Invertible matrix of function evaluations? Let $A := \{1, \dots, n\}$. Suppose there are maps $f_1, \dots, f_n: A \to \mathbb R$ that are linearly independent. Can one always choose $x_1, \dots, x_n \in A$ such that the following matrix is invertible?
$$ \begin{pmatrix} f_1(x_1) & \cdots & f_1(x_n) \\ \vdots & \ddots & \vdots \\ f_n(x_1) & \cdots & f_n(x_n) \end{pmatrix}$$
I think I can prove this statement by induction but I wonder if there is a more clever and elegant way to prove it? Maybe by writing down some suitable map or something like that?
 A: Yes.
We show the contrapositive, namely that if for all $x_1,x_2,\ldots,x_n\in A$, the matrix $ \begin{pmatrix} f_1(x_1) & \cdots & f_1(x_n) \\
\vdots & & \vdots \\
f_n(x_1) & \cdots & f_n(x_n) \end{pmatrix}$ is singular, then the $f_i$ are linearly dependent.
Assuming the hypothesis for the contrapisitive, if we take $x_1=1,x_2=2,\ldots,x_n=n$, the matrix 
$$ \begin{pmatrix} f_1(1) & \cdots & f_1(n) \\
\vdots & & \vdots \\
f_n(1) & \cdots & f_n(n) \end{pmatrix}$$
is singular.
Hence there exists a non-trivial linear combination of the rows of this matrix that equals the zero vector. That is, there exist $c_1,\ldots,c_n\in \Bbb{R}$, not all zero, such that
$$c_1 (f_1(1),\ldots, f_1(n)) + c_2(f_2(1),\ldots, f_2(n))+\cdots + c_n(f_n(1),\ldots,f_n(n)) = (0,\ldots,0).$$
Equating the $i$-th component for each $i$ gives $$c_1f_1(i)+c_2f_2(i) + \cdots + c_nf_n(i)=0$$ for all $i\in A$. This means that $c_1f_1+c_2f_2+\cdots + c_n f_n$ is the zero function. Since the $c_i$ are not all $0$, this means that the $f_i$ are linearly dependent, as required.
A: The space of maps $\{1, \dots, n\} = A \rightarrow \mathbb R$ is a vector space of dimension $n$, with a nice basis consisting of delta-functions:
$$
\delta_i(j) = \begin{cases}1, \quad if \quad  i=j \\ 0, \quad if \quad i \neq j\end{cases}
$$
Evaluating a map $f$ at some point $j$ just reveals it's $j$-th coordinate. The fact that $f_1, \dots, f_n$ are linearly independent means that, in any basis, the matrix of vector coordinates is invertible. Since coordinates in our basis are get by evaluation, this shows that
$$
\begin{pmatrix}
f_1(1) & \cdots & f_n(1) \\
\vdots & \ddots & \vdots \\
f_n(1) & \cdots & f_n(n)
\end{pmatrix}
$$
is invertible.
