I searched through literature but could not find any related topic for the question below. I hope some of you may be able to point me to the right direction.

Let $X: \mathbb{R}\rightarrow\mathbb{R}^{n\times n}$ be a family of $n\times n$ matrices $X(t)$ indexed by $t\in\mathbb{R}$. Assume that we know the determintant of $X(t)$ for all $t$. Is there a way to study the spectrum of matrix $X(0)$ at $t=0$ through the determinant function $\text{det}:t\mapsto\text{det}(X(t))$?

When $X(t) = A - tI$ for some matrix $A$ and identity matrix $I$ then $\text{det}(X(t))$ is the characteristic polynomial of $A$. Therefore knowing $\text{det}(X(t))$ means that we know its zeros or spectrum of $A=X(0)$. Thus, by studying $\text{det}(X(t))$ we can completely characterize the spectrum of $A$. For more general $X(t)$, under what conditions can we compute or infer about eigenvalues of $X(0)$ by using $\text{det}(X(t))$?

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    $\begingroup$ All you know is the value $\lambda_1(t) \cdots \lambda_n(t)$, so without extra information it is hard to see how one can derive any additional information? $\endgroup$ – copper.hat Nov 8 at 16:03
  • $\begingroup$ I assume $\lambda_i(t)$ are eigenvalues of $X(t)$. If I have access to $\lambda_i(t)$ then the problem is solved because that is all what I want. But I only have access to $\text{det}(X(t))$. The problem is easy when $X(t)= A-It$, so it may not be completely hopeless for more general case. $\endgroup$ – legon Nov 8 at 16:17
  • $\begingroup$ It is completely hopeless without further information. Consider the 2x2 case. As copper.hat rightfully points out, you know the product $\lambda_1(t)\lambda_2(t)$ and you want to extract from this the values of $\lambda_1(0)$ and $\lambda_2(0)$. But you cannot; for example, both $\lambda_1(t)=1, \lambda_2(t)=1$ and $\lambda_1(t)=\frac{1}{100}, \lambda_2(t)=100$ produce $\lambda_1(t)\lambda_2(t)=1$. The only thing you can do is that, if you know $\lambda_1$ and the determinant, you can compute $\lambda_2$. In arbitrary dimension, if you know $n-1$ eigenvalues, you can compute the missing one. $\endgroup$ – Giuseppe Negro Nov 8 at 16:22
  • $\begingroup$ If you want an explicit value, note that $$ X(t) = \pmatrix{s e^{t} & 0\\0& \frac{1}{se^{t}}} $$ will satisfy $\det X(t) = 1$ for all $t \in \Bbb R$, for any fixed $s \neq 0$. However, the eigenvalues of $X(0)$ will be $s,s^{-1}$. So, the fact that $\det X(t) = 1$ gives us no more information than the fact that $\det X(0) = 1$. $\endgroup$ – Omnomnomnom Nov 8 at 16:28
  • $\begingroup$ Thanks all for insightful comments. Yes, there are cases when it is hopeless to recover eigenvalues of $X(0)$, as rightly pointed out. But my main question is when recovery is possible, and under what condition on $X(t)$. I hope that there is a special case studied somewhere so I can try to emitate. $\endgroup$ – legon Nov 8 at 16:35

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