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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.

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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.

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  • $\begingroup$ 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

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