What is the dimension of eigenspaces in a function-based linear transformation?

  • Context: Graduate 
  • Thread starter Thread starter Tala.S
  • Start date Start date
  • Tags Tags
    Dimension
Click For Summary
SUMMARY

The discussion focuses on determining the dimension of eigenspaces in the context of the linear transformation f: C^∞(R) -> C^∞(R), defined by f(x(t)) = x'(t). The eigenvalue problem is set up with eigenvalues λ = -5 and λ = 0, leading to the eigenspaces E-5 = ke^-5t and E0 = k. It is concluded that each eigenspace is one-dimensional, as every solution can be expressed as a constant multiplied by the exponential function e^{\lambda t}.

PREREQUISITES
  • Understanding of linear transformations in functional spaces
  • Familiarity with eigenvalues and eigenspaces
  • Knowledge of exponential functions and their properties
  • Basic concepts of differential equations
NEXT STEPS
  • Study the properties of linear transformations in functional analysis
  • Learn about the spectral theorem for linear operators
  • Explore the concept of eigenfunctions in the context of differential equations
  • Investigate the implications of one-dimensional eigenspaces in various applications
USEFUL FOR

Mathematicians, physicists, and students studying linear algebra and differential equations, particularly those interested in functional analysis and eigenvalue problems.

Tala.S
Messages
43
Reaction score
0
Linear transformation f:C^∞(R) -> C^∞(R)

f(x(t)) = x'(t) a) I have to set up the eigenvalue-problem and solve it :

My solution : ke^λtb) Now I have to find the dimension of the single eigen spaces when λ is

-5 and 0. My solution :

Eigenspaces :

E-5 = ke^-5t

E0=k (because ke^0t = k)

But I don't know how to find the dimension of the single eigen spaces ?

I'm used to working with vectors but now it's functions and I'm not sure about the dimension.
 
Last edited:
Physics news on Phys.org


Solved !
 
Good! I presume that you realized that since every solution, Ce^{\lambda x} is a constant, C, times the single function e^{\lambda x}, the space is one dimensional.
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 6 ·
Replies
6
Views
4K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K