dimensionality of the involved subspaces.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants are examining the relationship between the bases of the subspaces and their implications in a five-dimensional space. Questions about linear independence and the nature of the vectors involved are raised, particularly regarding how six vectors in a five-dimensional space relate to the concept of spanning and basis.

Discussion Status

The discussion is active, with participants questioning assumptions about linear independence and exploring the definitions related to the dimensionality of the subspaces. Some guidance has been offered regarding the significance of linear dependence in the context of the proof.

Contextual Notes

There is an ongoing examination of the definitions and implications of linear independence and dependence, as well as the constraints imposed by the dimensionality of the space in question.

gavin1989
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Prove that if L and P are three dimensional 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?
 
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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 don't 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 ...?
 
they are not linearly independent? and thus not a basis for such space
 
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?
 
then they don't share common subspaces...
 
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).
 
matt grime said:
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?

gavin1989 said:
okay, if so, they will form a six dimentinal subspace, which don't span the vector spcce.


thanks a lot!

matt grime said:
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.
 

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