Orthogonal Diagonalization of a Symmetric Matrix

AI Thread Summary
The discussion focuses on the orthogonal diagonalization of a symmetric matrix with repeated eigenvalues. The matrix in question has eigenvalues of lambda = 2, but the participants clarify that these need to be recalculated correctly. The process involves finding eigenvectors after determining the correct eigenvalues, with a determinant equation provided for solving. The rational root theorem is suggested as a method to find potential rational roots for the polynomial derived from the determinant. The conversation emphasizes the importance of accurately identifying eigenvalues and eigenvectors for successful diagonalization.
aznkid310
Messages
106
Reaction score
1

Homework Statement



Orthogonally diagonalize the matrix:
| 2 1 1|
| 1 2 1|
| 1 1 2|


Homework Equations



Since this only has three of the same eigenvalues ( lambda = 2), how do i use
A = PD(P^t)? What is P?

After row reduction, I got x = y = z = 1. This would give me the first column of P, but what about the other two columns?


The Attempt at a Solution



This is a symmetric matrix and the eigen values are lambda = 2,2,2

solving (2I - A)x = 0 i get | 0 1 1 |
| 1 0 1 |
| 1 1 0 |

After row reduction: | 1 0 0 |
| 0 1 0 |
| 0 0 1 |

Which means x = y = z = 1?
 
Physics news on Phys.org
To start out with, the eigenvalues aren't 2,2,2. Try that again. What are they? Next you have to find the eigenvectors.
 
| (lambda - 2) 1 1 |
| 1 (2-lambda) 1 |
| 1 (2-lambda) 1 |

Taking the determinant, i get: -L^3 + 6L^2 - 9L + 4

How do i solve this?
 
Well, you don't "solve" a polynomial, you solve an equation. What you mean, of course is solve -\lambda^3 + 6\lambda^2 - 9\lambda + 4= 0

By the "rational root theorem", the only rational roots, if it has any, must be \pm 1, \pm 2, \pm 4, the integer factors of 4. Try those. It is always a good idea to try the simple things first!
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...

Similar threads

Back
Top