What are the eigenvalues of P(A)?

  • Thread starter Thread starter chocolatefrog
  • Start date Start date
  • Tags Tags
    Eigenvalue Matrix
Click For Summary
SUMMARY

The discussion focuses on proving that the polynomial expression αo + α1λ + α2λ2 + ... + αdλd is an eigenvalue of the matrix αoI + α1A + α2A2 + ... + αdAd, where A is an n x n matrix and λ is an eigenvalue of A. The key argument involves manipulating determinants and using properties of eigenvalues, specifically showing that the polynomial P(A) - μI is invertible if and only if all (A - r_iI) are invertible. This establishes a direct relationship between the eigenvalues of the polynomial and the eigenvalues of the matrix A.

PREREQUISITES
  • Understanding of eigenvalues and eigenvectors in linear algebra.
  • Familiarity with polynomial functions and their properties.
  • Knowledge of determinants and matrix invertibility.
  • Basic concepts of matrix algebra, specifically for n x n matrices.
NEXT STEPS
  • Study the properties of eigenvalues and eigenvectors in depth.
  • Learn about the Cayley-Hamilton theorem and its applications.
  • Explore the relationship between polynomials and matrices, particularly in the context of matrix functions.
  • Investigate determinant properties and their implications for matrix invertibility.
USEFUL FOR

Students and professionals in mathematics, particularly those studying linear algebra, matrix theory, and eigenvalue problems. This discussion is beneficial for anyone looking to deepen their understanding of polynomial matrices and their eigenvalues.

chocolatefrog
Messages
11
Reaction score
0

Homework Statement



αo, α1,..., αd [itex]\inℝ[/itex]. Show that αo + α1λ + α2λ2 + ... + αdλd [itex]\inℝ[/itex] is an eigenvalue of αoI + α1A + α2A2 + ... + αdAd [itex]\inℝ^{nxn}[/itex].


2. The attempt at a solution

If λ is an eigenvalue of A, then |A - Iλ| = 0. Also, λn is an eigenvalue An. So we basically have to somehow prove the following equation (after rearranging):

1(A - Iλ) + α2(A2 - Iλ2) + ... + αd(Ad - Iλd)| = O

3. Relevant equations

I can't seem to get my head around this one. I almost used the triangular inequality to prove it before I realized that these are determinants we are dealing with, not absolute values. :/
 
Physics news on Phys.org
So let [itex]P(z)[/itex] be a polynomial. We wish to prove that [itex]P(\lambda)[/itex] is an eigenvalue of [itex]P(A)[/itex].

For each [itex]\mu\in \mathbb{C}[/itex], we can write

[tex]P(z)-\mu=r_0(z-r_1)...(z-r_n)[/tex]

Thus

[tex]P(A)-\mu I = r_0(A-r_1 I)...(z-r_n I)[/tex]

So [itex]P(A)-\mu I[/itex] is invertible iff all [itex](A-r_i I)[/itex] is invertible. What does that imply for the eigenvalues?
 

Similar threads

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