Consider the matrix $V$ given by
$$ V = \phi_k(z_j) $$
with $\phi_0, \dots, \phi_n$ a basis for the space of polynomials of degree $n$ and $z_0, \dots, z_n$ pairwise distinct points. A basis is the following: ${1,x,x^2,\dots,x^n}$. Is the resulting matrix a Vandermonde matrix? The problem I am asked to solve is the following: is it true that, if the points are pairwise distinct, this generalized Vandermonde matrix is invertible (given $\phi_0, \dots, \phi_n$ a basis for the space of polynomials of degree $n$)? Do I have to consider a more general basis?
If $\psi_0,...,\psi_n$ and $\varphi_0,...,\varphi_n$ are two bases for the same space, then there is an invertible matrix $g\in GL_{n+1}(\mathbb{R})$ such that $g\bar {\varphi}=\bar{\psi }$. It then follows that $det(\psi_k (z_i))=\det(g)\det(\varphi_k(z_i))$, so that the first is invertible iff the second is invertible. Since you already know a basis for which this matrix is invertible (e.g. $1,x,...,x^n$, or even better construct polynomials $\varphi_i$ such that $\varphi_i(z_j)=\delta_{i,j}$), then for every basis of polynomials this matrix is invertible.
EDIT:
Some Examples: If we take the basis $1,x,x^2$, then the matrix we get is
$$\left(\begin{array}{ccc} 1 & 1 & 1\\ z_{1} & z_{2} & z_{3}\\ z_{1}^{2} & z_{2}^{2} & z_{3}^{2} \end{array}\right)$$ which is a Vandermonde matrix and therefore invertible. We can also choose the basis $$f_1(x)=\frac{(x-z_2)(x-z_3)}{(z_1-z_2)(z_1-z_3)}, f_2(x)=\frac{(x-z_1)(x-z_3)}{(z_2-z_1)(z_2-z_3)}, f_3(x)=\frac{(x-z_1)(x-z_2)}{(z_3-z_1)(z_3-z_2)}$$ and then the matrix will be the identity matrix.
Finally, let's take the basis $1,1+x,1+x+x^2$ which can be written as $$\left(\begin{array}{c} 1 \\ 1+x\\ 1+x+x^2 \end{array}\right)= \left(\begin{array}{ccc} 1 & 0 & 0\\ 1 & 1 & 0\\ 1 & 1 & 1 \end{array}\right) \left(\begin{array}{c} 1 \\ x\\ x^2 \end{array}\right)$$ Then the matrix is $$\left(\begin{array}{ccc} 1 & 1 & 1\\ 1+z_{1} & 1+z_{2} & 1+z_{3}\\ 1+z_1+z_{1}^{2} & 1+z_2+z_{2}^{2} & 1+z_3z_{3}^{2} \end{array}\right)= \left(\begin{array}{ccc} 1 & 0 & 0\\ 1 & 1 & 0\\ 1 & 1 & 1 \end{array}\right) \left(\begin{array}{ccc} 1 & 1 & 1\\ z_{1} & z_{2} & z_{3}\\ z_{1}^{2} & z_{2}^{2} & z_{3}^{2} \end{array}\right) $$ Because the Vandermonde matrix is invertible, and we multiply it by an invertible matrix, then the new matrix is also invertible.