Let $A$ and $B$ be two covariance matrices such that $AB=BA$. Is $AB$ a covariance matrix?
A covariance matrix must be symmetric and positive semi definite. The symmetry of $AB$ can be proved as follows: $$(AB)^T = B^TA^T = BA = AB$$
The question is, how to prove or disprove the positive semi definitive character of $AB$?
$\endgroup$ 03 Answers
$\begingroup$Two commuting matrices can be diagonalized by the same matrix. The positive semi definite follows immediately.
$\endgroup$ $\begingroup$Expanding the word "immediately" of Quang Hoang's answer:
$$AB = QD_AQ'QD_BQ' = QD_A D_B Q', $$
where $D_A$ and $D_B$ are the respective diagonal matrices of $A$ and $B$. The diagonal entries of these matrices are nonnegative, so are the diagonal entries of the product of $D_A D_B$.
$\endgroup$ $\begingroup$Your question about "the positive semi definitive character of $AB$" can be answered as follows: $A\cdot B$ is positive semi definite if and only if $A \cdot B$ is normal (i.e. $(A\cdot B)^T\cdot (A\cdot B) = (A\cdot B)\cdot (A \cdot B)^T$). Reference: Meenakshi, A. and C. Rajian (1999). On a product of positive semidefinite matrices. Linear Algebra and its Applications 295 (1-3), 3–6.
Using ist symmetry $(A \cdot B)^T = (A\cdot B)$ we can in fact show that $A \cdot B$ is normal: $(A \cdot B)^T \cdot (A \cdot B) = (A \cdot B) \cdot (A \cdot B) = (A \cdot B) \cdot (A \cdot B)^T$. So $A B$ is symmetric and positive semi definite and thus a covariance Matrix.
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