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I am trying to prove the following fact, which seems intuitive to me, yet I cannot find a way to show it.

Consider the set of functions $\{f_1,\ldots,f_n\}$, $f_i:\mathbb{R}\rightarrow\mathbb{R}$. Suppose that this set is linearly independent, i.e. $$\sum_i f_ic_i=0\implies c_i=0\;\;\; \forall i$$

Then there exists numbers $(x_i)_{i=1}^n$ such that the matrix

\begin{bmatrix} f_1(x_{1}) & f_2(x_{1}) & \dots & \dots & f_n(x_{1}) \\ f_1(x_{2}) & \dots & \dots & \dots & \dots \\ \dots & \dots & \dots & \dots & \dots \\ f_1(x_{n}) & \dots & \dots & \dots & f_n(x_{n}) \end{bmatrix} has a nonzero determinat.

I have tried to reason by contraposition, assuming that the determinant vanishes for any choice of the $(x_i)_{i=1}^n$ but I didn't get anywhere.

Lorenzo Stanca
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    Can you prove the $n=2$ case? That could be easier and may given insight for the general case. – Michael Dec 26 '17 at 18:25
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    Note also that it is enough to show (for the $n=2$ case) that there exist $x,y \in \mathbb{R}$ such that vectors $(f_1(x), f_1(y))$ and $(f_2(x), f_2(y))$ are linearly independent. Similar for the general case. – Michael Dec 26 '17 at 18:34
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    My post here answers your question. If you are satisfied with that, then I will go ahead and close this question. If not, be sure to explain what it is that you don't understand from the post. – Ben Grossmann Dec 26 '17 at 19:10
  • Thanks, this is what I was looking for. – Lorenzo Stanca Dec 26 '17 at 20:03

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