First $2n+1$ trigonometric functions and Chebyshev Sets Hopefully, this will be my last question on Chebyshev sets. I have a question on a statement at page 55 from The Approximation of Functions: Linear theory by John R. Rice.

Consider a set of $n$ functions $\{f_1(x),f_2(x),\cdots,f_n(x)\}$ on [a,b]. They are called the Chebyshev set on $[a,b]$ when all the linear combinations
$$\sum_{i=1}^n a_i f_i(x)$$
have at most $n-1$ distinct roots on $[a,b]$ for any $a_1,\cdots,a_n$.
Theorem: Given a function set $\{f_1(x),f_2(x),\cdots,f_n(x)\}$ on [a,b], pick $n$ distinct points $x_1,x_2,\cdots,x_n$ from $[a,b]$ to define an $n\times n$ matrix $T$:
\begin{equation*}
T = 
\begin{bmatrix}
f_1(x_1) & f_2(x_1)& \cdots & f_n(x_1) \\
f_1(x_2) & f_2(x_2)& \cdots & f_n(x_2) \\
\vdots & \vdots & \ddots & \vdots \\
f_1(x_n) & f_2(x_n)& \cdots & f_n(x_n) \\
\end{bmatrix}.
\end{equation*}
Then $T$ is invertible for every set of distinct points $x_1,\cdots,x_n$ in $[a,b]$ if and only if the function set forms a Chebyshev set in $[a,b]$.

Then the book explains two examples of Chebyshev sets:


*

*A function set $\{1,x,\cdots,x^n\}$ forms a Chebyshev set for any interval.

*The first $2n+1$ trigonometric function set $\{1, \cos x, \sin x, \cdots, \cos nx, \sin nx\}$ forms a Chebyshev set for the interval $[0, 2\pi]$. But the roots at $x=0$ and $x=2\pi$ count as a single root since trigonometric functions are periodic.
The first example is easy as this function set yields an $n\times n$ Vandermonde matrix. As long as this matrix consists of function evaluations at $n$ distinct points in any interval, this matrix has non-zero determinant and therefore invertible. It'll complete the proof that this function set forms a Chebyshev set.
But I'm not sure how to approach the second example of $2n+1$ trigonometric functions. Any hints or suggestions?
 A: Case 2. follows from case 1. together with the substitutions $n \leftarrow 2n$ and $x \leftarrow e^{\textrm {i} x}$. If $c_0, \ldots, c_{2n}$ are complex numbers, not all equal to zero, then the polynomial $$c_0+ c_1z+\ldots+c_{2n}z^{2n}$$ has at most $2n$ (complex) roots. The map $x\mapsto e^{\textrm{i} x}$ is a bijection between the strip $$\{x \in \mathbb{C} \mid 0 \leq \operatorname{Re}(z) < 2\pi \}$$ and $\mathbb{C}\setminus\{0\}$. Therefore $$c_0 + c_1e^{\textrm{i}x} + c_2e^{2\textrm{i}x}+\ldots+c_{2n}e^{2n \textrm{i} x}$$ has at most $2n$ roots in this strip and a fortiori at most $2n$ roots in the interval $[0, 2\pi)$. Now let $$\begin{eqnarray}c_n&=&a_0 \\ c_{n-k}&=&a_k + b_k \textrm{i}\\ c_{n+k}&=&a_k - b_k \textrm{i}\end{eqnarray}$$ for $k\in\{1,\ldots,n\}$ and some real $a_0, \ldots, a_n, b_1, \ldots b_n$ not all equal to zero. Then $$e^{-n\textrm{i}x}\left(c_0 + c_1e^{\textrm{i}x} + c_2e^{2\textrm{i}x}+\ldots+c_{2n}e^{2n \textrm{i} x}\right) = a_0 + 2\sum_{k=1}^n\left(a_k\cos(kx)+b_k\sin(kx)\right)$$ and therefore the right hand side also has at most $2n$ roots in the interval $[0,2\pi)$.
