Prove: Existence of Basis for Finite Vector Space with Linearly Independent Set

  • Thread starter Thread starter eyehategod
  • Start date Start date
  • Tags Tags
    Basis
Click For Summary
SUMMARY

The discussion centers on proving the existence of a basis for a finite-dimensional vector space V that includes a given linearly independent set S. It is established that if S spans V, then S is already a basis. If S does not span V, one must identify a vector not expressible as a linear combination of the vectors in S, and demonstrate that adding this vector to S maintains linear independence. This process continues until a basis for V is constructed.

PREREQUISITES
  • Understanding of linear independence in vector spaces
  • Familiarity with the concept of a basis in linear algebra
  • Knowledge of finite-dimensional vector spaces
  • Ability to perform linear combinations of vectors
NEXT STEPS
  • Study the properties of linear independence and spanning sets in vector spaces
  • Learn about the process of constructing bases from linearly independent sets
  • Explore the concept of dimension in finite-dimensional vector spaces
  • Investigate examples of basis construction in various vector spaces
USEFUL FOR

Students and educators in linear algebra, mathematicians focusing on vector spaces, and anyone interested in understanding the foundational concepts of linear independence and basis construction.

eyehategod
Messages
82
Reaction score
0
Let S be a linearly independent set of vectors from the finite dimensional vector space V. Prove that there exists a basis for V containing S. Can anyone help me out? I can't figure out how to approach this.
 
Physics news on Phys.org
What is a basis? What do you need to have one? What do you already have? Can you think of a way to construct the rest?
 
If the independent set already spans the space, you are done. If not then there exists a vector that cannot be written as a linear combination of the vectors in the independent set. Can you show that adding that new vector to the set still gives you an independent set of vectors? If that new set spans the vector space, you are done. If not ...
 

Similar threads

  • · Replies 15 ·
Replies
15
Views
2K
Replies
34
Views
4K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 30 ·
2
Replies
30
Views
2K
  • · Replies 18 ·
Replies
18
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
15
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K