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As far as I know, the matrix: $$M = \begin{pmatrix}1 & 0 & i\sqrt{3} \\ 0 & 2 & 0 \\ -i\sqrt{3} & 0 & 3 \end{pmatrix} $$

Is hermitian with eigenvalues: $\lambda_1 = 0, \lambda_2 = 2, \lambda_3 = 4$.

And corresponding eigenvectors: $V_{\lambda_1 = 0} = \begin{pmatrix}-i\sqrt{3} \\ 0 \\ 1\end{pmatrix} $, $V_{\lambda_2 = 2} = \begin{pmatrix}0\\ 1 \\ 0\end{pmatrix} $, $V_{\lambda_3 = 4} = \begin{pmatrix}\dfrac{i}{\sqrt{3}}\\ 0 \\ 1\end{pmatrix} $

Since these are eigenvectors of distinct eigenvalues they should be orthogonal but the first and third are not orthogonal.

Why is this?

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    $\begingroup$ Maybe they are hermitian orthogonal. $\endgroup$ – Charlie Frohman Nov 19 '18 at 16:55
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They are orthogonal. \begin{align}V_0^HV_2&=0\\ V_2^HV_4&=0\\ V_0^HV_4&=\begin{pmatrix}i\sqrt3&0&1\end{pmatrix}\cdot\begin{pmatrix}\frac i{\sqrt3}\\ 0\\ 1\end{pmatrix}=0\end{align}

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  • $\begingroup$ I did not check that the original eigenvectors are OK, but there is a sign difference in the top term of your vertical vector. $\endgroup$ – Andrei Nov 19 '18 at 17:00
  • $\begingroup$ @Andrei Like the one you do when you take the Hermitian of something? $\endgroup$ – Saucy O'Path Nov 19 '18 at 17:01
  • $\begingroup$ Ah! Thank you so much $\endgroup$ – Rzmwood Nov 19 '18 at 17:02
  • $\begingroup$ @SaucyO'Path thanks. I somehow missed that. $\endgroup$ – Andrei Nov 19 '18 at 17:07

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