I want to ask a determinant problem and want to prove that they are non-zero (even better if there is a display expression), where $m$ is a positive integer:
\begin{equation}\left| \begin{matrix} 1&{\cos \left( {\frac{{2\pi }}{{2m}} \cdot 1 \cdot 0} \right)}&{\sin \left( {\frac{{2\pi }}{{2m}} \cdot 1 \cdot 0} \right)}&{\cos \left( {\frac{{2\pi }}{{2m}} \cdot 2 \cdot 0} \right)}&{\sin \left( {\frac{{2\pi }}{{2m}} \cdot 2 \cdot 0} \right)}& \cdots &{\cos \left( {\frac{{2\pi }}{{2m}} \cdot \left( {m - 1} \right) \cdot 0} \right)}&{\sin \left( {\frac{{2\pi }}{{2m}} \cdot \left( {m - 1} \right) \cdot 0} \right)}&{\cos \left( {\frac{{2\pi }}{{2m}} \cdot m \cdot 0} \right)} \\ \vdots &{\cos \left( {\frac{{2\pi }}{{2m}} \cdot 1 \cdot 1} \right)}&{\sin \left( {\frac{{2\pi }}{{2m}} \cdot 1 \cdot 1} \right)}&{\cos \left( {\frac{{2\pi }}{{2m}} \cdot 2 \cdot 1} \right)}&{\sin \left( {\frac{{2\pi }}{{2m}} \cdot 2 \cdot 1} \right)}& \cdots &{\cos \left( {\frac{{2\pi }}{{2m}} \cdot \left( {m - 1} \right) \cdot 1} \right)}&{\sin \left( {\frac{{2\pi }}{{2m}} \cdot \left( {m - 1} \right) \cdot 1} \right)}&{\cos \left( {\frac{{2\pi }}{{2m}} \cdot m \cdot 1} \right)} \\ \vdots & \vdots & \vdots & \vdots & \vdots & \cdots & \vdots & \vdots & \vdots \\ 1&{\cos \left( {\frac{{2\pi }}{{2m}} \cdot 1 \cdot \ell } \right)}&{\sin \left( {\frac{{2\pi }}{{2m}} \cdot 1 \cdot \ell } \right)}&{\cos \left( {\frac{{2\pi }}{{2m}} \cdot 2 \cdot \ell } \right)}&{\sin \left( {\frac{{2\pi }}{{2m}} \cdot 2 \cdot \ell } \right)}& \cdots &{\cos \left( {\frac{{2\pi }}{{2m}} \cdot \left( {m - 1} \right) \cdot \ell } \right)}&{\sin \left( {\frac{{2\pi }}{{2m}} \cdot \left( {m - 1} \right) \cdot \ell } \right)}&{\cos \left( {\frac{{2\pi }}{{2m}} \cdot m \cdot \ell } \right)} \\ \vdots & \vdots & \vdots & \vdots & \vdots & \cdots & \vdots & \vdots & \vdots \\ 1&{\cos \left( {\frac{{2\pi }}{{2m}} \cdot 1 \cdot \left( {2m - 1} \right)} \right)}&{\sin \left( {\frac{{2\pi }}{{2m}} \cdot 1 \cdot \left( {2m - 1} \right)} \right)}&{\cos \left( {\frac{{2\pi }}{{2m}} \cdot 2 \cdot \left( {2m - 1} \right)} \right)}&{\sin \left( {\frac{{2\pi }}{{2m}} \cdot 2 \cdot \left( {2m - 1} \right)} \right)}& \cdots &{\cos \left( {\frac{{2\pi }}{{2m}} \cdot \left( {m - 1} \right) \cdot \left( {2m - 1} \right)} \right)}&{\sin \left( {\frac{{2\pi }}{{2m}} \cdot \left( {m - 1} \right) \cdot \left( {2m - 1} \right)} \right)}&{\cos \left( {\frac{{2\pi }}{{2m}} \cdot m \cdot \left( {2m - 1} \right)} \right)} \end{matrix} \right|\end{equation} and
