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Congruence transformation

  1. Apr 21, 2004 #1
    i was reading goldstein (oscillations)

    in it, it is said that

    Va = (lambda)*Ta

    where V is the potential energy matrix, T is the kinetic eneregy matrix, lambda is an eigenvalue and a is the corresponding eigenvector(of displacement from equilibrium)

    and it is said that the matrix of eigenvectors A diagonalises the matrix V through a congruence transformation A'VA = L
    where L is the diagonal matrix with its diagonal elements as eigenvalues. and A' is the transpose of A. this equation is solved by taking
    |V-LI| = 0

    here I is the identity matrix and 0 is the zero matrix.
    but this secular equation is allowed only when the matrix A diagonalises V through a similarity transformation, isn't it?

    it is the equation
    Va = (lambda)*a

    which will yield
    |V-LI| = 0 and

    not the equation Va = (lambda)*Ta isn't it?
    Last edited: Apr 21, 2004
  2. jcsd
  3. Apr 21, 2004 #2


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    Science Advisor

    Yes, that's true. IF a matrix has a "complete set of eigenvectors" (there exist a basis for the vetor space consisting of eigenvectors of the matrix), THEN the matrix A, having those eigenvectors as columns, diagonalizes V: A-1VA= L. Notice that I used A-1 rather than A'. Of course, if we can construct an orthonomal basis of eigenvectors, then A will be an orthogonal matrix and A-1= A'.
  4. Apr 22, 2004 #3
    it's true that A diagonalises V .

    but my question is:

    since a'Ta = 1 , and not a'a = 1 ,( so that a is not orthogonal)
    isn't A' not equal to (A inverse)??

    ......by the way, how did you write (A inverse )...the mathematical way?!
  5. Apr 23, 2004 #4
    can somebody help me? i have no one to discuss things with as i am studying physics on my own...that's why i put up the question here
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