Another question on Chebyshev set and Invertible Matrix This is a follow-up question on chebyshev sets. I'm trying to prove Theorem 4 in Parks and McClellan paper but I don't know how. Can anyone explain how to approach this proof?

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: Assume that $\{f_1(x),f_2(x),\cdots,f_n(x)\}$ is a Chebyshev set on [a,b]. Pick $n+1$ distinct points $x_1,x_2,\cdots,x_{n+1}$ from $[a,b]$ to define an $(n+1)\times (n+1)$ matrix $T$
\begin{equation*}
T = 
\begin{bmatrix}
f_1(x_1) & f_2(x_1)& \cdots & f_n(x_1) & \delta_1 \\
f_1(x_2) & f_2(x_2)& \cdots & f_n(x_2) & \delta_2 \\
\vdots & \vdots & \ddots & \vdots & \vdots \\
f_1(x_{n+1}) & f_2(x_{n+1})& \cdots & f_n(x_{n+1}) & \delta_{n+1} \\
\end{bmatrix}
\end{equation*}
where each $\delta_i$ is non-zero which alternates in sign.
Then $T$ is invertible.

One thing that pops up in my head was to show that the determinant of $T$ is non-zero. As the last column of $T$ alternate in sign, I may apply the Laplace expansion along the last column to obtain the determinant as the summation of minors. But each minor doesn't necessarily have the same sign and there's no guarantee that these summation is always non-zero. How can I prove this?

I will assume $x_1<x_2<\dots<x_{n+1}$. Let $(a_1, \dots, a_{n+1})$ be in the kernel of $T$. Then $a_1\,f_1 + \ldots + a_n\,f_n$ has a root in each of the $n$ intervals $[x_k, x_{k+1})$ for $k\in\{1, \dots, n\}$. (Consider the cases $a_{n+1} = 0$ and $a_{n+1} \neq 0$ separately.) Therefore $a_k=0$ for all indices $k$ since $\{f_1, \dots, f_n\}$ is a Chebyshev set. So the kernel of $T$ is trivial. 
