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I am new to linear algebra and have a problem, suppose a covariance matrix is given by $\Sigma$, which is invertible and can be written as $$\Sigma=U \Lambda U^*$$ where $U$ is a rotation matrix and $\Lambda$ is a diagonal matrix. I know that covariance matrix transpose is equal to the matrix itself, but does the same hold for conjugate transpose in case of invertible covariance matrix?

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  • $\begingroup$ Since $U$ is the rotation matrix, it is unitary. So the answer is yes $\endgroup$ – polfosol Sep 27 '16 at 12:30
  • $\begingroup$ if I am right than doing conjugate transpose will give $$(K_z)^*=(U diag U^*)^*$$ which gives $$U^* diag^* U$$ $$diag^*=diag $$ so you mean to say that $$U^*=U$$ then why we need to write $$K_z=(U diag U^*) and not K_z=(U diag U)$$ $\endgroup$ – Userhanu Sep 27 '16 at 12:47
  • $\begingroup$ First, please don't use this notation. Write $K_z=U\Sigma U^*$ which is nicer and more readable (upon saying that $\Sigma$ is diagonal). Second, it is sufficient for $\Sigma$ to be real, which is true for a positive definite matrix. Third, take a look at this $\endgroup$ – polfosol Sep 27 '16 at 13:06
  • $\begingroup$ But there U and V are two different rotation matrix, and here we have both same. Sorry but I still can't get you. $\endgroup$ – Userhanu Sep 27 '16 at 13:15
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    $\begingroup$ Covariance matrices have real entries. Conjugation does not matter. $\endgroup$ – Wintermute Sep 27 '16 at 14:50
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for arbitrary matrices with appropriate dimensions, $$ (ABC)^\star = C^\star B^\star A^\star $$

(I added this because you made the mistake in the comment and seemingly used $(ABC)^\star = A^\star B^\star C^\star$ which is incorrect)

therefore in particular:

$$ \begin{eqnarray} (U\Lambda U^\star)^\star &=& (U^\star)^\star \Lambda^\star U^\star &=& U\Lambda U^\star \end{eqnarray}$$

since the singular values are real and thus $\Lambda=\Lambda^\star$.

Note: as was commented, a covariance matrix has real entries so the hermitian transpose is actually just simply the transpose and you're back on the result that you knew from before.

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