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Homework Help: 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"

    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


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

    and i think

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

    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. "

    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?

  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


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    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!
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