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A matrix is positive semidefinite (PSD) if it's symmetric and all its eigenvalues are non-negative. There are many other equivalent definitions.

Suppose $A$, $B$ are two PSD matrices. $AB$ is PSD if and only if $AB$ is symmetric. This is well-known.

This can be extended to the product of three PSD matrices: Is the product of $3$ positive semidefinite matrices positive semidefinite?

However, it fails for products of four matrices and greater: If the product of $n$ positive definite matrices is symmetric, is it also positive definite?

Suppose I have $n$ PSD matrices $A_1, A_2, \dots, A_n$ and suppose the product $A_1 A_2 \dotsb A_n$ is symmetric. Is there any non-trivial condition on the $A_i$ or on the product that guarantees that the product is PSD?

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  • $\begingroup$ Related $\endgroup$ May 23, 2023 at 6:28
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    $\begingroup$ The tag numerical-linear-algebra should only be used to talk about numerical algorithms related to linear algebra, in contexts where those algorithms are actually being used or derived. The text of your question does not make it at all clear that this is the goal, hence the tag is not really appropriate here. $\endgroup$
    – Xander Henderson
    May 27, 2023 at 15:14
  • $\begingroup$ @XanderHenderson That is fine. I have no issue. Though these algorithms can often be used to show matrices are PSD so in that sense I thought it was worth the tag. $\endgroup$
    – Dan1618
    May 29, 2023 at 1:14

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One sufficient condition (along with the condition that $A_1A_2\cdots A_n$ is Hermitian), generalised from a proof for the case $n=3$, is that $B=A_2A_3\cdots A_n$ is a diagonalisable matrix over $\mathbb C$ with a real nonnegative spectrum. This condition is also necessary when $A_1$ is positive definite.

Given this sufficient condition, $B$ has a matrix square root that can be expressed as $g(B)$ for some polynomial $g$. Now, since $A_1A_2\cdots A_n$ is supposed to be Hermitian, we have $A_1B=B^\ast A_1$. So, inductively, we have $A_1B^k=(B^k)^\ast A_1$ for all nonnegative integer $k$. Hence $A_1g(B)=g(B)^\ast A_1$. Consequently, $A_1A_2\cdots A_n=A_1B=A_1g(B)^2=g(B)^\ast A_1g(B)$ is PSD.

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