# Orthogonal Diagonalization

1. Mar 3, 2008

### aznkid310

1. The problem statement, all variables and given/known data

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

2. Relevant 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?

3. 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?

2. Mar 3, 2008

### Dick

To start out with, the eigenvalues aren't 2,2,2. Try that again. What are they? Next you have to find the eigenvectors.

3. Mar 4, 2008

### aznkid310

| (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?

4. Mar 4, 2008

### HallsofIvy

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!