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

In summary, when proving that if L and P are three dimensional subspaces of R5, they must have a nonzero constant in common, it is important to consider whether the basis vectors of these subspaces are linearly independent. If not, then they may not span the space and therefore will not have a nonzero constant in common.
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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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  • #2
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?
 
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okay, if so, they will form a six dimentinal subspace, which don't span the vector spcce.


thanks a lot!
 
  • #4
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
they are not linearly independent? and thus not a basis for such space
 
  • #6
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
then they don't share common subspaces...
 
  • #8
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
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.
 

1. What does it mean for L and P to have a common nonzero constant?

Having a common nonzero constant means that there is a number that can be multiplied by all the elements in both L and P, resulting in the same value. This number is not equal to zero, as that would mean the two sets have no common elements.

2. How is this proof related to R5?

R5 refers to the 5-dimensional Euclidean space, which is the space in which the elements of L and P exist. This proof shows that despite being in a high-dimensional space, L and P still have a common nonzero constant, demonstrating a fundamental property of vector spaces.

3. What are the implications of this proof?

This proof has implications in linear algebra and other areas of mathematics. It shows that even in high-dimensional spaces, certain properties and relationships still hold true. It also allows for the use of techniques and methods from linear algebra in higher dimensions.

4. How does this proof relate to real-world applications?

Many real-world systems can be represented and analyzed using vectors in high-dimensional spaces. This proof provides a fundamental understanding of the properties and relationships of these vectors, which can be applied to various fields such as physics, engineering, and data analysis.

5. Is this proof applicable to all sets in R5?

Yes, this proof applies to all sets in R5 as long as the sets are linearly dependent. However, if the sets are linearly independent, then they cannot have a common nonzero constant. This proof only holds true for linearly dependent sets in R5.

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