# Positive definite matrix implies the **infimum** of eigenvalues are positive?

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.

## 1 Answer

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

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