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}.

  • $\begingroup$ No. Just take the $1x1$ matrices $P(x) = 1/sqrt(1+x^2)$ over $\Omega=\mathbb R$. $\endgroup$ – daw Dec 7 '18 at 13:04
  • $\begingroup$ Or for $n=2$, $P(x) =diag( sqrt(1+x^2),1/sqrt(1+x^2))$. $\endgroup$ – daw Dec 7 '18 at 13:05

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.


Originally, the text-part of the question asked whether we knew that, over $\Omega$, the determinant of $P$ is always positive. This is different from the case that the infimum of the determinant is positive (the infimum condition implies that the determinant is always bounded away from zero).

For the edited question:

It is enough for $\Omega$ to be compact and $P$ to be continuous. Suppose that $\Omega$ has the desired property and the infimum of the smallest eigenvalue goes to zero. In this case, there exists a sequence $\{x_i\}$ so that $P(x_i)\in\Omega$ for all $i$ and $\lambda_{\min}(P(x_i))\rightarrow 0$.

Since the determinant is bounded from below and the determinant is the product of the eigenvalues, it must be that the largest eigenvalue of $P(x_i)$ is growing (without bound). Since the magnitudes of the eigenvalues are bounded in terms of the maximum of the entries of the matrix, for an eigenvalue to grow without bound, the maximum of the entries of the matrix must also be growing.

We can find a subsequence $\{x_{i_j}\}$ so that in one particular entry of the matrix, the magnitude of that entry is growing without bound. By sequential compactness, there is some subsequence that converges in $\Omega$. However, since the magnitude of the entry is growing without bound, it follows that the corresponding entry of the limit must be $\pm\infty$, which is absurd.

  • 2
    $\begingroup$ You also need compactness. $\endgroup$ – daw Dec 7 '18 at 13:04
  • $\begingroup$ Thanks. My question has been edited now. $\endgroup$ – winston Dec 7 '18 at 13:53
  • $\begingroup$ Can you re-edit your answer? I do not quite understand what you are talking about, because your conclusion is not clear, though the description is easier to understand. $\endgroup$ – winston Dec 7 '18 at 13:56
  • $\begingroup$ I've edited the answer with more details. $\endgroup$ – Michael Burr Dec 7 '18 at 17:25

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