Number of linear functions from $\{0,1\}^n$ to $\{0,1\}$

Question

For $x,y \in \{0,1\}^n$, let $x \oplus y$ be the element of $\{0,1\}^n$ obtained by the component-wise exclusive or of $x$ and $y$. A boolean function $F:\{0,1\}^n \to \{0,1\}$ is said to be linear if $F(x \oplus y) = F(x) \oplus F(y)$, for all $x$ and $y$.
Find the number of linear functions from $\{0,1\}^n$ to $\{0,1\}$.

The abscissa can take $2^n$ values and the ordinate can be $0$ or $1$ for each value of abscissa. So, the number of linear functions should be $2^{2^n}$, but the answer is $2^n$?

Answer 1

The XOR operation between two string is the sum in $\mathbb{F}_2$, that is, the field with two element $\{0,1\}$.

That means you're actually looking for the cardinality of the linear function between the vectorial spaces $\mathbb{F}_2^n$ and $\mathbb{F}_2$. (Also called Dual of $\mathbb{F}_2^n$).

Those are determined on what are the images of a base in $\mathbb{F}_2^n$, but it has $n$ elements, and you can send each of them either to $0$ or $1$.

This means that there are $2^n$ linear functions.

Note: it is also a common knowledge that the dual of a finite-dimensional space is isomorphic to itself, so it has the same number of elements: $|\mathop{\rm Hom}(\mathbb{F}_2^n,\mathbb{F}_2)|=|\mathbb{F}_2^n|=2^n$