Proving (u,v,u+v) Cannot Be a Basis for <u,v,u+v>

  • Thread starter Thread starter sana2476
  • Start date Start date
  • Tags Tags
    Basis
Click For Summary
SUMMARY

The discussion centers on proving that the set (u, v, u+v) cannot form a basis for the vector space spanned by . The proof demonstrates linear dependence by showing that the equation αu + βv + γ(u+v) = 0 has non-trivial solutions for α, β, and γ. Specifically, the values α = -1, β = -1, and γ = 1 satisfy this equation, confirming that the vectors are linearly dependent and thus do not form a basis.

PREREQUISITES
  • Understanding of vector spaces and bases
  • Knowledge of linear dependence and independence
  • Familiarity with linear combinations of vectors
  • Basic algebraic manipulation skills
NEXT STEPS
  • Study the concept of vector spaces in linear algebra
  • Learn about the criteria for linear independence
  • Explore examples of bases in different vector spaces
  • Practice solving linear equations involving multiple variables
USEFUL FOR

Students of linear algebra, mathematics educators, and anyone interested in understanding vector spaces and linear dependence.

sana2476
Messages
33
Reaction score
0

Homework Statement



Prove (u,v,u+v) can not be a basis for <u,v,u+v>.


Homework Equations





The Attempt at a Solution



Let αu+βv+γ(u+v)=0
αu+βv=-γ(u+v)
α/γ(u)+β/γ(v)=-(u+v)
α/γ(u)+β/γ(v)+1*(u+v)=0

Since α/γ,β/γ,1 are not all zeros, therefore, (u,v,u+v) are linearly dependent. Hence it doesn't form basis for <u,v,u+v>.


Let me know if this is the right approach towards the proof.
 
Physics news on Phys.org
Showing that (u,v,u+v) are linearly dependent is the correct way to prove they don't form a basis. To show that they are linearly dependent, you only have to find an example of α, β, and γ (not all zero) for which αu+βv+γ(u+v)=0. I would just write down a specific choice of numbers that works; this is not difficult to do by guess and check.
 
So I guess we can achieve that by saying the following:

-1*(u) + (-1)*v + 1*(u+v) = 0

Since none of the constants are actually zero. Therefore, (u,v,u+v) are infact linealy dependent.
 

Similar threads

  • · Replies 19 ·
Replies
19
Views
3K
Replies
9
Views
1K
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
12
Views
2K