Let's say I have a Hadamard matrix $W$ and a vector $y$ such that $W*x = y$ where $y$ is fixed and supposed to be known, and $x$ is a vector too. my question, how many possible vectors of $x$ can be exist in order to get the same result $y$?
For example, let's have $W = [1, -1; -1, 1]$ the vector $x$ can be either [1; 1] or [-1; -1] and the result of both multiplication is unique which is $y = 0$.
Is there a general rule for that ?
The kernel $\ker(W) := \{x \in \mathbb{R}^n : Wx = 0\}$ is always a vector space. That means, if $Wx = 0$ has a solution that is different from $x = 0$, you will automatically have infinitely many solutions. Indeed, If $x$ solves $Wx = 0$, then also $W(\lambda x) = \lambda Wx = 0$ for all $\lambda \in \mathbb{R}$. Likewise, if $x$ is a solution of $Wx = y$, then $x' = x+v$ is also a solution for all $v \in \ker(W)$. So it all depends on the dimension of $\ker(W)$. If $\ker(W) = \{0\}$, then the solution is unique.