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?

  • $\begingroup$ 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
  • 1
    $\begingroup$ Covariance matrices have real entries. Conjugation does not matter. $\endgroup$ – 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)

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.