A basic property of vector spaces is that a vector space $V \subseteq \mathbb{F}_2^n$ of dimension $n-d$ can be characterized by $d$ linearly independent linear constraints - that is, there exist $d$ linearly independent vectors $w_1, \ldots, w_d \in \mathbb{F}_2^n$ that are orthogonal to $V$.From a Fourier perspective, this is equivalent to saying that the indicator function $1_V$ of $V$ has $d$ linearly independent non-zero Fourier coefficients. Note that $1_V$ has $2^d$ non-zero Fourier coefficients in total, but only $d$ of them are linearly independent.I am looking for an approximate version of this property of vector spaces. Specifically, I am looking for a statement of the following form:Let $S \subseteq \mathbb{F}_2^n$ be of size $2^{n-d}$. Then, the indicator function $1_S$ has at most $d\cdot\log(1/\varepsilon)$ linearly independent Fourier coefficients whose absolute value is at least $\varepsilon$.This question can be viewed from a Structure vs. Randomness perspective - Intuitively, such a claim says that every large set can be decomposed to a sum of a vector space and a small biased set. It is well known that every function $f:\mathbb{F}_2^n \to \mathbb{F}_2$ can be decomposed into a linear part of which has $\mathrm{poly}(1/\varepsilon)$ large Fourier coefficients, and a pseudorandom part that has small bias. My question asks whether the linear part has only a logarithmic number of linearly independent Fourier coefficients.
Linearly independent Fourier coefficients
co.combinatorics;linear algebra;boolean functions;fourier analysis
Isn't the following a counter-example?Let $f(x)$ be the majority of $x_1, \ldots, x_{1/\epsilon^2}$, which is an indicator of a set of size $2^n/2$, so $d = 1$.  However, $\hat{f}(\{i\}) = \Theta(\epsilon)$ for $1 \le i \le 1/\epsilon^2$, so you have $1/\epsilon^2$ linearly independent large Fourier coefficients.