1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Diagonalisation of matrix

  1. Mar 23, 2010 #1
    1. The problem statement, all variables and given/known data
    This is part of a larger problem of finding [tex] e^A [/tex], where [tex]


    A = \left[ \begin{array}{ccc} 0 & 0 & -2 \\ 1 & 2 & 1 \\ 1 & 0 & 3 \end{array} \right]

    [/tex]


    2. Relevant equations
    -


    3. The attempt at a solution


    I generally have no problems with diagonalising a matrix, however, in this case I just can't seem to get it to work. I've got the eigenvectors to be [tex] (-1, 0, 1)^T, (0, 1, 0)^T, (-2, 1, 1)^T [/tex]

    Once carried through the steps of finding [tex] P, A, P^{-2} [/tex] and finding the diagonal, I come out with

    [tex] D = \left[ \begin{array}{ccc} 2 & 0 & 0 \\ 0 & 2 & 2 \\ 0 & 0 & 1 \end{array} \right] [/tex]

    I've checked it on wolframalpha and on maple and both confirm that my inverse is correct and following through the multiplication both come out with the same result. Chances are I've just made a silly mistake somewhere. Can someone help me spot it?

    Thankyou
     
  2. jcsd
  3. Mar 23, 2010 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    You mean you did D=P^(-1)AP where P is the matrix whose columns are the eigenvectors, right? I did the same thing and got D=diag(2,2,1). What did you get for P^(-1)?
     
  4. Mar 23, 2010 #3
    Sorry, I did mean [tex] P^{-1} [/tex]

    Which I had as:

    [tex]

    \left[ \begin{array}{ccc} 1 & 0 & 2 \\ 1 & 1 & 1 \\ -1 & 0 & -1 \end{array} \right] [/tex]
     
  5. Mar 23, 2010 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Looks right.

    With P:
    [tex]


    \left[ \begin{array}{ccc} -1 & 0 & -2 \\ 0 & 1 & 1 \\ 1 & 0 & -1 \end{array} \right]
    [/tex]

    I get P^(-1)AP=diag(2,2,1). I do notice that if I do PAP^(-1) I get your off diagonal result D. That's not the right order.
     
    Last edited: Mar 23, 2010
  6. Mar 23, 2010 #5
    Oh, yes, it was down to wrong order. Very silly!

    Thanks for clearing that up
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook