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Suppose $f_1, f_2,...$ are a set of functions $\mathbb{R} \rightarrow \mathbb{R}$ so that each is the power of some non-constant function h. So $ f_ i=h^{n_i}$ for some natural number n.
Is it possible for such a set to be linearly dependent?
What additional conditions could ensure independence?
I was thinking about this because I need to show that a set in linearly independent. And I was not convinced that having the functions be powered lhjlj would be sufficient.
The following applies over an arbitrary field, with scalar valued functions on an arbitrary set (and was written before these were specified). I will make the simplifying assumption that in fact $f_i = h^i$, which results in no loss of generality.
The set $\{h,h^2,h^3,\ldots\}$ is linearly independent if and only if $h$ takes on infinitely many values. (In particular, with the assumption that $h:\mathbb R\to\mathbb R$, it would suffice for $h$ to be nonconstant and continuous.)
First suppose that $h$ takes on infinitely many values, and let $n$ be a positive integer. To see that $\{h,h^2,\ldots,h^n\}$ is linearly independent, suppose that $a_1,a_2,\ldots,a_n$ are scalars such that $a_1h+a_2h^2+\cdots a_nh^n=0$ identically. Let $x_1,\ldots, x_n$ be points in the domain where $h$ takes distinct nonzero values, say $y_i=h(x_i)$, with $y_i\neq y_j$ if $i\neq j$. Then the equations you get from plugging the $x_i$s into the identity $a_1h+a_2h^2+\cdots a_nh^n=0$ can be organized into the matrix equation $$\begin{bmatrix} y_1 & y_1^2 & \cdots & y_1^n\\ y_2 & y_2^2 & \cdots & y_2^n\\ \vdots & \vdots & \ddots & \vdots\\ y_n & y_n^2 & \cdots & y_n^n \end{bmatrix} \begin{bmatrix} a_1 \\a_2\\ \vdots \\ a_n\end{bmatrix}=0.$$
If $A$ is the square matrix in this equation, note that $A$ has the same determinant as the invertible Vandermonde matrix $\begin{bmatrix} 1&\vec 0\\ \vec1& A\end{bmatrix}$, where $\vec{1}$ is the column vector of all $1$s and $\vec{0}$ is the row vector of all $0$s. Thus, $A$ is invertible, which implies that $a_1=a_2=\cdots=a_n=0$.
Now suppose that $h$ takes on only finitely many values, say $k$ of them, and that the sets where $h$ takes on distinct values are $E_1,E_2,\ldots,E_k$. The vector space of functions that are constant on each $E_i$ is finite dimensional with dimension $k$, and each $h^i$ lies in this space. Therefore $\{h,h^2,\ldots,h^{k+1}\}$ is linearly dependent.
