1. Nov 27, 2011 #1

   thoughtgaze

   

   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?

    

   thoughtgaze, Nov 27, 2011
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3. Dec 3, 2011 #2

   jshtok

   

   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.

    

   jshtok, Dec 3, 2011

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