Find the $\det(I + A)$ where A is a trigonometric matrix

Question

Let $A$ be a matrix of order $ n $ with elements defined as $a_{st} = \sin((s + t)\theta)$, where $\theta = \frac{2\pi}{n}$ and $ 1 \leq s, t \leq n $. I am interested in computing the determinant of $I + A$, where $I$ is the identity matrix of the same order.

My approach is to use the Spectral Theorem and express $A$ in the form $VDV^{-1}$, where $V$ is an orthogonal matrix and $D$ is a diagonal matrix with the eigenvalues of $A$ on its diagonal. Utilizing the identity $\det(I + AB) = \det(I + BA)$, I rewrite $\det(I + A)$ as follows:

$ \det(I + A) = \det(I + VDV^{-1}) = \det(I + V^{-1}VD) = \det(I + D). $

Therefore, it seems that finding the eigenvalues of $A$ would be sufficient to solve the problem. Through some case analysis, I've observed that for $n > 2$, the eigenvalues of $A$ appear to be $\frac{n}{2}$, $-\frac{n}{2}$, and $0$ (with $0$ having a multiplicity of $n-2$). However, I am struggling to prove this conclusively. I only figure out that $0$ is indeed a eigenvalue because each row of $A$ sum to $0$.

Questions:

1. Is my approach to finding $\det(I + A)$ correct?
2. How can I rigorously prove the observed eigenvalues of $A$ for $n > 2$?
3. Are there any alternative methods or insights that could simplify the problem?

Answer 1

Your overall approach is correct. The only question that remains is how to find eigenvalues of $A$. Due to the following relation \begin{equation} A_{st} = \sin{((s+t)\theta)} = \sin{(s\theta)}\cos{(t\theta)} + \cos{(s\theta)}\sin{(t\theta)}, \end{equation} the matrix $A$ is in fact a matrix of rank two since the above relation allows us to write $A$ as \begin{equation} A = u v^T + v u^T, \end{equation} where $u,v$ are two $n$-dim vectors given by \begin{equation} u = (1, \cos(\theta), \cos(2\theta),\ldots,\cos(n\theta))^T, v = (1, \sin(\theta), \sin(2\theta),\ldots,\sin(n\theta))^T. \end{equation} From this rank-two form, following the procedure here (Eigenvalue of rank 2 matrix) together with $v^Tv=u^Tu=n/2$ and $v^Tu=0$, it is easy to see the only non-zero eigenvalues are $n/2$ and $-n/2$.

Answer 2

This is similar to Hanwen's answer. Instead of computing the eigenvalues one can use the identity $\det(I+XY) =\det(I+YX)$ as follows:

With the column vectors $$ \begin{align} u &= (1, \cos(\theta), \cos(2\theta),\ldots,\cos(n\theta))^T,\\ v &= (1, \sin(\theta), \sin(2\theta),\ldots,\sin(n\theta))^T \end{align} $$ is $$ \begin{align} \det(I + A) &= \det\left(I + \begin{pmatrix}u & v\end{pmatrix} \begin{pmatrix}v^T \\ u^T\end{pmatrix}\right) \\ &= \det\left(I + \begin{pmatrix}v^T \\ u^T\end{pmatrix}\begin{pmatrix}u & v\end{pmatrix} \right) \\ &= \det \begin{pmatrix} 1 + v^T u & v^T v \\ u^T u & 1+u^T v \end{pmatrix} \\ &= \det \begin{pmatrix} 1 & n/2 \\ n/2 & 1\end{pmatrix} \\ & = 1 - \frac{n^2}{4} \, . \end{align} $$