Solve Unique Decomposition Problem: Linear Algebra Vectors & Scalars

  • Context: Graduate 
  • Thread starter Thread starter arshavin
  • Start date Start date
  • Tags Tags
    Algebra
Click For Summary
SUMMARY

The discussion centers on the unique decomposition of vectors in a vector space V under a linear map ℓ: V → R. It establishes that if a vector z is not in the nullspace of ℓ, any vector x in V can be expressed uniquely as x = v + cz, where v belongs to the nullspace of ℓ and c is a scalar. The proof begins by considering two cases: when x is in the nullspace of ℓ and when it is not, leading to the conclusion that a nonzero scalar c exists such that ℓ(x) = cℓ(z).

PREREQUISITES
  • Understanding of vector spaces and linear maps
  • Familiarity with nullspaces and their properties
  • Knowledge of scalar multiplication in linear algebra
  • Basic proof techniques in mathematics
NEXT STEPS
  • Study the properties of linear maps and their nullspaces
  • Explore the concept of unique decomposition in linear algebra
  • Learn about scalar multiplication and its implications in vector spaces
  • Investigate proof techniques specific to linear algebra problems
USEFUL FOR

Mathematics students, educators, and professionals in fields requiring linear algebra knowledge, particularly those focusing on vector spaces and linear transformations.

arshavin
Messages
21
Reaction score
0
Let V be a vector space and ℓ : V → R be a linear map. If z ∈ V is not in the
nullspace of ℓ, show that every x ∈ V can be decomposed uniquely as x = v + cz ,
where v is in the nullspace of ℓ and c is a scalar.
 
Physics news on Phys.org
I'll give the start of the proof:

There are two situations:
1) x is in the nullspace of l, then the statement is trivial.
2) x is not in the nullspace of l, then l(x)\neq 0 and l(z)\neq 0. Thus, there exists a nonzero c, such that l(x)=cl(z). Try to continue the argument (hint: what happens to x-cz?
 

Similar threads

  • · Replies 19 ·
Replies
19
Views
4K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 23 ·
Replies
23
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K