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A well-known result for the product of two PD (or PSD) matrices is:

Proposition: Let $A$ and $B$ be positive definite (respectively positive semidefinite) Hermitian matrices of the same size. If $D:=AB$ is Hermitian, then $D$ is also positive definite (respectively positive semidefinite).

This can be extended to the product of three PD (or PSD) matrices: Is the product of three positive semidefinite matrices positive semidefinite

Proposition: Let $A$, $B$, and $C$ be positive definite (respectively positive semidefinite) Hermitian matrices of the same size. If $D:=ABC$ is Hermitian, then $D$ is also positive definite (respectively positive semidefinite).

Is there any extension of this to the product of $n$ PD (or PSD) matrices?

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No. Here is a random counterexample. Let $$ P=\pmatrix{3&6&8\\ 8&8&4\\ 2&4&5}, \quad\Lambda=\pmatrix{4\\ &8\\ &&12}, \quad D=\pmatrix{1\\ &-1\\ &&-1} $$ and $$ A=P\Lambda P^{-1} =\pmatrix{36&-3&-36\\ -64&0&112\\ 16&-2&-12}. $$ One may use a computer to verify that $AD$ has three different positive eigenvalues $4$ and $2(11\pm\sqrt{97})$. Since every diagonalisable real square matrix with a positive spectrum is a product of two positive definite matrices, we have $A^{-1}=S_1S_2$ and $AD=S_3S_4$ for some $S_1,S_2,S_3,S_4\succ0$. Now $S_1S_2S_3S_4=D$, which is indefinite.

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