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

Question

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?

Answer 1

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.