I'm stuck in the "$\Longrightarrow$" of the following question.
Exercise. Let $f_1,...,f_n\colon X\to\Bbb{K}$ functions defined on an arbitrary set $X$, where $\Bbb{K}$ is a field. Show that $(f_j)_{j=1}^n$ is a linearly independent family of $\Bbb{K}^X$ if and only if there exists $x_1,...,x_n\in X$ such that $$\det\left[\begin{array}{ccc}f_1(x_1)&\cdots&f_n(x_1)\\\vdots&\ddots&\vdots\\f_1(x_n)&\cdots&f_n(x_n)\end{array}\right]\ne0$$
The part "$\Longleftarrow$" is pretty simple, but the another implication... My colegues tells me that we can prove by induction on $n$, this seems ok to me. But, anyone there has some another idea? Thanks for any help!
Suppose that $(f_i)$ is a linearly independent family. That is, for any coefficients $c_1,\dots,c_n$ not all equal to zero, there exists an $x \in X$ such that $\sum_{i}c_i f_i(x) \neq 0$.
Let $F:X \to \Bbb K^n$ denote the map $x \mapsto (f_1(x),\dots,f_n(x))$ (which I will take to be a column-vector, for the purposes of computation). Now, suppose $(f_i)$ is a linearly independent family. That is, for every vector $c = (c_1,\dots,c_n) \in \Bbb K^n$, there exists an $x \in X$ such that $c^TF(x) \neq 0$. Thus, the span of the vectors $\{F(x) : x \in X\}$ satisfies $$ \operatorname{span}(\{F(x) : x \in X\})^\perp = \{0\}. $$ Thus, $\operatorname{span}(\{F(x) : x \in X\}) = \Bbb K^n$. Thus, there exist $x_1,\dots,x_n$ such that the vectors $F(x_1), \dots, F(x_n)$ are linearly independent. If we take these vectors $F(x_1), \dots, F(x_n)$, then we find that this matrix is square and must have a non-zero determinant, as desired.