Suppose we have two covariance matrices $A$ and $B$. They satisfy the condition $AB=BA$. Is $AB$ a covariance matrix?

My answers:

We can easily check that $(AB)'=B'A'=BA$, then $AB$ is symmetric. But I have no idea how to check it is positive semi-definite. I can't come up with an example showing it isn't a covariance matrix, either.

Any help would be appreciated.


Since $AB=BA$, then $A$ and $B$ can be simultaneously diagonalized by some matrix $U$. Hence it follows \begin{align} AB = UD_1U^{-1}UD_2U^{-1} = UD_1D_2U^{-1}. \end{align} Thus, the eigenvalues of $AB$ are product of eigenvalues of $A$ and $B$. Thus, it follows $AB$ is also positive semi-definite since the eigenvalues are nonnegative.

  • $\begingroup$ Product of the corresponding eigenvalues. $\endgroup$ – Batman Sep 25 '16 at 1:26
  • $\begingroup$ @Batman Ops. I have corrected the statement. Thanks. $\endgroup$ – Jacky Chong Sep 25 '16 at 1:27
  • $\begingroup$ @JackyChong, I can't understand why "Since AB=BA, then A and B can be simultaneously diagonalized by some matrix U". Could you elaborate it a little more in detail? Thanks again! $\endgroup$ – Pandaaaaaaa Sep 25 '16 at 2:28
  • $\begingroup$ $AB=(U_A \Lambda_A U_A^{-1}) (U_B \Lambda_B U_B^{-1}) = (U_B \Lambda_B U_B^{-1}) (U_A \Lambda_A U_A^{-1}) = BA$, I can't figure out what is the $U$ as mentioned in the answer, thanks again! $\endgroup$ – Pandaaaaaaa Sep 25 '16 at 2:29
  • $\begingroup$ @Pandaaaaaaa Commuting matrices have the same eigenbasis. $\endgroup$ – Jacky Chong Sep 25 '16 at 2:48

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.