Let $x_0, \dots , x_n$ be the roots of the Chebyshev polynomial $T_{n+1}(x)$.
We define:
$$A=\begin{pmatrix} \frac{1}{\sqrt2}T_0(x_0) & \cdots & \frac{1}{\sqrt2}T_0(x_n) \\ T_1(x_0) & \cdots & T_1(x_n) \\ \vdots & \vdots& \vdots \\ T_n(x_0) & \cdots & T_n(x_n) \\ \end{pmatrix}$$
Is $A$ invertible? If so, calculate $A^{-1}$.
I have tried to solve it for the roots of $T_2(x)$ and I found that the matrix is invertible.
How can I generalize this?
On
The orthogonality is an important property of the Chebyshev polynomials. However, the invertibility is true in much more general case. For if some linear combination of rows $\alpha_0 T_0 + \alpha_1 T_1 + \cdots + \alpha_n T_n$ has $n+1$ roots $x_i$, then the polynomial is $0$, and so all the $\alpha_i$ are $0$ ( the $T_i$ are linearly independent (being of different degrees)).
Related: Chebyshev systems.
Since $T_0=1$, the first row is a constant, and so just for the sake of symmetry, I'll write it as $$ \begin{pmatrix} \frac{1}{\sqrt2}T_0(x_0) & \cdots & \frac{1}{\sqrt2}T_0(x_n) \\ T_1(x_0) & \cdots & T_1(x_n) \\ \vdots & \vdots& \vdots \\ T_n(x_0) & \cdots & T_n(x_n) \\ \end{pmatrix}, $$ where the $T_0$'s in the first row are actually evaluated at all of the roots $x_0$ up to $x_n$ rather than evaluated all at $x_0$. In this case, the invertibility of the matrix follows directly from a discrete orthogonality condition for the $T$'s, which is $$ \sum_{k=0}^{N-1}T_i(x_k)T_j(x_k)=\begin{cases} 0 & i\neq j \\ N & i=j=0 \\ N/2 & i = j \neq 0 \end{cases}, $$ which holds provided that $N>\textrm{max}(i,j)$. Each of these orthogonality conditions is realized by taking the dot product of any two columns in the matrix above, and since each dot product is zero, the rows are mutually linearly independent, and hence the matrix is invertible.
Related: Prove that $T_n$ satisfy $ \sum_{k=0}^{N-1}{T_i(x_k)T_j(x_k)} = \begin{cases} 0 &: i\ne j \\ l\neq 0 &: i=j \end{cases} \,\! $