# Does the conjugate transpose of invertible covariance matrix is the matrix itself?

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?

• Since $U$ is the rotation matrix, it is unitary. So the answer is yes – polfosol Sep 27 '16 at 12:30
• 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 $$K_z=(U diag U^*) and not K_z=(U diag U)$$ – Userhanu Sep 27 '16 at 12:47
• 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 – polfosol Sep 27 '16 at 13:06
• But there U and V are two different rotation matrix, and here we have both same. Sorry but I still can't get you. – Userhanu Sep 27 '16 at 13:15
• Covariance matrices have real entries. Conjugation does not matter. – Wintermute Sep 27 '16 at 14:50

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)
$$\begin{eqnarray} (U\Lambda U^\star)^\star &=& (U^\star)^\star \Lambda^\star U^\star &=& U\Lambda U^\star \end{eqnarray}$$
since the singular values are real and thus $\Lambda=\Lambda^\star$.