Subspace proof

[gavin1989]

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?

 

[matt grime]

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]

okay, if so, they will form a six dimentinal subspace, which dont span the vector spcce.


thanks a lot!

 

[matt grime]

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 .....?

 

[gavin1989]

they are not linearly independent? and thus not a basis for such space

 

[Matterwave]

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?

 

[gavin1989]

then they dont share common subspaces...

 

[matt grime]

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).

 

[HallsofIvy]

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.