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Let $A$ and $B$ be two real matrices such that $AB$ is positive semi-definite where positive semi-definite is to mean that $x^TMx \geq 0$ for all $x$.

Let $D$ be a diagonal, full rank, square, positive definite, real matrix (i.e has only positive values along the diagonal).

Is $ADB$ positive semi-definite?

If not - are there any conditions on non-square $A$ and $B$ that do make $ADB$ positive semi-definite?

From Theorem 2 in https://cms.math.ca/openaccess/cjm/v15/cjm1963v15.0313-0317.pdf, we know that the product of three positive definite matrices is positive definite if the product is Hermitian. But I wasn't sure if that extended to this use case.

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  • $\begingroup$ Real or complex? $\endgroup$ – user251257 Jan 5 '18 at 2:51
  • $\begingroup$ Real in my case -- but if a theorem exists for the complex case, it would be great to know! $\endgroup$ – Robert Jan 5 '18 at 2:52
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    $\begingroup$ No. Take A, B as diagonal matrices with zero and positive entries and D with negative entries $\endgroup$ – jnyan Jan 5 '18 at 2:57
  • $\begingroup$ Oops -- you are right... I meant to add the condition that D is positive definite! $\endgroup$ – Robert Jan 5 '18 at 2:58
  • $\begingroup$ Does positive definite includes symmetric/hermitian? $\endgroup$ – user251257 Jan 5 '18 at 3:02
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For your first question, the answer is no in general. It should be easy to generate a random counterexample using computer by rejection method, e.g. $$ A=I_2,\ B=\pmatrix{2&0\\ 2&1},\ D=\pmatrix{1&0\\ 0&3} $$ so that $$ AB+(AB)^T=\pmatrix{4&2\\ 2&2},\ (ADB)+(ADB)^T=\pmatrix{4&6\\ 6&6}. $$

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  • $\begingroup$ I might be misunderstanding, and it might not matter, but $D$ in your example isn't diagonal. My feeling is that you are still correct that in general the answer is no - but at the very least the condition where $A = B^T$ satisfies that condition. $\endgroup$ – Robert Jan 5 '18 at 4:55
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    $\begingroup$ @Robert The answer is still no if $D$ is required to be diagonal. See my new edit. $\endgroup$ – user1551 Jan 5 '18 at 5:22
  • $\begingroup$ For posterity - the test @user1551 used here is that for $A$ to satisfy $x^TAx > 0$ for all $x$, $A + A^T$ must be positive definite - in his example $AB + (AB)^T$ is positive definite, but $(ADB) + (ADB)^T$ is not. See math.stackexchange.com/questions/387192/… $\endgroup$ – Robert Jan 5 '18 at 5:41

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