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$H + H^T$ is a positive definite matrix and $P$ is also a positive definite matrix.

Will $Q = PH + H^TP$ be a positive definite matrix?

In my calculations, it is not positive definite. But I read a paper saying that $Q$ should be positive definite. Is it so?

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    Nobody's ever done that before! @Fatima, the place you just posted this question is the _meta_ site, which is for discussion of the main site, not for discussion of mathematics.2011-08-09
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    @Qiaochu: If you mean nobody's posted a math question on meta, it happened once before: http://math.stackexchange.com/questions/33394/prime-numbers-which-solve-2s-1mod-p2011-08-09
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    @Jonas: my apologies. I should have been more precise: I've never seen anyone do that before!2011-08-09
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    Fatima: What calculations, and what paper?2011-08-09
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    @Jonas: paper is "A new approach to the LQ design from the viewpoint of the inverse regulator problem" by Takao Fujii2011-08-09
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    @Qiachu: I am extremely sorry if I posted my question in wrong place. I am not familiar of this system. This is my second question.2011-08-09

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You refer to your calculations; does that mean you already have a counterexample? My calculations seem to agree with yours, as seen in the example $H=\begin{bmatrix}1&0\\1&1\end{bmatrix}$ and $P=\begin{bmatrix}1&0\\0&5\end{bmatrix}$.

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    yes, I wanted to use the results of the mentioned paper for my work but I came across matrices which do not satisfy the results mentioned in the paper.2011-08-09
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    Fatima, thanks for giving the name of the article, but I do not know precisely what is claimed in the article, and I do not have access to it at the moment. Perhaps there are additional hypotheses there. If you would like to ask further questions on the particular claim made in the article, it would help to provide a little more context, and perhaps an excerpt for those like me who cannot view it.2011-08-09
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If

$$\mathbf H=\begin{pmatrix}15&9&7\cr-1&9&-8\cr-3&-9&11\cr\end{pmatrix}$$

and

$$\mathbf P=\begin{pmatrix}81&-5&30\cr-5&75&-54\cr30&-54&54\cr\end{pmatrix}$$

then $\mathbf Q$ isn't positive definite, having two positive and one negative eigenvalues. It should be easy to generate other counterexamples...

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    How have you generated this one? The numbers seem rather arbitrary.2011-08-10
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    Yes @Patrick, I just randomly generated some unsymmetric matrix $H$ and perturbed it so that $H+H^T$ is symmetric positive definite...2011-08-11
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    Oh =) Okay cool2011-08-11