# 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

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– 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

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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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## 2 Answers

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. – Fatima Tahir Aug 9 '11 at 5:54
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. – Jonas Meyer Aug 9 '11 at 6:02

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
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