# If the product of $n$ positive definite matrices is symmetric, is it also positive definite?

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?

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.