# Spectral properties via determinant function

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 $$\text{det}(X(t))$$?

• 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? – copper.hat Nov 8 at 16:03
• 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. – legon Nov 8 at 16:17
• 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. – Giuseppe Negro Nov 8 at 16:22
• 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 $\det X(0) = 1$. – Omnomnomnom Nov 8 at 16:28
• 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. – legon Nov 8 at 16:35