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Consider the matrix product $\prod_i^n A_i$, where each $A_i$ is a $2\times2$ matrix having the form $A_i = \left( \begin{smallmatrix} \lambda + \alpha_i & -\beta_i \\ 1 & 0\end{smallmatrix}\right)$ with

  • $\lambda \in \mathbb{C}$ and obeys $\Re \lambda \le 0$ and $\Im \lambda\ge 0$.
  • Each $\alpha_i\in\mathbb{R}$ and obeys $\alpha_i\ge 0$
  • Each $\beta_i \in \mathbb{R}$ and obeys $\beta_i \ge 0$.

It is easy to check that the determinant of the product obeys $\det \big(\prod_i^n A_i\big)=\prod_i\beta_i$. Are there any upper or lower bounds (or any other results) on the trace $\mathrm{tr} \prod_i^n A_i$ that can be written in terms of $\lambda$ and the entries $\alpha_i$ and $\beta_i$? Equivalently, I am interested in any results that say something about the two eigenvalues of the product $\prod A_i$, stated in terms of $\lambda$ and the entries $\alpha_i$ and $\beta_i$.

By the way, based on other considerations, I can constrain this trace to be real and strictly positive.

I am particularly looking for bounds that (1) depend on $\lambda$ and (2) become tight in the simple case when all of the $A_i$ are equal to each other (i.e., when $\mathrm{tr}(A^n) = \lambda_1^n + \lambda_2^n$, where $\lambda_1$ and $\lambda_2$ are eigenvalues of $A$). Bounds stated in terms of maximal/minimum elements of $\alpha_i$ or $\beta_i$ are fine.

I know some simple bounds can be derived from the determinant, but such bounds generally do not obey (1) and (2). I have also derived the bound $$ \left\vert \mathrm{tr}\Big[\prod_i^n A_i\Big]\right\vert \le \left\Vert \prod_i^n A_i\right\Vert \le \prod_i^n\left\Vert A_i\right\Vert, $$ where $\left\Vert \cdot \right\Vert $ is the trace norm. Unfortunately, I don't think this is tight when all the $A_i$ equal each other.

Note that each matrix can be written as a product of symmetric matrices $A_i = \left( \begin{smallmatrix} \lambda+\alpha_i & 1 \\ 1 & 0\end{smallmatrix}\right)\left(\begin{smallmatrix} 1 & 0 \\ 0 & -\beta_i\end{smallmatrix}\right)$. So if it is helpful, the problem can be treated as bounding the trace of a product of symmetric $2\times 2$ matrices.

As a side note, this problems arises in applying the transfer matrix method in physics.

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    $\begingroup$ @AlexRavsky - The second equality is not true. The trace is commutative on symmetric matrices when $n \leq 3$ or when the matrices are permuted cyclicly, but not in general. Choosing $\lambda = 0$, $\alpha_i = \beta_i = i$ for $1 \leq i \leq n = 2$ gives the left trace equal to $-1$ and the right trace equal to $5$. $\endgroup$ Commented May 16, 2020 at 9:47

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