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Suppose $P(x): \mathbb{R} \to \mathbb{R}^{n \times n}$ is always a positive definite matrix, does it imply that the infimum (over $\mathbb{R}$) of the minimum eigenvalue of $P(x)$ is always positive?, that is, $\inf_{x \in \mathbb{R}} \{\lambda_{{\rm min}}(P(x)) \} > 0$ In a mathematical way:

Is the following conclusion correct? \begin{equation} \forall x \in \mathbb{R},\;P(x) \succ 0 \implies \inf_{x \in \mathbb{R}} \{\lambda_{{\rm min}}(P(x)) \} > 0 \end{equation}

Please be careful with the infimum.

Please go to the other similar question here, which is exactly the real (and more difficult in my opinion) question that I want to ask.

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No. Consider $n=1$ and $P(x)=e^x$ or in general, $P(x)=e^{xI_n}$.

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  • $\begingroup$ I think it should be $P(x)=e^{-x}$. $\endgroup$ – winston Dec 7 '18 at 10:59
  • $\begingroup$ @winston $e^x$ is OK. It is always positive and $e^x\to0$ when $x\to-\infty$. $\endgroup$ – user1551 Dec 7 '18 at 11:00
  • $\begingroup$ Yes,you are right. $\endgroup$ – winston Dec 7 '18 at 11:01
  • $\begingroup$ Actually my intended question is more difficult. Can you go to this math.stackexchange.com/questions/3029778/… $\endgroup$ – winston Dec 7 '18 at 11:07

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