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 ...?
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.
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.
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.
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.
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.