Finding the eigenvalues of maps of polynomial vector spaces

In summary, the conversation discusses finding the eigenvalues and geometric multiplicities for two maps from a vector space of real-coefficient polynomials with degree strictly less than five. The first map involves differentiation and the second map involves multiplying by 2+x^3 and crossing out terms of degree five or greater. The conversation goes on to mention different approaches to solving for eigenvalues, such as considering two cases where the eigenvalue is 0 or not 0, and the use of systems of equations and ordinary differential equations.
  • #1
cloverforce
3
0

Homework Statement


Let V be the vector space of all real-coefficient polynomials with degree strictly less than five. Find the eigenvalues and their geometric multiplicities for the following maps from V to V:

a) G(f) = xD(f), where f is an element of V and D is the differentiation map.

b) F(f) is obtained by multiplying f by 2+x^3 and crossing out all terms of degree five or greater


Homework Equations


None I can think of.


The Attempt at a Solution


For a), I wrote out that if v is an eigenvector, then for it there exists some h such that G(v) = h*v = h(a4*x^4+...+a1*x+a0) = 4*a4*x^4 + 3*a3*x^3+...+a1*x and from this tried to set up equations such as h*a4 = 4*a4,.., h*a1=a1, h*a0= 0, and so on. But from here I'm not sure how to solve for the eigenvalues. I've never done this before and there has been no explanation either in lecture or in the book of how to approach this kind of problem.

For b), I did the same and came up with F(v) = h*v = (2*a4+a1)*x^4 + (2*a3+a0)*x^3 + 2*a2*x^2 + 2*a1*x + 2*a0 (this is what happens when one simplifies the polynomial and crosses out the summands with terms of degree five or higher). Again, not sure what to do from here.
 
Physics news on Phys.org
  • #2
Given an operator f, I think it's usually easier to consider two cases: 1. the eigenvalue is 0 (considered when Ker f =/= {0}), 2. the eigenvalue is not 0.

For instance, in part a), you can easily check which polynomials are eigenvectors with eigenvalue 0. Then just suppose that the eigenvalue h is not 0, and then using the systems of equations you've written out it's pretty easy to see what h could be. Note the obvious reinterpretation of problem a) as a first order separable ordinary differential equation, which leads to slick solution.
 

1. What are eigenvalues?

Eigenvalues are a mathematical concept used to describe the behavior of linear transformations, such as maps of polynomial vector spaces. They represent the scalar values that, when multiplied by a vector, result in the same vector but possibly scaled. In other words, they represent the directions in which a transformation stretches or compresses a vector.

2. How do you find the eigenvalues of a map of polynomial vector spaces?

To find the eigenvalues of a map of polynomial vector spaces, you need to solve the characteristic equation. This equation is obtained by setting the determinant of the map's matrix representation equal to zero. The solutions to the characteristic equation are the eigenvalues of the map.

3. Can a map of polynomial vector spaces have complex eigenvalues?

Yes, a map of polynomial vector spaces can have complex eigenvalues. This is because the characteristic equation may have complex solutions, and the eigenvalues are the roots of this equation.

4. How do eigenvalues relate to eigenvectors?

Eigenvectors are the corresponding vectors to eigenvalues. They represent the directions in which a transformation is applied without changing direction. In other words, they are the vectors that are only scaled by the transformation. The eigenvector associated with an eigenvalue can be found by solving the system of equations formed by the map's matrix representation and the eigenvalue.

5. Why are eigenvalues important in the study of maps of polynomial vector spaces?

Eigenvalues are important in the study of maps of polynomial vector spaces because they provide valuable information about the behavior of the transformation. They can determine if the transformation stretches or compresses vectors, if it has a fixed point, and if it is invertible. Additionally, eigenvalues can be used to diagonalize a matrix, simplifying calculations and making it easier to study the transformation.

Similar threads

  • Calculus and Beyond Homework Help
Replies
24
Views
806
  • Calculus and Beyond Homework Help
Replies
5
Views
533
  • Calculus and Beyond Homework Help
Replies
9
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
2K
  • Calculus and Beyond Homework Help
2
Replies
58
Views
3K
  • Calculus and Beyond Homework Help
Replies
3
Views
237
  • Calculus and Beyond Homework Help
Replies
5
Views
833
  • Calculus and Beyond Homework Help
Replies
12
Views
981
  • Calculus and Beyond Homework Help
Replies
8
Views
478
  • Calculus and Beyond Homework Help
Replies
2
Views
165
Back
Top