1. Not finding help here? Sign up for a free 30min 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!

Matrix Diagonalisation eigen?

  1. Sep 7, 2011 #1
    1. The problem statement, all variables and given/known data

    i have matrix A which is diagonalisable

    by doing an example on wiki under the section " how do diagonalise a matrix"
    http://en.wikipedia.org/wiki/Diagonalizable_matrix

    i realise that

    A x is not equal to [itex]\lambda[/itex] x , where x are eigen vectors of A , [itex]\lambda[/itex] is eigen values

    instead

    x A = [itex]\lambda[/itex] x

    and i think

    A x = x [itex]\lambda[/itex]

    QUESTION 1)
    if this is so, why do they always write the hamiltonian as H |x> = E |x> ??? shouldn't it be H |x> = |x> E ?

    if i remembered correctly, for matrix multiplication AB =/= BA right?

    but i read wiki and it says something like (i can't remember the exact phrasing)

    "it is equal if both A and B are diagonalisable matrix , and are both n by n matrix. "


    QUESTION 2)
    also, for PT A P = [itex]\lambda[/itex]

    if i want to "bring over" the P, is it like this

    A = P [itex]\lambda[/itex] PT

    but why is it like this?



    thanks!
     
  2. jcsd
  3. Sep 7, 2011 #2
    You are right that matrices don't commute, however the multiplication of a matrix with a scalar (real or complex number) always commutes. So λx = xλ. What you write above is also not correct. Ax is not at all the same as xA, the wiki page is correct.

    Question 2):

    this works for so called orthogonal matrices where P^T * P = P*P^T = 1 (the identity matrix). It is not true for general matrices.
     
  4. Sep 7, 2011 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    You do not have to choose orthonormal vectors for eigenvalues but you can. If you do then the matrix P is "orthogonal"- in particular [itex]P^T= P^{-1}[/itex]. Without P being orthogonal, that is not true but it is still true that [itex]P^{-1}AP= D[/itex] so, multiplying on the left by P and on the right by [itex]P^{-1}[/itex],
    [tex]P(P^{-1}AP)P^{-1}= PDP^{-1}[/tex]
    [tex](PP^{-1})A(PP^{-1})= IAI= A= PDP^{-1}[/tex]

    In the case that P is "orthogonal", You can replace [itex]P^{-1}[/itex] with [itex]P^T[/itex].
     
  5. Sep 7, 2011 #4
    oh shucks ... for question 1 i realise i multiplied the P ,matrix of eigen vectors, with the diagonal matrix eigen values

    but in fact i think they are talking about the individual eigen value and eigen functions , not the combined P.

    so yup i get it thanks everyolne!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Matrix Diagonalisation eigen?
  1. Diagonalisable matrix (Replies: 4)

Loading...