I feel like self-adjoint / Hermitian operators are the "best" operators, since an operator that is self-adjoint can be orthogonally diagonalized, according to the Spectral Theorem (over the complex number field, an operator only needs to be normal to be orthogonally diagonalizable).

I know that self-adjoint implies normal, but does self-adjoint also imply orthogonal / unitary: $AA^T = A^TA = I$?


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    $\begingroup$ No. Consider $\begin{bmatrix} 2 & 0 \\ 0 & 1\end{bmatrix}$. This matrix is self-adjoint but not orthogonal. It is also untrue that Hermitian matrices or necessarily unitary. $\endgroup$ – user137731 Dec 2 '14 at 3:31
  • $\begingroup$ Ok, thanks so much, @Bye_World :) $\endgroup$ – User001 Dec 2 '14 at 4:06

Definitely not. A self-adjoint operator just need $A^T=A$. Note the matrix of $T^*$ is the transposed of the matrix of $T$.

  • $\begingroup$ Got it - thanks@user193702. $\endgroup$ – User001 Dec 2 '14 at 4:06

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