Diagonalize operator A by matrix S

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To diagonalize operator A using matrix S, it is not necessary for S to be formed from normalized eigenvectors. While S can be constructed from non-normalized eigenvectors, this will still yield a diagonal matrix. However, the diagonal elements of this matrix will not correspond to the eigenvalues of A if S is not normalized. Normalization of eigenvectors ensures that the diagonal elements accurately represent the eigenvalues. Thus, for correct diagonalization, using normalized eigenvectors is recommended.
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Suppose, i want to diagonalize operator A by matrix S (A'= S^\\dagger A S). Do i need to form S from "NORMALIZED" eigenvectors? I checked and found that even S formed from not-normalized eigenvectors works.
 
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If S isn't normalized, you will get a diagonal matrix. However, the diagonal elements of the matrix will not be the eigenvalues of A.
 

Thread 'Two equivalent statements of time reversal symmetric Hamiltonian'

Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?
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