# About positive-definiteness of a matrix Q depending on matrices H and P

$H + H^T$ 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. –  Qiaochu Yuan Aug 9 '11 at 3:18
@Qiaochu: If you mean nobody's posted a math question on meta, it happened once before: math.stackexchange.com/questions/33394/… –  Jonas Meyer Aug 9 '11 at 3:46
@Jonas: my apologies. I should have been more precise: I've never seen anyone do that before! –  Qiaochu Yuan Aug 9 '11 at 3:48
Fatima: What calculations, and what paper? –  Jonas Meyer Aug 9 '11 at 4:01
@Jonas: paper is "A new approach to the LQ design from the viewpoint of the inverse regulator problem" by Takao Fujii –  Fatima Tahir Aug 9 '11 at 4:43
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## migrated from meta.math.stackexchange.comAug 9 '11 at 3:17

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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. –  Patrick Da Silva Aug 10 '11 at 12:53
Yes @Patrick, I just randomly generated some unsymmetric matrix $H$ and perturbed it so that $H+H^T$ is symmetric positive definite... –  Ｊ. Ｍ. Aug 11 '11 at 1:40
Oh =) Okay cool –  Patrick Da Silva Aug 11 '11 at 2:21
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 $P=\begin{bmatrix}1&0\\0&5\end{bmatrix}$.