Distribution of weight of special type of random-matrix vector product? Let $G$ be a matrix of dimension $k \times n$ sampled uniformly randomly from $F_2^{k \times n}$. It is a well known fact that $y = xG$ is uniformly distributed in $F_2^n - \{0\}$ for all $x \in F_2^k$. It follows that distribution of Hamming weight of $\textbf{Y}$ follows a binomial distribution $B(n, \frac{1}{2})$.
If one imposes additional restriction over $G$ that all of $n$ columns must be distinct ( assuming $ n < 2^k)$, what will be the distribution of Hamming weight of $\textbf{Y} = x \textbf{G}$ for all $x \in F_2^k - \{0\}$ now?
Will it be still a binomial distribution $B(n, \frac{1}{2})$?
In other words, will $y = xG$ be still uniformly distributed in $F_2^n$ for all $x \in F_2^k - \{0\}$?
 A: $\newcommand{\de}{\delta}$Let $Y:=\mathbf Y$ and $G:=\mathbf G=[G_{ij}]$, with values in $F_2^{k\times n}$. It  appears that the equality $Y=xG$ is understood as
$$Y_j=\bigoplus_{i\in[k]}x_i G_{ij}$$
for $j\in[n]:=\{1,\dots,n\}$, with $Y=[Y_1,\dots,Y_n]$ with values in $F_2^{1\times n}$ and $x=[x_1,\dots,x_k]\in F_2^{1\times k}$, where $\bigoplus$ denotes the addition in $F_2$.
If so, the answer to the posted question is no. E.g., take $k=2$, $n=2$, and $x=[1,0]$. Then $G$ takes $4\cdot3=12$ equally likely values in $F_2^{2\times2}$, each with probability $\frac1{12}$, and the Hamming weight of $xG$ takes values $0,1,2$ with respective probabilities $\frac2{12},\frac8{12},\frac2{12}$, which differ from the corresponding binomial probabilities $\frac14,\frac24,\frac14$.

The OP has asked how to extend this counterexample to arbitrary $n$ (and, apparently, to arbitrary $k$ such that $n\le2^k$).
To do so, take indeed any natural $k$, any natural $n\le2^k$, and also take $x=[1,0,\dots,0]\in F_2^{1\times k}$. Then $Y_j=G_{1j}$ for $j\in[n]$.
The number of matrices in $F_2^{k\times n}$ with pairwise distinct columns is
\begin{equation*}
    N_{k,n}=\prod_{i=0}^{n-1}(2^k-i). 
\end{equation*}
So, the distribution of the Hamming weight of $xG$ is given here by the formula
\begin{equation*}
\begin{aligned}
    p_{k,n}(m)&:=P\Big(\sum_{j\in[n]}Y_j=m\Big) \\ 
    &=P\Big(\sum_{j\in[n]}G_{1j}=m\Big) \\ 
    &=\binom nm \frac{N_{k-1,m}N_{k-1,n-m}}{N_{k,n}}
\end{aligned}
\tag{1}\label{1}
\end{equation*}
for $m\in\{0,\dots,n\}$.
In particular,
\begin{equation*}
\begin{aligned}
    p_{k,n}(0)=\frac{N_{k-1,n}}{N_{k,n}}
    =\prod_{i=0}^{n-1}\frac{2^{k-1}-i}{2^k-i}
=\frac1{2^n}\prod_{i=0}^{n-1}\frac{2^k-2i}{2^k-i}<\frac1{2^n}
\end{aligned}
\end{equation*}
if $n\ge2$, so that $p_{k,n}(0)$ differs from the corresponding binomial probability $\frac1{2^n}$.

Actually, the distribution of the Hamming weight of $xG$ is the same for all $x\in F_2^k\setminus\{0\}$. Indeed, rewrite \eqref{1} as
\begin{equation*}
\begin{aligned}
    P\Big(\sum_{j\in[n]}Y_j=m\Big)
    &=\binom{2^{k-1}}m\binom{2^{k-1}}{n-m}\Big/\binom{2^k}n,  
\end{aligned}
\tag{2}\label{2}
\end{equation*}
again for $m\in\{0,\dots,n\}$. So, $\sum_{j\in[n]}Y_j$ has the hypergeometric distribution with parameters $2^k,2^{k-1},n$ -- if $x=[1,0,\dots,0]\in F_2^{1\times k}$.
Now take any $x\in F_2^k\setminus\{0\}$. By symmetry, the distribution of the random matrix $G$ is the same as that of the random submatrix, say $H=[H_{ij}]\in F_2^{k\times n}$, of the nonrandom matrix $M=[M_{i,p}]\in F_2^{k\times2^k}$, where
\begin{equation*}
    M_{i,p}:=\de_p^i
\end{equation*}
for $(i,p)\in[k]\times[2^k]$; $\de_1,\dots,\de_{2^k}$ is any fixed enumeration of the set $\{0,1\}^k$; $\de_p^i$ is the $i$th coordinate of $\de_p$, so that $\de_p=(\de_p^1,\dots,\de_p^k)$;
\begin{equation*}
    H_{ij}:=M_{i,P_j}
\end{equation*}
for $(i,j)\in[k]\times[n]$, and $(P_1,\dots,P_n)$ is a uniformly distributed random element of the set of all $n$-tuples $(p_1,\dots,p_n)\in[2^k]^n$ with pairwise distinct $p_1,\dots,p_n$.
Informally, the random matrix $H$ is obtained by selecting $n$ columns of the nonrandom matrix $M$ at random without replacement, with the preservation of the order in which the selections were made. However, because the addition of real numbers is commutative, the order of the selections will not affect the distribution of the Hamming weight $\sum_{j\in[n]}Z_j$
of $xH$, where
\begin{equation*}
    Z_j:=\bigoplus_{i\in[k]}x_i H_{ij}=\bigoplus_{i\in X}H_{ij} 
    \;=\;\bigoplus_{i\in X}M_{i,P_j} 
\end{equation*}
and
\begin{equation*}
    X:=\{i\in[k]\colon x_i=1\}\ne\emptyset. 
\end{equation*}
By symmetry,
\begin{equation*}
|E_1|=|E_0|=\frac{2^k}2=2^{k-1}, 
\end{equation*}
where $E_t:=\{p\in[2^k]\colon\bigoplus_{i\in X}M_{i,p}=t\}$ and $|\cdot|$ denotes the cardinality.
So, the Hamming weight $\sum_{j\in[n]}Z_j$ of $xH$ will take a value $m$ if and only if in the corresponding sample without replacement of $n$ columns out of the $2^k$ columns of the matrix $M$ there will be exactly $m$ columns with indices $p\in E_1$, with the remaining $n-m$ columns with indices $p\in E_0$.
Therefore and because $(Z_1,\dots,Z_n)$ equals $(Y_1,\dots,Y_n)$ in distribution, we arrive at the hypergeometric distribution formula \eqref{2}.
