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.

  • $\begingroup$ Do you mean $f_i=h^{n_i}$ for some natural number $n_i$? $\endgroup$ Jul 5, 2011 at 0:14
  • $\begingroup$ Yes. I was not thinking of n as fixed, but I think that would make it more clear. $\endgroup$ Jul 5, 2011 at 0:16
  • 1
    $\begingroup$ Sorry to be pesky, but I want people on this site to be able to help. You have $N$ in there, I assume you mean $n_i$. And what does "set of functions in R2" mean? What is R2? What is the domain of these functions? Is it $\mathbb{R}^2$? Are these functions to the reals? To $\mathbb{R}^2$? $\endgroup$ Jul 5, 2011 at 0:26

1 Answer 1


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.

  • $\begingroup$ Another way to see that $A$ is invertible using Vandermonde matrices is to note that $$A=\begin{bmatrix} y_1 & 0 & \cdots & 0\\ 0 & y_2 & \cdots & 0\\ \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & \cdots & y_n \end{bmatrix} \begin{bmatrix} 1 & y_1 & \cdots & y_1^{n-1}\\ 1 & y_2 & \cdots & y_2^{n-1}\\ \vdots & \vdots & \ddots & \vdots\\ 1 & y_n & \cdots & y_n^{n-1} \end{bmatrix}.$$ $\endgroup$ Jul 5, 2011 at 1:06

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