Matrix rank and linear independence of functions

The following proposition is easy to verify:

Given $n$ functions $f_1,...,f_n$, they are linearly independent in an interval $I$ if there exists a set of $n$ points in $I$, namely $x_1,...,x_n$, such that the matrix \begin{pmatrix} f_1(x_1) & \cdots & f_n(x_1) \\ \vdots & \ddots & \vdots \\ f_1(x_n) & \cdots & f_n(x_n) \end{pmatrix}

has full rank.

My question is: can I replace the if by if and only if? That is, if $n$ functions are linearly independent, is it true that there exists a set of $n$ points such that the above matrix has full rank? I am pretty convinced so, but I am not being able to prove it.

Solution 1:

Let's try to prove this by contrapositive. Define $$ M(x_1,\dots,x_n) = \begin{pmatrix} f_1(x_1) & \cdots & f_n(x_1) \\ \vdots & \ddots & \vdots \\ f_1(x_n) & \cdots & f_n(x_n) \end{pmatrix} $$ We have supposed, then, that for every collection $x_1,\dots,x_n$, the matrix $M$ fails to have full rank. That is, there exists a function $g:\Bbb R^n \to \Bbb R^n \setminus \{0\}$ such that $$ M(x_1,\dots,x_n)g(x_1,\dots,x_n) = 0 $$ We wish to conclude that $g$ may be taken to be a constant function. Suppose for contradiction that it is impossible to do so. That is, there exists a collection $\mathbf x^1,\dots,\mathbf x^m \in \Bbb R^n$ such that $\bigcap_{i=1}^m \ker M(\mathbf x^i) = \{0\}$. Equivalently, the block matrix $$ \tilde M = \pmatrix{M(\mathbf x^1) \\ \vdots \\ M(\mathbf x^n)} $$ has full rank. This means, however, that an $n \times n$ submatrix of $\tilde M$ has full rank. However, this submatrix is necessarily of the form $M(\mathbf x^{i_1}_{k_{i_1}},\dots,\mathbf x^{i_n}_{k_{i_n}})$. Thus, we have contradicted the supposition that $M$ never has full rank.

Thus, there exists a constant $g = (c_1,\dots,c_n)$ such that $Mg = 0$. In other words, $$ \sum_{i=1}^n c_i f_i(x) = 0 $$ for any $x \in \Bbb R$. That is, the functions $f_i$ are linearly dependent.