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Congruence transformation

  1. Nov 27, 2011 #1
    If A diagonalizes V under a congruence transformation (instead of a similarity transformation), are the diagonal elements, the eigenvalues of V?

    If A is orthogonal, it is obvious that this is true, but what if A is not orthogonal?
     
  2. jcsd
  3. Dec 3, 2011 #2
    First glance:

    There are some conditions on A to allow diagonal of D=A'VA be composed of eigenvalues of V. If this is the case, V and D have the same determinant and trace (first is a product of eig.values and the second is the sum). Same determinant implies
    det(A'VA) = det(V),
    which means det(A) =1 or det(A)=-1. Same trace implies
    tr(V)=tr(A'VA)=tr(AA'V).
    I don't immediately see what does this condition mean for A.
     
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