It is known (see e.g. https://www.math.ucdavis.edu/~bremer/classes/fall2018/MAT128a/lecture9.pdf) that any continuous function $f: [-1,1] \to \mathbb{C}$ admits a Chebyshev expansion $$ f(x) = \sum_{n=0}^{\infty} a_n T_n(x) $$ where $$ a_n = \frac{2}{\pi} \int_{-1}^1 f(x) T_n(x) \frac{dx}{\sqrt{1-x^2}}, \quad T_n(x) = \cos(n \cdot \arccos(x)) $$ If one considers the point-wise approximation of $f$ by a finite expansion: for $f$ that is $k$-times continuously differentiable then $|a_n| = O(n^{-k})$, and since in addition $|T_n(x)| \leq 1$ for $|x| \leq 1$ then $$ \left| f(x) - \sum_{n=0}^N a_n T_n(x) \right| \leq \sum_{n=N+1}^{\infty} |a_n| \to 0 $$ for $k\geq 2$. So let us consider the expansion of the trigonometric function $f(x) = \cos(m x)$ for $x\in [-1,1]$. Since $f$ is $k$-times continuously differentiable for any $k$ then $f(x)$ can be approximated by a linear combination of Chebyshev polynomials of degree at most $m$ for error $m^{-c}$ for any $c>0$.
Let us write for any fixed $x\in [-1,1]$ a set of $N$ linear equations:
$$
\bar{c} = A \cdot \bar{t}
$$
where for each $i\in [N]$: $\bar{c}_i = \cos(i x)$ and $\bar{t}_i = T_i(x)$ and $A$ is an $N\times N$ matrix where $A_{i,j}$ is the $j$-th coefficient in the Chebyshev expansion of $\cos(i x)$.
By the above we have:
$$
\left\| \bar{c} - A \cdot \bar{t} \right\| = o(1)
$$
where the norm here is the $\ell_1$-norm. What can be said of $A$? Under what conditions is it invertible, and in such cases, could it be that in addition we have:
$$
\left\| A^{-1} \bar{c} - \bar{t} \right\| = o(1)
$$
namely that $A$ is "well-conditioned" in the sense of the $\ell_1$-norm ?