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Orthogonal Diagonalization

  1. Mar 3, 2008 #1
    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. jcsd
  3. Mar 3, 2008 #2

    Dick

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    Science Advisor
    Homework Helper

    To start out with, the eigenvalues aren't 2,2,2. Try that again. What are they? Next you have to find the eigenvectors.
     
  4. Mar 4, 2008 #3
    | (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?
     
  5. Mar 4, 2008 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Well, you don't "solve" a polynomial, you solve an equation. What you mean, of course is solve [itex]-\lambda^3 + 6\lambda^2 - 9\lambda + 4= 0[/itex]

    By the "rational root theorem", the only rational roots, if it has any, must be [itex]\pm 1[/itex], [itex]\pm 2[/itex], [itex]\pm 4[/itex], the integer factors of 4. Try those. It is always a good idea to try the simple things first!
     
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