Conditions for nonsingularity of a matrix with trigonometric elements

Question

Let $q$ be a positive integer and define $Q\triangleq \{1,2,...,q\}$. For all $i\in Q$, and all $k\in\mathbb{N}$, define \begin{align} f_{i,k}\triangleq \begin{bmatrix}\cos k\theta_i&\sin k\theta_i\end{bmatrix}\in\mathbb{R}^{1\times 2}, \end{align} where $\theta_i>0$. Define \begin{align} P_{q,k}\triangleq \begin{bmatrix}f_{1,k}&f_{2,k}&...&f_{q,k}\\ f_{1,k+1}&f_{2,k+1}&...&f_{q,k+1}\\ \vdots&\vdots&~&\vdots\\ f_{1,k+2q-1}&f_{2,k+2q-1}&...&f_{q,k+2q-1}\end{bmatrix}\in\mathbb{R}^{2q\times 2q} \end{align}

In the case, where $q=1$, it follows that \begin{align} \det~ P_{1,k}&=\det\begin{bmatrix}\cos k\theta_1&\sin k\theta_1\\\cos (k+1)\theta_1&\sin (k+1)\theta_1\end{bmatrix}\\ &=\cos k\theta_1\sin (k+1)\theta_1-\sin k\theta_1\cos (k+1)\theta_1\\&=\sin \theta_1. \end{align} Thus, if $\sin\theta_1\neq 0$, then $P_{1,k}$ is nonsingular for all $k\in\mathbb{N}$.

For $q\in\{2,3,4,...\}$, what are the conditions on $\theta_i$, where $i\in Q$, such that for all $k\in\mathbb{N}$, $P_{q,k}$ is nonsingular?

Answer 1

Hint: Computing the derivative of the determinant of your matrix with respect to $k$ we see at once that it is independent of $k$, so we just take $k=1$.

Next we know that $$\cos(n\theta)=T_n(\cos\theta),$$ with $T_n$ the $n$th Chebyshev polynomial of the first kind. Smilarly, $$\sin(n\theta)=U_n(\cos(\theta))/\sin(\theta),$$ with $U_n$ the Chebyshev polynomial of the second kind now.

$U_n$ and $T_n$ are polynomials of degree $n$. Using these facts and row operations you can considerably simplify your matrix until it becomes something similar to a Vandermonde matrix, up to some factors, and using that you should find the formula I wrote in a comment above for the determinant.