Diagonalization of a matrix with repeated eigenvalues

Click For Summary

Homework Help Overview

The discussion revolves around the diagonalization of matrices, specifically focusing on matrices with repeated eigenvalues. Participants are exploring the conditions and methods for diagonalization in this context.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to understand the process of diagonalizing a matrix with repeated eigenvalues and questions whether identical eigenvectors are involved. Others suggest trying specific examples, such as the identity matrix, to clarify the concept of eigenvalues and their multiplicities.

Discussion Status

The discussion is ongoing, with participants raising questions about the nature of eigenvalues and the implications of multiplicity. Some guidance has been offered regarding the relationship between eigenvalues and the determinant of the matrix, but no consensus has been reached on the overall approach to diagonalization.

Contextual Notes

There appears to be some confusion regarding the calculation of eigenvalues, particularly in relation to the identity matrix and its eigenvalue multiplicity. Participants are navigating the foundational concepts of eigenvalues and their significance in diagonalization.

bemigh
Messages
29
Reaction score
0
Hey guys,
I know its possible to diagonalize a matrix that has repeated eigenvalues, but how is it done? Do you simply just have two identical eigenvectors??
Cheers
Brent
 
Physics news on Phys.org
Well, try an example -- the simplest you can think of. How about the identity matrix? Can you diagonalize it? What are its eigenvectors?
 
Ok, well... i would get an eigenvalue of 1, with a multiplicity of 3. Alright, so when solving for my eigenvalues, by plugging 1 into my matrix A-(lamda)I ; i would just be left with 0's. How can i get eigenvalues from this?? I appreciate your help by the way...
 
Am I misunderstanding something? If you know what eigenvalues are, then you must know how to find them in this simple case. If \lambda is an eigenvalue of matrix A, then A-\lambda I is NOT invertible (since the equation A- \lambda I= 0 does not have a unique solution) and so det(A- \lambda I)= 0. Since you are finding the eigenvalues, you do NOT yet know that \lambda= 1 so you are NOT "left with 0's. In this case, that equation is (1- \lamba)^3= 0 which obviously has 1 as a triple root.
 

Similar threads

  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 4 ·
Replies
4
Views
1K
Replies
8
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
4
Views
2K