I have a question that I'm trying to solve which is as following:
If $x$ is a fixed vector determine how many binary vectors $y$ of length $n$ do exist such that $\langle x,y\rangle = 0$
Inner product of two vectors is as following: $$\langle x,y\rangle = x_1y_1 + x_2y_2 + x_3y_3 + \cdots + x_ny_n$$
I tried and I got as following: $wt(x)2^{n-wt(x)}$ where $wt(x)$ is the number of $1$s in vectors and $n-wt(x)$ is the number of non-zero in $x$. And $2^{n-wt(x)}$ is the number of zeroes in x. It is correct for some values, but not for general! my way to do is to write two vectors of x and y and suppose x is fixed with different number of 0s and 1s. Then, if $x_i=0$, then $y_i=0$ or $y_i=1$ to get $\langle x,y\rangle=0$ So we have two possible values for y if $x_i=0$ and if $x_i=1$, then y must be equal 0. Thus, I calculate the number of zeros in $x$ as ($n-wt(x)$) and powered by $2$ since it has two possible values and times $wt(x)$ because if $x$ has $k$ number of $1$s, then we need $k$ number of $1$s for $y$ to make $\langle x,y\rangle=0$. Thus, we get $wt(x)2^{n-wt(x)}$.
Thank you!
Vectors orthogonal to $x$ always form an $n-1$ dimensional subspace if $x \neq 0$, and if $x=0$ then the while $n$-dimensional space is orthogonal to it.
An $m$-dimensional space over field of size $q$ has $q^m$ elements. In your case, $q=2$, and