\begin{equation}\left| \begin{matrix} 1&{\cos \left( {\frac{{2\pi }}{{2m + 1}} \cdot 1 \cdot 0} \right)}&{\sin \left( {\frac{{2\pi }}{{2m + 1}} \cdot 1 \cdot 0} \right)}&{\cos \left( {\frac{{2\pi }}{{2m + 1}} \cdot 2 \cdot 0} \right)}&{\sin \left( {\frac{{2\pi }}{{2m + 1}} \cdot 2 \cdot 0} \right)}& \cdots &{\cos \left( {\frac{{2\pi }}{{2m + 1}} \cdot m \cdot 0} \right)}&{\sin \left( {\frac{{2\pi }}{{2m + 1}} \cdot m \cdot 0} \right)} \\ 1&{\cos \left( {\frac{{2\pi }}{{2m + 1}} \cdot 1 \cdot 1} \right)}&{\sin \left( {\frac{{2\pi }}{{2m + 1}} \cdot 1 \cdot 1} \right)}&{\cos \left( {\frac{{2\pi }}{{2m + 1}} \cdot 2 \cdot 1} \right)}&{\sin \left( {\frac{{2\pi }}{{2m + 1}} \cdot 2 \cdot 1} \right)}& \cdots &{\cos \left( {\frac{{2\pi }}{{2m + 1}} \cdot m \cdot 1} \right)}&{\sin \left( {\frac{{2\pi }}{{2m + 1}} \cdot m \cdot 1} \right)} \\ \vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 1&{\cos \left( {\frac{{2\pi }}{{2m + 1}} \cdot 1 \cdot \ell } \right)}&{\sin \left( {\frac{{2\pi }}{{2m + 1}} \cdot 1 \cdot \ell } \right)}&{\cos \left( {\frac{{2\pi }}{{2m + 1}} \cdot 2 \cdot \ell } \right)}&{\sin \left( {\frac{{2\pi }}{{2m + 1}} \cdot 2 \cdot \ell } \right)}& \cdots &{\cos \left( {\frac{{2\pi }}{{2m + 1}} \cdot m \cdot \ell } \right)}&{\sin \left( {\frac{{2\pi }}{{2m + 1}} \cdot m \cdot \ell } \right)} \\ \vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 1&{\cos \left( {\frac{{2\pi }}{{2m + 1}} \cdot 1 \cdot 2m} \right)}&{\sin \left( {\frac{{2\pi }}{{2m + 1}} \cdot 1 \cdot 2m} \right)}&{\cos \left( {\frac{{2\pi }}{{2m + 1}} \cdot 2 \cdot 2m} \right)}&{\sin \left( {\frac{{2\pi }}{{2m + 1}} \cdot 2 \cdot 2m} \right)}& \cdots &{\cos \left( {\frac{{2\pi }}{{2m + 1}} \cdot m \cdot 2m} \right)}&{\sin \left( {\frac{{2\pi }}{{2m + 1}} \cdot m \cdot 2m} \right)} \end{matrix} \right|\end{equation}
I've checked that all the cases where m equals $1,2$ are true. But I have no idea about the general situation. Here's where the questions came from and what I found.
The source of the problem is: consider the data of the $T$-period, which may be denoted as $f(0), f(1),... f(T-1)$. I want to fit it completely with the linear combination function $g(n)$ of $1$, $\cos(2k\pi n/T)$, $\sin(2k\pi n/T)$, i.e. $f(n)=g(n)$, where $n=0,...,T-1$. This is equivalent to a discrete version of Weierstrass's second approximation theorem. I guess that the order of the finite-order trigonometric function $g$ could be no more than $T/2$ (where $T$ is even), or $(T-1)/2$ (where $T$ is odd). In equivalent terms, the elements of the coefficient matrix are the elements of the determinant. I want to prove that the determinants are non-zero. Interestingly, the constant term of $g$ is exactly the mean of the periodic data (which is immediately proved by adding all rows together using the trigonometric property), which is similar to the mean of the function on the torus $T^d$ for the Fourier series of the periodic function. Many thanks!