# 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? $$$$\forall x \in \mathbb{R},\;P(x) \succ 0 \implies \inf_{x \in \mathbb{R}} \{\lambda_{{\rm min}}(P(x)) \} > 0$$$$

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.

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