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I think the simplest way to reduce $$A=M-\omega I$$ to a $3\times 3$ matrix is to use the Schur complement with respect to the $(\bar 2,\bar 2)$-elements of $A$, \begin{align} C = A/A_{\bar 2,\bar 2} …

This determinant can be calculated using the block transfer matrix method of Molinari (Linear Algebra and its Applications 429 (2008) 2221-2226, https://arxiv.org/abs/0712.0681), if $w_i \neq 0$. You …

This is a partial answer, now with added material. Assume that $n$ is a multiple of 3, the other two cases should be similar. I'll denote $C_1\mapsto a$ and $C_2\mapsto b$, such that for, e.g., $n=6$, …

OK, second try, equation references are pointing to my other answer: Using the definitions (3) for $\alpha_j$ and $\beta_j$, we can proof the OPs conjecture in the following way: As the determinants $ …

The characteristic polynomial of $\mathrm A_n$ can be calculated exactly in the limit $n\to\infty$: As pointed out by @Gottfried, the characteristic polynomial $c_n(x)=\det(\mathrm A_n-x 1)$ can be wr …

I guess that the determinant only factors if there is a stronger relation between the $z_j$. As example one can give a closed expression for $D_k(N)$ if $z_j = \exp(\iota (j-1) x)$. For $N$ even and …

Let $A(\theta)$ be a real $n\times n$ matrix. If $n$ is odd, there must be at least one real eigenvalue due to complex conjugate pairs, see the argument of @Carlo. If $n$ is even, you can discuss the …

Defining the $2 \times 2$ transfer matrix \begin{align}\tag{1} Q = \begin{pmatrix} -\lambda & 1 \\ -1 & 0 \end{pmatrix}, \end{align} the characteristic polynomial (CP) of the $M \times M$ matrix $A_M$ …

A quick numerical investigation for $N$ up to 512 leads to a conjectured exact large-$N$ expansion of the largest eigenvalue, $$ \lambda_0(N) = 4 - \frac{2\pi}{N} + \frac{\pi^2}{2 N^2} - \frac{\pi^3}{ …

OK, let's call the block matrix above $M$. First eliminate $N$ by a substitution $c\mapsto d N$. Then substitute $z_i \mapsto d y_i$ to eliminate $d$. Then you can construct the Schur complement w.r.t …

Based on your comment to my posting "Conjecture for a certain Cauchy-type determinant", I found that in addition to the formula above (I write $q$ instead of $x$, as all this is related to $q$-series, …

The "closest" form you can expect is the well known determinant formula for tridiagonal matrices, which in your case can be written as $$ \Delta = \det \left[ \begin{pmatrix} -x_n^{-n} & 0 \\ 0 & 1 …

I’ll try to explain it in physicists words: A matrix with all eigenvalues equal is proportional to the identity matrix. The eigenvectors are maximally degenerate, as every arbitrarily oriented orthono …