I asked a similar question in here, but actually what I want to ask is more difficult as described below:
Suppose $P(x): \mathbb{R} \to \mathbb{R}^{n \times n}$ is always a positive semi-definite matrix. Now if there is a set $\Omega \subset \mathbb{R}$ such that we know the infimum of the determinant of $P(x)$ over $\Omega$ is always positive, then does it imply that the infimum (over $\Omega$) of the minimum eigenvalue of $P(x)$ is always positive? In a mathematical way:
Is the following conclusion correct?
\begin{equation}
\inf_{x \in \Omega}\{\det(P(x))\}>0 \implies \inf_{x \in \Omega} \{\lambda_{{\rm min}}(P(x)) \} > 0
\end{equation}.
Best Answer
No. Consider e.g. $P(x)=\operatorname{diag}(x,\frac1x)$ over $\Omega=[1,+\infty)$.
It is true, however, that if $\Omega$ is compact, $P$ is continuous and $P(x)$ is positive definite over $\Omega$, then $\inf_{x\in\Omega}\lambda_\min(P(x))>0$. This is because the eigenvalues of a matrix vary continuously with the matrix's entries and every continuous function attains its minimum on a compact set.